The structure of digraphs with excess one

A digraph G $G$ is k $k$ ‐geodetic if for any (not necessarily distinct) vertices u,v $u,v$ there is at most one directed walk from u $u$ to v $v$ with length not exceeding k $k$ . The order of a k $k$ ‐geodetic digraph with minimum out‐degree d $d$ is bounded below by the directed Moore bound M(d,k)=1+d+d2+⋯+dk $M(d,k)=1+d+{d}^{2}+\cdots +{d}^{k}$ . The Moore bound can be met only in the trivial cases d=1 $d=1$ and k=1 $k=1$ , so it is of interest to look for k $k$ ‐geodetic digraphs with out‐degree d $d$ and smallest possible order M(d,k)+ϵ $M(d,k)+{\epsilon }$ , where ϵ ${\epsilon }$ is the excess of the digraph. Miller, Miret and Sillasen recently ruled out the existence of digraphs with excess one for k=3,4 $k=3,4$ and d≥2 $d\ge 2$ and for k=2 $k=2$ and d≥8 $d\ge 8$ . We conjecture that there are no digraphs with excess one for d,k≥2 $d,k\ge 2$ and in this paper we investigate the structure of minimal counterexamples to this conjecture. We severely constrain the possible structures of the outlier function and prove the nonexistence of certain digraphs with degree three and excess one, as well closing the open cases k=2 $k=2$ and d=3,4,5,6,7 $d=3,4,5,6,7$ left by the analysis of Miller et al. We further show that there are no involutary digraphs with excess one, that is, the outlier function of any such digraph must contain a cycle of length ≥3 $\ge 3$ .


Introduction
The undirected degree/girth problem asks for the smallest possible order of a graph that has degree d and girth g.If g = 2k + 1 is odd, then the order of any such graph is bounded below by the undirected Moore bound 1 + d + d(d − 1) + • • • + d(d − 1) k−1 .A graph that has order exceeding the Moore bound by ǫ has excess ǫ; a graph that meets the Moore bound is called a Moore graph.It was shown in [1,11,16] that, with the trivial exception of cycles and complete graphs, for odd g = 2k + 1 the Moore bound can be met only for diameter k = 2 and degrees d = 3 (the Petersen graph), d = 7 (the Hoffman-Singleton graph) and possibly d = 57.Interestingly, it is even harder for a graph to have excess one than to meet the Moore bound; Bannai and Ito showed in [2] that there are no graphs with excess one and girth 2k + 1 ≥ 5.
A directed graph G is k-geodetic if for any pair of (not necessarily distinct) vertices u, v of G there is at most one directed walk from u to v in G with length ≤ k.The following analogue for directed graphs of the degree/girth problem was raised by Sillasen in [23]: It was shown in [28] that for any d, k ≥ 1 there exists a diregular k-geodetic digraph with degree d, so this problem is well-posed.
A lower bound on the order of a k-geodetic digraph G with minimum out-degree d can be derived as follows.Fix a vertex u of G; this will be Level 0 of a directed Moore tree of depth k.There are at least d out-neighbours of u; draw these at Level 1 of the tree with arcs from u at Level 0. In general once we have completed Level t of the tree, where 0 ≤ t ≤ k − 1, we draw arcs from each vertex of Level t to each of its out-neighbours at Level t + 1.It follows by induction that for 0 ≤ t ≤ k there are at least d t vertices at Level t of the Moore tree.Moreover by k-geodecity all of the vertices in this tree must be distinct.It follows that the order of G is bounded below by the directed Moore bound It was shown by Bridges and Toueg in [7] that the only digraphs that meet the directed Moore bound are directed cycles and complete digraphs.Hence for d, k ≥ 2 we are interested in k-geodetic digraphs with minimum out-degree d and order M (d, k) + ǫ for some small excess ǫ.A smallest such digraph is a geodetic cage.
In this paper we will be interested in digraphs with excess ǫ = 1; we will call such a digraph a (d, k; +1)-digraph, which we will typically denote by the letter G.For every vertex u of such a digraph there is a unique vertex o(u) such that the distance from u to o(u) in G satisfies d(u, o(u)) ≥ k + 1; we call the associated function o : V (G) → V (G) the outlier function of G.The following results on the structure of (d, k; +1)-digraphs were proven in [19,23].
• The outlier function o of G is an automorphism.
Further results on the problem and examples of geodetic cages can be found in [25,26,27,28].As no non-trivial digraphs with excess one have been found, we make the following conjecture.
It is instructive to compare this problem with the state of the art of the directed degree/diameter problem (a survey of which can be found in [20]), which asks for the largest digraphs with given maximum degree d, diameter k and order M (d, k) − δ for some small defect δ.A digraph with defect δ = 1 is a (d, k; −1)-digraph or an almost Moore digraph.
The plan of this paper is as follows.Firstly in Section 2 we use counting arguments to deduce some strong conditions on digraphs with excess one and a high level of symmetry.In Section 3 we investigate the structure of digraphs with degree three and excess one.Then in Section 4 we use the approach of Sillasen in [24] to analyse the set of vertices fixed by an automorphism of a (d, k; +1)digraph.As the outlier function o is an automorphism of G these results tell us quite a lot about the structure of o as a permutation.In the final part of this paper, Section 5, we use a spectral approach to exploit the results of Section 4 to prove the non-existence of certain (d, k; +1)-digraphs.

Vertex-transitive digraphs with excess one
All of the known Moore graphs are vertex-transitive.This suggests that it is of interest to look for digraphs with order close to the directed Moore bound that have a high degree of symmetry.In her thesis [22] Sillasen uses this approach on digraphs with defect one; we now emulate this approach for digraphs with excess one.
In [22] as a basis for her counting arguments Sillasen divides the vertices of an almost Moore digraph into two types, Type I and Type II.Adapting this notation, we make the following definition.
The type of a vertex is preserved by any automorphism of G.This leads us to the following observation.
If every vertex of G is Type I, then we can obtain a strong divisibility condition on d and k.The reason for this is that all directed (k + 1)-cycles of G are arc-disjoint.

Lemma 2.
No arc of G is contained in more than one directed (k + 1)-cycle.
Proof.Suppose that an arc (u, v) is contained in two (k + 1)-cycles.Then there are two distinct k-paths from v to u, which contradicts k-geodecity.
Proof.Let (u, v) be such an arc.As u = o(v) there is a path of length k in G from v to u, so the arc (u, v) is contained in a (k + 1)-cycle, which is unique by Lemma 2.

Corollary 1. Suppose that every vertex of
Proof.As all vertices of G are Type I, by Lemma 2 we can partition the arcs of Computer search shows that for k between 2 and 10000, the following values of d and k satisfy this condition:  3, 5, 7, 11, 13, 23, 55, 91, 163, 236, 1235, 1356.We see that the condition in Corollary 1 is quite strong.On the other hand, if G contains a Type II vertex then these vertices group themselves into cycles in a natural way.Lemma 4. Suppose that G contains a Type II vertex u.Then u has a unique Type II out-neighbour, namely o − (u).
Proof.Applying the automorphism o − to the arc (o(u), u), we see that (u, o − (u)) is also an arc.By inspection, o − (u) is a Type II vertex.Suppose that u ′ is an arbitrary Type II out-neighbour of u.
It follows immediately that in a vertex-transitive (d, k; +1)-digraph all vertices must be Type I.
Proof.Suppose that G contains a Type II vertex; by vertex-transitivity, every vertex is Type II.But this contradicts Lemma 4.
In particular, if G is vertex-transitive, then it must satisfy the divisibility condition in Corollary 1.
In fact, we can significantly extend Corollary 1 using the fact that for any vertex u of a vertextransitive (d, k; +1)-digraph the vertex o − (u) cannot be close to u.
. However, as o − is an automorphism of G, there is an arc o − (u) → o − (u 2 ), so that o − (u 2 ) also lies in T (u 1 ).As o − (u 2 ) appears twice in the Moore tree, k-geodecity is violated.Hence t ≥ k.
Proof.Let u be any vertex of G and draw the Moore tree of depth k rooted at u.By Lemma 6 all vertices in T (u) have a path of length ≤ k to u.For 1 ≤ t ≤ k − 1, let us say that a vertex v at Level t of the Moore tree is All vertices of N + (u) must be long by k-geodecity.It is easily seen that for 1 ≤ t ≤ k − 2 every vertex at Level t has one short out-neighbour and d − 1 long out-neighbours at Level t + 1.By induction for 2 ≤ t ≤ k − 1 there are d t−1 short vertices in Level t of the tree.Therefore for 2 ≤ t ≤ k − 1 the vertex u is contained in d t − d t−1 closed walks of length k + t and these walks must be cycles by k-geodecity.Equation 1 follows by double-counting pairs (u, Z), where u is a vertex of G and Z is a directed (k + t)-cycle containing u.
Using the strengthened divisibility condition in Theorem 1, computer search shows that the only values of d and k in the range 3 ≤ d ≤ 12 and 2 ≤ k ≤ 10000 for which there can exist a vertex-transitive (d, k; +1)-digraph are k = 2: d = 3, 6, 9, 12, k = 3: d = 6, 10.The results of [19] show that in practice there are no (d, 3; +1)-digraphs for d ≥ 2 and no (d, 2; +1)digraphs for d ≥ 8, so the only remaining values of d and k in this range are (d, k) = (3, 2) and (6,2).In Section 7 we will see that such digraphs also do not exist.This scarcity of vertex-transitive (d, k; +1)-digraphs can be taken as evidence in favour of Conjecture 1.
There are some simple number-theoretic conditions on d and k that force G to contain a Type II vertex, so that G cannot be vertex-transitive.
Theorem 2. If d ≥ 3 and k ≥ 2 satisfy any of the following conditions then any (d, k; +1)-digraph contains a Type II vertex: • iv) there is an odd prime p such that p|(k + 1) and d ≡ 2 (mod p).
Proof.For part i), suppose that d and k are odd, but G contains only Type I vertices.By Corollary 1, (k + 1) divides d(M (d, k) + 1), so M (d, k) must be odd.However contains an even number of odd summands and hence is even.
Finally, for part iv) suppose that p is an odd prime such that p|(k + 1) and d ≡ 2 (mod p); then if every vertex is Type I we must have implying that p is even, a contradiction.These results can be extended to larger values of the excess ǫ under the assumption of arctransitivity.Proof.Write N + (u) = {u 1 , u 2 , . . ., u d }.Suppose that an element v of O − (u) lies at distance t ≤ k from u.We can assume that v ∈ T (u 1 ).By arc-transitivity, there are d − 1 automorphisms φ i of G, i = 2, 3, . . ., d, that map the arc u → u 1 to the arc u → u i .Each image φ i (v) of v under these automorphisms must also belong to O − (u), so each branch contains an element of O − (u).As ǫ < d, one of these elements must be repeated in the Moore tree rooted at u, violating k-geodecity.Therefore we must have O − (u) ⊆ O(u) and the result follows.
Proof.Again we denote the number of distinct directed r-cycles in G by Z r .Using the same reasoning as in the proof of Theorem 1, we see that each vertex of G is contained in These results may suggest that the permutation digraphs are smallest possible arc-transitive k-geodetic digraphs [28].

Diregular digraphs with excess one and degree three
Sillasen has shown that there are no (2, k; +1)-digraphs [23]; therefore a reasonable next step is to ask whether there are any (3, k; +1)-digraphs.It was proven in [5] that there are no (3, k; −1)digraphs; the strategy of the proof is to show that any two distinct vertices of a (3, k; −1)-digraph can have at most one common out-neighbour (and, conversely, at most one common in-neighbour), classify vertices u according to the distance from u to its repeat r(u) and then count the different types of vertices in two different ways to arrive at a contradiction.In this section we show that the first main result of [5], that any two vertices have at most one common out-neighbour, continues to hold in the setting of digraphs with degree three and excess one.
The fact that a (d, k; +1)-digraph G must be diregular leads to a 'duality principle'.This phenomenon has been observed for almost Moore digraphs; in [5] it is shown that taking the converse of a (d, k; −1)-digraph yields another (d, k; −1)-digraph.The proof for digraphs with excess one is trivial and we omit it.The Duality Principle will allow us to interchange out-neighbourhoods and in-neighbourhoods in our results.Recall that o − is the inverse function of the outlier function, i.We begin with a lemma that holds generally for (d, k; +1)-digraphs; it describes the situation of vertices with identical out-neighbourhoods.
Draw the Moore tree of depth k rooted at u; by k-geodecity, u appears only at Level 0, the root position, of this tree.As the Moore tree rooted at v differs from the Moore tree rooted at u only at Level 0, and u = v, it follows that v cannot reach u by a path of length ≤ k, so o(v) = u and, by symmetry, o(u) = v.The result for in-neighbourhoods follows by the Duality Principle.
For the remainder of this section let G be a diregular (3, k; +1)-digraph with outlier function o.Our first goal is to show that no pair of distinct vertices can have identical out-neighbourhoods; to achieve this we need a lemma for pairs of vertices with exactly two common out-neighbours.Proof.Let u and v be as described.This configuration is shown in Figure 1.By k-geodecity u 3 ∈ T (u 1 ) ∪ T (u 2 ).Hence there are three possible positions for the vertex u 3 in the Moore tree rooted at v: i)

Lemma 10. Let u, v be distinct vertices of G with exactly two common out-neighbours, i.e. |N
).The vertex w has three out-neighbours w 1 , w 2 and w 3 .By kgeodecity none of these out-neighbours can lie in T (v 3 ).At most two of the out-neighbours can lie in {v, o(v)}, so it follows that w has an out-neighbour, say w 1 , that lies in T (v 1 )∪T (v 2 ) = T (u 1 )∪T (u 2 ); without loss of generality w 1 ∈ T (u 1 ).Hence there is a path of length ≤ k from u to w 1 via u 1 .There is also a path from u to w 1 with length ≤ k formed from the arc u → u 3 , followed by the path from u 3 to w with length k − 1 − ℓ and the arc w → w 1 .This violates k-geodecity.It follows that option iii) must hold, i.e.
The setup is shown in Figure 2. By Lemma 9 we know that v = o(u) and u = o(v).For i = 1, 2, 3 denote the in-neighbour of u i that does not lie in {u, v} by u * i .We cannot have u * 1 = u * 2 = u * 3 , for otherwise by Lemma 9 we would have o Then in G − the pair of vertices u i , u j has exactly two common outneighbours, so that by Lemma 10 we obtain o Applying Lemma 10 to the pairs u 1 , u 3 and u 2 , u 3 in turn, we deduce that o − (u 1 ) = o − (u 2 ) = u * 3 , again a contradiction, as o is a permutation.
Having ruled out identical out-neighbourhoods, we can complete the proof of our desired result.
Theorem 3. Any two distinct vertices of a (3, k; +1)-digraph G have at most one common outneighbour and at most one common in-neighbour.
Observe that now u 3 and v 3 have two out-neighbours in common, namely u 10 and u 11 , so by Lemma 10 we have o(u 3 ) = u and o(v 3 ) = v.Applying the outlier automorphism to the arcs incident with u, we deduce that o(u) = v 3 has arcs to o(u 3 ) = u and o(u 1 ) and o(u 2 ), so {o(u 1 ), o(u 2 )} = {u 10 , u 11 }.By 2-geodecity u 10 can have arcs only to N + (u 1 ) and N + (u 2 ).As u 10 has three out-going arcs and cannot have the same out-neighbourhood as u 1 or u 2 by Corollary 4, it follows that u 10 must have two common out-neighbours with either u 1 or u 2 ; without loss of generality N + (u 10 ) = {u 4 , u 5 , u 7 }.Applying Lemma 10 to the pair u 1 , u 10 we see that o(u 1 ) = u 7 .As we have already determined that o(u 1 ) ∈ {u 10 , u 11 }, this is a contradiction.The last part of the theorem follows by the Duality Principle.
Theorem 3 allows us to prove our first non-existence result for degree three, namely that there are no (3, 2; +1)-digraphs.
Proof.Suppose that G is a diregular (3, 2; +1)-digraph.Fix an arbitrary vertex u of G with N + (u) = {u 1 , u 2 , u 3 } and draw the Moore tree rooted at u as shown in Figure 4. We set At least one of the vertices c 1 , c 2 is not equal to o(u), say c 1 = o(u).By 2-geodecity we can assume that c 1 = u 4 .By Theorem 3 c 1 has no arcs to T (u) − {u 3 }, at most one arc to N + (u 2 ) and Figure 3: Configuration for Theorem 3

Automorphisms of digraphs with excess one
In [24] Sillasen uses counting arguments to deduce information on the form of the subdigraph of an almost Moore digraph that is induced by the set of vertices fixed by an automorphism.Using the same approach we can deduce a strong result on the action of automorphisms of a digraph with excess one.This will later help us to analyse the structure of the outlier function of a (d, k; +1)digraph.
Let G be a (d, k; +1)-digraph and φ ∈ Aut(G) a non-identity automorphism of G. Denote by F ix(φ) the set of vertices of G that are fixed by φ and let F IX(φ) be the subdigraph induced by F ix(φ).
Firstly we show that the fix-set F ix(φ) is closed under the action of the outlier automorphism.
Iteration of o implies the result.
As the outlier automorphism is fixed-point-free, it follows from Lemma 11 that any fix-set F ix(φ) cannot consist of a single vertex and, if φ fixes just two vertices u, u ′ of G, then these two vertices are outliers of each other, i.e. o(u) = u ′ and o(u ′ ) = u.
We will now assume that F ix(φ) contains at least three vertices.
Lemma 12.If u, v ∈ F ix(φ) and P is a path of length ≤ k from u to v, then all vertices of P are contained in F ix(φ).
Proof.Let u, v and P be as described.Suppose that there is a vertex u ′ ∈ V (P ) that is not fixed by φ.Then P and φ(P ) are distinct ≤ k-paths from u to v, contradicting k-geodecity.
Proof.For any vertex u ∈ F ix(φ) we will denote the out-degree and in-degree of u in the subdigraph F IX(φ) by d + φ (u) and d − φ (u) respectively.We will show that for any two (not necessarily distinct) vertices u, v ∈ F ix(φ) we have d + φ (u) = d − φ (v); this implies the desired result.For this pair u, v we will write Assume that v ∈ o(N + (u)).Then for 1 ≤ i ≤ d there is a unique ≤ k-path P i from u i to v. Suppose that u → v.By k-geodecity, none of the paths P i pass through the same in-neighbour of v, so without loss of generality there is a then repeating this reasoning for the out-neighbours of u other than v shows that we still have then for 2 ≤ i ≤ d we can assume that there is a ≤ (k − 1)-path from u i to v i and as before u i ∈ F ix(φ) if and only if v i ∈ F ix(φ) for 2 ≤ i ≤ d.There is an arc u → u 1 , so as o is an automorphism there exists an arc o(u) → o(u 1 ) = v, giving o(u) ∈ N − (v).There are ≤ k-paths from u to each v i for 2 ≤ i ≤ d, so we must have o(u) = v 1 .As o(u) ∈ F ix(φ) by Lemma 11 we have v 1 ∈ F ix(φ).Also by Lemma 11 we have o − (v) = u 1 ∈ F ix(φ), so again we see that By Lemma 13, F IX(φ) is diregular with degree ≥ 1 (as we are assuming that |F ix(φ Proof.As a subdigraph of G, F IX(φ) is k-geodetic.By Lemma 13, F IX(φ) is diregular with degree d ′ .We are assuming that F ix(φ) contains at least three vertices, so by Lemma 12 F IX(φ) contains a path and d ′ ≥ 1.We are also assuming that φ is not the identity automorphism, so φ does not fix all vertices of G and d ′ ≤ d − 1.By diregularity and Lemma 14, it follows that F IX(φ) As there are no diregular (2, k; +1)-digraphs [24], we have the following result.

Structure of the outlier function
We will now make use of some of the results from the preceding sections to deduce useful information on the permutation structure of the outlier function of a digraph with excess one.Let G be a (d, k; +1)-digraph.By Lemma 1 G is diregular and the outlier function of G is an automorphism.Therefore every vertex u of G has an associated order ω(u), which is the smallest integer such that o ω (u) = u.Cholily et al. have used the vertex orders of the repeat function to successfully analyse the structure of digraphs with defect one in such papers as [3] and [8].We can immediately apply the method of [8] to make a connection between the vertex orders for the outlier function and the existence of short paths in G.
Proof.Suppose that for some 1 ≤ i ≤ r − 1 the order of u i does not divide t.Then o t (u i ) = u i , so we obtain two ≤ k-paths u 0 , u 1 , . . ., u i , . . ., u r and o t (u 0 ), o t (u 1 ), . . ., o t (u i ), . . ., o t (u r ) = u 0 , o t (u 1 ), . . ., o t (u i ), . . ., u r from u 0 to u r , a contradiction.Corollary 8.If p is the minimum vertex order of G and W is a walk of length ≤ k between two vertices u, v with order p, then every vertex on W has order p.
Proof.Suppose that there is a vertex w on W such that ω(w) > p. Then W and o p (W ) are two distinct walks of length ≤ k between u and v, contradicting k-geodecity.
We now make two definitions that will help us to analyse the structure of the permutation o.As o is an automorphism it follows that any power o r of o is also an automorphism of G.In Section 4 we classified the possible fixed sets of any non-identity automorphism of G.We therefore record the following implication of Corollary 7.
Corollary 9.For any integer r ≥ 2, the set of vertices of G with order dividing r induces one of the following: • the empty digraph, • a pair of vertices that form a transposition in o, Conjecture 1 claims that there are no non-trivial (d, k; +1)-digraphs; one approach to proving this conjecture is to study the properties of a minimal counterexample.Let k ≥ 2 and k = 3, 4. Suppose that there exists a (d, k; +1)-digraph with d ≥ 2 and let d ′ be the smallest possible value of d ≥ 3 such that there exists a (d, k; +1)-digraph; then we will refer to a (d ′ , k; +1)-digraph as a minimal (d, k; +1)-digraph.For a fixed k, Corollary 9 strongly restricts the structure of the outlier automorphism of a minimal (d, k; +1)-digraph.
Corollary 10.A minimal (d, k; +1)-digraph G satisfies one of the following: • the outlier function o of G contains a unique transposition, or • the vertices of G with order ω(G) form a directed (k + 2)-cycle.
Proof.By Corollary 9 the automorphism o ω(G) fixes either i) every vertex of G, in which case every vertex of G has order ω(G) and G is outlier-regular, ii) two vertices that are outliers of each other, so that ω(G) = 2 and o contains a unique transposition, or iii) a (k + 2)-cycle.
If a minimal (d, k; +1)-digraph is not outlier-regular, then Corollary 10 allows us to deduce the subdigraph induced by the set of vertices with smallest order.Proof.Suppose that G is a non-outlier-regular minimal (d, k; +1)-digraph with outlier function that does not contain a unique transposition.Then by Corollary 10 the vertices with order equal to the index ω(G) of G induce a directed (k + 2)-cycle C. For any vertex u in the cycle its outlier o(u) also has order ω(G), so the outlier of u must be the vertex preceding u on the cycle C; it follows that ω(G) = k + 2.
We will find the following classification of this behaviour convenient.According to this definition, every minimal (d, k; +1)-digraph is either Type A, Type B or outlierregular.Let us now return to the problem of digraphs with degree three and excess one; if any such digraph exists it is minimal.
Proof.By Lemma 16 we have ω(G) = k+2 and the vertices with order k+2 induce a (k+2)-cycle C. Pick a vertex u on C and write N + (u) = {u 1 , u 2 , u 3 }, where u 1 also lies on C. The automorphism o (k+2) fixes u and u 1 , but not u 2 and u 3 , so o (k+2) transposes u 2 and u 3 .Thus o 2(k+2) fixes every vertex in T 1 (u) and by Corollary 9 every vertex of G has order either k + 2 or 2(k + 2).If there are r cycles in o with length 2(k + 2), then we obtain Lemma 18.Let G be a (3, k; +1)-digraph of Type B. Then k ≡ 3, 5 (mod 6).If k ≡ 0, 2 (mod 6), then G contains two vertices of order two, with all other vertices of G having order six.
Proof.Assume that G is a non-outlier-regular (3, k; +1)-digraph with outlier function o containing a unique transposition.Let u and o(u) be the vertices of G with order two, where N + (u) = {u 1 , u 2 , u 3 }.The automorphism o 2 fixes u, but fixes no vertex in {u 1 , u 2 , u 3 }.We can thus assume that o 2 permutes u 1 , u 2 , u 3 in a 3-cycle (u 1 u 2 u 3 ).
By Theorem 3 u and o(u) have at most one common out-neighbour.Suppose that N + (u) ∩ N + (o(u)) = ∅; we can assume that u 1 is the common out-neighbour of u and o(u).However, applying the automorphism o 2 to G shows that o 2 (u 1 ) is a common out-neighbour of u and o(u) and, since G is Type B we have o 2 (u 1 ) = u 1 , thereby violating Theorem 3. It follows that o contains the 6-cycle (u ) and hence o 6 fixes every vertex of G, so that the order of every vertex apart from u and o(u) is either 3 or 6.
Suppose that there is a vertex with order 3. Then o 3 fixes a (k + 2)-cycle.If there are r cycles in o of length 6, then Hence 6|(M (3, k) − k − 3).This implies that k ≡ 1 (mod 3).On the other hand, suppose that all vertices of G have order six, with the exception of the two vertices with order two.Then 6|(M (3, k) − 1), which implies that k is even.
Proof.Assume that G is a (3, k; +1)-digraph such that k satisfies each of these conditions.By Lemmas 17 and 18, G is neither Type A nor Type B and hence must be outlier-regular.As the order of G is prime, it follows that its outlier function o consists of a single cycle of length M (3, k)+1; thus G is vertex-transitive.However, any vertex-transitive digraph with prime order is a circulant digraph, which is not k-geodetic for k ≥ 2. It follows that there is no (3, k; +1)-digraph for such k.
The first k for which Corollary 11 applies are k = 3, 15 and 63.This provides an independent proof of the non-existence of (3, 3; +1)-digraphs [19], as well as ruling out the existence of (3, k; +1)digraphs for some larger k.

Spectral results
Now that we have more information about the permutation structure of o, we can apply some more powerful spectral results developed in [19].If A is the adjacency matrix of a (d, k; +1)-digraph G with order n, J is the n × n all-one matrix and P is the permutation matrix associated with the permutation o, then counting the paths of length ≤ k we have that We will now exploit the connection in Equation 2 between the permutation structure of the outlier function o of a (d, k; +1)-digraph G and the spectrum of G.We will use the following concise description of the permutation structure of the outlier function o from [19].
Definition 5.For any (d, k; +1)-digraph G and 1 ≤ j ≤ M (d, k)+1, the number of cycles of length j in the permutation o will be denoted by is the number of odd cycles in the permutation o with length divisible by j, • m ′′ (j) is the number of even cycles in the permutation o with length divisible by j, and is the total number of cycles in o with length divisible by j.
Note that as the outlier function is fixed-point-free we always have m 1 = 0. We will also need the following family of polynomials derived from the cyclotomic polynomials.
Definition 6.For n, k ≥ 1 the polynomial F n,k (x) is defined by where Φ n (x) is the n-th cyclotomic polynomial.
In [19] Miller et al. derive the following relation between the characteristic polynomial of J − P and the permutation vector of G.
Proof.Assume that G is a 2-outlier-regular (d, k; +1)-digraph with order n = M (d, k) + 1, i.e. the outlier function o of G contains only transpositions.Thus m 2 = n 2 and m i = 0 for i = 2. Therefore by Lemma 19 the characteristic polynomial of J − P is It follows from Equation 2 that the spectrum of G consists of , and For any integer r ≥ 0 we define As for all vertices u of G we have o − (u) = o(u), the reasoning of Theorem 1 shows that each vertex of G is contained in d directed (k + 1)-cycles.By k-geodecity, any closed (k + 1)-walk must be a cycle, so it follows by counting walks of length and Each eigenvalue λ i satisfies 1 summing this geometric series and rearranging we obtain λ i (λ k i − 1) = 0, so each λ i is either zero or a k-th root of unity.Hence for 1 ≤ r ≤ k we have Λ k+r = Λ r .Similarly each eigenvalue µ i satisfies µ k+1 i = 2 − µ i , so that for 1 ≤ r ≤ k we have In particular Subtracting this from Equation 3 with r = 1 yields For any prime p the polynomial p + x + x 2 + • • • + x k is irreducible over Q [15].As each µ i is a solution of 2 + x + x 2 + • • • + x k = 0, it follows that the roots of 2 + x + x 2 + • • • + x k must appear with equal multiplicity among the µ i .Therefore k must divide n 2 − 1.The sum of the roots of 2 Rearranging, we obtain In particular, it follows from Lemma 7 and Theorem 5 that no (d, k; +1)-digraph with d, k ≥ 2 can be arc-transitive.

2-geodetic digraphs with excess one
In [19] the authors use spectral techniques to show that there are no 2-geodetic digraphs with excess one and degree d ≥ 8. Sillasen's first paper on the subject [23] proves that there are no (2, 2; +1)-digraphs.Theorem 4 of the present work further showed that there are no (3, 2; +1)digraphs.This leaves open the existence of (d, 2; +1)-digraphs for d = 4, 5, 6 and 7. We will see that no (d, 2; +1)-digraphs exist for these values of d.We first rule out the existence of outlier-regular (d, 2; +1)-digraphs, then use an inductive approach to deal with the remaining cases.
Accordingly we shall now assume that any (d, 2; +1)-digraph is outlier-regular.Then the index ω(G) of the digraph G must be a non-unit divisor of the order 2 + d + d 2 of the (d, 2; +1)-digraph.These divisors are displayed in Table 1.Theorem 5 shows that there are no 2-outlier-regular digraphs, so we have already dealt with the divisors in red.Furthermore, if the index ω(G) of the digraph is equal to the order 2 + d + d 2 of the digraph G, then G is vertex-transitive and by Corollary 2 the size 2d + d 2 + d 3 of G must be divisible by 3; this precludes the existence of outlier-regular digraphs with the divisors written in blue in Table 5.For the remaining possible structures of the outlier function we will need the exact factorisation of the characteristic polynomial of a (d, 2; +1)-digraph from [19].Lemma 20. [19] The characteristic polynomial of a (d, 2; +1)-digraph factorises in Q[x] as m ′ (j) 2 j≥6,j even where a 1 and a 2 are non-negative integers that satisfy the simultaneous equations Lemma 21.An outlier-regular (d, 2; +1)-digraph cannot have odd index ω(G).
Lemma 21 disposes of all of the green entries in Table 1.
Lemma 22 disposes of the pink divisors in Table 1.
which is not an integer for ω ∈ {4, 8, 16}.This gets rid of all of the orange entries in Table 1.
The only remaining option for an ω-outlier-regular (d, 2; +1)-digraph G is that d = 6 and ω = 4. Then m(4) = m ′′ (1) = 44/4 = 11 and d 2 − d + 1 = 31.Equation 5 becomes a 1 + a 2 = 11 and a 1 − a 2 = 9, which has solution a 1 = 10 and a 2 = 1.It follows from Lemma 20 that the spectrum of G is {6 (1) , 0 (10) , −1 (1) , , i (11) , (−i) (11) } where multiplicities are indicated in round brackets.Summing the third powers of the eigenvalues, it follows that T r(A 3 ) = 240.It follows by Lemma 16 that any minimal (d, 2; +1)-digraph must be either Type A (with a directed 4-cycle C of vertices with order 4 and all other vertices with order greater than 4) or Type B. We will take an inductive approach.Theorem 4 shows that there is no (3, 2; +1)-digraph, so we can take any (4, 2; +1)-digraph to be minimal, which allows us to show that (4, 2; +1)-digraphs do not exist, so that any (5, 2; +1)-digraph is minimal and so on.We make the following two observations from Lemma 20 and Corollary 7 respectively.Proof.By Lemma 23, a minimal (d, 2; +1)-digraph is either Type A, in which case the smallest non-zero entry of the permutation vector is m 4 = 1, or Type B, in which case the smallest entry is m 2 = 1.By Corollary 7, for each r ≥ 1 the automorphism o r has fix-set of size 0, 2 or 4, or else fixes every vertex of G. Therefore if for some r ≥ 2 the automorphism o r fixes 3 or ≥ 5 vertices of G, then o r is the identity automorphism and every vertex of G has order dividing r.Suppose that the or ii) (u 2 u 3 u 4 )(u 5 u 6 ).In case i) every vertex of G has order 4, 5, 10 or 20; by Lemma 25 m 5 = 0.If m 20 > 0, then by Lemma 25 m 4 = 1, m 10 = 0, m 20 = 2.By Lemma 20, this yields a 1 + a 2 = 3 and a 1 − a 2 = 25, which would imply that a 2 is negative.Therefore m 10 = 4, m 20 = 0, giving a 1 + a 2 = 5, a 1 − a 2 = 29, which again is impossible.
Adding up the third powers of these eigenvalues, we see that the trace of A 3 is T r(A 3 ) = 378.A vertex lies in 7 directed triangles if it is Type I and 6 if it is Type II; therefore if there are α Type I vertices in G and β Type II vertices, then counting paths we see that 7α + 6β = 378, α + β = 58.Solving these equations gives α = 30, β = 28.As a vertex has the same type as its outlier, it follows that the subdigraph B induced by the Type II vertices consists of 7 directed 4-cycles and there is a set A of 28 vertices v such that d(v, o − (v)) = 2, the other two Type I vertices being u and o(u).
B has size 28 and (B, V (G) − B) = (V (G) − B, B) = 168.This means that the subdigraph induced by A ′ = A ∪ {u, o(u)} has size 42.Fix a vertex v ∈ A ′ .The outlier o(v) of v also lies in A ′ , so v can reach every vertex of B by ≤ 2-paths.The vertex v has 7 out-going arcs.Suppose that the out-degree of v in the subdigraph induced by A ′ is ≤ 2. Each arc from v to B allows v to reach 2 vertices of B; therefore the largest possible number of vertices of B that v could reach by ≤ 2-paths would be achieved if v has two out-neighbours in A ′ , each of which has out-neighbourhood contained in B ; however, this would still only allow v to reach 24 of the 28 vertices of B. It follows that the minimum out-degree in A ′ is ≥ 3, which implies that the size of the subdigraph induced by A ′ is ≥ 90, a contradiction.This completes the remaining cases in the classification of (d, 2; +1)-digraphs from [19].Combined with the results of [19], we see that to find a digraph with excess one, we must look at digraphs that have degree at least three and are at least 5-geodetic.

Lemma 6 .
If G is a vertex-transitive (d, k; +1)-digraph, then for any vertex u of G we have d(u, o − (u)) ≥ k.

Lemma 9 .
Let u and v be vertices of a (d, k; +1)-digraph such that N + (u) = N + (v), where u = v.Then v = o(u) and u = o(v), i.e. the outlier function o transposes u and v.The same result holds

Figure 1 :
Figure 1: Vertices with two common out-neighbours

Figure 4 :
Figure 4: The Moore tree for Theorem 4

Definition 2 .
The index ω(G) of a (d, k; +1)-digraph G is the value of the smallest vertex order in G, i.e. ω(G) = min{ω(u) : u ∈ V (G)}.Definition 3. A (d, k; +1)-digraph is outlier-regular if its outlier function o is a regular permutation.If each vertex of G has order ω, then G is ω-outlier-regular.

Lemma 16 .
If a minimal (d, k; +1)-digraph G is not outlier-regular, then either its outlier function o contains a unique transposition, or else ω(G) = k + 2 and the vertices with order k + 2 induce a directed (k + 2)-cycle.

Definition 4 .
A minimal (d, k; +1)-digraph G such that the vertices of G with order ω(G) form a directed (k+2)-cycle is Type A, whereas if the outlier function o of G contains a unique transposition, then G is Type B.

A
vertex is contained in 6 directed triangles if it is Type I and 5 directed triangles if it is Type II.If there are α Type I vertices and β Type II vertices in G, it follows that 6α + 5β = 240 and α + β = 44.Solving these equations, we have α = 20 and β = 24.Let A be the subdigraph of G induced by the Type I vertices and B the subdigraph induced by the Type II vertices.By Lemma 4 it follows that B consists of a collection of 6 directed 4-cycles.Thus |(A, B)| = |(B, A)| = 120.Thus each vertex in A has out-neighbourhood entirely contained in B. Each vertex in B has just one out-neighbour in B and so each arc from a vertex u of A allows it to reach just twelve vertices of B by paths of length ≤ 2, which is impossible.