On the number of spanning trees in random regular graphs

Let $d \geq 3$ be a fixed integer. We give an asympotic formula for the expected number of spanning trees in a uniformly random $d$-regular graph with $n$ vertices. (The asymptotics are as $n\to\infty$, restricted to even $n$ if $d$ is odd.) We also obtain the asymptotic distribution of the number of spanning trees in a uniformly random cubic graph, and conjecture that the corresponding result holds for arbitrary (fixed) $d$. Numerical evidence is presented which supports our conjecture.


Introduction
In this paper, d denotes a fixed integer which is at least 2 (and usually at least 3). All asymptotics are taken as n → ∞, with n restricted to even integers when d is odd.
The number of spanning trees in a graph, also called the complexity of the graph, is of interest for a number of reasons. The complexity of a graph is an evaluation of the Tutte polynomial (see for example [14]). The Merino-Welsh conjecture [10] relates the complexity of a graph with two other graph parameters, namely, the number of acyclic orientations and the number of totally cyclic orientations of a graph. (Noble and Royle [13] recently proved that the Merino-Welsh conjecture is true for series-parallel graphs.) The complexity of a graph also plays a role in the theory of electrical networks (see for example [12]).
We are interested in the number of spanning trees in random regular graphs. The first significant result in this area is due to McKay [7], who proved that for d 3, the nth root of the number of spanning trees of a random d-regular graph with n vertices converges to as n → ∞, with probability one. An alternative proof of this was later given by Lyons [6,Example 3.16].
McKay [8,Theorem 4.2] gave an asymptotic expression for the expected number of spanning trees in a random graph with specified degrees, up to some unknown constant. His result holds when the maximum degree is bounded and the average degree is bounded away from 2 (independently of n). When specialised to regular degree sequences, [8,Theorem 4.2] states that the expected number of spanning trees in G n,d is asymptotic to for some unknown constant c d .
Other work on asymptotics for the number of spanning trees has focussed on circulant graphs, grid graphs and tori (see for example [2] and the references therein).
Our first result, Theorem 1.1, provides the value of the constant c d from (2), proving that For our second result we investigate the distribution of the number of spanning trees in random d-regular graphs using the small subgraph conditioning method, and obtain the asymptotic distribution in the case of cubic graphs, presented in Theorem 1.2. We provide partial calculations for arbitrary fixed degrees, which lead us to conjecture that the corresponding result holds in general (see Conjecture 1.3). In order to precisely state our main results we must introduce some notation and terminology.

Notation and our main results
Let N denote the natural numbers (which includes 0). For integers n, k let (n) k denote the falling factorial n(n − 1) · · · (n − k + 1). Square brackets without subscripts denote the electronic journal of combinatorics 21(1) (2014), #P1.45 extraction of coefficients of a generating function. We use 1(·) to denote both the indicator variable of an event and the characteristic function of a set (the particular set will appear as a subscript). We use standard asymptotic notation throughout, with the exception that indicates convergence in distribution of a sequence of random variables. (We use this notation rather than d → to avoid overloading the symbol "d", which we use for the degree of the graph.) Let G n,d denote the uniform model of d-regular simple graphs on the vertex set {1, . . . , n}. Define the random variable Y G to be the number of spanning trees in a random G ∈ G n,d .
Clearly Y G is identically zero if n 3 and d < 2. A 2-regular graph has a spanning tree if and only if it is connected (that is, forms a Hamilton cycle), in which case it has exactly n spanning trees. Hence the distribution of Y G can be inferred from [15,Equation (11)]. For the remainder of the paper we assume that d 3.
Our first result gives an asymptotic expression for the expectation of Y G .
This theorem is proved at the end of Section 3. Next, for fixed d 3 and for each positive integer j, define Our second theorem gives the asymptotic distribution of the number of spanning trees in the case of cubic graphs. Theorem 1.2. Let Z j ∼ Poisson(λ j (3)), with each Z j independent. Consider the number of spanning trees in a random cubic graph, normalized by the expectation given in Theorem 1.1 for d = 3. The asymptotic distribution of this quantity is given by This theorem is proved in Section 4. We conjecture that an analogous result holds for arbitrary (fixed) d 3.
where Z j ∼ Poisson(λ j (d)) and each Z j is independent.
We present numerical evidence which supports this conjecture in Section 4.1.

Plan of attack
From now on, we omit explicit mention of d in the constants ζ j = ζ j (d) and λ j = λ j (d) from (3).
As is standard in this area, most of our calculations will be performed in the uniform probability space P n,d of pairings (also called the configuration model [1,9,15]. Let d and n be positive integers such that dn is even. Consider a set of dn prevertices distributed evenly into n sets, called buckets. (We prefer the terminology "prevertices" to "points".) A pairing is a partition of the prevertices into dn/2 sets of size 2, called pairs. Then using Stirling's formula. By contracting the prevertices in each bucket to a vertex, each pairing projects to a labelled d-regular multigraph, with loops permitted. Let Ω n,d denote the set of such multigraphs, and denote the projection of a pairing P by G(P ). (We will occasionally informally refer to "partial" pairings, where only a subset of the prevertices are paired. The projection of a partial pairing is defined in the same way.) Each G ∈ G n,d is the projection of (d!) n different pairings (permuting the prevertices in each bucket), so we can recover the uniform model G n,d from P n,d by conditioning on the event that the projected multigraph of a random pairing is simple.
Moreover, this and the convergence in (A1) hold jointly.
We now define the random variables X j and Y to which these results will be applied. For each j 1, let γ j : Ω n,d → N give the number of cycles of length j in a multigraph. (A loop is a 1-cycle, and a pair of edges on the same two vertices is a 2-cycle.) Then the random variable X j = γ j •G is the number of j-cycles in the projection of a random pairing P ∈ P n,d . Write X = (X j ) j 1 for the sequence of all cycle counts. It is well known [1] that for any positive integer m, the random variables X 1 , . . . , X m are asymptotically independent Poisson random variables, and that the mean of X j tends to the quantity λ j = λ j (d) given in (3). Hence Condition (A1) of Theorem 1.4 holds.
Let τ : Ω n,d → N be the function which counts spanning trees in d-regular multigraphs. Define Y G as the restriction of τ to G n,d , and define Y = τ • G. Then, Y G is the number of spanning trees in a random G ∈ G n,d , as in Section 1.1. Y is accordingly the number of spanning trees in the projection of a random pairing P ∈ P n,d . We will investigate the asymptotic distribution of Y G through analysis of Y .
In Section 2 we obtain an asymptotic formula for the expected value of Y . In Section 3 we analyse the interaction of the number of spanning trees with short cycles, establishing that (A2) holds for λ j and ζ j as given in (3). This enables us to prove Theorem 1.1 and to prove that (A3) holds. Then in Section 4 we investigate the second moment of Y . We can prove that Condition (A4) holds when d = 3, leading to a proof of Theorem 1.2. Using our partial calculations for general degrees, we provide numerical evidence that strongly supports Conjecture 1.3.

Expected number of spanning trees
In this section we compute EY . Let T n denote the set of labelled trees on n vertices, so that |T n | = n n−2 by Cayley's formula (see for example, [11] where M T,P is the number of ways to embed T into the multigraph G(P ). (When G(P ) is simple, M T,P is zero or one). Now, we want to condition on the degree of each of the n vertices in T . Define the set of possible degree sequences D n = δ ∈ N n : n j=1 δ j = 2(n − 1) .
We can decompose |P n,d | EY as where T ∼ δ denotes the event that vertex j has degree δ j in T , for all j = 1, . . . n.
To evaluate the innermost sum in (7), fix some δ ∈ D n and some T ∈ T n with T ∼ δ. We need to count the number of pairings that include T , with the embedding of T identified. That is, if for some pairing P , the tree T can be embedded in G(P ) in multiple ways, then we count each different way separately. Now, exactly δ j of the prevertices in bucket j must contribute to T , and there are (d) δ j ways to choose and order these prevertices. So, there are n j=1 (d) δ j ways to pair up the n − 1 edges corresponding to a copy of T . Then, there are prevertices remaining, which can be paired in #P ((d − 2)n + 2) ways. This yields The inner sum in (8) is the number of trees with degree sequence δ, which is the multinomial n − 2 It follows that the total number of ways to choose a spanning tree on n vertices and choose a partial pairing that projects to that tree is Hence, by Stirling's approximation and (4) we conclude that It follows that Hence for d 3 and n sufficiently large, we have EY = 0.

Interaction with short cycles
Recall that X j = γ j • G is the number of cycles of length j in the projection of a random pairing P ∈ P n,d . For some fixed m 0, ρ ∈ N m , let X ρ = m j=1 (X j ) ρ j . In this section we will compute an asymptotic formula for E[Y X ρ ]/EY , in the form required by Condition (A2 ).
We have Note that m j=1 (γ j (G(P ))) ρ j the electronic journal of combinatorics 21(1) (2014), #P1.45 is the number of ways to choose, for each j ∈ {1, . . . , m}, an ordered set of ρ j cycles of length j. This will result in an ordered set of ρ is the number of ordered sets of cycles in which each cycle is disjoint, and γ ρ is the number of ordered sets in which some vertices are shared between multiple cycles. We can further decompose γ ρ by the structure of the interaction between the cycles. That is, according to the multigraph that is the union of the cycles, and the specification of which edges of this union belong to which cycle. This expresses γ ρ as a sum of terms γ (j) ρ . The number of terms J in this decomposition depends on ρ, but is O(1) as n → ∞. Define . We proceed to calculate E (0) . As is standard when applying this method (see for example, [5, Theorem 9.6]), E[Y X ρ ] is asymptotically dominated by E (0) (the contribution due to disjoint cycles). See Lemma 3.1 for some justification for this fact.
Let C n,j be the set of all j-cycles on the vertex set {1, . . . , n}, and define the Cartesian product Each R ∈ C n,ρ is an ordered set of |ρ| cycles. We use the notation R j,k for the kth cycle of length j in R. Next, define Similarly to (6), we have where M (T,R),P is the number of ways to embed the tree T and the cycles in R into the multigraph G(P ).
We further condition on the edge intersection between the embedding of T and the cycles in R. We use a binary sequence of length j to encode the intersection of an j-cycle the electronic journal of combinatorics 21(1) (2014), #P1.45 with a spanning tree. Picking an arbitrary start vertex and direction for the cycle, if the kth edge of the cycle is to be included in the intersection, then the kth element of the corresponding sequence is one; otherwise it is zero. All sequences q ∈ {0, 1} j represent possible intersections, except the sequence (1, . . . , 1), because a tree contains no cycles.
Define the set of all possible intersection sequences for a cycle of length j, by Also, define the Cartesian product So, for each R ∈ R n,ρ , specifying some Q ∈ Q ρ fully specifies the intersection between the cycles in R and a tree T .
We have where M (T,R,Q),P is the number of ways to embed T and the cycles in R in P , such that the intersection between the embedding of T and the cycle R j,k is consistent with Q j,k , for j = 1, . . . , m and k = 1, . . . , ρ j . Fixing Q ∈ Q ρ , we will now evaluate the innermost triple sum in (13). Consider the following process: 1. Choose some R ∈ R n,ρ .
2. Choose a partial pairing that projects to R.
3. Extend this to a pairing of a spanning tree consistent with Q.

Pair the remaining prevertices arbitrarily.
We will find that the number of ways to complete each step is independent of the other steps. Then, E (0) is a product of the number of ways to complete each step, summed over all Q ∈ Q ρ .
First let n ρ = m j=1 jρ j be the total number of vertices in each R ∈ R n,ρ . The number of ways to choose the vertices for some R ∈ R n,ρ is n n ρ and the number of different arrangements of disjoint cycles on those vertices is where jχ(j) is the size of the automorphism group of a j-cycle: That is, the number of ways to complete Step 1 is Next, the number of ways to complete Step 2 is Note for future reference that we have Next, we count the number of ways to extend this pairing to a tree T consistent with Q. We do this by constructing a new irregular pairing model P n,ρ from the prevertices still unpaired after Step 2. Recall that Q describes a union of disjoint paths; for each of these paths, combine the unpaired prevertices remaining in each constituent vertex of the path to form a super-bucket. If the path has k vertices then the resulting super-bucket has k(d − 2) prevertices. Let |Q| be the number of super-buckets formed in this way, so the total number of buckets in P n,ρ is n := n − n ρ + |Q|. Now, consider an extension of a pairing of cycles from Step 2, to a (partial) pairing of the edges of a tree T consistent with Q, as per Step 3. The pairs from this extension correspond uniquely to a (partial) pairing P in the pairing model P n,ρ . By the construction of P n,ρ , the projection T = G(P ) of this pairing is simply T with some subpaths contracted to single vertices. Since contracting edges of a tree cannot create cycles, G(P ) is itself a (spanning) tree. Similarly, every pairing of a tree in P n,ρ corresponds to an extension of a pairing of cycles to a pairing of a tree in P n,d consistent with Q. So the number of ways s (0) 3 to complete Step 3 equals the number of ways to choose and pair up a spanning tree in P n,ρ .
We will perform this count as in Section 2, by conditioning on the degree in T of each bucket in P n,ρ . Put an arbitrary ordering on the |Q| super-buckets, and let d j be the number of prevertices in the jth super-bucket. For a degree sequence δ, let |δ| be its degree sum. Define the sets D Q = δ ∈ N |Q| : δ j d j for all j , The set D Q contains all possible degree-in-T sequences for the |Q| super-buckets. For some δ ∈ D Q , the set D δ contains all possible degree sequences for the n − n ρ remaining ordinary buckets. So, we have where T ∼ (δ , δ) denotes the event that the super-buckets have degree-in-T sequence δ and the remaining vertices have degree-in-T sequence δ. Proceeding as before, after fixing some (T , δ , δ), there are ways to pair the edges of T . There are (9). So, we have where Finally, for Step 4 there are dn − 2n ρ − 2(n − 1) = (d − 2)n − 2(|Q| − 1) prevertices remaining, which can be paired in ways.
Proof. We can estimate general E (j) with some slight modifications to the above calculations. We would need a different R n,ρ ⊆ C n,ρ that contains all possible ways to embed an ordered set of cycles with a particular union U into the vertex set {1, . . . , n}. The intersection between a spanning tree and the cycles in some R ∈ R n,ρ would then be a subforest of U , so it would be more complicated to explicitly define a set Q ρ that encodes all possibilities for the intersection. However, the number of possible intersections is still independent of n.
Using the same 4 steps, the decomposition E (j) = s We conclude that E (j) /E (0) = O n |U |− U . Any non-disjoint union between distinct cycles has more edges than vertices, so the lemma is proved.
We now want to to express (18) in the form required by (A2 ). So, we consider each cycle independently. For a sequence q ∈ I j , let q[k] be the number of paths with k vertices in the intersection encoded by q. So, for Q ∈ Q ρ we have Also, let |q| = j k=1 q[k] be the total number of paths in the intersection encoded by q. Now, recall that if the jth super-bucket was collapsed from a path of length k, then d j = k(d − 2). Also recall that n ρ = m j=1 jρ j and note that As a result, we have .
Note that takes the role of δ j for the jth super-bucket. We have proved that Condition (A2 ) is satisfied. It remains to simplify our expression for λ j . We have Now, recall that (1, . . . , 1) / ∈ I j . So, to evaluate the sum over q, we may identify a particular element in the sequence to be zero. By symmetry, we arbitrarily choose the last. We also condition on |q|: define Note that Λ j,1 = jµ for all j, because the only sequence q ∈ I j with |q| = 1 and q j = 0 is (1, . . . , 1, 0). For |q| > 1, the first path in (the intersection encoded by) a sequence can contain anywhere between 1 and j − 1 vertices. Ranging over the possibilities, we have for t > 0. To solve this recurrence, define the generating function We have If q ∈ I j , then there are j positions to place a zero, and if |q| = t then there are t zeros in q. Hence ).
To complete this section we will establish that conditions (A2) and (A3) of Theorem 1.4 hold, and prove Theorem 1.1.
Proof. The calculations of this section show that Condition (A2 ) of Lemma 1.5 is satisfied with λ j given by (20). Then Lemma 1.5 guarantees that (A2) is satisfied. Using the Taylor expansion of log(1 − z), it follows from (3) that Taking the exponential of both sides and rearranging establishes the stated expression for exp ∞ j=1 λ j ζ 2 j , which is finite for d 3. Hence Condition (A3) holds, as required.
Proof of Theorem 1.1. We know that condition (A2) holds (as proved above), and hence Substituting using (3) and (12)  We now want to calculate EY 2 . As a first step we transform this problem into one of evaluating the coefficient of a certain generating function.
Lemma 4.1. Let d 3 be fixed and define where M (T 1 ,T 2 ),P is the number of ways to embed the ordered pair of trees (T 1 , T 2 ) into the multigraph G(P ). We will estimate this sum by choosing some T 1 , T 2 ∈ T n and counting the ways to pair up their edges, then counting the ways to complete the pairing. We break up this process in a similar way to Section 3: 1. Choose b ∈ {1, . . . , n}, which will be the number of connected components in the intersection of the embeddings of T 1 and T 2 . (As we will see later, when d = 3 we must restrict to b ∈ {1, . . . , n/2 + 2}.) 2. Choose a partition (ν 1 , . . . , ν b ) of n into positive parts. That is, ν j is a positive integer for j = 1, . . . , b and b j=1 ν j = n. Here ν j will be the number of vertices in the jth connected component of the interesection. (We should divide by b! to account for our assumption that the connected components are labelled).
3. Choose a partition of the n vertices into b groups, where the size of the jth group is ν j .
4. In each group, choose a spanning tree on that group and choose a partial pairing that projects to that tree. This specifies a component of the intersection. Now, collapse the buckets in each group into a single super-bucket, giving exactly b super-buckets. The jth super-bucket has dν j − 2(ν j − 1) unpaired prevertices. We now want to pair up two pair-disjoint spanning trees T 1 , T 2 in the collapsed pairing model. These will extend to T 1 and T 2 using the intersection subtrees chosen in Step 4. 5. For j = 1, . . . , b, choose δ 1 j and δ 2 j , the degree of vertex j in T 1 and T 2 respectively, in such a way that δ 1 j + δ 2 j dν j − 2(ν j − 1). These must also satisfy b j=1 as they are the degree sequence of a spanning tree on b vertices.
6. Choose two trees T 1 , T 2 on the b vertices that are consistent with the degree sequences chosen in Step 6.
7. Pair up these two trees in a pair-disjoint way.
8. Pair all remaining prevertices to complete a d-regular pairing.
Given ν, the number of ways to complete Step 3 is By (10), the number of ways to complete Step 4 is The number of ways to complete Step 6 is by (9), and the number of ways to complete Step 7 is Finally, for prevertices remaining, so the number of ways to complete Step 8 is For this construction to make sense, the quantity (d − 2)n − 2(b − 2) must be nonnegative. This is certainly true when d It will be convenient to work with the nonnegative variables defined for j = 1, . . . , b. Let be the set of possible sequences ν from Step 2 and let be the set of sequences arising from Step 5 using (21). Combining all of the above gives the electronic journal of combinatorics 21(1) (2014), #P1.45 It follows that This is equal to the expression in the lemma statement, by definition of S 2 (b).
We now seek to evaluate By Stirling's approximation and the ratio test, the radius of convergence of the series ∞ j=1 is analytic in the disk {z : |z| < 1}. Define β = b/n and let r β ∈ (0, 1) be fixed for each β (we will determine this later). Then, with the contour Γ : [−π, π] → C defined by θ → r β e iθ , we have by Cauchy's coefficient formula. Let be the sample space for pairs (β, θ), and define a n : X n → C by Finally, let We now apply the saddle point method to estimate the sum in (24) in the case that d = 3. Our proof is adapted from that of [3,Theorem 2.3].

It follows that
Next suppose that (β, θ) ∈ F (4) n ∩ X n . Then we have 1 2 < β It remains to consider F (1) n . Define (β, θ) = βn n , θ , so that F n = n Xâ n ( x )dx. For any y ∈ R 2 , define x y = x * + y/ √ n and B n = y : x y ∈ X (1) , so that we can make the change of variables Note that for y ∈ B n , so that B n is approximately a ball of radius O n 1/6 . Next, a first-order Taylor uniformly. Now, for each y ∈ R 2 , we have x y → x * . For n large enough so that y ∈ B n , we have ψ( x y ) → ψ(x * ) by continuity and e nφ(xy) → e 1 2 y T Hy by (29). We therefore have 1 Bn (y)â n ( x y ) → ψ(x * )e 1 2 y T Hy for all y.
We now pull these calculations together to prove the following.
We can now complete the proof of Theorem 1.2.
Hence Conjecture 1.3 is a consequence of (A4).

Support for Conjecture 1.3
Let p d (n) denote the quotient This plot supports our conjecture that p d (n) ∼ 1 for all d 4. Indeed, the rate of convergence to 1 appears to increase as d increases.
We now give an asymptotic result which is equivalent to Conjecture 1.3. Combining