Counting trees in graphs

Erdős and Simonovits proved that the number of paths of length t in an nvertex graph of average degree d is at least (1 − δ)nd(d − 1) · · · (d − t + 1), where δ = (log d)−1/2+o(1) as d → ∞. In this paper, we strengthen and generalize this result as follows. Let T be a tree with t edges. We prove that for any n-vertex graph G of average degree d > t, the number of labelled copies of T in G is at least (1− ε)nd(d− 1) · · · (d− t+ 1) where ε = (4t)5/d2. This bound is tight except for the term 1 − ε, as shown by a disjoint union of cliques. Our proof is obtained by first showing a lower bound that is a convex function of the degree sequence of G, and this answers a question of Dellamonica et. al.


Introduction
If H and G are graphs, let N H (G) denote the number of labelled copies of H in G.More precisely, N H (G) is the number of injections φ : V (H) → V (G) such that φ(u)φ(v) ∈ E(G) for every edge uv of H. Let N G (v) denote the neighborhood of vertex v in a graph G and let d G (v) = |N G (v)|.Let (a) b = a(a − 1)(a − 2) . . .(a − b + 1) when a b 1 and a is real and b is an integer, and let (a) 0 = 1.Let T be a t-edge tree and G = (V, E) be a graph with minimum degree at least t.In this paper, we obtain lower bounds on N T (G).This is a basic question in combinatorics, for example, the simple lower bound v∈V (d(v)) t in the case when T is a star is the main inequality needed for a variety of fundamental problems in extremal graph theory.Counting paths and walks in graphs has numerous applications, such as finding bounds on the spectral radius of the graph and the energy of a graph (see Täubig and Weihmann [6] and the references therein), as well as in standard graph theoretic reductions of combinatorial problems such as counting words over alphabets with forbidden patterns.Blakley and Roy [2] used elementary linear algebraic methods to show that if A is any symmetric pointwise non-negative matrix and v is any pointwise non-negative unit vector, then for any t ∈ N, A t v, v Av, v t .If t is even, this is a simple consequence of the fact that A is diagonalizable and Jensen's Inequality applied to a convex combination of the t-th powers of the eigenvalues of A. When A is the adjacency matrix of an n-vertex graph of average degree d, this shows that the number of walks of length t is at least nd t .There are now a number of proofs of this result, for instance another approach to counting walks was used by Sidorenko, using an analytic method and the tensor power trick [5], as well as generalizations to various other inequalities involving walks (see [6] and the references therein).
Alon, Hoory and Linial [1] showed that the number of non-backtracking walks in an nvertex graph G of average degree d and minimum degree at least two is at least nd(d−1) t−1 , which is again tight for d-regular graphs.A key fact in their proof is that if one considers a random walk of length t on the graph, starting with a uniformly randomly chosen edge, then the distribution of every edge of the walk is also uniform.The method of Alon, Hoory and Linial [1] was generalized in Dellamonica et.al. [3] to show that the number of homomorphisms of a t-edge tree T in G is at least nd v∈V d(v) (t−1)d(v)/nd .By Jensen's Inequality, this is at least nd t when G has no isolated vertices.
In Problem 4 of [3], the question of finding similar lower bounds for N T (G) when T is a tree is raised.In the case T is a path with three edges, it was shown in [3] that N T (G) nd v∈V (d − 2) 2d(v)/dn when G has minimum degree at least three.If P is a t-edge path, and G is a d-regular graph, then clearly N P (G) n(d) t with equality for d t if and only if every component of G is a clique.Erdős and Simonovits [4] showed that if G is any graph of average degree d, then N P (G) n(d − o(d)) t as d → ∞.In this paper, we answer Problem 4 of [3], extend the result in [4] to arbitrary trees and strengthen their lower bound to n(d − O(1/d)) t as d → ∞: Theorem 1.If T is any t-edge tree and G is an n-vertex graph of average degree d t, and This result is best possible up to the factor 1 − ε d in view of the graph comprising n/(d + 1) cliques of order d + 1 t + 1 when d + 1 is a positive integer dividing n, since these graphs have exactly n(d) t copies of T .Note that since Since we do not have an example of a graph G with n vertices and average degree d for which N T (G) < n(d) t , we propose the following conjecture: Conjecture 2. For every t > 1, there exists d 0 = d 0 (t) such that for any tree T with t edges and any n-vertex graph G with average degree d d 0 , N T (G) n(d) t .
Once more the example of disjoint complete graphs of order d+1 shows that this conjecture if true is best possible.Perhaps it is even true that if d is an integer, then this is the only example where equality holds.
This paper is organized as follows.In the next section, we determine a lower bound on N T (G) for a t-edge tree T , as a function of the degree sequence of G.This answers Problem 4 of [3].The proof of this result extends the method of [1] by carefully controlling the distribution of the edges in a random embedding of a tree in a graph of average degree d.
The function of degrees is "almost" convex, and in Section 3 we give a suitable approximate form of Jensen's Inequality, which is used in Section 4 to derive Theorem 1.

Degree sequences and counting trees
As a stepping stone to Theorem 1, we are going to prove the following theorem, which gives a lower bound for N T (G) in terms of the degree sequence of G, and addresses Problem 4 in [3]: Let T be a t-edge tree, and let G = (V, E) be an n-vertex graph with average degree d and minimum degree k > t > 1. Then where for and for Proof.Let Ω be the set of all labelled copies of T in the graph G. Fix a breadth-first search ordering x = (x 1 , x 2 , . . ., x t+1 ) of the vertices of T , starting with a leaf x 1 of T .For 2 i t + 1, let a(i) be the unique h < i such that x h is the parent of x i in the ordering x.Note that a(2) = 1 and a(3) = 2. Let v 1 v 2 be a randomly and uniformly chosen (oriented) edge of G and define φ(x 1 ) = v 1 and φ(x 2 ) = v 2 .Having defined φ(x j ) = v j for all j < i, let x i be mapped to a uniformly chosen vertex of One can always define φ in this manner since d(v a(i) ) k t + 1 i and therefore Here we noted that v a(i) ∈ {v 1 , v 2 , . . ., v i−1 }.Then φ gives a probability measure P on the sample space Ω.Let v denote a vector (v 1 , v 2 , . . ., v t+1 ) of vertices of G.We write φ( x) = v to denote the event that φ( By the inequality of arithmetic and geometric means: In the next two claims, we fix v ∈ V .For i ∈ [t + 1], define We observe that g 1 (v) = g 2 (v) = d G (v)/dn.For 3 i t + 1, we determine upper and lower bounds on To bound the quantities g i (v), let T i be the subtree of T consisting of the vertices x 1 , x 2 , . . ., x i in order.Now consider a breadth-first search ordering y 1 , y 2 , . . ., y i of T i such that y 1 = x i and y i = x 1 .For 2 j i, let b(j) be the unique h < j such that y h is the parent of y j in the ordering y 1 , y 2 , . . ., y i and let Proof of Claim 1.The number of choices of an embedding of T i with x i mapping to v is 1. (7) Here we are counting the images of T i according to the breadth-first search ordering y 1 , y 2 , . . ., y i of T i with y 1 = x i and y i = x 1 , so that the image of y j is u j for j ∈ [i].At the stage where y j is to be embedded, we have to select a vertex u j ∈ N G (u b(j) ) such that u j is not equal to any of v = u 1 , u 2 , . . ., u j−1 , which have already been embedded.By definition of N + (u b(j) ), this is equivalent to Suppose we fix such an embedding with vertices u 1 , u 2 , . . ., u i .Let v 1 , v 2 , . . ., v i be the re-ordering of u 1 , u 2 , . . ., u i such that x j is mapped to v j for j ∈ [i].This fixed embedding has probability We now prove {v a(j) : 3 j i} and {u b(j) : 3 j i} are equal as multisets.For example, if T i is a path, then these multisets are sets, and {v a(j) : 3 i − 1, the number of times u appears in the multiset {u b(j) : 3 j i} is the number of j with 3 j i such that b(j) = .In other words, it is the number of times u is a parent of some vertex u j in the ordering u 1 , . . ., u i .This is precisely d T (y ) − 1.Similarly, the number of times u ∈ {v 2 , v 3 , . . ., v i−1 } appears in the multiset {v a(j) : 3 j i} is the the electronic journal of combinatorics 23(3) (2016), #P3.39 number of times u is a parent of some vertex v j in the ordering v 1 , . . ., v i , and this is again d T (y ) − 1.We conclude {v a(j) : 3 j i} and {u b(j) : 3 j i} are equal as multisets.Therefore by (9), Replacing the summand 1 in (7) with P(φ( x) = v), we obtain Claim 1.
For each v ∈ V and 3 i t + 1, define θ i (v) by Note that since v∈V g (v) = v∈V P(φ(x ) = v) = 1 for all t + 1, we have and this verifies (2).The goal of the next claim is to verify (3).
Proof of Claim 2. We only prove the upper bound, since the lower bound is similar.For convenience, for 3 i, and v = u 1 = u b(2) , define This shows f (i) . Continuing in this way we obtain the electronic journal of combinatorics 23(3) (2016), #P3.39 For any j such that 3 Inserting this in the preceding upper bound for g i (v), we find for any j with 3 j i t + 1, This also holds for i = 2 as Replacing i with a(i) 2 and selecting j = i, By the definition (10), this gives θ i (v) (k/(k − t)) t .The lower bound is similar: we use the lower bound on g i (v) from Claim 1, together with the observation Then we use for 3 j t + 1, and as above this gives the required lower bound on θ i (v).Now we finish the proof of Theorem 3: (5) becomes where θ i (v) satisfies ( 2) and (3), by Claims 1 and 2. This proves (1).
).If n is even, y i = γ for all i n and the electronic journal of combinatorics 23(3) (2016), #P3.39 This shows the bound in Lemma 4 is asymptotically tight as δ → 0. Note that if δ = 0, then n i=1 x y i /γn i γ directly from Jensen's Inequality.

Proof of Theorem 1
We prove Theorem 1 by induction on n.Let G be an n-vertex graph with average degree , then 1 − ε d 0 and so Theorem 1 is true in this case.This also proves the theorem when n < 16t 2 .Suppose d 16t 2 .If k is the minimum degree of G, we consider the two cases k < d/8 and k d/8 separately.
Case 1. k < d/8.In this case, remove from G a vertex of degree k to get a graph H with n − 1 vertices and average degree at least From the second to the third line, we used 3d/4n > 9d 2 /16n 2 since d < n, and 15d 2 /16n < d 2 /n.Recall ε d = (4t) 5 /d 2 .Note also d > d t, so by induction, and since 1−ε The proof is complete in this case.

Concluding remarks
• Theorem 1 can be improved if structural information on G is given.For instance, if G is an n-vertex triangle-free graph of average degree d, and T is the t-edge path, then one can adapt the proof of Theorem 1 to obtain A disjoint union of complete bipartite graphs K d,d shows this is tight up to the factor 1 − O( t 5 d 2 ).It is plausible as in Conjecture 1 that this construction has the fewest copies of T amongst all triangle-free graphs of average degree d, when d is large enough.
• Fix a tree T with t edges and degrees 1, d 1 , d 2 , . . ., d t in a breadth-first labelling starting with a leaf.A local isomorphism of T in a graph G is a neighbourhood preserving embedding of T in G i.e. a map φ : V (T ) → V (G) such that if φ(x) = v for some x ∈ V (T ), then φ(y) = φ(z) whenever y, z ∈ N T (x) are distinct.Using the method of Theorem 3, one can show that the number of local isomorphisms of T in G is at least If G has minimum degree say at least 2t, and average degree d, it is possible to show that the above expression is at least If T is a path, then we recover the main result of [1] stating that the number of nonbacktracking walks of t edges in G is at least nd(d − 1) t−1 .Furthermore, the bound is tight since equality is achieved for any d-regular n-vertex graph.Finally, if G has girth at least diam(T ), then every local isomorphism of T in G is an isomorphism, so we obtain the lower bound Again, this is tight for any d-regular graph of sufficiently large girth.