Asymptotic enumeration of sparse uniform linear hypergraphs with given degrees

Asymptotic enumeration of sparse uniform linear hypergraphs with given degrees Vladimir Blinovsky Instituto de atemática e Estatística Universidade de São Paulo, 05508-090, Brazil Institute for Information Transmission Problems Russian Academy of Sciences, oscow 17994, Russia vblinovs@yandex.ru Catherine Greenhill School of athematics and Statistics UNSW Australia Sydney NSW 05, Australia c.greenhill@unsw.edu.au September 014 Abstract A hypergraph is simple if it has no loops and no repeated edges, and a hypergraph is linear if it is simple and each pair of edges intersects in at most one vertex. n 3, let r = rn) 3 be an integer and let k = k 1,..., k n ) be a vector of nonnegative integers, where each k j = k j n) may depend on n. Let = n) = n j=1 k j for all n 3, and define the set I = {n 3 rn) divides n)}. We assume that I is infinite, and perform asymptotics as n tends to infinity along I. Our main result is an asymptotic enumeration formula for linear r-uniform hypergraphs with degree sequence k. This formula holds whenever the maximum degree k max satisfies r 5 k 4 max = o). A new asymptotic enumeration formula for simple uniform hypergraphs with specified degrees is also presented, which holds when r 4 k 3 max = o). The proof focuses on the incidence matrix of a hypergraph, interpreted as a bipartite graph, and combines known enumeration results for bipartite graphs with a switching argument. Research supported by the Australian Research Council. For 1

1 Introduction Hypergraphs are combinatorial structures which can model very general relational systems, including some real-world networks [5, 6, 9]. Formally, a hypergraph or a set system is defined as a pair V, E), where V is a finite set and E is a multiset of multisubsets of V. We refer to elements of E as edges.) Note that under this definition, a hypergraph may contain repeated edges and an edge may contain repeated vertices. Any -element multisubset of an edge e E is called a link in e. If a vertex v has multiplicity at least in the edge e, we say that v is a loop in e. So every loop in e is also a link in e.) The multiplicity of a link {x, y} is the number of edges in E which contain {x, y} counting multiplicities). A hypergraph is simple if it has no loops and no repeated edges: that is, E is a set of edges, and each edge is a subset of V. Here it is possible that distinct edges may have more than one vertex in common. This definition of simple hypergraph appears to be standard, and matches the definition of simple hypergraphs given by Berge [1] in the case of uniform hypergraphs.) A hypergraph is called linear if it has no loops and each pair of distinct edges intersect in at most one vertex. Note that linear hypergraphs are also simple.) For a positive integer r, the hypergraph V, E) is r-uniform if each edge e E contains exactly r vertices counting multiplicities). Uniform hypergraphs are a particular focus of study, not least because a -uniform hypergraph is precisely a graph. We seek an asymptotic enumeration formula for the number of linear r-uniform hypergraphs with a given degree sequence, when the maximum degree is not too large the sparse range), and allowing r to grow slowly with n. To state our result precisely, we need some definitions. Write [a] = {1,,..., a} for all positive integers a. Given nonnegative integers a, b, let a) b denote the falling factorial aa 1) a b + 1). We are given a degree sequence k = kn) = k 1,..., k n ) with sum = n) = n i=1 k i, and we are also given an integer r = rn) 3, for each n 3. For each positive integer t, define n t = t n) = k i ) t. Then 1 = and t k max t 1 for t. Let H r k) denote the set of simple r-uniform hypergraphs on the vertex set [n] with degree sequence given by k = k 1,..., k n ), and let L r k) be the set of all linear hypergraphs in H r k). Note that H r k) and L r k) are both empty unless r divides. Our main theorem is the following. i=1

Theorem 1.1. For n 3, let r = rn) 3 be an integer and let k = k 1,..., k n ) be a vector of nonnegative integers, where each k j = k j n) may depend on n. Let = n) = n j=1 k j for all n 3, and suppose that the set I = {n 3 rn) divides n)} is infinite. Let k max = k max n) = max n j=1 k j for all n 3. If and r 5 k 4 max = o) as n tends to infinity along elements of I, then! L r k) = /r)! r!) /r n i=1 k i! exp r 1) r 1) + O r 5 k 4 max/ ) ) 4. When r =, our result is weaker than ckay and Wormald [15] as their expression has smaller error term and more significant terms. However, we believe that our result is the first asymptotic enumeration for r-uniform linear hypergraphs with r 3, and the first enumeration result for sparse r-uniform hypergraphs which allows r to grow with n. In order to improve the accuracy of Theorem 1.1, a much more detailed analysis of double links is required. We will present such an analysis in a future paper. A brief survey of the relevant literature is given in the next subsection. In Section we outline our approach, which is to use the incidence matrix of a hypergraph and prior enumeration results for bipartite graphs in order to enumerate linear hypergraphs. First we show that some undesirable substructures are rare in random bipartite graphs with the appropriate degrees. As a corollary of this, we obtain a new enumeration result for sparse simple uniform hypergraphs. Theorem 1.1 then follows from a switching argument for bipartite graphs which is used to remove subgraphs isomorphic to K,, as these correspond to double links in the hypergraph. This switching argument is presented in Section 3. 1.1 History In the case of graphs, the best asymptotic formula in the sparse range is given by ckay and Wormald [15]. See that paper for further history of the problem. The dense range was treated in [1, 14], but there is a gap between these two ranges in which nothing is known. An early result in the asymptotic enumeration of hypergraphs was given by Cooper et al. [3], who considered simple k-regular r-uniform hypergraphs when k, r = O1). Recently, Dudek et al. [4] proved an asymptotic formula for simple k-regular r-uniform hypergraphs with k = on 1/ ) and r = O1). An asymptotic enumeration formula for sparse simple uniform hypergraphs with irregular degree sequences was given in []. We restate this result below. 3

Theorem 1.. [, Theorem 1.1] Let r 3 be a fixed integer and define 1 if r {3, 4, 5, 6}, η = ηr) = 0 otherwise. Let k, and k max be defined as in Theorem 1.1. Assume that r divides for infinitely many values of n, and and kmax +η = o) as n tends to infinity along these values. Then ) H r k) =! /r)! r!) /r n i=1 k i! exp r 1) + Ok+η max/) For an asymptotic formula for the number of dense simple r-uniform hypergraphs with a given degree sequence, see [10]. As far as we are aware, there are no prior asymptotic enumeration results for uniform linear hypergraphs when r 3.. Hypergraphs, incidence matrices and bipartite graphs Suppose that G is an r-uniform hypergraph with degree sequence k which has no loops but may have repeated edges). Let A be the n /r) incidence matrix of G, where the rows of the incidence matrix correspond to vertices 1,,... n in that order, and the columns correspond to the edges of the hypergraph, in some order. Then A is a 0-1 matrix as G has no loops), the row sums of A are given by k and each column sum of A equals r. If G is simple that is, if G H r k)) then all columns of A are distinct, and hence there are precisely /r)! possible distinct) incidence matrices corresponding to G. Conversely, every 0-1 matrix with rows sums given by k, column sums all equal to r and with no repeated columns can be interpreted as the incidence matrix of a hypergraph in H r k). It will be convenient to work with the bipartite graphs whose adjacency matrices are the incidence matrices of hypergraphs. Let B r 0) k) be the set of bipartite graphs with vertex bipartition {v 1,..., v n } {e 1, e,..., e /r }, such that degree sequence of v 1,..., v n ) is k and every vertex e j has degree r. We sometimes say that a vertex v j is on the left and that a vertex e i is on the right. An example of a 3-uniform hypergraph, its incidence matrix with edges ordered in lexicographical order) and corresponding bipartite graph is shown in Figure 1. Double links will be of particular interest: there are two double links in the hypergraph in Figure 1, and each corresponds to a subgraph of the bipartite graph which is isomorphic to K,. One is induced by {v 1, v, e 1, e } and the other by {v 5, v 6, e 3, e 4 }.) 4

v 1 e 1 1 3 4 5 6 1 1 0 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 1 1 v v 3 v 4 v 5 e e 3 v 6 e 4 Figure 1: A hypergraph, its incidence matrix and corresponding bipartite graph. It follows from [8, Theorem 1.3] that B 0) r =! r!) /r n j=1 k j! exp r 1) ) + Or kmax/).1) whenever 1 rk max = o 1/ ). In fact, the result of [8] is more accurate but we are unable to exploit the extra accuracy here, so we state a simplified version.) Next, let B r k) denote the set of all bipartite graphs in B r 0) k) such that no two vertices e i1, e i on the right) have the same neighbourhood. These bipartite graphs correspond to 0-1 matrices with no repeated columns, which in turn can be viewed as incidence matrices of simple) hypergraphs in H r k). Hence /r)! H r k) = B r k)..) To work towards linear hypergraphs, we identify some desirable properties of the corresponding bipartite graphs. Given positive integers a, b, say that the bipartite graph B has a copy of K a,b if B contains K a,b as an induced subgraph. Define N = 3 max { log, r 1) / } and let B r + k) denote the set of all bipartite graphs B B r 0) properties: which satisfy the following i) B has no copy of K 3,. ii) B has no copy of K,3. 5

iii) No two copies of K, have a vertex e j on the right) in common. This implies that any two copies of K, in B are edge-disjoint.) iv) Any three copies of K, in B involve at least five vertices on the left. That is, at most two of the three copies of K, can share a common vertex.) v) There are at most N edge-disjoint copies of K, in B. To motivate this definition, note that B B r + k) if and only if the corresponding hypergraph G = GB) satisfies the following properties: i) ii) iii) iv) The intersection of any two edges of G contains at most two vertices. Any link has multiplicity at most two in G. That is, the intersection of any three edges of G contains at most one vertex.) No edge of G contains more than one double link. That is, if e 1 and e are edges of G which share a double link then e 1 is not involved in any other double link in G, and similarly for e.) No vertex can belong to three double links, and if a vertex v belongs to two double links say {v, x} and {v, y} are both double links) then both x and y belong to precisely one double link. v) There are at most N double links in G. In particular, as r 3, any hypergraph G = GB) with B B r + k) has no loops and) no repeated edges, so is simple. ckay [11] proved asymptotic formulae for the probability that a randomly chosen bipartite graph with specified degrees contains a fixed subgraph, under certain conditions. We state one of these results below, which will be use repeatedly. In fact the statement below is a special case of [11, Theorem 3.5a)], obtained by taking J = L and H = in the notation of [11], and with slightly simplified notation.) Lemma.1. [11, Theorem 3.5a)]) Let Bg) denote the set of bipartite graphs with vertex bipartition is given by I = {1,..., n} {1,..., m } and degree sequence g = g 1,..., g n ; g 1,..., g m ). Here vertex i has degree g i for i = 1,..., n, and vertex j has degree g j for j = 1,..., n.) Let L be a subgraph of the complete bipartite graph on this vertex bipartition, and let Bg, L) 6

be the set of bipartite graphs in Bg) which contain L as a subgraph. Write E g = n i=1 g i and E l = n i=1 l i, where l = l 1,..., l n ; l 1,..., l m) is the degree sequence of L. Finally, let g max and l max denote the maximum degree in g and l, respectively, and define Γ = g max g max + l max 1) +. If E g Γ E l then Bg, L) Bg) i I g i) li E g Γ) El. Using this lemma, we now analyse the probability that a uniformly random element of B r 0) k) satisfies properties i) v). Theorem.. Under the conditions of Theorem 1.1, B r +! k) = r!) /r n i=1 k i! exp r 1) ) + Or5 kmax/) 4. Proof. Throughout this proof, consider a uniformly random element B B r 0) k). We will apply Lemma.1 several times with g = k 1,..., k n ; r,..., r). In each application, L is a subgraph with constant maximum degree. Hence g max = max{k max, r} and Γ = g max g max + l max 1) + = Or + k max). For i), let v j1, v j, v j3 [n] be distinct vertices on the left, and let e i1, e i be distinct vertices on the right. Applying Lemma.1 with L = K 3,, we find that the probability that B has a copy of K 3, on the vertices {v j1, v j, v j3 } {e i1, e i1 } is at most r r 1) r ) + Or + k max)) 6 k j1 ) k j ) k j3 ). By assumption, k max + r = o). ultiplying this by the number of choices for {e i1, e i } and summing over all choices of j 1, j, j 3 ) with 1 j 1 < j < j 3 n shows that the expected number of copies of K 3, in B is at most ) /r ) r 6 k j1 ) k j ) k j3 ) O 6 j 1 <j <j 3 ) r 4 3 = O 4 = Or 4 k 3 max/)..3) 7

Hence property i) fails with probability Or 4 k 3 max/). For future reference, we note that.3) still holds under the weaker condition r 4 k 3 max = o). Next we prove that property ii) holds with high probability. Arguing as above using Lemma.1 with L = K,3 shows that the expected number of copies of K,3 in B is at most ) r 3 3 O = Or 3 k max/). 4.4) 3 To show that property iii) is likely to hold, consider the subgraph L which consists of a copy of K, on the vertices {v j1, e i1, v j, e i } and a copy of K, on the vertices {v j3, e i, v j4, e i3 }, where {j 1, j } and {j 3, j 4 } are two pairs of distinct vertices on the left, and e i1, e i, e i3 are distinct vertices on the right. If {j 1, j } {j 3, j 4 } = 1 then the two copies of K, share a common edge. By Lemma.1, the expected number of such subgraphs in B is ) r 4 3 O = Or 4 kmax/), 4.5) 4 since L has 7 edges in this case. However, when {j 1, j } {j 3, j 4 } = the subgraph L has 8 edges and the expected number of such subgraphs is, by Lemma.1, ) r 5 4 O = Or 5 k max/). 4.6) 5 This case can only arise if r 4, but it does not hurt to use this error bound when r = 3 since then it is Ok 4 max/), matching the bound from.4).) We now show that iv) holds with high probability. By iii), we need only consider sets of three edge-disjoint copies of K, with at most 4 vertices on the left. By considering cases and applying Lemma.1, the expected number of such sets is ) r 6 kmax 8 O = Ork max/), 4.7) which is sufficiently small. As a representative example, suppose that the three copies of K, all share one common vertex on the left, with no other vertex coincidences. The expected number of sets of three copies of K, with this property is Or 6 3 6 / 6 ) = Or 6 k 8 max/ ), as claimed. It remains to observe that there are only a constant number of cases to consider.) Now we turn to v). Let Q 1 = max { log, r 1) / } and define a = Q 1 + 1. We first show that the expected number of sets of a vertex-disjoint copies of K, in B is O1/). Fix j 1,..., j a ) [n] a such that k jl 8 for l = 1,,..., a

and j l 1 j l for l = 1,,..., a. Let i 1,..., i a ) {1,..., /r} a be a a)-tuple of distinct) edge labels. The probability that there is a copy of K, on {v jl 1, v jl } {e il 1, e il } for l = 1,..., a is rr 1)) a O 4a ) a l=1 k jl ), by Lemma.1. There are at most /r) a choices for i 1,..., i a ), and for an upper bound we can sum over all possible values of j 1,..., j a ). This counts each set of a vertex-disjoint copies of K, precisely 4 a a! times. From any suitable pair j 1,..., j a ), i 1,..., i a )), the two vertices on the left of each K, may be permuted and the two vertices on the right of each K, may be permuted; furthermore the a copies of K, may be permuted.) It follows that the expected number of sets of a vertex-disjoint copies of K, in B is a ) ) r 1) a 1 r 1) a ) k jl ) O = O 4 a a! a a! 4 j 1,...,j a ) [n] a l=1 ) e r 1) a ) = O 4a by choice of a. Next, let = O e/8) a ) Q = max { log, r 1) 4 4 / 4 } = O1/).8) and define b = Q + 1. From iii) and iv), we can assume that any copy of K, in B is either vertex-disjoint from all other copies of K, in B, or shares one vertex on the left with precisely one other copy of K, in B. Call such a pair of copies of K, a fused K, pair. A fused K, pair involves 7 vertices 3 on the left, 4 on the right) and 8 edges. Arguing as above, the expected number of sets of b vertex-disjoint) fused K, pairs involving 8b edges) is, using Lemma.1, at most ) 1 r 1) 4 b e ) ) 4 r 4 b 4 = O b! 8 + Or + kmax)) 4 8 b 4 = O e/8) b) = O1/),.9) by choice of b. It follows that the expected number of edge-disjoint copies of K, in B is at most Q 1 + Q with probability O1/). Since r 1) 4 4 4 r 1)4 k 3 max 3 r 1), 9

we have Q 1 + Q 3Q 1 = N. Hence v) holds with the required probability. Recalling.3).7), we conclude that B r + k) / B r 0) k) = 1 + O r 5 kmax/), 4 and the result follows using.1). As a by-product, we obtain a new asymptotic enumeration formula for sparse simple uniform hypergraphs with given degrees. The result below allows r to grow slowly with n, whereas [, Theorem 1.1] restated earlier as Theorem 1., for ease of comparison) only considered fixed r 3. However for 7 r = O1), the result of [] has wider applicability and a better error bound, as it assumes only k max = o) and achieves error bound Ok max/). For r {3, 4, 5, 6}, Corollary.3 precisely matches [, Theorem 1.1].) Corollary.3. For n 3, let r = rn) 3 be an integer and let k = kn) = k 1,..., k n ) be a vector of nonnegative integers, where each k j = k j n) may depend on n. Let = n) = n j=1 k j for all n 3, and suppose that the set I = {n 3 rn) divides n)} is infinite. If and r 4 kmax 3 = o) as n tends to infinity along elements of I, then! H r k) = /r)! r!) /r n i=1 k exp r 1) ) i! + Or4 kmax/) 3. Proof. Since r 3, it follows from.3) that B r k) / B r 0) k) = 1 + Or 4 kmax/), 3 and as noted earlier,.3) holds when r 4 k 3 max = o). Combining this with.1) and.) completes the proof. 3 Double links For nonnegative integers d, let C d be the set of bipartite graphs in B + r k) with exactly d edge-disjoint copies of K,. The corresponding hypergraph has exactly d double links.) The sets C d partition B + r k), and so B + r k) = N We estimate this sum using a switching operation which we now define. d=0 C d. 3.1) Given B C d, a d-switching is described by an 8-tuple u 1, u, w 1, w, f 1, f, g 1, g ) of vertices of B with u 1, u, w 1, w {v 1,..., v n } and f 1, f, g 1, g {e 1,..., e /r }, such that 10

B has a copy of K, on {u 1, u } {f 1, f }, w 1 g 1 and w g are edges in B. The corresponding d-switching produces a new bipartite graph B with the same vertex set as B and with edge set EB ) = EB) \ {u 1 f 1, u 1 f, w 1 g 1, w g }) {u 1 g 1, u 1 g, w 1 f 1, w f }. 3.) The d-switching operation is illustrated in Figure below. Note that in the hypergraph setting, the d-switching replaces the four edges f 1, f, g 1, g of the original hypergraph with the edges f 1, f, g 1, g defined by f j = f j \ {u 1 }) {w j }, g j = g j \ {w j }) {u 1 } for j = 1,.) w 1 g 1 w 1 g 1 u 1 f 1 u 1 f 1 u f u f w g w g Figure : A d-switching Let dist G x, y) denote the length of the shortest path from x to y in a graph G. The bipartite graph B produced by the d-switching belongs to C d 1 if the following conditions hold in B: I) Neither of g 1 or g belong to a copy of K,. II) For j = 1, we have dist B u 1, g j ) 5 and dist B w j, f j ) 5. Hence in particular, all vertices in the 8-tuple are distinct.) If I) fails then the d-switching may destroy more than one copy of K,, while if II) fails then the d-switching may create a new copy of K,, or may result in a double edge.) 11

A reverse d-switching is the reverse of a d-switching. A reverse d-switching from a bipartite graph B C d 1 is described by an 8-tuple of vertices u 1, u, w 1, w, f 1, f, g 1, g ) with u 1, u, w 1, w {v 1,..., v n } and f 1, f, g 1, g {e 1,..., e /r } such that u 1 g 1, u 1 g, w 1 f 1, u f 1, u f, w f are all edges of B. The reverse d-switching produces the bipartite graph B defined by 3.). This operation is depicted in Figure by following the arrow in reverse. The bipartite graph B produced by the reverse d-switching belongs to C d if the following conditions hold in B : I ) None of f 1, f, g 1, g belong to a copy of K,. II ) For j = 1, we have dist B u 1, f j ) 5 and dist B w j, g j ) 5. Hence in particular, all vertices in the 8-tuple are distinct.) We will analyse d-switchings to obtain an asymptotic expression for C d / C d 1, and then combine these to find an expression for L r k) = C 0 //r)!, which is the quantity of interest. The following summation lemma from [8] will be needed. The statement has been adapted slightly from that given in [8], without affecting the proof given there.) Lemma 3.1 [8, Corollary 4.5]). Let N be an integer and, for 1 i N, let real numbers Ai), Ci) be given such that Ai) 0 and Ai) i 1)Ci) 0. Define A 1 = min N i=1 Ai), A = max N i=1 Ai), C 1 = min N i=1 Ci) and C = max N i=1 Ci). Suppose that there exists a real number ĉ with 0 < ĉ < 1 such that max{a/n, C } ĉ for all A [A 3 1, A ], C [C 1, C ]. Define n 0,..., n N by n 0 = 1 and n i = 1 i Ai) i 1)Ci)) n i 1 for 1 i N. Then where Σ 1 N n i Σ, i=0 Σ 1 = exp A 1 1 A 1C ) eĉ) N, Σ = exp A 1 A C 1 + 1 A C 1) + eĉ) N. First we analyse one d-switching. 1

Lemma 3.. Suppose that d is an integer such that 1 d N. Then C d = C d 1 r )) 1) kmax d + r k 1 + O max). 4d Proof. Suppose that B C d. There are d ways to choose a copy of K,, and four ways to order the two vertices u 1, u ) on the left and the two vertices f 1, f ) on the right. Then there are 4d) ways to select distinct edges w 1 g 1, w g, avoiding edges which lie in a copy of K,. Hence there are at most 8-tuples giving a d-switching in B. 4d 4d) = 4d 1 + Od/) ) However, some of these 4-tuples are illegal and must be removed. All tuples counted above satisfy condition I). There are Odr+k max )) 8-tuples in which two of the vertices coincide, and Odrk max ) 8-tuples in which all eight vertices are distinct but an edge exists in B from u 1 to g j or from w j to f j, for some j {1, }. There are Odr k max) 8-tuples for which dist B u 1, g j ) = 3 or dist B w j, f j ) = 3, for some j {1, }. For example, given a copy of K, on {u 1, u } {f 1, f }, there are Or k max ) choices for w 1 within distance three of f 1, and then Ok max ) choices for a neighbour g 1 of w 1.) Hence condition II) fails for Odr k max) 8-tuples, and number of d-switchings which can be applied to B is )) d + r 4d 1 kmax + O. Next, given B C d 1, there are ) ways to select u 1, g 1, g ) and u, f 1, f ) such that u 1 g 1, u 1 g, u f 1 and u f are all edges of B. Then there are r 1) ways to choose w 1, w ) such that w j f j is an edge of B and w j u, for j = 1,. Some of these choices are illegal: condition I ) fails for Odr k max ) 8-tuples, while in Or k max r + k max ) ), two of the vertices coincide. In Or 3 k max ) there is an edge from w j to g j or an edge from u 1 to f j, for some j {1, }, and the number of 8-tuples with dist B u 1, f j ) = 3 or with dist B w j, g j ) = 3 for some j {1, } is Or 4 k 3 max ). Hence condition II ) fails for Or 4 k 3 max ) 8-tuples, and the number of reverse d-switchings which can be applied to B is r 1) 1 + O kmax d + r k max) )). Taking the ratio completes the proof: the error term from the reverse d-switchings dominates since 1/ k max /. 13

We can now prove our main result. The proof is similar to those in related enumeration results such as [7]. We present the proof in full in order to demonstrate how the factors of r arise in the error bounds since previous results only dealt with r =, or assumed that r was constant). Proof of Theorem 1.1. First we prove that N d=0 r 1) C d = C 0 exp 4 + O )) r 4 kmax 4. 3.3) Let d be the first value of d N for which C d =, or d = N + 1 if no such value of d exists. We saw in Lemma 3. that any B C d can be converted to some B C d 1 using a d-switching. Hence C d = for d d N. In particular, 3.3) holds if C 0 =, so we assume that d 1. By Lemma 3., there is some uniformly bounded function α d such that for 1 d N, where C d C 0 = 1 C d 1 d C 0 Ad) d 1)Cd)) 3.4) Ad) = r 1) α d r 4 kmax 3, Cd) = α d r k max 4 4 for 1 d < d, and Ad) = Cd) = 0 for d d N. We wish to apply Lemma 3.1. It is clear that Ad) d 1)Cd) 0, from 3.4) if 1 d < d or by definition, if d d N. If α d 0 then Ad) Ad) d 1)Cd) 0 by 3.4), while if α d < 0 then Ad) is nonnegative by definition. Now define A 1, A, C 1, C by taking the minimum and maximum of Ad) and Cd) over 1 d N. Let A [A 1, A ] and C [C 1, C ] and set ĉ = 1 0. Since A = r 1) /4 + o1) and C = o1), we have that max{a/n, C } ĉ for sufficiently large, by the definition of N. Hence Lemma 3.1 applies and gives an upper bound N d=0 C d r 1) C 0 exp 4 + O Since e/10) N e/10) 3 log 1, this gives N d=0 C d r 1) C 0 exp 4 )) r 4 kmax 4 + O e/10) ) N. + O )) r 4 kmax 4. 14

In the case that d = N + 1, the lower bound given by Lemma 3.1 is the same within the stated error term, which establishes 3.3) in this case. This leaves the case that 1 d N. Considering the analysis of the reverse switchings from Lemma 3., this case can only arise if But then = Od k max + r k 3 max) = O k max r k max + log ) ). r 1) 4 ) r k = O maxr kmax + log ) = O ) r 4 kmax 4, so the trivial lower bound of 1 matches the upper bound within the error term. Hence 3.3) also holds when 1 d N. Therefore 3.3) holds in both cases. Applying Theorem. and 3.1) gives L r k) = C 0 /r)! = B+ r k) /r)! = completing the proof. exp r 1) 4! /r)! r!) /r n j=1 k j! exp )) r 4 kmax 4 + O r 1) r 1) 4 + O )) r 5 kmax 4, Acknowledgement We are very grateful to Brendan ckay [13] for suggesting that use of the incidence matrix of the hypergraph would simplify the calculations. References [1] C. Berge, Hypergraphs: Combinatorics of Finite Sets, Elsevier, Amsterdam, 1989. [] V. Blinovsky and C. Greenhill, Asymptotic enumeration of sparse uniform hypergraphs with given degrees, Preprint 013). arxiv:1306.01 [3] C. Cooper, A. Frieze,. olloy and B. Reed, Perfect matchings in random r-regular, s-uniform hypergraphs, Combinatorics, Probability and Computing 5 1996), 1 14. 15

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