Asymptotic determination of edge-bandwidth of multidimensional grids and Hamming graphs

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1 Asymptotic etermination of ege-banwith of multiimensional gris an Hamming graphs Reza Akhtar Tao Jiang Zevi Miller. Revise on May 7, 007 Abstract The ege-banwith B (G) of a graph G is the banwith of the line graph of G. More specifically, for any bijection f : E(G) {,,..., E(G) }, let B (f, G) = max{ f(e ) f(e ) : e an e are incient eges of G}, an let B (G) = min f B (f, G). We etermine asymptotically the ege-banwith of -imensional gris P n an of the Hamming graph K n, the -fol Cartesian prouct of K n. Our results are as follows. () For fixe an n, B (P n) = c()n + O(n 3 ), where c() is a constant epening on, which we etermine explicitly. () For fixe even n an, B (K n) = ( + o()) π n (n ). Our results exten recent results by Balogh et al. [5] who etermine B (P n) asymptotically as a function of n an B (K ) asymptotically as a function of. Introuction Let G = (V (G), E(G)) be a simple graph on n vertices. A labeling f is a bijection of V (G) to {,..., n}. When there is no ambiguity, we will simply write B(f) for B(f, G). The banwith of f is B(f, G) := max{ f(u) f(v) : uv E(G)}. The banwith B(G) of G is B(G) := min{b(f, G)}. f Dept. of Mathematics an Statistics, Miami University, Oxfor, OH 45056, USA, reza@calico.mth.muohio. eu. Dept. of Mathematics an Statistics, Miami University, Oxfor, OH 45056, USA, jiangt@muohio.eu. Research supporte in part by the National Security Agency uner Grant Number H Dept. of Mathematics an Statistics, Miami University, Oxfor, OH 45056, USA, millerz@muohio.eu

2 The notion was introuce by Harper in his influential paper [3] in which he etermine the banwith of the -imensional hypercube by solving the corresponing vertex isoperimetric problem in the hypercube. There are several motivations for stuying the banwith problem: sparse matrix computations, representing ata structures by linear arrays, VLSI layouts, mutual simulations of interconnection networks an minimizing the effects of noise in the multi-channel communication of ata (see [0,, 4, 7]). The banwith problem is NPhar an is inappromaximable by any multiplicative constant even for trees [8]. Banwiths are known only for a few families of graphs incluing hypercubes [3], multiimensional gris [8,, 3], complete trees [6], an various mesh-like graphs (see [6, 7, 0]). The ege-banwith was introuce by Hwang an Lagarias [8]. Here we label the eges instea of the vertices, an the banwith of an ege-labeling f of a graph G is B (f, G) := max{ f(uv) f(vw) : uv, vw E(G)}. In other wors, it is the maximum ifference of labels between a pair of incient eges. When there is no ambiguity, we will write B (f) for B(f, G). The ege-banwith of a graph G is B (G) := min B (f, G). f Naturally, B (G) = B(L(G)), where L(G) is the line graph of G. In [9], Jiang et al. reintrouce the notion of ege-banwith an stuie the relationship between B(G) an B (G). They etermine the ege-banwith of caterpillars, the complete graph K n, an the balance complete bipartite graph K n,n. A. Gupta [] pointe out that the inequality B(T ) B (T ) B(T ) for a tree T obtaine in [9] together with Unger s inappromixation result [8] for banwith imply that etermining the ege-banwith is also NP-har. Recently, there has been an increase of interest in the stuy of ege-banwith. Calamoneri et al [9] obtaine tight bouns on the ege-banwith of complete k-ary trees an bouns on the ege-banwith of the hypercube an butterfly graphs. Balogh et al [5] subsequently obtaine asymptotically tight bouns on the ege-banwith of two imensional gris an tori, the Cartesian prouct of two cliques an the hypercube. Sharpening the result of Balogh et al [5] on the two imensional gris an tori while confirming a conjecture of Calamoneri et al [9], Pikhurko an Wojciechowski [7] showe that the ege-banwith of an m by n gri, where m n, is n. They also showe that the ege-banwith of an m by n torus, where m n, is between 4n 5 an 4n. In an unpublishe manuscript, Akhtar, Jiang, an Pritikin have inepenently shown that the ege-banwith of an m by n gri, where m n, is between n an n, an that the ege-banwith of an m by n torus, where m n, is between 4n 5 an 4n. In this paper, we etermine the ege-banwith of the -imensional gris Pn asymptotically when is fixe an n, an we obtain lower an upper bouns on the egebanwith of the Hamming graph Kn. When n is a fixe positive even integer an our lower an upper bouns match asymptotically. The Cartesian prouct of graphs G an H, enote by G H, is the graph with vertex set V (G H) = {(u, v) : u V (G), v V (H)} specifie by putting (u, v) ajacent to (u, v ) if an only if () u = u an vv e(h), or () v = v an uu E(G). The -fol Cartesian prouct G G G is enote by G. Let P n enote a path on n vertices.

3 The -fol Cartesian prouct P n, enote by P n, is also known as the -imensional gri. Equivalently, we can view P n as a graph whose vertices are vectors u, u,..., u of length where i, u i {0,,..., n } an any two vertices x, x,..., x an y, y,..., y are ajacent if an only if there exists i {,,..., } such that x i y i = an x j = y j for all j i. Given a vertex x = x, x,..., x, we efine the weight of x by wt(x) = x +x + +x. For each r {,,..., n}, let L(n,, r) = {x V (P n) : wt(x) = r}. Let l(n,, r) = L(n,, r) an l (n, ) = max{l(n,, r) : 0 r (n )}. We etermine B (P n) asymptotically as a function of n. Theorem. Let be a fixe positive integer. We have B (P n) = ( + o())l (n, ) = c()n + O(n 3 ), ( )!. Further- where c() is a constant epening on, given by c() = Σ / j=0 ( ) j( j more, c(). ) ( j) A simple calculation shows that c() =. Hence B (Pn) = ( + o())n, which was obtaine by Balogh et al[5]. For another example, c(3) = 3, yieling 4 B (Pn) 3 = ( + o()) 9 4 n. Theorem. etermines asymptotically the ege-banwith of gris of arbitrary imension. The -fol Cartesian prouct of the complete graph K n, enote by Kn, is also calle the Hamming graph. We obtain lower an upper bouns on B (Kn) for fixe n as a function of. When n is even, our lower an upper bouns match asymptotically, yieling Theorem. Let n be a fixe positive even integer. We have B (Kn) = ( + o()) n (n ). π General bouns The stanar techniques for obtaining lower bouns on banwith use isoperimetric inequalities. Many vertex an ege isoperimetric problems have been consiere in the literature. Given a graph G an a set S V (G), let (S) = {v V (G) S : u S such that uv E(G)}. We call (S) the (vertex) bounary of S. In other wors, (S) = N G (S) S. Given an optimal numbering f of V (G), let S be the set of vertices receiving labels,,, k. Then the highest label assigne to a vertex in (S) is at least k + (S). Let v be the vertex with the highest label in (S). It has a neighbor u in S, whose label is at most k. So f(v) f(u) (S), which implies B(G) = B(f, G) (S). Similarly, consier a vertex x in (V S) with the smallest label. Its label is at most k (V S) +. It has a neighbor y in V S, whose assigne label is at least k +. So f(y) f(x) (V S). Thus, B(G) = B(f, G) (V S). Therefore, we have B(G) max{ (S), (V S) }. This yiels the following 3

4 Proposition. [3] Let G be a graph an k an integer, where 0 k V (G). Then B(G) min max{ (S), (V S) }. S V (G), S =k For each k, let L k (G) = min S V (G), S =k (S). By Proposition., B(G) L k (G). Since this hols for each k with 0 k V (G), we have B(G) max k L k (G). This lower boun max k L k (G) for B(G) is often referre to as the Harper boun. In general, the Harper boun nees not be sharp an calculating it is ifficult (NP-har). When the Harper boun is not very useful, it is sometimes useful to consier the iterate bounary (shaow) instea. Given a nonnegative integer q, let ( q) (S) = {v V (G) S : v is at istance at most q from some vertex in S}. Hence, in particular, (S) = ( ) (S). Consier an optimal numbering f of V (G). Let S be the set of vertices receiving labels,,..., k. Let q be any integer such that q n. Let v be a vertex in q (S) with the highest label. Then f(v) k + q (S). Vertex v is at istance at most q from some vertex u in S. Note that f(u) k. Hence for some ege on a shortest u, v-path, the ifference between the f-labels of its two enpoints is at least f(u) f(v) /q q (S) /q. This yiels Proposition. Let G be a graph an k an integer, where 0 k V (G). Then B(G) min max S V (G), S =k q q n ( q) (S) Our iscussions above apply similarly to a set of eges. We efine the bounary of a set S E(G) of eges in G by (S ) = {e E(G) S : e S such that e an e are incient}. The iterate bounary for S is then given, for q, by ( q) (S ) = {e E(G) S : e is at istance at most q from some ege of S }. We see that (S ) = ( ) (S) an ( q) (S ) = ( q ) (S) ( ( q ) (S )). The ege analogues of Propositions. an. are then obtaine by replacing B(G) by B (G) an V (G) by E(G).. 3 The weight function in multiimensional gris Recall that V (P n) = { x, x,..., x : x i {0,,..., n } for all i =,,..., }. Two vertices x, x,..., x an y, y,..., y are ajacent if they iffer by in one coorinate an agree in all other coorinates. Again, the weight wt(x) of a vertex x = x, x,..., x is efine by wt(x) = x + x + + x. Given positive integers n, an an integer r with 0 r (n ), let L(n,, r) = {x V (P n) : wt(x) = r}. Let l(n,, r) = L(n,, r). Let l (n, ) = max{l(n,, r) : 0 r (n )}. It is easy to see that l(n,, r) is the number of integer solutions to the equation x + x + + x = r, where 0 x i n for each i []. By consiering the value of x one can easily erive the following recurrence relation on l(n,, r), which also appeare in [6]. 4

5 Proposition 3. Let n, be positive integers an r an integer. We have l(n,, r) = if 0 r n an l(n,, r) = 0 otherwise. For all an r with an 0 r (n ), l(n,, r) = Σ n j=0 l(n,, r j), an for other values of r we have l(n,, r) = 0. It is straightforwar to erive an exact formula for l(n,, r) using a generating function. This was one in [6], we inclue a short proof for completeness. Proposition 3. Let n,, r be positive integers. We have ( )( ) l(n,, r) = Σ r j=0( ) r jn + j. j Proof. By prior iscussion, l(n,, r) is the number of integer solutions to x +x + +x = r with 0 x i n for each i []. For fixe n,, the generating function for l(n,, r) is ( x g(x) = ( + + x n ) n ) = = ( x n ) ( x x ). As l(n,, r) equals coefficient of x r in the above expansion, we have ( )( ) ( )( ) r jn + l(n,, r) = Σ j=0( ) j = Σ r r jn + j j=0( ) j. j The function l(n,, r) is well-stuie in the theory of posets as the rank number in the poset of ivisors of a number. Consier a positive integer m with prime factorization m = p k p k... p k, where the p i are istinct primes an k i for each i. The rank of m is K = k + k an N r (m) is the number of ivisors of m of rank r. It is easy to see that in the case k = k = = k = n, N r (m) is precisely l(n,, r). Chapter 4 of Anerson [3] gives a etaile iscussion about the function N r (m). In particular, we have Proposition 3.3 ([3] Chapter 4) Let K = (n ). Then. l(n,, i) = l(n,, K i) for each i with 0 i K.. For fixe n an, l(n,, i) is strictly increasing in i for i K an strictly ecreasing in i for i K. By Proposition 3.3, for fixe n an, l(n,, r) is a symmetric an unimoal function of r on {0,,..., (n )} with maximum value at r = (n ) an at r = (n ). Thus, l (n, ) = l(n,, (n ) ). Setting r = (n ) in the formula in Proposition 3. for l(n,, r) then gives us a formula for l (n, ). When is fixe an n tens to infinity, the leaing term is a multiple of n whose leaing coefficient can be expresse exactly as a sum. We then give a close form estimate of this leaing coefficient using earlier results of Anerson [] given below. 5

6 Theorem 3.4 ([]) Let k, k,, k be nonnegative integers. Let K = k + + k. Let s enote the number of integer solutions to the equation x + x + + x = K, where i 0 x i k i. Let A = 3 Σ i=k i (k i + ) an τ = Π i=( + k i ). Then there exist positive constants C an C such that τ τ s C. A A C Furthermore, we can take C = an for any small ɛ > 0 we can take C = 3 ɛ when K is sufficiently large. Corollary 3.5 Let n, be positive integers, where n. There exist positive constants C an C such that n C l n (n, ) C. Furthermore, we can take C = an for any small ɛ > 0 we can take C = ɛ when n is sufficiently large. Proof. By our earlier iscussion, l (n, ) = l(n,, (n ) ) is the number of integer solutions to the equation x + x + + x = (n ), where 0 x i n for each i. We apply Theorem 3.4. Here, k i = n, for each i []. So A = 3 (n ) an τ = n. Also, s = l (n, ). The claim follows. We now give the exact formula for l (n, ) together with an asymptotic formula for it. There is an exact formula for the coefficient of the leaing term in the form of a summation; this coefficient can be boune using Corollary 3.5. Before we procee, we nee the following routine estimation of binomial coefficients. Recall that if x is a real number an k is an integer then ( ) x k := x(x ) (x k+). Also, it is straightforwar to check that e x x for all reals k! x an x e x for x with 0 < x <. Lemma 3.6 Let k be a positive integer an N, m nonnegative real numbers such that N > max{k, mk}. Then N k k ( ) ( ) N+m k! N k N k mk ( + ). So, ) k! N ( N+m k N k (m + )N k. k! Proof. For the lower boun, we have ( ) N+m k / N k > (N+m k)k / N k > (N k)k / N k = ( k k! k! k! k! k! N )k. Since 0 < k <, ( k N N )k e k N k k ). N / N k For the upper boun, we have ( N+m k ( + x )x < e when x > 0, we have ( + m) N m < e an hence ( + m N N )k < e mk k! < (N+m)k k! / N k k! = ( N+m N )k = ( + m N )k. Since N. Since mk < N an e x < + x when 0 < x <, we have e mk N < + mk. Therefore, ( ) N+m N k / N k < + mk. k! N The last statement follows reaily from the lower an upper bouns. 6

7 Theorem 3.7 Let n, be positive integers. Then l (n, ) = Σ j=0( ) j ( )( ( j)n j + We have l (n, ) =, l (n, ) =. For fixe 3, as n, we have l (n, ) = c()n + O(n ), where c() is a constant epening on, given by Also, c(). ( ) c() = Σ / ( j) j=0( ) j j ( )!. Proof. The exact formula for l (n, ) given above is obtaine by setting r = (n ) in the formula for l(n,, r) in Proposition 3.. For fixe, we erive an asymptotic formula for l (n, ) as a function of n. Fix j with j > 0. Let N = ( j)n an m = ( ( j)n + ) N. Then m /. For sufficiently large n, since is fixe, we have N > max{( ), m( )}. Applying Lemma 3.6, we have ( ( j)n + ) ( j) ( )! n (m + )n ( + 4)n. ). So, l = Σ j=0( ) j ( j = Σ / j=0 ( ) j )( (n jn ( ) j + ) ( j) ( )! n + O(n ). Let c() = Σ / j=0 ( ) j( ) ( j). We have j ( )! l (n, ) = c()n + O(n ). Note that c() is a constant epening only on an by Corollary 3.5, c(). Remark 3.8 When n is fixe an, using Laplace s metho one can show that l (n, ) = 6 π ( n n ) n + o(n ). See the concluing remarks section for further iscussion. Next, we show that for all r relatively close to (n ) an (n ), l(n,, r) is close to the maximum value l (n, ). This property is important to establishing our lower boun on the ege-banwith of Pn in the next section. 7

8 Lemma 3.9 Let n,, r be integers such that n,. If r (n ) = t, where 0 t n, then l(n,, r) ( t n )l (n, ). Proof. Let M = (n ). By Proposition 3.3, l (n, ) = l(n,, M). For convenience, let f(j) = l(n,, j). By Proposition 3.3, f(j) is symmetric on [0, (n )( )] an it is unimoal with peak(s) at M = (n )( ) an M = (n )( ). By Proposition 3., l (n, ) = l(n,, M) = Σ M i=m (n )f(i). Consier the interval of integers I = [M (n ), M]. Assume first that n is o. In this case M = M = (n )( ) is the only element at the center of I. The function f(j) is symmetric an unimoal on I with a single peak at (n )( ). Let A enote the sum of f(j) over the first t elements of I, B the sum of f(j) over the last t elements of I, an C the sum of f(j) over the mile n t elements of I. The symmetry an unimoality of f(j) imply that A = B an C n t. So, l (n, ) = A + B + C ( + n t )A, which implies A t t A = B t n l (n, ). Now, by Proposition 3., l(n,, M + t) is the sum of f(j) over I + t, the translate of I by t. So clearly, l(n,, M + t) l(n,, M) A ( t n )l (n, ). Similarly, l(n,, M t) l(n,, M) B ( t n )l (n, ). When n is even, we consier the subcases epening on whether is even or o. In each subcase, similar analysis shows that A, B ( t n )l (n, ) an hence l(n,, M + t) ( t n )l (n, ) an l(n,, M t) ( t n )l (n, ). 4 The ege-banwith of a multiimensional gri Bollobás an Leaer [8] solve the vertex isoperimetric problem in gris. We will use their result to obtain asymptotically tight bouns on the ege-banwith of a multiimensional gris. Our result extens that of Balogh et al. on -imensional gris to gris of any imension. Definition 4. The simplicial orer on V (P n) is efine by x < y if either wt(x) < wt(y) or wt(x) = wt(y) an x s > y s, where s = min{t : x t y t }. Bollobás an Leaer [8] showe that for any k, with 0 k V (P n), the initial segment of length k in the simplicial orer has the smallest bounary among all sets of k vertices. For each r with 0 r (n ), let B(n,, r) = {x V (P n) : wt(x) r}. In other wors, B(n,, r) = r j=0 L(n,, j). Note that each ege in P n has one enpoint in L(n,, r) an the other enpoint in L(n,, r + ) for some r. Theorem 4. (Vertex isoperimetric inequality in the gri) [8] Let A V (P n) an let C be the initial segment of length A in the simplicial orer on V (P n). Then N(A) N(C). In particular, if A B(n,, r) then N(A) B(n,, r + ). Also, for all q we have N ( q) (A) B(n,, r + q). 8

9 Definition 4.3 Let G be a graph on n vertices. Let σ : x < x < < x n be a linear orer on V (G). The labeling f of V (G) satisfying f(x i ) = i is the vertex labeling of V (G) inuce by σ. Lemma 4.4 Let G = P n. Let f be the labeling of V (G) inuce by the simplicial orer on G. For every uv E(G), we have f(u) f(v) l (n, ) + l (n, ). Proof. Without loss of generality, suppose wt(u) = r an wt(v) = r +. Let A = {x L(n,, r) : f(x) > f(u)}, B = {x L(n,, r + ) : f(x) < f(v)} an C = {x L(n,, r + ) : f(x) > f(v)}. Let a = A, b = B, c = C. Then f(u) f(v) = a + b + an b + c + = l(n,, r + ). Suppose u = u, u,..., u an v = v, v,..., v. Since uv E(G) an wt(v) = wt(u) +, there exists j [] such that v j = u j + an v i = u i for all i [] j. By our efinition of A, vertices in A all have their first coorinate at most u. Let A = {x A : x = u or u } an let A = {x A : x u }. We have A = A A. Since there are at most l (n, ) vertices other than u that have u in the first coorinate an there are at most l (n, ) vertices that have u in the first coorinate, A l (n, ). For each x A, let g(x) = x +, x,..., x. It is easy to see that g is an injection of A into L(n,, r + ). Furthermore, for each x A, since x + < u v, we have f(g(x)) > f(v) by the efinition of the simplicial orer. So g(x) C. So, g is an injection of A into C. Thus, A c an a = A = A + A l (n, ) + c. Now, we have f(u) f(v) = a + b + l (n, ) + c + b + = l (n, ) + l(n,, r + ) l (n, ) + l (n, ). Theorem 4.5 Let be a fixe positive integer. For all positive integers n we have B (P n) [l (n, ) + l (n, )] = c()n + O(n ), where c() is efine as in Theorem 3.7. Proof. First, we efine a igraph H from G = Pn by orienting each ege xy from x to y if x < y in the simplicial orer. For each vertex x, let E + (x) enote the set of out-eges from x. Note that E + (x) for each x. We efine a labeling g of E(H) (an of E(G)) using,,..., E(H) as follows. Suppose the vertices are u, u,... where u < u < in the simplicial orer. Starting with we assign the first E + (u ) consecutive labels to E + (u ), then the next E + (u ) consecutive labels to E + (u ), an so on. Let e = u i u j an e = u j u k be two incient eges in G at the vertex u j. We consier three cases epening on how e an e are oriente. Case. i < j < k or i > j > k. 9

10 By symmetry, we may assume i < j < k. Then we have wt(u i ) = r, wt(u j ) = r, wt(u k ) = r + for some r. By Lemma 4.4, j i l (n, ) + l (n, ). Note that e E + (u i ) an e E + (u j ). By our efinition of g, we have g(e ) g(e) j t=i E + (u t ) (j i + ) (l (n, ) + l (n, )). Case. i < j an j > k. In this case, we have wt(u i ) = wt(u k ) = r an wt(u j ) = r for some r. In particular, k i l(n,, r ) l (n, ). Also, e E + (u i ) an e E + (u k ). Without loss of generality, suppose i < k. By our efinition of g, we have g(e ) g(e) k t=i E + (u t ) k i + l (n, ) (l (n, ) + l (n, )). Case 3. i > j an k > j. In this case, e, e E + (u j ), an g(e) g(e ) < (l (n, ) + l (n, )). We have shown that g(e) g(e ) (l (n, ) + l (n, )) for every pair of incient eges e an e in G. This yiels B (G) B (g, G) (l (n, ) + l (n, )). Using l (n, ) = c()n + O(n ), we get B (G) c()n + O(n ). We now erive a lower boun on B (Pn) that matches the upper boun in Theorem 4.5 asymptotically when is fixe an n. Our proof is base on the metho use by Calamoneri et al. [9] an Balogh et al. [5]. We nee an easy lemma. Lemma 4.6 Let n an be positive integers. Let A V (P n). Then there are at least A n eges incient to A in P n. Also, E(P n) = n n. Proof. The graph Pn is a spanning subgraph of Cn, the -fol cartesian prouct of C n. We can think of obtaining Cn from Pn by aing eges of the form uv, where u = u, u,..., u an v = v, v,..., v satisfy that u i = 0, v i = n or u i = n, v i = 0 for some i [] an u j = v j for all j [] {i}. It is easy to see that there are n such eges. Since Cn is -regular, we have E(Pn) = E(Cn) n = n n. In Cn, since each vertex has egree, there are at least A / = A eges incient to A. So in Pn, there are at least A n eges incient to A. Theorem 4.7 Let be a fixe positive integer. Then we have as n B (P n) l (n, )( o()) = c()n + Ω(n 3 ). Proof. Throughout the proof, whenever necessary, we assume that n is sufficiently large. Let g be an ege-labeling of G = Pn with B (g, G) = B (G). Let S enote the set of eges receiving labels,..., E(G) /. We color the eges in S re an the rest of the eges white. Let us call a vertex re if all of its incient eges are re, a vertex white if all of its incient eges are white, a vertex mixe if it is incient to both re eges an white eges. Let R enote the set of re vertices, W the set of white vertices, an M the set of mixe vertices. We consier two cases. For convenience, let l = l (n, ). Case. M 5l. 0

11 Call a vertex v ba if (v) 3. Let D enote the set of ba vertices in G. If v = v, v,..., v is ba, then it has a 0 or n in at least three of the coorinates. So D ( ) 3 3 n 3 < 3 n 3. By Corollary 3.5, when n is sufficiently large, we have l n > 3 n 3 > D. Hence, M D 5l l = 4l. Each vertex in M D has egree at least. So the total number of eges incient to M is at least ( ) M D / = ( ) M D 4( )l l, since. Let E(M) enote the set of eges in G incient to M an let E = E(G). We have E(M) l. So, either E(M) (E S) l or E(M) S l. Note that E(M) (E S) (S) an E(M) S (E S). Hence, we have either (S) l or (E S) l. By Proposition., we have B (g, G) l = c()n + O(n ). Case. M < 5l. We have R + M + W = n. Hence, either R < n / or W < n /. Without loss of generality, we may assume that R < n /; otherwise we switch S with E S an hence R with W. Let H = G[R M] enote the subgraph of G inuce by R an M. Note that S E(H). Hence, S E(H) ( R + M )/ = ( R + M ). On the other han, S = E(G) / (/)(n n ). So, we have ( R + M ) S n n. From this we get So, R + M n n. R n n M n n 5l. By Corollary 3.5, for sufficiently large n we have l n. Hence, n l an R n l 5l n ( + 5)l. We have n ( + 5)l < R < n. () Since l(n,, j) is symmetric on [0, (n )] an Σ (n ) j=0 l(n,, j) = V (G) = n, we have Σ (n ) j=0 l(n,, j) n Σ (n ) j=0 l(n,, j). () For convenience, let M = (n ). By Lemma 3.9, l(n,, M t) ( t n )l, for t [0, n ]. Let n be sufficiently large so that n > ( ) +7. We have ( ) Σ +6 t= l(n,, M t) Σ +6 t= ( t +7 n )l = ( + 6)l n l ( + 5)l. (3) By () an (3), we have Σ M 7 j=0 l(n,, j) R Σ M j=0l(n,, j). In other wors, B(n,, M 7) < R < B(n,, M + ). (4)

12 Therefore, we have B(n,, r) R < B(n,, r + ) for some r [M 7, M + ]. Let q = n. We may assume that n is sufficiently large so that q + 7. Let A be a subset of R with A = B(n,, r). By Theorem 4., N ( q) (A) B(n,, r + q). Since N ( q) (A) N ( q) (R), we have N ( q) (R) B(n,, r + q). Hence, ( q) (R) = N ( q) (R) R B(n,, r + q) B(n,, r + ) = Σ q j=l(n,, r + j). (5) For each j [r +, r + q], we have j [M 7, M + q + ]. Since j M q +, by Lemma 3.9, l(n,, j) ( q + n )l ( n n )l ( )l. n Hence, ( q) (R) Σ q j=l(n,, r + j) (q )( n )l. Let E( ( q) (R)) enote the set of eges incient to ( q) (R). By Lemma 4.6, E( ( q) (R)) ( q) (R) n. Finally, note that E( ( q) (R)) ( q+) (S). Applying Proposition. to the line graph L(G), we have B (g, G) ( q+) (S) /(q + ) E( ( q) (R)) /(q + ) ( (q) (R) n ) /(q + ) [ (q )( ] ) l /(q + ) n /(q + ) n ( ) ( n = l ) + O(n 3 ) (using q = n ) n n = c()n + O(n 3 ). (by Theorem 3.7) 5 Ege-banwith of the Hamming graph I: upper boun In this section we erive an upper boun on B (K n). We view any vertex x in K n as an n-ary -string x = x, x,..., x, where i x i {0,,..., n }. Two vertices x = x, x,..., x an y = y, y,..., y are ajacent in K n if an only if the two strings iffer in precisely one coorinate. As in Section 4, for each x = x, x,..., x, we efine the weight of x as wt(x) = x + x x. Note that when n =, K n is just the -imensional hypercube Q an for each vertex x, wt(x) is the number of s in the binary string that represents x. The ege-banwith B (Q ) was asymptotically etermine by Balogh, Mubayi, an Pluhár

13 [5] while the vertex banwith B(Q ) was completely etermine by Harper in his paper [3]. We will combine the labelings use in these results to esign a labeling that yiels an upper boun on B (K n). Let us recall the labelings use in [5] an [3]. Definition 5. The vertex-hales numbering of V (Q ) is a bijection h : V (Q ) {,,..., } such that h(x) < h(y) if either wt(x) < wt(y) or wt(x) = wt(y) an x s s = min{t : x t y t }. > y s, where Note that in Section 4, we calle a similar orering on V (P n) the simplicial orer. Harper showe that the vertex-hales numbering achieves the vertex banwith for Q. Theorem 5. [3] Let h be the vertex-hales numbering of Q. Then ( ) ( ) k B(Q ) = B(h) = Σ k=0 = ( + o()). k/ / The last equality use stanar estimates of binomial coefficients. Next we escribe the labeling use by Balogh, Mubayi, an Pluhár in establishing an asymptotically tight upper boun on B (Q ). For convenience, we will call this numbering the ege-hales numbering. Our escription of the numbering is equivalent to the one use in [5]. Definition 5.3 [5] The ege-hales numbering of Q is a bijection f : E(Q ) {,,..., } such that for any two eges vw an xy where wt(w) = wt(v) + an wt(y) = wt(x) + we have f(vw) < f(xy) if either () h(v) < h(x) or () v = x an h(w) < h(y). Theorem 5.4 [5] Let f enote the ege-hales labeling of Q. Then ( ) ( ) ( ) ( ) + o() B (Q ) B (f) / + o(). / Now, we combine the two numberings mentione above to obtain a total numbering on V (Q ) E(Q ), which we will call the mixe-hales numbering of Q. This numbering is prouce by the following algorithm. Algorithm 5.5 (The mixe-hales numbering m of Q ) Input: The -imensional hypercube Q. Output: A bijection m: V (Q ) E(Q ) {,,..., + }. Initialization: () Denote the eges of Q by e i, i, accoring to the ege-hales numbering f of Q. That is, e i is the ege e with f(e) = i. () Let m(0 ) =, where 0 enotes the all 0 string of length. (3) For all i, i, let m(e i ) = i +. (4) Set i =. 3

14 Iteration: (5) Suppose e i = xy, where wt(y) = wt(x) +. If m(y) is not yet efine, then (5a) Let m(y) = m(e i ) + ; (5b) For all j > i, let m(e j ) = m(e j ) +. (6) Let i = i +. (7) If i = +, terminate; otherwise go to step (5) Figure. The mixe-hales numbering of Q 3. 8 Intuitively speaking, to obtain the mixe-hales numbering we process the eges one by one in increasing orer of ege-hales label. The algorithm gives a vertex y an m-label at the earliest opportunity, as soon as we process the first ege in the ege-hales numbering incient to y. See Figure for the mixe-hales labeling of Q 3. Next, we summarize some useful facts about mixe-hales numbering in the following proposition. In particular, we see that the orering on V (Q ) an the orering on E(Q ) inherite from the mixe-hales numbering m of Q are precisely the vertex-hales numbering h an the ege-hales numbering f of Q, respectively. Proposition 5.6 Let h, f, m enote the vertex-hales, ege-hales, an mixe-hales numberings of Q, respectively. Let 0 enote the all 0 string of length an the all string of length. For each vertex x V (Q ) {0, }, let x enote the neighbor of x with smallest h label an x + the neighbor of x with largest h label. Then. For each vertex x V (Q ) {0, }, the string representing x is obtaine from the string for x by flipping the rightmost to a 0 an the string representing x + is obtaine from the string for x by flipping the rightmost 0 to a.. For each vertex x V (Q ) {0, }, among the eges incient to x, xx has the smallest m label an xx + has the largest m label. Hence, in particular, m(x) = m(xx ) +. 4

15 3. For any two eges e, e in Q, if f(e) < f(e ) then m(e) < m(e ). 4. For any two eges vw, xy in Q, where wt(w) = wt(v) + an wt(y) = wt(x) +, m(vw) < m(xy) if an only if either h(v) < h(x) or v = x an h(w) < h(y). 5. For any two vertices x, y V (Q ) {0, }, if h(x) < h(y), then h(x ) h(y ) an h(x + ) h(y + ). 6. For any two vertices, x, y in Q, if h(x) < h(y) then m(x) < m(y). Proof. Part follows immeiately from the efinition of the vertex-hales numbering (Definition 5.). Parts an 3 follow immeiately from Algorithm 5.5. Part 4 follows from the efinition of the ege-hales numbering f (Definition 5.3) an Part 3. To prove Part 5, suppose h(x) < h(y). Since h(x) < h(y), we have wt(x) wt(y). If wt(x) < wt(y), then wt(x ) < wt(y ) an h(x ) < h(y ) hol trivially. So we may assume that wt(x) = wt(y). By Part, the string representing x is obtaine from the string representing x by flipping the rightmost in x to 0 an the string for y is obtaine from the string for y by flipping the rightmost in y to 0. Let j enote the smallest coorinate in which x an y iffer. Since h(x) < h(y), we have x j = an y j = 0. Since wt(x) = wt(y), if x has k many s in coorinates j +, j +,..., then y shoul have exactly k + many s in coorinates j +, j +,...,. If k, then clearly h(x ) < h(y ). If k = 0, then x = y an hence h(x ) = h(y ). By a very similar argument, we have h(x + ) h(y + ). To prove Part 6, we may assume without loss of generality that x, y / {0, }. Suppose h(x) < h(y). Since m(x) = m(xx ) + an m(y) = m(yy ) +, to prove m(x) < m(y) it suffices to prove that m(xx ) < m(yy ). By Part 5, h(x ) h(y ). Thus, we have either h(x ) < h(y ) or x = y an h(x) < h(y). By Part 4, we have m(xx ) < m(yy ). This completes the proof. In the next proposition, we boun the number of vertices an the number of eges whose m-labels lie between the m-labels of two incient eges. Proposition 5.7 Let h, f, m enote the vertex-hales, ege-hales, an mixe-hales numberings of Q, respectively. Let e, e be two incient eges in Q, where m(e) < m(e ). Then there are at most B(h) + vertices z with m(e) m(z) m(e ) an there are at most B (f) + eges e with m(e) m(e ) m(e ). Proof. By Proposition 5.6 part, an ege e satisfies m(e) m(e ) m(e ) if an only if f(e) f(e ) f(e ). Hence, since e an e are incient, there are at most f(e ) f(e) + B (f) + such eges e. Next, suppose e an e are both incient to x. Let z be a vertex with m(e) m(z) m(e ). If x = 0 or, then it is easy to see that there are at most < B(h) such vertices z. Hence, we may assume that x / {0, }. By Proposition 5.6 Part, m(xx ) m(z) m(xx + ). Since m(x) = m(xx ) + an m(z) > m(xx ), we have m(z) m(x). We show that also m(z) m(x + ). Suppose first that h(z ) > h(x). Then by Proposition 5.6 Part 4, m(zz ) > m(xx + ), an hence m(z) = m(zz ) + > m(xx + ), a contraiction. So, we must have h(z ) h(x). By Proposition 5.6 Part 5, h((z ) + ) h(x + ). 5

16 Since h(z) h((z ) + ), we have h(z) h(x + ) an thus m(z) m(x + ). So, any vertex z satisfying m(e) m(z) m(e ) must satisfy h(x) h(z) h(x + ). There are at most h(x + ) h(x) + B(h) + such vertices z. Lemma 5.8 Let p, q, t be positive integers. Let G be a graph obtaine from Q by replacing each vertex x of Q by a t-vertex graph G x having p eges an each ege xy of Q by a set of q cross eges between V (G x ) an V (G y ). Then B (G) p(b(h) + ) + q(b (f) + ), where h, f enote the vertex-hales an ege-hales numberings of Q, respectively. Proof. Apply Algorithm 5.5 to obtain the mixe Hales numbering m of Q. List elements of V (Q ) E(Q ) in the orer etermine by m, call this list L. We prouce a labeling g of E(G) as follows. Start with label. As we scan L, each time we encounter a vertex x, we allocate the next p consecutive labels to the eges of G x, an each time we encounter an ege e = xy, we allocate the next q consecutive labels to the set of q cross eges between V (G x ) an V (G y ). Consier any pair of incient eges e an e in G with g(e) < g(e ). Suppose both are incient to vertex w, an w lies in G x. If x {0, }, then it is easy to see that g(e) g(e ) p + q < p(b(h) + ) + q(b (f) + ). Hence we may assume that x / {0, }. By our labeling scheme g(e) g(e ) is maximum when e is among the set of q eges of G associate with ege xx in Q an e is among the set of q eges of G associate with ege xx + in Q. By Proposition 5.7 an the efinition of g, g(e) g(e ) p(b(h) + ) + q(b (f) + ). Now, we apply Lemma 5.8 to get an upper boun on B (Kn). For convenience, we consier only even n. For o n, we can upper boun B (Kn) by B (Kn+). We can view Kn as being obtaine from Q by replacing each vertex of Q with a copy of Kn/ an replacing each ege of Q by the set of eges between two neighboring copies of Kn/ in K n. More specifically, for each x = x,..., x V (Q ), let O(x) enote the subgraph of Kn inuce by the set of vertices {w = w,..., w : 0 w i n/ if x i = 0, n/ w i n if x i = }. Then each O(x) is a copy of Kn/. For each ege xy E(Q ), let E(O(x), O(y)) enote the set of eges in Kn having on enpoint in O(x) an the other enpoint in O(y). It is easy to see that E(O(x), O(y)) = ( n ) ( n ) = ( n )+ for all xy E(Q ). We enote this quantity by q(n, ). Applying Lemma 5.8 with p = e(kn/ ) an q = q(n, ), we have the following. Lemma 5.9 Let be a positive integer an n a positive even integer. We have B (K n) e(k n)(b(h) + ) + q(n, )(B (f) + ). Theorem 5.0 Let n be a fixe positive even integer. Let be a positive integer. We have as. B (K n) ( + o()) ( / ) (n ) (n ) = ( + o()) π n (n ), 6

17 Proof. Using Lemma 5.9, Theorem 5., an Theorem 5.4, e(kn/ ) = ( ) n ) ( n/, q(n, ) = ) ( ) ( n )+, an Σk=0 = ( + o()) = ( + o()) π, we have B (K n) ( k k/ / [ ( ) ( )] [ ( )] n ( ) [( ) ( )] n/ k n + + Σ k=0 + k/ + o() / ( ) [ ( ) ( )] n n/ = ( + o()) + ( + o()) ( ) ( ) n + / / ( ) ( ) (( ) n n/ = ( + o()) + ( ) ) n / ( ) ( ) + = ( + o()) n (n ) / = ( + o()) n (n ). π On a historic note, the iea of blowing up each vertex of Q to an orthant subgraph O(x) of Kn was use by Harper in [5] in his constructive upper boun for B(Kn). It was inepenently iscovere by one of us while supervising an M.A. thesis [4]. To evelop a constructive upper boun on B (Kn), we use this iea by labeling eges internal to orthants in blocks in the orer of the vertex-hales numbering of Q which achieve B(Q ) [3]. Our next step was to label cross eges between neighboring orthants in blocks in the orer of the ege-hales labeling of Q achieving B (Q ) asymptotically, escribe earlier an originating in [5]. In this way both classes of eges, either internal to orthants or crossing between orthants, are labele. The final step was to merge these two labelings efficiently. The key iea here is to notice that the ege-hales labeling of Q is in some natural sense inuce by the vertex-hales numbering. 6 Ege-banwith of the Hamming graph II: lower boun In this section we establish a lower boun for B (Kn) which matches the upper boun of the previous section asymptotically when n is even. Our technique employs a theorem of Harper giving a solution to the isoperimetric problem in [0, ]. The approach is to look at Kn geometrically as a -imensional box having sie length an so containing n many -imensional cells of length /n in each imension. More specifically, we consier the following mapping from V (Kn) to [0, ]. For each i {0,,..., n }, let I i enote the interval [ i, i+] of real numbers. For each vertex x = x n n, x,, x in Kn, where each x i {0,,..., n }, let g(x) = I x I x I x. Thus g(x) is a -imensional cell of length /n in each imension. Uner this mapping, a subset W of k vertices in V (Kn) then correspons to a collection S of k of these cells. S is a set of measure k/n in [0, ], the -fol prouct of the unit interval, equippe with the Lebesque measure. Given two points x = x, x,..., x an y = y, y,..., y in [0, ], we say that x an y are neighbors if there exists k {,,..., } such that x k y k but x i = y i for all i k. We 7

18 write x y if x an y are neighbors. Given a measurable subset S of [0, ], we efine the shaow Φ(S) to be Φ(S) = {y [0, ] S : x S x y}. We efine iterate shaows of S recursively as following. Let Φ () (S) = Φ(S), an inuctively efine Φ (l+) (S) = Φ(Φ (l) (S)). For a measurable subset S of [0, ], let S enote its measure. The following proposition is clear from our efinitions an iscussions above. Proposition 6. Let g be the mapping from V (K n) to [0, ] efine above. Let W be a subset of V (K n) an l a positive integer with l. We have g( ( l) (W )) = Φ ( l) (g(w )). Hence, ( l) (W ) = Φ ( l) (g(w ) n. To establish a lower boun on B (K n) we will nee to establish a lower boun on (W ) for any subset W of V (K n) of a given size k. By Proposition 6., it suffices to consier the problem of minimizing Φ(S) over all measurable subsets S of [0, ] of a given measure. In [4], Harper showe that the problem of minimizing Φ(S) over all measurable subsets S of [0, ] of a given measure reuces to that of minimizing Φ(S) over all suitably compresse such subsets. He then showe using variational methos that for any given v, 0 v, the smallest value of Φ(S) over all such subsets of measure v is achieve by a Hamming ball, which we efine as follows. Definition 6. Let t be real number with 0 t. For any binary -tuple x of Q, let K(x, t) be the subset of [0, ] efine by K(x, t) = {y [0, ] : 0 y i t if x i =, an t < y i if x i = 0, i }. Define the subset HB(, k, t) of [0, ] calle a Hamming ball by HB(, k, t) = wt(x) k K(x, t). Note that if x has weight i, then ( ) the measure of K(x, t) is t i ( t) i. So the measure of HB(, k, t) is v = v(, k, t) = Σ k i=0 i t i ( t) i an the size of its shaow is Φ(HB(, k, t)) = ) t k+ ( t) k. For convenience, we use α(, m, t) to enote ( ) t m ( t) m. Then ( k+ Φ(HB(, k, t)) = α(, k +, t). We can now state Harper s theorem. Theorem 6.3 ([4] Theorem ) For any given v with 0 v, the minimum of Φ(S) over all subsets S of [0, ] of measure v is achieve by some HB(, k, t) for suitable k an t. To make effective use of Theorem 6.3, we nee to analyze Φ(HB(, k, t)) = α(, k +, t). Observe that if X( is ) a ranom variable rawn from the binomial istribution BIN(, t), then v(, k, t) = Σ k i=0 i t i ( t) i = P r(x k) an α(, k +, t) = ( ) k+ t k+ ( t) k = P r(x = k + ). 8 m

19 Our general approach is again the one use in [9], [5], an the proof of Theorem 4.7. Given an optimal ege labeling of E(Kn), we consier the set S E(Kn) of size about E(Kn) / receiving the smallest labels. We then lower boun (S ) or ( q) (S) for an aequate q. By an argument similar to the one use the proof of Theorem 4.7, lower bouning (S ) is reuce to lower bouning (S) for a corresponing set S V (Kn) with S near n. This then reuces to lower bouning Φ(S) where S [0, ], with S having measure near. By Theorem 6.3, we nee to estimate Φ(HB(, k, t)) = α(, k +, t) when v(, k, t) is near. In light of our earlier observation, this means lower bouning P r(x = k + ) when P r(x k) is close to, where X is a ranom variable rawn from BIN(, t). One coul in principle obtain such a lower boun by approximating BIN(, t) using a normal istribution. But the error analysis in such an approximation is quite involve. Further, when the expecte value t is either too close to 0 or too close to, a normal istribution approximation becomes infeasible. Here we use a self-containe an completely combinatorial approach to obtain our estimates. We think our approach is of inepenent interest. We nee a lemma from []. Lemma 6.4 ([] Lemma B.7.) Let n be a positive integer an p a real number such that 0 p. Let X be a ranom variable rawn from the binomial istribution B(n, p) (where n is the number of inepenent trials an p is the probability of success of each trial). Then P r(x np ) / P r(x np ). Lemma 6.4 suggests that if P r(x k) is close to then k is close to the expecte value np. Thus, we nee to lower boun P r(x = k + ) for those k close to np. Lemma 6.5 If z be a real number with 0 < z <, then z e z. Let x, y, a be positive real numbers such that x, y a. Then ( + a x )x ( a y )y e a x a y. Proof. For any real number w with 0 < w, we have ln( + w) = w w + w3 3 w4 4 + w w > w w, ln( w) = w w w3 3 w4 4 w w. If 0 < z <, then ln( z) z z z. So, z e z. Since x, y a, 0 < a x, a y. Let K = ( + a x )x ( a y )y. It suffices to show that ln K a a x y. Inee, we have ln K = x ln( + a x ) + y ln( a y ) x(a x a x ) + y( a y a y ) = a x a y. Our next lemma says that for if X is a ranom variable rawn from BIN(n, p) where np is not too small or too large an k is near np then P r(x = k) is lower boune by ( o()) πn as n. We postpone its somewhat technical proof to the Appenix. 9

20 Lemma 6.6 Let n be a positive integer. Let p be a real number such that 0 < p <. Suppose 5 ln n np n 5 ln n. Let k be a nonegative integer. Let X be a ranom variable rawn from BIN(n, p). If k np ln n, then P r(x = k) πn e n when n is sufficiently large. Now we use Lemma 6.6 to give lower bouns on Φ(S), for subsets S of [0, ] having measure near. Theorem 6.7 Let be a sufficiently large positive integer. Let S be a subset of [0, ] with measure v = S. If S π ln e 0 ln, then Φ(S) π e Proof. By Theorem 6.3, Φ(S) is minimum when S is a Hamming ball. Hence, we may assume S = HB(, k, t) for some k an t, where k is an integer with 0 k an t is a real number with 0 t. We have S = v(, k, t) = Σ k i=0α(, i, t) an Φ(S) = α(, k +, t). As observe earlier, if X is a ranom variable rawn from BIN(, t), then S = P r(x k) an Φ(S) = P r(x = k + ). By our assumption,. P r(x k) π ln e 0 ln. (6) By Lemma 6.4, P r(x t ) P r(x t ). (7) We consier several cases. Case. 5 ln t 5 ln. By Lemma 6.6, for each integer m with m t ln, we have P r(x = m) π e when is large. Hence, for sufficiently large, we have P r(x t + ln 3 ) + ( ln 3) π e + π ln e 0 ln P r(x k), where the last inequailty follows from (6). So k t + ln 3. By a similar argument, we have k t ( ln 3). It follows that (k + ) t ln. Since 5 ln t 5 ln, by Lemma 6.6, we have Φ(S) = P r(x = k + ) π e 0.,

21 Case. t < 5 ln. In this case, we have t < 5 ln. Suppose t 4. Then P r(x k) P r(x = 0) = ( t) ( 4 ) e 4 e > 0.6, contraicting (6) for large. So, t >. 4 If k > 3t, then by Markov s inequality we have P r(x > k) < an hence P r(x k), 3 3 contraicting (6) for large. So, we have k 3t 5 ln. Let m = k +. Then m 6 ln. We have (recalling that t 5 ln ) 4 ( ) Φ(S) = P r(x = k + ) = P r(x = m) = t m ( t) m m ( ) m ( ) m ( ) t m t m ( t) e t e 0 ln. m m 4m = e m ln 4m 0 ln e 4 ln π e, for large. 4 Case 3. t > 5 ln. In this case, we apply almost ientical reasoning as in Case, but with t playing the role of t. First we show t. Then we apply Markov s inequality to X to show 4 that k 5 ln an therefore m 6 ln. Then we establish the lower boun on Φ(S) as in Case, with t playing the role of t an m playing the role of m. We omit the etails. Corollary 6.8 Let be a sufficiently large positive integer. Let S be a subset of [0, ] with measure v = S. Suppose Then either π ln e or there exists an integer l such that Φ(S) π e 0 ln S 0 ln,. Φ ( l) (S) Φ(S) l + π e 0 ln.

22 Proof. For each i, let S i = Φ (i) (S). Let l enote the smallest m such that S S S m ln e 0 ln π. Let i {,..., l}. Let S = S S S i. Then S = S + S S i + π ln e Since also S S ln e 0 ln π, by Theorem 6.7, we have Since this hols for all i l, we have If l > ln 3, then S i = Φ(S ) π e. S S l (l ) π e. 0 ln. S S l > ( ln 4) π e > π ln e 0 ln, contraicting our choice of l. So, we have l ln 3. Now, if S = Φ(S) 0 e ln π, then we are one. Otherwise, we have Φ ( l) (S) Φ(S) = S S l S Thus since l + ln 3, we get π ( ln ) e 0 ln. Φ ( l) (S) Φ(S) l + π e 0 ln. The next two corollaries are just the equivalents of Theorem 6.7 an Corollary 6.8 for K n, uner the corresponence between [0, ] an K n escribe in Proposition 6.. Corollary 6.9 Let n an be positive integers, where is sufficiently large. Let S be a subset of V (K n). If S n π ln e 0 ln n, then Φ(S) π e n.

23 Corollary 6.0 Let n an be positive integers, where is sufficiently large. Let S be a subset of V (K n). Suppose π ln e 0 ln n S n. Then either (S) π 0 e ln n or there exists an integer l such that ( l) (S) (S) l + π 0 e ln n. Corollary 6.9 reaily yiels the following lower boun on the vertex banwith B(K n). Theorem 6. Let n be a fixe positive integer. We have B(K n) ( o()) π n, as. Proof. Consier a labeling f of V (Kn) that achieves the banwith. Let S enote the set of vertices receiving labels,,..., n /. By Corollary 6.9, (S) ( o()) π n. Thus, B(Kn) = B(f) (S) ( o()) π n. In [5], Harper obtaine the same lower boun on B(K n) when both n an go to. For even n, the lower boun for B(K n) in Theorem 6. matches asymptotically the upper boun for B(K n) given in [5] as. Recently, Balogh et al [6] obtaine improve upper an lower bouns on B(K 3 n). We now establish a lower boun on B (K n) that asymptotically matches the upper boun given in Theorem 5.0 when n is even. We follow the approach use in [9] an [5]. Theorem 6. Let n be a fixe positive integer. We have B (Kn) ( o()) π n (n ), as. Proof. Let f be an optimal labeling of E(Kn) using labels,..., ( ) n n. Let S enote the set of eges receiving ( ) the first half of the labels. That is, S is the set of the eges receiving labels,,..., n n. Let us call the eges in S re an the rest of the eges white. For a vertex x V (Kn), let E(x) enote the set of eges incient to x. A vertex x is calle re if all the eges in E(x) is re, an white if all the eges in E(x) are white; otherwise it is calle mixe. Let R, W, an M enote the set of re, white, an mixe vertices, respectively. We have R + W + M = n. (8) For x M, let r(x) enote the number of re eges in E(x). Hence r(x) (n ). By (ouble) counting the re eges an the white eges, we have 3

24 R (n ) + Σ x M r(x) = ( ) n n = W (n ) + Σ x M ((n ) r(x)). (9) It reaily follows from (9) that R, W < n. Note that the white eges incient to M belong to (S). Similarly, the re eges incient to M belong to (E(K n) S). Therefore, we have (S) Σ x M((n ) r(x)) an (E(K n) S) Σ x Mr(x) (0) Combining these two inequalities we obtain B (K n) max{ (S), (E(K n) S)} If M π n e 0 ln, then by (), we have M (n ). () 4 B (K n) M (n ) 4 π n (n ) e an we are one. Hence, we may assume that 0 ln = ( o()) π n (n ), Either M < π n e 0 ln. () Σ x M r(x) M (n )/ or Σ x M ((n ) r(x)) M (n )/. Without loss of generality, let us assume that the first inequality hols, since otherwise we coul switch the roles of re an white vertices. Combining this with (9) an (), we have ( ) n n R (n ) + M (n )/ R (n ) + n (n ). (3) π This yiels the lower boun n R < π n. By Corollary 6.0, we have either (R) π 0 e ln n, (4) or there exists an integer l such that ( l) (R) (R) l + π 0 e ln n. (5) 4

25 Now, clearly (R) M. So if (4) hols then we have M π n e 0 ln, contraicting (). Hence, we may assume that (5) hols instea. Let C = ( l) (R) (R). Let E(C) enote the set of eges incient to C. Recalling that S is the set of re eges, we have E(C) ( l+) (S). Note that E(C) C (n )/. Applying Proposition. to the line graph L(K n), we have B (K n) B (f) ( l+) (S) (l + ) E(C) (l + ) = ( l) (R) (R) (n ) (l + ) π 0 e (n ) ln n = ( o()) π n (n ). C (n ) (l + ) (by (5)) 7 Concluing Remarks. We thank one of the referees for pointing us to the work of Pellegrini [6]. There the weight function l(n,, r) is expresse as an exponential sum, using the recurrence in Proposition 3. an stanar results in Fourier series. This le to ([6] Proposition 3) l(n,, r) = π π 0 ( ) sin(nx) cos(x(r (n ))) x. sin x Recall that l (n, ) = l(n,, n( ) ). So letting r = n( ) an aapating Pellegrini s application of Laplace s metho to estimate this integral for fixe n an, we obtain l (n, ) = 6 π ( n n ) n + o(n ), when n is fixe an. (6) We note that in Pellegrini s application it is important to have n fixe an. Since we are concerne with the case when is fixe an n, we cannot use formula (6) in our result. Formula (6) will be useful when one stuies B (P n) when n is fixe an.. We have asymptotically etermine B (Kn) for fixe even n an, showing that B π n (n ). In the upper boun labeling, the evenness of n mae it possible to partition Kn into many orthants, each isomorphic to Kn/. 5

26 When n = m + is o, a similar partition is possible. Here, an orthant corresponing to a weight k vertex of Q is the Cartesian prouct of k copies of K m+ an k copies of K m. Applying the same labeling scheme, an noting that a maximum ilation occurs at a pair of cross eges incient to a vertex in an orthant associate with a hypercube noe of weight ( ), we obtain an upper boun (m+) m+ B (Kn) m + m m m + π n (n )( + o()), when n = m + is fixe an. This upper boun is within a factor of m+ m from the lower boun in Theorem 6.. m m+ Note that this factor is close to an tens to as m. It woul be interesting to etermine B (Kn) asymptotically when n is o an. We suspect that the upper boun given here is closer to the truth. 8 Appenix: Proof of Lemma 6.6 Proof. We recall Stirling s formula that for all integers n, n! = πn (n/e) n e θ n for some θ = θ(n) with 0 < θ <. We have for some θ, θ, θ 3 (0, ) P r(x = k) = = ( ) n p k ( p) n k k πn( n e )n πk( k π e )k π(n k)( n k pk ( p) n k e e )n k ( ) np k ( n np k n k n k(n k) θ n θ k θ 3 (n k) ) n k e k (n k). (7) Let a = np k. Then np = k + a an a ln n by our assumption. We have ( ) np k k ( ) n np n k ( = + a ) k ( a ) n k e a k a n k e ln n( k + n k ), (8) n k k n k where the first inequality follows from Lemma 6.5. By (7) an (8), we have for large n P r(x = k) n π k(n k) n+ e (ln )( k + n k ) π n k(n k) ln n( e k + n k ) (9) Note that k(n k) is unimoal an is maximize at k = n/. We consier two cases. Case. k n 3 4 or n k n 3 4. In this case, k(n k) n 3 4 n. So n n k(n k) n 4 ln n(n 4 ln n) k(n k) n 3 4 ln n. By (9), we have. Also, 4 n k 4 ln n. So, P r(x = k) π n 3 8 e ln n ln n = π n 3 8 ln n πn e n, for large n. 6