The minimum number of spanning trees in regular multigraphs

In a recent article, Bogdanowicz determines the minimum number of spanning trees a connected cubic multigraph on a ﬁxed number of vertices can have and identiﬁes the unique graph that attains this minimum value. He conjectures that a generalized form of this construction, which we here call a padded paddle graph , would be extremal for d -regular multigraphs where d ≥ 5 is odd. We prove that, indeed, the padded paddle minimises the number of spanning trees, but this is true only when the number of vertices, n , is greater than 9 d +68 . We show that a diﬀerent graph, which we here call the padded cycle , is optimal for n < 9 d +6 8 . This fully determines the d -regular multi-graphs minimising the number of spanning trees for odd values of d . We employ the approach we develop to also consider and completely solve the even degree case. Here, the parity of n plays a major role and we show that, apart from a handful of irregular cases when both d and n are small, the unique extremal graphs are padded cycles when n is even and a diﬀerent family, which we call ﬁsh graphs , when n is odd.


Introduction
The celebrated Matrix-Tree Theorem establishes a link between the number of spanning trees of a (multi-)graph and linear algebra, thereby providing an efficient way to obtain the number of spanning trees of a given graph through the computation of the determinant of a certain matrix.
It tells us little, however, about the extremal values taken by this number over graph classes.
Following the extremal graph theory tradition, a number of works pursued this line of research.
In this context, a very natural class of graphs to consider is that of regular graphs.In particular, the question has been well studied for regular simple graphs and there are some results on asymptotic values for the minimum and maximum number of spanning trees a connected France.Email: sereni@kam.mff.cuni.cz.
‡ Carnegie Mellon University in Qatar.Email: zyilma@qatar.cmu.edu.Supported by the Qatar Foundation through Carnegie Mellon Qatar's Seed Research program. 1 where the infimum and supremum are taken over all connected d-regular n-vertex simple graphs.
McKay [10] showed that while Alon [1] proved that 1/(d+1) . ( For the case d = 3, Kostochka [8] showed, in a strong sense, that δ d = 2 3/4 , by proving that τ (G) ≥ 2 3(n+2)/4 for all cubic simple graphs G on n ≥ 5 vertices (this result was shown to hold for the class of 2-connected cubic graphs in earlier work by Valdes [12]).The value δ 3 = 2 3/4   matches the upper bound given by (1).To our knowledge, the exact value of δ d is not yet determined for d ≥ 4 and, as underlined by Alon [1], it should indeed be a difficult question.
Alon [1] also proposed to study the question on multigraphs, noticing that loops should not be allowed -for otherwise, for odd d, there is always a d-regular n-vertex multigraph with a unique spanning tree.Let δ d be defined as δ d except that the infimum is taken over all connected d-regular n-vertex multigraphs (without loops).Alon sketched a neat argument proving that δ d has order exactly √ d, where the lower bound is obtained by a slight modification of his Theorem 1.1.Further, he explained that the conclusion of van der Waerden's conjecture, which had been already established by then [3,4] (nowadays the reader can also consult Gurvits's proof [6] for an elementary and totally different argument) implies that any d-regular n-vertex loopless multigraph actually contains (Ω( √ d)) n linear forests -that is, forests such that each connected component is a path.
We shall provide an exact formula for δ d for all values of d, and actually even the exact minimum value of τ over the class of connected d-regular n-vertex loopless multigraphs for all values of d and all possible values of n.In addition, we explicitly provide all graphs attaining this minimum value.Apart from a few exceptional cases, the extremal graphs belong to one of the three families of graphs which we introduce here.
Definition 1.1.Let d be an integer greater than 2 and n an integer greater than 3.
• If n is even, then P C d,n is the padded cycle graph, illustrated in Figure 1, left: it is the d-regular cycle graph on n vertices with edges of alternating multiplicities 1 and d − 1.
We will mention here that the case d = 3 and n ≥ 6 of Theorem 1.2 was recently solved by Bogdanowicz [2].Indeed, our work was motivated by his conjecture that the padded paddle graph is the unique extremal structure that minimises the number of spanning trees in connected odd-regular graphs.The existence of two competing structures, contrary to the conjecture, differentiates the odd degree case and makes inductive approaches more difficult.
While our proofs of Theorems 1.2, 1.3 and 1.4 share the same general approach, some technicalities due to the parities of d and n and some irregular cases when d is even and n is small necessitate separating the exposition.We will start by collecting, in Section 2, the common notation and preliminary results (many of which we recall from the literature) which will be applicable in all cases.The proof of Theorem 1.2, which completely solves the odd degree case, is presented in Section 3. The even degree case is discussed in Section 4. This contains the proofs of Theorems 1.3 and 1.4 as well as all exceptional cases not covered in those theorems.

Preliminaries
As the majority of our work considers graphs that contain multiedges, we will use the terms "graph" and "multigraph" interchangeably.We use the term simple graph if we want to forbid multiedges (except where explicitly stated, all graphs in this paper are loopless).
Given a multigraph G and any two vertices i, j ∈ V (G), we define w G (i, j) to be the number of edges of G between i and j.We write i ∼ G j if w G (i, j) ≥ 1, that is, if i and j are adjacent in G.If w G (i, j) = 1 then we may speak unambiguously of the edge ij of G.For convenience, if f is an edge of G then we define the multiplicity of f (in G) to be the number of edges of G with the same end-vertices as f .

Concavity of τ
Let L(G) be the Laplacian matrix of the graph G; that is, for all i, j ∈ V (G).The Matrix-Tree Theorem, which applies to multigraphs also, gives the relationship between Laplacian matrices and the number of spanning trees.
with equality only if A = cB for some c > 0. If we take A and B to be the (n − 1) × (n − 1) matrices obtained from L(H 1 ) and L(H 2 ) by deleting their first rows and first columns, thus creating positive definite matrices, we obtain (using also Theorem 2.1) 1) , and the result follows.
We will often apply the following consequence of Proposition 2.

Lifts
It is well known that if H is a graph and f is an edge in H, then where H/f is the graph obtained by contracting the edge f (that is, deleting all edges between the endpoints of f and then identifying the endpoints).As, in general, both deletion and contraction of an edge in a regular graph result in a graph which is not regular, we regain regularity by employing a 'lifting' operation similar to one used by Ok and Thomassen [11].Let x, y 1 and y 2 be three distinct vertices in a graph H, and suppose that f i is an edge in H between x and y i , for i ∈ {1, 2}.Lifting f 1 and f 2 means deleting the two edges f 1 and f 2 and adding an edge between y 1 and y 2 .If x is a vertex of degree 2m in H, a complete lift of x is the process of first performing a sequence of m lifts of pairs of edges incident with x and then deleting the vertex x (which is, by then, isolated), thereby producing a multigraph H x .Observe that if there exists a vertex y such that w H (x, y) ≥ m + 1, then it is not possible to perform a complete lift of x since it will not be possible to pair up the edges incident with x such that edges in a same pair span three different vertices, as required in our definition of lift.Conversely, it is possible to perform a complete lift of x as soon as w H (x, y) ≤ m for all vertices y.It is possible to produce a connected multigraph H x via a complete lift of x if, in addition, H is connected and H − x has at most m + 1 components.
Theorem 2.4 (Ok and Thomassen [11]).Let H be a graph with a vertex x of degree 2m.Let H x be a graph obtained from H by a complete lift of x.Then where where the minimum is taken over all sequences of positive integers d 1 , . . ., d k with varying length k such that k i=1 d i = 2m, and over all connected k-vertex graphs X with degree sequence d 1 , d 2 , . . ., d k such that d i ≤ d i for each i ∈ {1, . . ., k}. [11] have determined the values c 1 = 1, c 2 = 2, c 3 = 8/3, and c 4 = 18/5.

Ok and Thomassen
We observe that while in the definition of c m the minimum is taken over all graphs X with degree sequence d i ≤ d i (and, therefore, potentially fewer than m edges), it is enough to consider only graphs with exactly m edges, as we formalise and show next.This has the benefit of reducing the search space for an exhaustive computational search to determine more values of c m and also allows us to combine Theorem 2.4 with a (direct extension to multigraphs of a) theorem of Grone and Merris [5] to find a general lower bound for c m .Proposition 2.5.Using the notation of Theorem 2.4, , As long as there are at least two indices i = j such that d i < d i and d j < d j , we can add a new edge between the two corresponding vertices of X to form a new connected graph X .
The number of spanning trees of X is larger than that of X, and thus X along with the sequence D show that c m is not attained by D and X.We may thus assume that there exists a unique index i ∈ {1, . . ., k} such that d i < d i .It then follows that d i ≤ d i − 2 as both k i=1 d i and k i=1 d i are even.Consider the sequence (s j ) 1≤j≤k+1 defined by Let X be the connected graph obtained from X by adding a new vertex of degree 1 joined to the vertex with degree d i .Then X has degree sequence s 1 , . . ., s k+1 , where and therefore c m is not attained by D and X, which concludes the proof.
Taking into consideration the result of Proposition 2.5, we have calculated a few more values of c m using exhaustive computer search.The values, and the graphs attaining them, are given in Table 1.(For m = 5, the diamond graph, the graph obtained from K 4 by removing one edge, also attains the value c 5 = 9/2.) Knowing that X has exactly m edges also allows us to use the following result of Grone and Merris [5], which was originally stated for simple graphs but the proof of which, using linear algebra, applies equally to multigraphs.

Optimal Substructures
One may observe that all three of our extremal graph families contain long paths with edges of alternating multiplicity 1 and d − 1.Additionally, two of these families contain what we call pendant triangles.To be precise, let G be a connected multigraph with at least 4 vertices, and suppose that {u, v, w} is a set of three vertices inducing a triangle T in G.We say that T is a pendant triangle if at most, and hence exactly, one vertex in T has a neighbour not in T .This vertex is then an articulation point of G, and the other two vertices in T are the terminal vertices of T .The following two statements hint at the structural relevances of long alternating paths and of pendant triangles.Proof.The degree condition and the number of edges imply that almost every vertex in H has degree exactly d.Let x be a vertex of degree at most d − 1.Let B 1 , . . ., B r be the maximal 2-connected blocks of H with x ∈ B 1 .Let n i be the number of vertices in B i .Observe that Pick vertices s i , t i ∈ B i where s 1 = x, s i is an articulation point for all i ≥ 2, and t i = s i is arbitrary.Let s i = y i,1 , y i,2 , . . ., y i,n i = t i be an st-labeling (also known as a bipolar orientation) of B i .That is, for each j ∈ {2, . . ., n i − 1}, the vertex y i,j has at least one neighbour in B i that comes before it and at least one neighbour in B i that comes after it in the ordering.It is well known that there is such an ordering for any pair of vertices (s, t) in a 2-connected graph [9].
We can build a spanning tree of B i by selecting, for each vertex y i,j with j ≤ n i − 1, one of its incident edges leading to a vertex that comes after y i,j in the ordering in B i .For j ∈ {1, . . ., n i − 1}, let d i,j be the number of edges in B i between y i,j and {y i,s : s > j}.Observe that τ (B i ) ≥ n i −1 j=1 d i,j with equality only if B i has exactly two vertices (otherwise, B i contains a cycle and other spanning trees exist -where some vertex y i,j has more than one neighbour y i,s with s > j).Then, ( The product of the rightmost side of (2) contains r i=1 (n i − 1) = n − 1 terms.Note also that the number of edges in B i is exactly n i −1 j=1 d i,j , and hence the number of edges in H is r for all i, j, the product is minimised when n/2 of the terms are d − 1 and the remaining n/2 − 1 terms are 1.Therefore, τ (H) Note also that if the block tree of H is not a path, it will have at least three leaves.This would imply the existence of at least one terminal vertex t i,n i of degree d.This means that B i has at least three vertices and the above inequality is again strict.Therefore, equality is attained only if H is the path graph as specified.
Given a subset X ⊆ V (G), let ∂X be the number of edges with exactly one endpoint in X. Proof.We are going to build a connected d-regular n-vertex multigraph G that has fewer spanning trees than G unless the conclusion of the statement holds in G.To this end, let H be the subgraph of G induced by {u, v, w}, and note that H is necessarily a triangle.Let G H be obtained from G by replacing u, v, w by a single vertex x joined to each vertex having a , as every spanning tree of G decomposes into a spanning tree of T and a spanning tree of G H = G − y − z.
We now observe that τ (T ) ≤ τ (H) with equality if and only if H and T are isomorphic.

The Odd Degree Case
We here prove Theorem 1.2, which completely solves the odd degree case.We will start by determining the number of spanning trees in padded paddles and padded cycles.It should be noted here that a misprint appears in the expression for τ (P P d,n ) given by Bogdanowicz 1.For any integer n ≥ 6, 2. For any integer n ≥ 4, We find it convenient to define the terms and at times use the following equivalent reformulation of Theorem 1.2.

Theorem 3.2. Let G be a connected d-regular n-vertex multigraph where d ≥ 3 is odd and n
Furthermore, the only graphs attaining the minimum are the padded cycle graph if n < 9d+6 8 and the padded paddle graph if n > 9d+6 8 .
The proof proceeds by induction on the number n of vertices.We show in our base case that the theorem holds for n ∈ {4, 6}.The inductive step makes use of the observation that 2. For all odd d ≥ 5, the padded cycle has the fewest spanning trees of all connected d-regular multigraphs on 6 vertices.

The padded paddle has the fewest spanning trees of all connected cubic multigraphs on 6
vertices.
Proof.By Proposition 2.2, any graph that minimises the number of spanning trees cannot be the convex combination of other graphs.In particular, Corollary 2.3 implies that if such a graph contains C 4 as a subgraph, then there cannot be a path, edge-disjoint from the cycle, that connects opposite vertices in the cycle.This implies, for one, that the underlying simple graph cannot contain the diamond graph.Note also that every vertex in the underlying simple graph must have degree at least 2.
For n = 4, it follows that the underlying simple graph must be C 4 .Then, by Corollary 2.3, the multigraph minimising the number of spanning trees is the padded cycle.
For n = 6, we will show first that the only graphs that are not convex combinations of other graphs are the four graphs in Figure 3 graphs with an equal number of edges but that is not the case).

Case 2:
The underlying simple graph contains C 4 .
Suppose that v 1 v 2 v 3 v 4 is a 4-cycle (necessarily chordless).We consider how to extend the graph creating neither a new path between v 1 and v 3 nor one between v 2 and v 4 . .We have Then, H is a path graph between u and x with edges of alternating multiplicities d − 1 and 1.
with strict inequality unless H is the alternating path as described.
One immediate consequence of Lemma 3.4 is the following.
Moreover, for every vertex x, the graph G − x has at most We proceed to show that every edge of G has one of the extreme multiplicities 1 or d − 1 unless it is contained in a pendant triangle.For each of the three cases we then show necessary structural properties fully characterizing the only two possible structures for G.
We now consider edges of multiplicity 1 and d − 1, starting with d − 1. thus applies.We make the following observations: Plugging 1 and 2 into the last equality given by 3, we obtain We observe that equality holds only if all inequalities written are equalities, which cannot hold  Proof.Assume, on the contrary, that w G (u, v) = 1, the graph G − uv is connected, u and v are not part of a pendant triangle and yet G is not the padded cycle.We will show that τ (G) is too large by writing and bounding from below each of the two terms in the right side.In both cases, we will perform a complete lifting operation so we first argue that it is possible to obtain connected graphs at the completion of the respective complete lifting operations.We start by establishing some facts on the multiplicities of edges incident with u or v.It turns out that (A) actually leads to a stronger statement.

(B).
Neither u nor v is incident to an edge of multiplicity greater than d−1 2 .
Proof.Suppose, on the contrary, that w G (u, u ) ≥ d+1 2 .Since d+1 2 ≥ 2 as d ≥ 3 and u is not incident to an edge of multiplicity d − 1 by (A), we deduce from Lemma 3.6 the existence of a vertex w such that T = {u, u , w} induces a pendant triangle.Since {u, v} is not part of a pendant triangle, it follows that w = v and ( We now turn our attention to τ (G − uv).In G − uv, both u and v have even degree d − 1.
We will first completely lift u in G − uv and then proceed to completely lift the vertex v in the resulting graph.The assertion (B) guarantees that we can indeed sequentially completely lift the vertices u and v in G − uv.It directly follows from the definitions that a graph produced by this process will be d-regular and have n − 2 vertices.What remains is to show that this process can lead to a connected graph.
Let G uv be a graph obtained by first completely lifting u in G − uv (yielding a connected graph G u ) and next completely lifting v in G u .Suppose the lifts are performed so that G uv has the smallest possible number of connected components among all graphs constructed in this way.
We will show that G uv is connected.
To this end, let E u be the set of edges created by lifting u, that is, the edges in G u but not in G. Similarly, let E v be the set of edges created by lifting v, that is, the edges in G uv but not in G u .We build an auxiliary multigraph L (possibly containing loops) as follows.For Consequently, at most one vertex of L has degree 1, all the others having degree at least 2.
Therefore, every connected component of L contains a cycle (where a loop is considered to be a cycle).
Suppose now that G uv , and hence L, is not connected.Then L contains an edge yz that is not in the same connected component as x and belongs to a cycle.Without loss of generality, assume that yz ∈ E u .Let uu 1 and uu 2 be the two edges of G − uv that were lifted to create yz.
By the definition of x, there exists an edge xx that belongs to E u .Similarly, let uu 3 and uu 4 be the two edges of G − uv that were lifted to create xx .Now, if we rather lift the pairs (uu 1 , uu 3 ) and (uu 2 , uu 4 ) instead of (uu 1 , uu 2 ) and (uu 3 , uu 4 ), and keep all other lifts the same, we obtain an auxiliary graph L that has fewer connected components than L. Indeed, the edge yz belongs to a cycle in L, meaning that the two different connected components of x and y in L will become one in L .This contradicts the definition of G uv , and thus implies that L, and hence G uv , is connected.Therefore, we can apply Theorem 2.4 to obtain Combining ( 3) and ( 4), we have and, noting that β(n) ≤ 4 3 β(n−2) for n ≥ 8, we infer that G fails to be optimal if c d−1 For d ≥ 7, we apply Proposition 2.7 and obtain Let us now gather the implications of the lemmas above.One consistent theme is that the padded cycle is a candidate to be the graph with the fewest spanning trees.If G is not the padded cycle, then each edge must either disconnect the graph (and have multiplicity 1 or d − 1) or belong to a pendant triangle.So the structure of G must be a tree with edges of multiplicity 1 or d − 1 and some pendant triangles.However, by Lemma 3.4, the tree structure cannot have two adjacent edges of multiplicity 1.Therefore, G must be a long, alternating path with pendant triangles at either end (internal vertices cannot support a pendant triangle).It only remains to show that there is exactly one pendant triangle at each of the two ends of the long path.Then Simplifying the above, we obtain f (a) = −da 2 + 2a(d 2 + d − 1) + (−2d 2 + 3d − 1), and so f (a) is a quadratic polynomial which achieves its maximum value at a = (d 2 + d − 1)/d.As

The Even Degree Case
The even degree case introduces some technical challenges not present in the odd degree case.
For one, while the number of vertices, n, has to be even when d is odd, its parity is not restricted when d is even.As seen in the statements of Theorems 1.3 and 1.4, the extremal graphs are different depending on the parity of n.
In addition, the d even case contains a few irregular extremal graphs, specifically when the pair (d, n) belongs to {(4, 6), (4, 7), (4,8), (4,9), (4,10), (4,11), (4, 13), (6,7), (6,9), (6,11), (8, 9)}, that are not covered in the statements of Theorems 1.3 and 1.4.To ease readability, the degree-4 case is given alone in Theorem 4.1, being peculiar in the sense that it is the only case where a version of the padded paddle graph can be optimal.Finally, the remaining cases excluded from With a slight abuse of notation, for n ≥ 5, let P P 4,n be the degree 4 padded paddle graph illustrated in Figure 4: we start from an (n − 4)-vertex path with end-vertices u 1 and u 2 and all edges of multiplicity 2, and then for each i ∈ {1, 2} we add a triangle with edge multiplicities 1, 1 and 3 and identify its vertex of degree 2 with u i (notice that u 1 = u 2 if n = 5, and then P P 4,5 is isomorphic to the 4-regular fish graph F 4,5 ).Here, τ   (d, n) belongs to {(6, 7), (6,9), (6,11), (8, 9)}, then with equality if and only if G is isomorphic to F * d,n (see Figure 5 for illustration).Proof.First, suppose that there is no vertex x such that {u, v, x} forms a triangle in G. Let G u be any connected multigraph obtained by completely lifting u in G. Then for every neighbour w of v in G different from u, the multiplicities of the edge vw in G and in G u are the same.In addition, for any new neighbour z that v might gain when lifting u, the multiplicity of the edge vz in G u cannot exceed the multiplicity of the edge vu in G, and hence is at most d 2 .It follows that v can be completely lifted in G u , as required.Now suppose that there is a vertex x such that {u, v, x} forms a triangle T in G. Since G does not contain a diamond by Corollary 2.3, the vertices u and v have no common neighbour other than x.Recall also that by Lemma 2.9, if {u, v, x} induces more than d edges, then the triangle T must be pendant.As a pendant triangle contains at most one vertex not incident to an edge of multiplicity greater than d 2 , we deduce that

Base Cases and Preliminary Results
Consequently, when lifting u, we may chose to pair at least min{w G (u, v), w G (v, x)} edges between u and v with edges incident to u but not to x.Then, after the complete lift of u, the multiplicity of the edge vx in the obtained graph G u is at most max{w(u, v), w(v, x)} ≤ d/2, and therefore we can completely lift v in G u .

Proof of Theorem 1.3
We prove Theorem 1.3 by induction on the number n of vertices.Throughout this section, the integer n is assumed to be even.
A key element of our inductive proof is the observation that, for n ≥ 6, In other words, u and v are terminal vertices of a (same) pendant triangle.
Proof.First note that the hypothesis of the lemma cannot hold for d = 4.So suppose that d ≥ 6 and set m = w G (u, v).Suppose there is no vertex z such that w(u, z)+w(v, z) ≥ d−m+1.Let G be a connected d-regular multigraph on n − 2 vertices obtained from G by deleting all m edges between u and v and identifying the two vertices u and v into a new vertex x.
where we used that d ≥ 6 for the third line and m for the second line.Therefore, G fails to be optimal unless there is a vertex z such that and hence Lemma 2.9 yields that {u, v, z} induces a pendant triangle in G.As w G (u, v) ≥ d/2+1, the vertex z must be the articulation point.This concludes the proof.
It follows from Lemma 4.7 that every vertex of G that is not a terminal vertex of a pendant triangle either is incident to an edge of multiplicity d−1, or is incident only to edges of multiplicity at most d/2.We call a vertex exceptional if all its incident edges have multiplicity at most d/2., and it thus follows that If d = 8, then we use the exact value of c 4 to deduce the following: For d = 6, as c 3 = 8/3 we have c 2 3 = 64/9 < 15/2 = 3(d − 1)/2.However, 64/9 > 4(d − 1)/3 ≥ τ (P C 6,n )/τ (P C 6,n−2 ) for n ≥ 8 and our assertion will hold if we show that P C 6,6 is optimal for n = 6.Indeed, let G u be a connected 6-regular multigraph obtained from G by completely lifting u.As τ (F 6,5 ) = 220 ≤ τ (G u ) by Lemma 4.4, it follows that which conclude the proof.
either has an edge of multiplicity d − 1 or belongs to a pendant triangle.If all vertices are of the former type, then G must be the padded cycle P C d,n .Otherwise, since G is connected, edges of alternating multiplicities 1 and d − 1 must form odd cycles.Consequently, if d ≥ 6, then Lemma 4.8 implies that G must be a collection of odd cycles (and pendant triangles) all sharing one vertex.However, this requires G to have an odd number of vertices, a contradiction.This establishes the statement of Theorem 1.3 whenever d ≥ 6.

Proof of Theorem 1.4 and Proposition 4.3
In this section, the integer n is assumed to be odd.We will simultaneously prove Theorem 1.4 and Proposition 4.3 by first identifying a family of graphs which includes the stated extremal structures for the two statements.Next, we show that, within this family, the graphs F d,n and F * d,n are the minimisers of τ for the stated values of d and n.We then establish the desired result by proving that, for d ≥ 6, any minimiser of τ over the whole class of connected d-regular multigraphs must belong to the above-identified family.
Let H d,n be the class of all connected d-regular n-vertex multigraphs that consist of pendant triangles and padded odd cycles (i.e., containing only edges of multiplicity 1 or d − 1) all sharing the same vertex.This vertex is called the centre of the graph.Since the centre is a cut-vertex, the number of spanning trees of such a graph is the product of the number of spanning trees of the odd cycles and triangles composing it.Let H d,n be a minimiser of τ over the class H d,n .

Optimising within H d,n .
We start by analyzing edge multiplicities within the pendant triangles.Proof.Suppose, on the contrary, that T 1 = {u, v, w} and T 2 = {u, x, y} are two pendant triangles in H d,n sharing the vertex u and containing no edges of multiplicity 1.Consider an even walk W = uvwuxyu that starts with u then traverses both triangles (arbitrarily choosing any of the (at least two) edges between consecutive vertices) and let M 1 = {uv, wu, xy} and M 2 = {vw, ux, yu} be the complementary sets of alternating edges in W .We note that both since each of the three fractions is less than 1 by convexity.Consequently, τ (H t d,n ) is minimised either at t = 1 or at t = t * = min{d/2 − 1, (n − 3)/2}.All that remains is to compare these two configurations.Note that for d = 4, the two configurations are identical and therefore optimal within H d,n .Observe also that if n = 5, then necessarily t * = 1, and hence H 1 d,5 = H t * d, 5 .We now consider the case where n ≥ 7.
Proof.First note that Consider the function Let us first look at the cases d = 6 and d = 8.We can simplify the above expressions to obtain    Let us start with the case where x ≥ d + 1. Observe that In the numerator as well as in the denominator, the two terms of the product sum to 2dx − , convexity implies that the quotient is less than 1 and, therefore, that f d (x + 2) < f d (x).
The last inequality holds because d ≥ 10, and inequality (5) uses the fact that for every This concludes the proof.
We gather here our observations about the ratios τ (H d,n )/τ (H d,n−2 ).
Observation 4.13.For all even d ≥ 6 and odd n ≥ 7 we have , we then deduce from Observation 4.13 that τ (G) > τ (H d,n ), a contradiction.
We deduce from the above statement that any vertex that is incident to an edge of multiplicity between d/2 and d − 2 must be a terminal vertex in a pendant triangle -recalling that d ≥ 6.Now consider vertices that are not terminal vertices in a pendant triangle.It follows that such vertices are either exceptional or incident to an edge of multiplicity d − 1.We next argue that an optimal graph cannot have more than one exceptional vertex.Before proving this, let us immediately show how this property allows us to conclude the proof.Suppose that the optimal graph must have at most one exceptional vertex and all remaining vertices must either be terminal vertices in a pendant triangle or incident to an edge of multiplicity d − 1.Then a path with edges of alternating multiplicities 1 and d − 1 must eventually close and form an odd cycle.Therefore, the optimal graph must be a member of H d,n .
It thus remains to prove the announced property.Suppose, for the sake of contradiction, that u and v are both exceptional vertices.By Proposition 4.6, we can obtain a connected d-regular (n − 2)-vertex multigraph G by sequentially lifting the two vertices u and v.Then, .   .
These contradictions establish that G has at most one exceptional vertex unless d = 6 and n = 9.
We now deal with the special case where d = 6 and n = 9, for which we have to use a different, more structural, approach to prove that G cannot have more than one exceptional vertex.One fact we need is that H 2 6,7 is the unique connected 6-regular 7-vertex with the fewest number of spanning trees.Indeed, the last two paragraphs do imply that H 6,7 ∈ H 6,7 and then Lemma 4.12 yields the statement.
We already know that all edges of multiplicity 3 or 4 are part of pendant triangles -actually, edges of multiplicity d/2 can never be part of a pendant triangle in a d-regular multigraph on more than 3 vertices, so that G has no edges of multiplicity 3. Now suppose that u and v do not belong to a same pendant triangle, and yet are linked by exactly 2 edges in G.If G − uv is disconnected, then let G 1 and G 2 be the two components of G − uv where G 1 has fewer vertices than G 2 .Without loss of generality, assume that u ∈ G 1 .Note that G 1 has at most 4 vertices and each of its vertices, except possibly u, has at least 2 neighbours in G 1 .
Suppose first that G 1 has 4 vertices.Then u has a unique neighbour u in G 1 for otherwise G 1 , and therefore G, would contain a diamond as a subgraph, which is impossible.Since u and v are linked by 2 edges, It follows that u and u are linked by 4 edges, and hence u, u is part of a pendant triangle, which is a contradiction since G 1 has 4 vertices.We deduce that G 1 has 3 vertices, and hence is a triangle with edge multiplicities 2, 2, 4. In particular, τ (G 1 ) = 20.
In G 2 , the vertex v has degree 4 (and no edge of multiplicity more than 2).Let G be a connected 6-regular 5-vertex multigraph obtained by completely lifting v in G 2 .Now, which is a contradiction.Consequently, G − uv is connected.In particular, every spanning tree of G − uv is a spanning tree of G that does not use any of the two edges between u and v. Let G be the connected 8-vertex multigraph obtained by deleting all edges between u and v and next identifying u and v into a single vertex x.Note that every spanning tree of G corresponds in a natural way to two spanning trees of G both containing an edge between u and v, and differing only on this edge.

We infer that
We now perform some complete lifts in G − uv and in G to obtain back connected 6-regular multigraphs.(The complete lifts we are going to make are possible since u and v are exceptional and do not belong to a same pendant triangle.) Let G be a connected 6-regular 7-vertex multigraph obtained from G − uv by subsequently lifting u and v.We have Let G x be a connected 6-regular 7-vertex multigraph obtained from G by completely lifting the new vertex x, so • τ (H 2 6,7 ).( 8) Combining ( 6), ( 7) and ( 8), we obtain τ (G) ≥ 4 + 36 5 τ (H 6,7 ) = 56 5 • τ (H 6,7 ) > τ (H 6,9 ), where the last inequality follows from Observation 4.13.This contradiction finishes to establish that edges of multiplicity 2 also belong to pendant triangles.
In what follows, most assertions implicitly rely on the fact that every edge of multiplicity greater than 1 and less than 5 is contained in a pendant triangle.Suppose now that G contains an exceptional vertex u that is not a cut-vertex.Then u must have 6 distinct neighbours, all in the same 2-connected block.Otherwise, u would be incident to an edge of multiplicity 2 (recalling that G has no edge of multiplicity 3), which has to be contained in a pendant triangle.
Consequently, u would have to be a terminal vertex of this pendant triangle, contradicting that u is exceptional.Similarly, because u is not a cut-vertex, the subgraph H induced in G by the neighbours of u is simple.Moreover, H has maximum degree 1, for otherwise a vertex of degree at least 2 in H along with two of its neighbours and u would induce a subgraph of G containing a diamond, a contradiction to the optimality of G.But then u and all its neighbours induce at most 9 edges, and the two remaining vertices are incident to 12 edges for a total of at most 21 edges in G, a contradiction to the 6-regularity of G.
As a result, every exceptional vertex of G is a cut-vertex.We now establish some properties of the 2-connected blocks of G. First, we note that all edges inside a 2-connected block must have multiplicity 1 or 5.This in particular implies that every 2-connected block of G contains at multigraph G uv by completely lifting v in G u .Furthermore, we note that G uv cannot be isomorphic to P P 4,n−2 .Indeed, note that P P 4,n−2 contains n − 7 ≥ 6 edges of multiplicity 2.
However, as each of u and v has 4 different neighbours in G, and G itself has no edge of multiplicity 2, the number of edges of multiplicity 2 in G uv is at most 4.This implies that n ≥ 17, and it follows that This provides a contradiction, because as reported earlier τ (F 4,n ) < 4 • τ (F 4,n−2 ) when n ≥ 11.
We thus proved that if G contains no edge of multiplicity 2, then either n is even and G = P C 4,n , or n is odd and at least 13 and then G = F 4,n .
Summing-up both situations (i.e G has an edge of multiplicity 2 or not), we see that it only remains to compare the cases for which we have found two possibilities: when n is either 13 we need to compare τ (P P 4,13 ) and τ (F 4,13 ), and when n is even and at most 14, we need to compare P P 4,n and P C 4,n .We see that τ (P P 4,13 ) < τ (F 4,13 ), and also that τ (P P 4,2k ) < τ (P C 4,2k ) if k ≤ 5 while the inequality is reversed if k ≥ 6, which concludes the proof.
d-regular n-vertex simple graph can have.Let us write τ (G) for the number of spanning trees of a graph G. Let δ d = lim inf n→∞ (τ (G)) 1/n and η d = lim sup n→∞ (τ (G)) 1/n

Figure 2 :•Figure 1 :
Figure 1: The padded cycle graph P C d,n (left) and the fish graph F d,n (right).

Proposition 2 . 8 .
Let H be a connected multigraph on n vertices with nd/2 − 1 edges and maximum degree d.Then τ (H) ≥ (d − 1) n/2 with equality only if H is a path graph with edges of alternating multiplicities d − 1 and 1.

Lemma 2 . 9 .
If G is a connected d-regular multigraph minimising the number of spanning trees and ∂{u, v, w} ≤ d − 2, then, without loss of generality, N G (v) ∪ N G (w) ⊂ {u, v, w}.
w}, which must have the same parity as d.Since d ≤ d − 2, we deduce that d − d is a positive and even integer.We can thus create a connected d-regular n-vertex multigraph G from G H by adding two vertices y and z such that fixed and the degree conditions a + b, a + c, b + c ≤ d, the quantity τ (H) is minimised when, up to symmetry, a = b = s − d and c = 2d − s, that is, when H and T are isomorphic -and, in particular, all edges of G in ∂{u, v, w} are incident to the same vertex of H.The conclusion follows.

[ 2 ,Lemma 3 . 1 .
Theorem 4], which should rather have been the expression given below.(Maybe this misprint is what caused the formulation of the incomplete conjecture, the incorrect expression being smaller than both τ (P P d,n ) and τ (P C d,n ).)Let d ≥ 3 be an odd integer.

Lemma 3 . 3 (
Base cases).Theorem 1.2 holds for all odd d and n ∈ {4, 6}.More precisely, 1.For all odd d ≥ 3, the padded cycle has the fewest spanning trees of all connected d-regular multigraphs on 4 vertices.

Lemma 3 . 4 .
and τ (P P d,6 ) = (9d 4 − 12d 3 − 2d 2 + 4d + 1)/16, with P C d,6 attaining the minimum for d ≥ 5.The graphs Gc d,6 and P P d,6 , which are identical for d = 3, are optimal for that case.Now fix n ≥ 8 and suppose that Theorem 1.2 holds for all d-regular multigraphs on at most n − 2 vertices.Let G be a connected d-regular n-vertex multigraph minimising the number of spanning trees.We will prove statements on the structure of G which will eventually show that G must be isomorphic to P C d,n or P P d,n as required.We will start by showing that the existence of multiple bridges implies the presence of a structure common to both the padded paddle and the padded cycle, namely, a path with edges of alternating multiplicity 1 and d − 1. Suppose uv and xy are two distinct bridges (i.e., cutedges of multiplicity one) in G, such that v = y and with u and x in the same component H of G − uv − xy (possibly u = x).

Lemma 3 . 6 .
If u ∼ G v, then one of the following holds: 1. w G (u, v) ∈ {1, d − 1}; or 2. u and v are part of a pendant triangle.Proof.The statement being trivial for d = 3, we assume that d ≥ 5. Recall that, by Lemma 2.9, if u and v have a common neighbour z such that {u, v, z} induces at least d + 1 edges in G, then the triple {u, v, z} forms a pendant triangle.Let us then set m = w G (u, v) and suppose, contrary to the statement, that 2 ≤ m ≤ d − 2 and yet there is no vertex z such that {u, v, z} induces at least d + 1 edges.Let G be obtained from G by first deleting all edges between u and v, and next identifying the vertices u and v into a new vertex x, which has thus degree 2(d − m) in G .Observe that τ (G) ≥ m • τ (G ).Note that there is no vertex z in G with w G (x, z) ≥ d − m + 1.In addition, Corollary 3.5 implies that G − u − v = G − x has at most d − m components.As observed earlier, it follows that it is possible to produce a connected graph by performing a complete lift of x in G .Let G x be a connected graph obtained from G by performing a complete lift of x.By Theorem 2.4 and Proposition 2.7,

Lemma 3 . 7 .
If w G (u, v) = d − 1 for two vertices u, v ∈ V (G), then one of the following holds: 1. G − u − v is disconnected; 2. u and v are part of a pendant triangle; or 3. G is the padded cycle.Proof.Suppose that G − u − v is connected.If u and v have a common neighbour w, then the triangle {u, v, w} is a pendant triangle.So suppose that a ∼ G u and b ∼ G v with v = a = b = u.We need to show that G is the padded cycle graph.Let G = G − u − v + f where f is an edge joining a and b.Note that G is a connected d-regular graph with n − 2 vertices, to which the induction hypothesis thus applies, and G − f = G − u − v is a connected graph with n − 2 vertices, maximum degree d and nd/2 − (d + 1) = (n − 2)d/2 − 1 edges, to which Proposition 2.8

Lemma 3 . 8 . 3 .
If w G (u, v) = 1 for two vertices u, v ∈ G, then one of the following must hold: 1. G − uv is disconnected; 2. u and v are part of a pendant triangle; or 13 G is the padded cycle.

(A ) .
Every edge of G incident to u or v has multiplicity less than d − 1. Proof.Suppose, on the contrary, that w G (u, u ) = d − 1, and hence u and v are the only two neighbours of u.Let v be the only neighbour of u different from u.First notice that v = v because u and v are not part of a pendant triangle.Second, G − u − u must be connected because G − uv is, and hence G − uv contains a path from v to u that avoids u, and hence contains v .Consequently, Lemma 3.7 implies that G is the padded cycle, a contradiction.The same reasoning applies to v by symmetry.
a contradiction.Let us now consider τ (G/uv).For simplicity of notation, let H = G/uv be obtained by contracting u and v into a single vertex w.Note that w has degree 2(d − 1) in H.As observed previously, it is possible to obtain a connected graph by completely lifting the vertex w if H − w has at most d components and w is not incident to an edge of multiplicity at least d.The first condition holds by Corollary 3.5 as H − w = G − u − v has at most d − 1 components.Second, w is not incident to an edge of multiplicity greater than d − 1 ≥ 2 in H, for otherwise u or v would be incident to an edge of multiplicity greater than d−1 2 in G, contradicting (B).Let H w be a connected d-regular (n − 2)-vertex graph obtained by completely lifting the contracted vertex w in the graph H = G/uv.By Theorem 2.4 and the induction hypothesis, we have each connected component of G − u − v we create an associated vertex in L. For each edge e in E u ∪ E v , we add an edge between the vertices associated to the end-vertices of e in L. (This may create loops if two edges leading to the same connected component of G − u − v were lifted together.)It follows that G uv and L have the same number of connected components.To lighten the writing, we shall canonically identify the edges of L with those in E u ∪ E v .Because G is connected, each connected component of G − u − v contains a neighbour of u or a neighbour of v. Consequently, these connected components can be partitioned into C u , C v , and C uv , depending on whether they have an edge only to u, only to v, or to both u and v, respectively.Furthermore, C uv is not empty because G − uv is connected; let x be a vertex of L associated to a connected component in C uv .By Lemma 3.4, at most one connected component in C u ∪ C v has exactly one edge to {u, v}.

Lemma 3 . 9 .
A vertex in G belongs to at most one pendant triangle.Proof.Suppose that T = {x, u, v} and T = {x, y, z} induce distinct pendant triangles.Let w be a vertex such that xw is a bridge.Such a vertex must exist because x cannot be incident to an edge of multiplicity d − 1, and at least one edge incident to x does not belong to a pendant triangle, since d is odd.As G is not a convex combination of other graphs, we may assume that w G (x, u) = w G (x, v) = 1, up to swapping T and T .Let w G (x, y) = w G (x, z) = a.Then τ (G) = (2d − 1)(a 2 + 2a(d − a))τ (G − u − v − y − z).Now consider the graph G obtained from G by deleting u and v, altering the triangle induced by T such that w G (x, y) = w G (x, z) = a + 1, deleting the edge xw, and introducing vertices u , v such that w G (x, u ) = w G (x, v ) = 1 and w G = (u , v ) = d − 1 (in other words, we remove one of the triangles at x and extend the path in G − u − v − y − z by 2).

Theorem 1 .
4 are presented in Proposition 4.3.Together these four parts, namely Theorem 1.3, Theorem 1.4, Theorem 4.1, and Proposition 4.3, give the unique minimiser graphs for each setting of n and d where d is even.

2 . 4 . 2 .
Definition For n odd and d even, let F * d,n be the d-regular multigraph on n vertices obtained from the disjoint union of d 2 − 1 triangles and one odd cycle by identifying one vertex from each into a single central vertex.Note that the central vertex has exactly d distinct neighbours (hence it is incident only to edges of multiplicity 1) and all the other edge multiplicities alternate between 1 and d − 1 on each cycle.

Proposition 4 . 3 .
Let G be a connected d-regular n-vertex multigraph.If the pair

Figure 5 :
Figure 5: The unique minimiser F * d,n in each of the four sporadic cases when d ≥ 6. Parallel edges are represented by a single edge with the multiplicity written next to it.

Lemma 4 . 4 .Proposition 4 . 6 .
For all even d ≥ 4, the unique connected n-vertex d-regular multigraph minimising the number of spanning trees is 1. for n = 2: an edge of multiplicity d; 2. for n = 3: a triangle with all edges of multiplicity d/2; Suppose G is a connected d-regular multigraph minimising the number of spanning trees and let u and v be two vertices in G not incident to edges of multiplicity more than d/2.Then it is possible to sequentially completely lift u and v to produce a connected d-regular graph.
1).Then, whenever P C d,n−2 minimises the number of spanning trees among connected d-regular multigraphs on n − 2 vertices, we can argue that a connected d-regular multigraph G on n vertices cannot be optimal if τ (G) > 3 2 (d − 1)τ (G ) for some connected d-regular multigraph G on n − 2 vertices.Fix n ≥ 6 and suppose that τ (G ) ≥ τ (P C d,n−2 ) for every connected d-regular multigraph G on n − 2 vertices.Let G be a connected d-regular n-vertex multigraph minimising the number of spanning trees.

Lemma
we can completely lift the vertex x in G , which produces a connected d-regular multigraph G x on n − 2 vertices.It follows from Theorem 2.4 and Proposition 2.7 that

Lemma 4 . 9 .
All but at most one pendant triangle in H d,n contain edges of multiplicities 1 and d − 1.
We now study the function x → f d (x) for x ∈ [5 , ∞), for any fixed even value of d ≥ 10.The typical shape for f d (x) is given in Figure6, where f d (5) = 1 and the curve has a local minimum at x = d − 1 and a local maximum at x = d + 1.Indeed, we will prove that f d (x) is decreasing on the two intervals [5 , d − 1] and [d + 1 , ∞).Then, establishing that f d (d + 1) < 1 completes the proof of the lemma.

least 3
vertices, for if a 2-connected block is be composed of only two vertices, then the 1 or 5 edges joining them would form an odd-cut in G, which is impossible in a regular multigraph of even degree.Second, because G has 9 vertices, no 2-connected block of G can contain more than 3 cut-vertices, since to each cut-vertex x of a block B, we can associate a distinct component of G − x not intersecting B, and such a component contains at least 2 vertices by degree regularity of G.Moreover, if a 2-connected block B of G contains exactly 3 cut-vertices {u, v, w}, then necessarily G consists of the simple triangle {u, v, w}, with pendant triangles at all three vertices.It follows that τ (G) = 3 • 20 3 > 10285 = τ (H 6,9 ), a contradiction.We thus proved that each 2-connected block of G contains at most 2 cut-vertices.It now follows that a 2-connected block with 2 cut-vertices must contain at least 4 (and at most 5) vertices, for if B is a 2-connected block with 3 vertices containing exactly two cut-vertices of G, then they must induce a triangle that is not pendant and yet contain edges of multiplicities greater than 1 and less than 5, a contradiction.As a consequence, if G contains at least 2 exceptional vertices, which in particular must be cut-vertices, then we can find two of them, u and v, that belong to the same 2-connected block B. So B has 4 or 5 vertices, and u and v are the only cut-vertices of G in B. Since B contains no pendant triangle, all vertices of B except u and v (which are exceptional) are incident to an edge of multiplicity 5 in B. This is a contradiction, as the only possibility is that B be a path u, u , v , v with edge multiplicities 1, 5, 1, which is not 2-connected.It follows that G contains at most one exceptional vertex, as needed.This concludes the proof.
the minimum is taken over all sequences of positive integers d 1 , . . ., d k with varying length k such that k i=1 d i = 2m, and over all connected k-vertex graphs X with degree sequence d 1 , d 2 , . . ., d k .Proof.Consider a positive integer m, a sequence D = d 1 , . . ., d k for some positive integer k such that k i=1 d i = 2m and a connected graph X with degree sequence D = d 1 , . . ., d k where d i ≤ d i for each i ∈ {1, . . ., k}.Our goal is to show that if D = D then c m is not attained by D and X.

Table 1 :
Values of the Ok-Thomassen c m term, the lower bound from Proposition 2.7, and graphs attaining the c m values.
6ecall that the minimum degree in the underlying simple graph is at least 2. Now, if either v 5 or v 6 has two or more neighbours among v 1 , v 2 , v 3 , v 4 , it is not possible to avoid creating a diamond or a path joining opposite vertices of the 4-cycle.Let us, therefore, suppose that both v 5 and v 6 have exactly one neighbour each among v 1 , v 2 , v 3 , v 4 .If they have a common neighbour, say v 1 , then the resulting graph cannot be extended to a regular multigraph (v 1 is forced to have degree d inside the C 4 because v 2 , v 3 and v 4 do).Otherwise, the simple graph must be the domino graph (C 6 with a chord forming two copies of C 4 ).Applying Corollary 2.3 to the two copies of C 4 , we see that Gb d,6is the only candidate with this underlying simple graph.Case 2b: v 5 and v 6 are not adjacent.Then both v 5 and v 6 have exactly two neighbours each among v 1 , v 2 , v 3 , v 4 .The only way to avoid creating a diamond or a path edge-disjoint from the 4-cycle and connecting two opposite vertices on it is, without loss of generality, to have edges v 1 v 5 , v 2 v 5 , v 3 v 6 , and v 4 v 6 (thus forming the co-domino graph).But this graph contains a C 6 and the graph remains connected after the removal of either maximum matching of the C 6 , contradicting Corollary 2.3.
1 2 Figure 3: The graphs P C d,6 , Gb d,6 , Gc d,6 , P P d,6 .Case 2a: v 5 and v 6 are adjacent.Therefore, we need only compare the number of spanning trees of P C d,6 , Gb d,6 , Gc d,6 , and P P d,6 7, and recalling that m ∈ {2, . . ., d − 2} with d being odd, there is equality if and only if d = 5 and m = 2.
1.By Proposition 2.8, this happens only if G − f is a path with edges of alternating multiplicities d − 1 and 1.This implies that G must be the padded cycle graph, as required.
contradicting the optimality of G.It follows that if G is not the padded cycle graph, then it must be the padded paddle graph.Comparing the values α d and β d (n), we see that τ (P C d,n ) < τ (P P d,n ) if and only if n ≤ (3d + 1) 2 /8d.The parity of d allows us to slightly simplify this expression to the one that appears in the statement of Theorem 1.2.
connected, since every edge induced by T 1 ∪ T 2 has multiplicity at least 2. Furthermore, H 1 , H 2 ∈ H d,n and 2H d,n = H 1 + H 2 .Proposition 2.2 then applies and contradicts the optimality of H d,n .There is at most one cycle in H d,n that is not a triangle.Proof.Suppose, on the contrary, that C 1 and C 2 are cycles in H d,n of lengths 2k + 1 and 2 + 1, respectively, where 2 ≤ k ≤ .Recall that each edge in C 1 ∪ C 2 has multiplicity either 1 or d − 1 so contradicting the optimality of H d,n in H d,n .For every integer t such that 1 ≤ t ≤ min{d/2 − 1, (n − 3)/2}, let us define the d-regular multigraph H t d,n ∈ H d,n as one having t − 1 triangles with edge multiplicities 1 and d − 1 one odd cycle of length n − 2t and one triangle with edge multiplicities at the centre v both being (d − 2t)/2.Note that H 1 d,n is isomorphic to the d-regular fish graph F d,n .Lemmas and 4.10 show that H d,n must be isomorphic to H t d,n for some integer t.In fact, we can make the following stronger statement.We have H d,n ∈ {H 1 d,n , H t * d,n } where t * = min{d/2 − 1, (n − 3)/2}).
1).Suppose G is a minimiser of τ over the class of connected d-regular n-vertex multigraphs, where d ≥ 6.Then G ∈ H d,n .Proof.Note first that if n = 5, then the lemma holds.We now assume that n ≥ 7 and that the lemma holds for graphs on n − 2 vertices.We begin by showing that edges of multiplicity m where 3 ≤ m ≤ d − 2 must be part of pendant triangles.Suppose that 3 ≤ m = w G (u, v) ≤ d − 2 and the vertices u and v do not belong to a same pendant triangle.This implies there is no vertex z such that {u, v, z} induce a triangle with more than d edges.Let G be a connected graph obtained by deleting all edges between u and v, identifying the vertices u and v into a new vertex x and completely lifting the vertex x.Then G is a connected d-regular (n − 2)-vertex multigraph and (8,9))The statement follows by direct computation, recalling thatH d,n = H t * d,n if (d, n) ∈ {(6, 7),(6,9),(6,11),(8,9)} and H d,n = H 1 d,n otherwise.One needs to note that for every fixed d, the function x → (d(x−1)−2) (d(x−3)−2) is decreasing with x. 4.3.2A Minimiser Must Belong to H d,n .Lemma 4.14.