The average number of spanning trees in sparse graphs with given degrees

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1 The average number of spanning trees in sparse graphs with given egrees Catherine Greenhill School of Mathematics an Statistics UNSW Australia Syney NSW 05, Australia Matthew Kwan Department of Mathematics ETH Zürich Zürich, 809, Switzerlan Mikhail Isaev Research School of Computer Science Australian National University Canberra, ACT 60, Australia Moscow Institute of Physics an Technology Dolgopruny, 700, Russia Brenan D McKay Research School of Computer Science Australian National University Canberra, ACT 60, Australia brenanmckay@anueuau 0 February 07 Abstract We give an asymptotic expression for the expecte number of spanning trees in a ranom graph with a given egree sequence =,, n, provie that the number of eges is at least n + max, where max is the maximum egree A key part of our argument involves establishing a concentration result for a certain family of functions over ranom trees with given egrees, using Prüfer coes Introuction The number of spanning trees τg in a graph G also calle the complexity of G is an important graph parameter that has connections to a wie range of topics, incluing the stuy of electrical networks, algebraic graph theory, statistical physics an number theory Research supporte by the Australian Research Council, Discovery Project DP0059

2 see for example [, 5, 7, 8] These connections are largely relate to the matrix tree theorem, which says that τg is equal to any cofactor of the Laplacian matrix of G There is a large boy of existing work concerning the approximate value of τg for graphs with given egree sequences, an ranom graphs with given egree sequences, especially in the regular case Let =,, n be a vector of positive integers with even sum, an let Γ enote the set of all graphs on the vertex set {,,, n} with egree sequence If every entry of equals then we write Γ n, for the set of all -regular graphs on {,,, n} Let G be the ranom graph with egree sequence, chosen uniformly at ranom from Γ, an let G n, be the ranom -regular graph on vertex set {,,, n}, chosen uniformly at ranom from Γ n, Unless otherwise state, all asymptotics in this paper hol as n, possibly along some infinite subsequence of N The number of spanning trees in a graph is strongly controlle by its egree sequence Let = n /n j, ˆ = j n j enote the arithmetic an geometric means of the egree sequence The best uniform upper boun for regular graphs is ue to McKay [], who prove that when 3, n log n τg = O / n log for all G Γ n, This was prove sharp within a constant by Chung an Yau [3] Kostochka [8] prove that ˆ ε n τg ˆn n for any connecte G Γ, where ε = εδ > 0 tens to zero as δ = min j j This lower boun extene a result of Alon [] on τg in the case of -regular graphs To iscuss ranom graphs, efine the ranom variables τ = τg an τ n, = τg n, That is, τ is the number of spanning trees in G, an τ n, is the number of spanning trees in G n, McKay [] prove that for fixe, τ /n n, / with probability An alternative proof in a much more general framework was given by Lyons in [9, Example 36] McKay [0] gave the expecte value E τ to within a constant factor, in the case that j = O for all j an the average egree is at least + ε, for some ε > 0 Specifically,

3 McKay prove that uner these conitions, the expecte number of spanning trees is ˆ n E τ = Θ n / / Greenhill, Kwan an Win [6] recently foun the asymptotic value of this Θ factor, for ranom -regular graphs with 3 = O Specifically, they prove that the Θ in is asymptotic to the constant / exp 3/ This is about e 3/ / for large They also gave the asymptotic istribution of the number of spanning trees in a ranom cubic graph In this paper, we obtain an asymptotic expression for E τ for a wier range of sparse egree sequences than in any of the above ranom graph results Our main result is the following Theorem Let = n =,, n be a vector of positive integers with even sum, for every n in some infinite subsequence of N Define max = max j, j = n n j, n /n ˆ = j, R = n n j an let H = / ˆ n 3/ n / / Suppose that max n Then the sequence is graphical for sufficiently large n, an the expecte number of spanning trees in G is given by R E τ = H exp R + O max 3 n + η, where { η = min max n, 3 max log n = O max log n5/ + n n / Some remarks about this result are given below } n, max Due to the Erős-Gallai Theorem, uner the conitions of Theorem the sequence is always graphical without any requirement for n to be large Since this fact is not require for our asymptotic formula, we omit the proof 3

4 Since max, the conition max n implies that > Other than the relative error term, the expression given by Theorem matches in the regular case, showing that the formula obtaine in for regular graphs with constant 3 also hols for -regular graphs with slowly growing in particular, it hols when = on /3 Uner our assumptions, the relative error term may not be vanishing, though it is always boune Let m = n j be the number of eges in any graph in Γ The conition max n is equivalent to the conition that m n + max, or in other wors, that there are at least max + more eges in any graph in Γ than in a tree on n vertices For example, when max = 3, our result hols with a boune error if the number of eges excees n by at least In particular, we have the following corollary when is close to Corollary Suppose that = + x/n where max x n / This correspons to graphs with n + x eges Then E τ = n e x n x 3/+x n ˆ 6 + R + R exp + 3x 6 n + O max x + x3 n Proof We estimate the various terms in Theorem First note that an / = + x/n / = e Ox/n 3/ ˆ / Next, a series expansion yiels log = + x/n log so we have / Then, we can compute / = log + log x n + 3 n = ˆ n 3/+x n x + x + x/n log + x n n x x 3 + O n n n e x 3x = n exp n + O = 3 x + O, n x 3 n

5 So, noting that R max, we have, 3 = x + O n + = x 6 + O n R R = Finally, the error term from Theorem is at most O max n + max = O 6 + R + R 6 max x + O max x n 3 Since this error term ominates the error from 3 uner our assumptions, the result follows From Corollary we see that when the average egree is close to but above, an the geometric mean ˆ is strictly greater than, then E τ tens to infinity This can be true even for egree sequences where the probability of connectivity tens to zero, even when Corollary oes not apply For example, consier the egree sequence with n/ vertices of egree 5 an n/ vertices of egree restricte to even n Here = 3 an ˆ = 5 > From Theorem it follows that the expecte number of spanning trees in G is n/ 80 Θ/n 7 which tens to infinity as n However, the probability that G is connecte tens to zero To see this, we work in the configuration moel [] For ease of notation, write n = t an let S be the set of configurations with t cells containing 5 points an t cells containing point If a configuration in S gives rise to a connecte graph then every point in a cell of size is paire with a point from a cell of size 5 There are at most 5t t t! t t! such configurations, an the probability that a ranom configuration in S is simple, conitione on connecteness, is at most The total number of simple configurations in S is 6t! Θ 5 3t 3t! where the Θ factor is the probability that a ranom configuration in S is simple: this tens to a constant boune away from zero, by [3, Theorem 6] Diviing by 5 gives the upper boun 5 5 n/ Θ = o

6 on the probability that a ranom element of G is connecte The case of ense irregular egree sequences will be treate in a separate paper Outline of our approach Let a k enote the falling factorial aa a k + We say that a sequence x = x,, x n of positive integers is a tree egree sequence if the entries of x sum to n We say that a tree egree sequence x is a suitable egree sequence if x j j for all j {,,, n} The intene meaning is that x is suitable as a egree sequence for a spanning tree of a graph with egree sequence For a suitable egree sequence x, let T x be the set of all trees with egree sequence x = x,, x n an T be the set of all trees with vertex set {,,, n} It is well-known that n T x = 6 x,, x n See for example [6, Theorem 3] Let τ x enote the number of spanning trees of G with egree sequence x, an enote by P, T the probability that the ranom graph G has T as a subgraph, for all T T Then the expecte number of spanning trees with egree sequence x in G can be written as E τ x = T T x P, T 7 an furthermore, the expecte number of spanning trees of any egree sequence in G is E τ = x E τ x where the sum is over all suitable egree sequences x We will estimate the summan in 7 using a theorem by McKay [3, Theorem 6], which we will restate below, incluing some necessary terminology, an with some minor reworing for consistency Theorem 3 [3, Theorem 6] Let g = g,, g n be a sequence of non-negative integers with even sum m, an let g max = max{g,, g n } Let X be a simple graph on the vertex set {,,, n} with egree sequence x = x,, x n, where x max = max{x,, x n } Suppose that g max an ˆ ɛ m, where ɛ < /3 an ˆ = + g max 3 g max + x max + Define λ = m n g i an µ = m g i g j ij X 6

7 Let Ng, X enote the number of simple graphs with egree sequence g an no ege in common with X Then uniformly as n Ng, X = m! m! m n g i! exp λ λ µ + O ˆ /m Given a suitable egree sequence x an a tree T T x, efine the parameters λ 0 = n λx = µt = n j, n + n + n j x j, {i,j} ET Using Theorem 3, we may prove the following i x i j x j Lemma Suppose that x is a suitable egree sequence an let T T x With notation as above, provie max n, P, T = n/ n n n j xj exp λ 0 + λ 0 λx λx µt + O max n n n Proof There is a bijection between the set of graphs with egree sequence which contain T, an those with egree sequence x which contain no eges of T Therefore, we can write N x, T P, T = 8 N, an Theorem 3 to estimate the numerator an enominator First, consier N x, T Let ˆ = + g max 3 g max + x max + where g max = max,,n j x j We require that ˆ ε m, where ε < is a constant 3 an m = n + is the number of eges in a graph with egree sequence x By assumption, m = n + + max Since g max, x max max an max 3 which follows since >, we have ˆ m + max 5 max + + max

8 which is strictly less than 3 Observe also that ˆ = O max Hence Theorem 3 applies an says that N x, T = n +! n/ +! n/+ n j x j! exp λx λx µt + O max/ n Similarly, we obtain N, = n! n/! n/ n j! exp λ 0 λ 0 + O max/ n, 9 noting that the value of the ˆ is smaller than in the previous application of Theorem 3, while the parameter m is larger Substituting these expressions into 8 completes the proof Observe that the only term in the argument of the exponential in Lemma which epens on the structure of T rather than just the egree sequence of T is µt For any suitable egree sequence x an any tree T T x, efine an let fx = λ 0 + λ 0 λx λx 0 βx = T x be the average value of e µt over all T T x T T x e µt Combining 6, 7 an Lemma, for any suitable egree sequence x we have E τ x = e O max / n n/ n n n j n n Now efine = e O max/ n n/ n n ˆn n n n! µx = T x n j xj e fx µt T T x n j e fx βx x j T T x µt, 3 the average value of µt over T x By proving that βx is close to e µx for each suitable egree sequence x, an evaluating µx, we will establish the following 8

9 Theorem 5 Suppose that the conitions of Theorem hol an that x is a suitable egree sequence Then with η, R an H as efine as in Theorem, E τ x = H n n n j R + exp λx x j λx n x j j x j + O max n n + η The structure of the rest of the paper is as follows In Section 3 we generalise the function µ an prove a concentration result for trees with given egrees This proof will involve a martingale concentration result of McDiarmi [] which we iscuss in Section The results of Section 3 are applie in Section to prove that the average of e µt over T T x is close to exp µx, an hence to prove Theorem 5, for any suitable egree sequence x Finally, Theorem is prove in Section 5 Before we begin, note that we use the following conventions in our summation notation: i j will always enote a sum over all orere pairs i, j with i j over some appropriate range which will be clear from the context, usually i, j =,, n On the other han, if i is fixe an we wish to sum over all j i for example, over all j {,,, n} \ i then we will write j:j i Concentration results Let P = Ω, F, P be a finite probability space A sequence F 0,, F n of σ-subfiels of F is a filter if F 0 F n A sequence Y 0,, Y n of ranom variables on P is a martingale with respect to F 0,, F n if i Y j is F j -measurable an has finite expectation, for j = 0,, n; ii EY j F j = Y j for j =,, n An important example of a martingale is mae by the so-calle Doob martingale process Suppose X, X,, X n are ranom variables on P an fx, X,, X n is a ranom variable on P of boune expectation Let σx,, X j enote the σ-fiel generate by the ranom variables X,, X j Define the martingale {Y j } with respect to the filter {F j }, where for each j, F j = σx,, X j an Y j = EfX, X,, X n F j In particular, F 0 = {, Ω} an Y 0 = E fx, X,, X n In this section we state some concentration results for martingales See McDiarmi [] for further backgroun an for any efinitions not given here Following McDiarmi [], 9

10 for j =,, n we efine the conitional range of Y j as rany j F j = ess supy j F j + ess sup Y j F j Here essential supremum may be replace by supremum, as in [], if the probability istribution is positive over Ω Our main tool is the following result from McDiarmi [] The tail boun on the probability is given by [, Theorem 3] The upper estimate on the moment generating function Ee hyn is an intermeiate step of McDiarmi s proof, see [, Section 35] The lower boun on K is ue to Jensen s inequality Theorem [] Suppose that P = Ω, F, P is a finite probability space Let Y 0, Y,, Y n be a martingale on P with respect to a filter F 0, F, F n, where F 0 = {, Ω}, such that n rany j F j ˆr as for some real ˆr Then E e Yn = e Y 0+K where 0 K 8 ˆr Furthermore, for any real t > 0, Pr Y n Y 0 t exp t /ˆr As a corollary, we obtain a concentration result for functions of sets of a given size Corollary Let [N] r be the set of r-subsets of {,, N} an let h : [N] r R be given Let C be a uniformly ranom element of [N] r Suppose that there exists α 0 such that ha ha α for any A, A [N] r with A A = r Then E e hc = exp E hc + K where K is a real constant such that 0 K min{r, N 8 r}α Furthermore, for any real t > 0, t Pr hc E hc t exp min{r, N r}α 0

11 Proof Let S N enote the set of permutations of {,, N} an τ = τ,, τ N be a uniform ranom element of S N Note that the set {τ,, τ r } is a uniformly ranom element of [N] r Define h : S N R by hω = h{ω,, ω r } for all ω S N Then hρ hρ α for all permutations ρ, ρ S N such that ρ ρ is a transposition Given ω = ω,, ω N S N, for k = 0,, N let h k ω = E hτ τj = ω j for j =, k Clearly, h 0 τ,, h N τ forms a martingale: it is the result of the Doob martingale process for hτ It follows from Frieze an Pittel [5, Lemma ] that ran hk τ στ,, τ k α Moreover, for any ω S N an k {r,, N}, we have E hτ τj = ω j, j k = h{ω,, ω r } Therefore ran hk τ στ,, τ k = 0 for all k > r Applying Theorem to the martingale h 0 τ,, h N τ, we conclue that hols with 0 K 8 rα, an that Pr hc E hc t exp t rα If r N r then we are finishe Otherwise we repeat the above argument using the bijection between subsets an their complements 3 Trees with given egrees In this section we consier sums of the form F T = {j,k} ET for a given function φ : {,,, n} [φ min, φ max ] R φj φk 3 Let F x be the average value of F over all trees with a given egree sequence x: F x = T x T T x F T

12 ξf T The goal of this section is to prove the following theorem, showing that the average of e over T x is close to e ξ F x, for ξ {, } We will measure this istance using the seminorm of φ given by φ m = min c R n φj c 3 Here the minimising value of c is any meian of {φj : j =,, n} Theorem 3 Let F satisfy 3 Then for any tree egree sequence x an for ξ {, }, T x T T x e ξf T = exp ξ F x + K for some real constant K which satisfies 0 K 8 L φ, where L φ = φ max φ min 3 min { φ max φ min n, φ m ln n + } Furthermore, if T is a uniformly ranom element of T x then for any real constant t > 0, Pr F T F x > t exp t /L φ First we give some explicit formulae which we will nee later Lemma 3 Let x be a tree egree sequence an consier the set T x of all trees with egree sequence x i Let S be a isconnecte forest with vertex set {,, n} an egree sequence s,, s n, where s j x j for j =,, n Let S,, S r be the components of S Then the probability that a uniform ranom tree in T x contains S is r i= j V S i x j s j n x j sj, n n r where x j sj = x j if s j = 0 In particular, for istinct j, k {,,, n}, the fraction of trees in T x in which vertices j, k are ajacent is x j + x k n ii The average value of F over T x is n n F x = φk x j φj n k= n n x j φj

13 Proof Define x = x,, x r, where x i = j V S i x j s j for i =,, r If x i = 0 for any i then T cannot contain S, as S is isconnecte Hence the result hols trivially in that case, an for the remainer of the proof we may assume that all entries of x are positive Next, observe that the entries of x sum to r, an hence x is a tree egree sequence Each tree in T x that contains S can be forme uniquely by the following process: Take any tree T on the vertex set {,, r} with egree sequence x For i =,, r, replace vertex i of T by S i an istribute the eges of T that were incient with i amongst the vertices of S i, so that each vertex j V S i has egree x j in the resulting tree By 6, the number of choices for T in Step is r x, while the number of ways r to istribute eges in Step is,,x r i= x i! j V S i x j s j! The first statement of i is proven by multiplying these expressions together, iviing by 6 an simplifying Then taking S to be the ege jk together with n trivial components completes the proof of i Now using linearity of expectation, 3, an part i, we calculate that establishing ii n F x = j<k x j + x k φjφk 33 = j kx j φjφk 3 = n n x j φj φk φj k= n n n = φk x j φj x j φj, k= We complete this section with the proof of Theorem 3, which involves the process use to construct the Prüfer coe of a labelle tree The Prüfer coe of a tree T T is a sequence b = b,, b n {,,, n} n Given T, fin the unique neighbour b of the lowestlabelle leaf a Then b becomes the first entry in the Prüfer coe for T We fin the next entry recursively by consiering the tree T a with the first leaf elete The process stops when a single ege remains: this ege is etermine by the egree sequence an oes not nee to be recore in the coe b We will refer to this process as the Prüfer process with 3

14 , 7,, 7, Figure : A tree an its corresponing Prüfer coe input T See Figure for an example The corresponence between trees an Prüfer coes is a bijection: see for example Moon [6, pp 5-6] This provies a proof of Cayley s formula an of 6 The following useful property of the Prüfer process may be prove by inuction on j Lemma 33 Let x be a tree egree sequence an let T T x Suppose that the Prüfer process with input T prouces the Prüfer coe b an the sequence a,, a n of leaves For any j =,, n, the initial sequence a,, a j is uniquely etermine by x an b,, b j When there is more than one tree uner consieration we will write a j T, b j T for the vertices ientifie at step j of the Prüfer process for the tree T To prove Theorem 3 we work with a martingale efine using the Prüfer coe of a tree A martingale construction base on the Prüfer coe was given by Cooper, McGrae an Zito [], for all labelle trees Our martingale is restricte to trees with a given egree sequence an we stuy a function for which it is more ifficult to boun the conitional ranges Proof of Theorem 3 Suppose that T an T are trees on {,,, n} with the same egree sequence For j = 0,, n 3, say that T an T are j-equivalent, an write j j T T, if b i T = b i T for i =,,, j By Lemma 33, if T T then a i T = a i T for i =,, j The j-equivalence relation inuces a partition of T x into equivalence classes C j,,, C j,rj, say, with r j l= C j,l = T x Let Y j,l equal the average of F T over T C j,l, an efine the function Y j on T x by Y j T = Y j,l if T C j,l Finally, efine the ranom variable Y j = Y j T, where T is a uniformly ranom element of T x Then Y 0 is the constant function which takes the value E F T everywhere, an Y n 3 = F T, since each equivalence class C n 3,l is a set of size Observe that Y 0,, Y n 3 is a martingale with respect to the filter F 0,, F n 3, where, for each j, F j is generate by the sets C j,,, C j,rj In fact, this is the Doob martingale process for the function F T of the ranom variables b T,, b n 3 T, which etermine T uniquely To apply Theorem we must calculate a value for ˆr Suppose that T an T are j -equivalent, where T, T T x an j {,, n 3} Then a j T = a j T, again by

15 Lemma 33 For ease of notation, write a i instea of a i T or a i T for i =,, j, an write b i instea of b i T or b i T for i =,, j For s =, let T s be the tree with n j vertices obtaine by eleting the vertices a,, a j from T s Both T an T have vertex set V j = {,,, n} \ {a,, a j } If b j T = b j T then T an T have the same egree sequence, since in this case precisely the same eges have been elete from T an T In this case, Y j T = Y j T Otherwise, the egree sequences of T, T iffer only for the two vertices b j T an b j T Specifically, vertex b j T has egree in T which is equal to its egree in T minus, while vertex b j T has egree in T which is equal to its egree in T plus Hence T an T have the same egree on all vertices in the set U j T, T = V j \ {b j T, b j T } For s =,, let y s be the egree sequence of T s on the vertex set V j an let T s enote the set of all trees on the vertex set V j with egree sequence y s Observe that y s an T s epen only on b,, b j, b j T s an x By relabelling the equivalence classes if necessary, we may assume that T s C j,s for s =, The map ϕ : C j,s T s which sens a tree T C j,s to T \ {a,, a j } is a bijection To see this, observe that the inverse map ϕ takes a tree in T s, as the vertices a,, a j an the eges {{a, b },, {a j, b j }, {a j, b j T s }} giving a tree in C j,s Therefore, for s =,, F T \ {a,, a j } = C j,s T T C s j,s i= T T s F T Combining this with 3 an the efinition of Y j,s, we see that for s =,, Y j,s = φkφl C j,s T C j,s {k,l} ET j = φa i φb i + φa j φb j T s + F T \ {a,, a j } C i= j,s T C j,s j = φa i φb i + φa j φb j T s + F T T s Applying Lemma 3ii gives Y j, Y j, = φb j T φb j T 5 T T s φa j n j l U j T,T φl

16 Note that if T = φb jt φb j T n j l U j T,T φaj φl j T then b j T = b j T an the above equality also hols Recall the efinition of φ m from 3, an let c R be the minimising value in this efinition By the triangle inequality, n j φa j φl φa j c + n j c φl l U j T,T φa j c + φ m n j l U j T,T 35 since U j T, T has n j elements an j n 3 Therefore, for any equivalence class C j,l, we have sup Y j T + sup Y j T T C j,l T C j,l φ max φ min n j sup T,T C j,l φ max φ min 3 min l U j T,T φa j φl { φ max φ min, φa j c + φ } m 36 n j Here we take the minimum of two possible upper bouns: the first arises from taking the worst case summan for both factors in the line above, while the secon arises by applying 35 to one of the factors Now let C j T enote the ranom set which is the equivalence class with respect to j which contains T It follows from 36 that rany j F j = sup Y j T + T C j T φ max φ min 3 min sup T C j T Y j T { φ max φ min, φa j T c + φ } m n j Using the efinition of c, the stanar upper boun on the harmonic series an the fact that each vertex is chosen as a j T at most once uring the Prüfer process, we get that n 3 rany j F j φ max φ min 3 min {φ max φ min n, φ m ln n + } Observe that the left han sie oes not change if F is replace by F an hence, the same boun is obtaine whether ξ = or ξ = Since Ee Y n 3 = Ee F T an Y 0 = EF T, applying Theorem completes the proof 6

17 Proof of Theorem 5 First we note the following corollary of Theorem 3 Recall the efinition of βx an µx from, 3, respectively Lemma Uner the conitions of Theorem, βx = exp µx + O min { max n, 3 max ln n n } Proof Set φj = j x j n + for j {,,, n}, an let ξ = We can take φ min = 0 an φ max = max / n + Next, we boun n j x j φ m = n + n + Finally, observe that 3 max φ m + φ max ln n + n + 3/ Now the result follows from Theorem 3 3 = O max ln n n + max ln n = O n 3 max ln n n We can now prove Theorem 5, giving an asymptotic expression for the expecte number E τ x of spanning trees in G with egree sequence x Proof of Theorem 5 Firstly note that, by 3, µx = We rewrite as E τ x = n n n + x j j x j k x k j k n x j j x j + O = e O max / n n/ n n ˆn n n max n n n n j e fx βx 3 x j 7

18 with fx as efine in 0 Applying Stirling s approximation gives n/ n n ˆn = H + O n n n where H is efine in the statement of Theorem Combining Lemma an gives { } βx = exp µx + O min max n, 3 max ln n n = exp n x j j x j n { } + O max n + min max n, 3 max ln n n In some cases, when is small, we can obtain a smaller error boun by a ifferent argument Observe that e µx βx, 5 using Jensen s inequality for the lower boun It follows from that µx = O max n + max Hence we can replace the upper boun on βx in 5 by e µx if we inclue an error term of this magnitue, leaing to βx = exp µx + O max n + max = exp n x j j x j + O max n n + max 6 using We may choose to use either this expression or, whichever gives the smaller boun Finally, observe that λ 0 + λ 0 = R + 7 Combining this with 0, 3, an 6 completes the proof 5 Proof of Theorem In this section we prove Theorem by summing the expression from Theorem 5 over all suitable egree sequences x Given a suitable egree sequence x, efine gx = fx µx = R + λx λx µx, 5 8

19 using 7 By an Theorem 5 we have n n j E τ = H n x x j exp gx + O max n + η 5 where the sum is over all suitable egree sequences x We now interpret this sum as an expecte value of a function of a nonuniform istribution on suitable egree sequences Lemma 5 Fix a partition A,, A n of {,,, n} such that A j = j for j =,, n, an let B be a uniformly ranom subset of {,,, n} of size n Define the ranom vector X = XB = X,, X n by X j = A j B + Then E τ = H exp O max n + η E e gx Proof Let x be a suitable egree sequence Since the sets A j are isjoint, there are n j x j ways to choose a subset of {,, n} with precisely xj elements in A j, for j =,, n It follows that n n j PrX = x = n x j Substituting this into 5 completes the proof Next, we prove that E e gx can be approximate by e E gx by applying Corollary We say that two suitable egree sequences x an x are ajacent if x an x iffer in precisely two entries, say in entries j an k, such that x j = x j + an x k = x k Ajacent egree sequences correspon to subsets A, A of {,,, n} of size n which have n 3 elements in common In orer to apply Corollary to g we must boun the amount by which gx can iffer from gx when x an x are ajacent Lemma 5 Suppose that x, x are two suitable egree sequences which are ajacent Then gx gx = O max n Proof Recall the efinition of g in 5 Firstly, observe that λx λx = λx λx λx + λx = O max λx λx since for any suitable x we have λx = O n max j x j = O max n 9

20 Next we calculate that λx λx = k x k j x j n + max = O n Therefore λx + λx λx λx = O max n Now we consier µ Suppose that y is a vector which isagrees with x in precisely one position, say y i = x i + ζ where ζ {, } Then using most conveniently in the form in 33, µy µx j x j i x i x i + x j + ζ n n + j:j i max = O = O max n n Applying this twice gives a boun of the same magnitue on µx µx, completing the proof Now we apply Corollary to prove the following Lemma 53 Uner the conitions of Theorem, E e gx = exp E gx + O max n Proof We will apply Corollary to hb = gxb, where the ranom set B is efine in Lemma 5 We set N = n an r = n Lemma 5 says that h changes by at most α = O max/ n if two entries of the vector change by one increasing an one ecreasing The value of the error term given by Corollary also epens on min{r, N r} = min{n, n + } We consier two cases If n n + then Corollary gives Ee gx = exp E gx + O max n n = exp E gx + O max n The secon equality follows since in this case n Otherwise it hols that n + < n, an here Corollary says that Ee gx = exp E gx + O max n + n 0

21 = exp E gx + O max, n as require To approximate E gx, we nee to be able to compute joint moments of the form E X j s X k t, where X = XB = X,, X n The ranom vector X,,, = X,, X n has a multivariate hypergeometric istribution, an from this it follows that the entries X,, X n of X = XB satisfy EX i s X j t = i s j t for i j See for example [7, Equation 396] We now fin an asymptotic expression for EgX Lemma 5 Uner the conitions of Theorem, EgX = R R + O 3 Proof First we estimate E µx using By 53 we have n s+t n s+t 53 3 max n Now EX j j X j k X k n n + = j k EX j n n + j EX j X k k EX j + EX j X k = j k n n n + n n + n 3 n n 3 = j k n + n 3 = j k 3 n + O 3 n 3 j k = j k n j n j = R + n + O maxn

22 Hence the expecte value of µx is given by E µx = 3 n + O 3 n 3 j k j k = 3 n + O R + n + O 3 n maxn 3 = R + + O max 5 n Next, recall that Applying 53 shows that λx = n + n j X j E j X j n + = n + j j EX j + EX j j n = n + n + n n = j n + n = j n + O n Therefore EλX = n j n + O = R + + O max n n 55 The same approach works for EλX but the etails are a little messier Observe that λx = n + j X j k X k + j k n j X j j X j 56

23 Applying 53 to the off-iagonal summans gives E j X j k X k n + = n + j k j k EX k j k EX j + j EX k + j k EX j X k + k EX j j EX j X k k EX j X k + EX j X k = j k n n + n + 6n n 3 + n n n 3 n = j k n + 3 n n + = j k n + O 3 n 3 Next, calculate j k = R + n + O 3 maxn j k Therefore the contribution to λx from the off-iagonal summans is n + O 3 n 3 j k j k = n + O R + 3 n 3 n + O 3 = R + + O 3 max n maxn The contribution to EλX from the iagonal terms of 56 that is, the secon summation in 56 is n E n + j X j j X j = O max EλX n 3 = O max, n using 55 Therefore EλX = R + + O 3 max 57 n The result follows by combining 7, 5, 55 an 57, after some rearranging 3

24 Now we may easily prove our main theorem Proof of Theorem The number of graphs with egree sequence is positive when n is sufficiently large, by 9 That is, is graphical for sufficiently large n The claime asymptotic expression for E τ then follows immeiately from Lemmas 5, 53 an 5 We also briefly justify the boun { max η = min n, 3 max log n = O max log n5/ + n n / } n, max Note that 3 max log n/ n max/ n if max log n When max log n, take the geometric mean of 3 max log n/ n an max Acknowlegements We woul like to thank the referees for their helpful comments References [] N Alon, The number of spanning trees in regular graphs, Ranom Structures Algorithms 990, no, 75 8 [] B Bollobás, A probabilistic proof of an asymptotic formula for the number of labelle regular graphs, Europ J Combin 980, 3 36 [3] F Chung an S-T Yau, Coverings, heat kernels an spanning trees, Electron J Combin 6 999, #R [] C Cooper, A R A McGrae an M Zito, Martingales on trees an the empire chromatic number of ranom trees, in Funamentals of Computation Theory 009 M Kuty lowski, M Gȩbala an W Charatonik, es, LNCS vol 5699, Springer, Heielberg, 009, pp 7 83 [5] A Frieze an B Pittel, Perfect matchings in ranom graphs with prescribe minimal egree, in Mathematics an Computer Science III: Algorithms, Trees, Combinatorics an Probabilities M Drmota, P Flajolet, D Gary an B Gittenberger, es, Birkhäuser, Basel, 00, pp 95 3 [6] C Greenhill, M Kwan an D Win, On the number of spanning trees in ranom regular graphs, Electron J Combin 0 #P5

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