On the Generalized Spectral Characterizations of Eulerian Graphs

A graph G is said to be determined by its generalized spectra (DGS for short) if, for any graph H, graphs H and G are cospectral with cospectral complements imply that H is isomorphic to G. In Wang [16] (J. Combin. Theory, Ser. B, 122 (2017) 438-451), the author gave a simple method for a graph to be DGS. However, the method does not apply to Eulerian graphs. In this paper, we give a simple method for a large family of Eulerian graphs to be DGS. Numerical experiments are also presented to illustrate the effectiveness of the proposed method. Mathematics Subject Classifications: 05C50


Introduction
The spectrum of a graph encodes a lot of combinatorial information about the given graph and thus has long been a useful tool in spectral graph theory.
A fundamental question in this area is: "Which graphs are determined by their spectra (DS for short)?".The problem was first raised in 1956 by Günthard and Primas [8], which relates the theory of graph spectra to Hückel's theory [9] from chemistry.It is also closely related to a famous question of Kac [10]: "Can one hear the shape of a drum?".Fisher [6] modelled the drum by a graph, and the frequency of the sound was characterized by the eigenvalues of the graph.Thus, the two problems are essentially the same.
Another motivation for the above question comes from complexity theory.It is a long standing open question whether the graph isomorphism problem is an easy or a hard problem, despite the recent breakthrough result of Babai [1].Since the spectrum can be computed in polynomial time, the focus is checking isomorphism for cospectral graphs.However, as is generally known, proving graphs to be DS is more challenging than constructing cospectral graphs.Up to now, the techniques for proving DS graphs heavily depend on some special properties of the spectra of these graphs, which cannot be extended to general graphs.For the background and some known results about this problem, we refer the reader to [4,5] and the references therein.
In recent years, Wang and Xu [13,14] and Wang [15,16] considered the above problem in the context of the generalized spectra.A graph G is determined by its generalized spectra (DGS for short) if, for any graph H, graphs H and G are cospectral with cospectral complements imply that H is isomorphic to G.In Wang [15,16], the author gave a simple method for determining whether a graph G is DGS, which works for a large family of general graphs.To describe the result, let G be a graph on n vertices with adjacency matrix A. The walk-matrix of G is defined as W = W (G) := [e, Ae, • • • , A n−1 e] (e is the all-one vector).In [15,16], Wang proved the following Theorem 1 (Wang [15,16]).If det W (G)/2 n/2 (which is always an integer) is odd and square-free, then G is DGS.
It is noticed, however, the above theorem fails for Eulerian graphs.(Recall that a graph is Eulerian if it admits an Eulerian tour, which traverses each edge exactly once; or equivalently, if it is connected and the degree of every vertex is even).This is because for an Eulerian graph, every entry (except for the ones in the first column) of the walk-matrix is divisible by 2, and hence 2 n−1 divides det W and det W/2 n/2 can never be odd and square-free.
This paper is devoted to investigating whether an Eulerian graph is DGS.The main contributions of the paper are as follows: • We show that for an Eulerian graph G with det W (G)/2 3n− 3 2 being odd and squarefree, G is either DGS or there exists a regular rational orthogonal matrix Q with level two (see Section 2 for the definitions) such that Q T A(G)Q is a (0,1)-matrix; • Based on the above result, we give a simple sufficient condition for an Eulerian graph with the above property to be DGS, by constructing a digraph associated with G and using a simple dimension argument.
It turns out that Eulerian graphs are among the most difficult family of graphs for which the existing method in [15,16] does not work well.The main reason is that the exponent of 2 in the prime factorization of det W is usually too high for Eulerian graphs, and "excluding" the prime 2 constitutes the most difficult part in showing a graph G to be DGS, as we shall see later.
We would like to mention that the proof of the main result of the paper follows the line of that in [15,16].However, several new ingenious ideas are needed to make the proof work.
The rest of the paper is organized as follows.In Section 2, we give some preliminaries that will be needed later in the paper.In section 3, we show that for Eulerian graphs G with det W (G)/2 3n− 3 2 being odd and square-free, the rational orthogonal matrices Q the electronic journal of combinatorics 26(1) (2019), #P1.9 such that Q T A(G)Q is (0,1)-matrix, must have level 1 or 2. In Section 4, an effective and novel method is provided to determine whether an Eulerian graph is DGS.In Section 5, we give some numerical results for illustrations.Conclusions and future work are given in Section 6.

Preliminaries
For convenience of the reader, we present some preliminary results that will be needed later in the paper.Throughout, let G = (V, E) be a simple graph with (0, 1)-adjacency matrix A = A(G).The spectrum of G consists of all the eigenvalues (together with their multiplicities) of the matrix A(G).The spectrum of G together with that of its complement will be referred to as the generalized spectrum of G in the paper (for some notions and terminologies in graph spectra, see e.g.[2]).
Two graphs are cospectral if they have the same spectrum.For a given graph G, we say that G is determined by its spectrum (DS for short), if any graph having the same spectrum as G is necessarily isomorphic to G. (Of course, the spectrum concerned should be specified.) The walk-matrix of a graph G, denoted by W (G) or simply W , is defined as where e denotes the all-one vector (we use this notation henceforth).There is a wellknown combinatorial interpretation of W , that is, the (i, j)-th entry of W is the number of walks of G starting from vertex i with length j − 1.
A graph G is called controllable if W is non-singular (see also [7]).Denoted by G n the set of all controllable graphs on n vertices.It was conjectured by Godsil that almost all graphs are controllable.Recently, O'Rourke and Touri [11] confirmed that the conjecture is true.
A rational orthogonal matrix Q is an orthogonal matrix with all entries being rational numbers, and it is called regular if Qe = e.
The following theorem gives a simple characterization of cospectral graphs with respect to the generalized spectra.
Lemma 2 (c.f.Wang and Xu [13]).Suppose G ∈ G n .Then G and H are cospectral with respect to the generalized spectra if and only if there exists a unique regular rational orthogonal matrix Q such that Define where O n (Q) denotes the set of all orthogonal matrices with rational entries.
By Theorem 3, in order to determine whether a given graph G ∈ G n is DGS or not, we only need to check whether all the Q's in Q(G) are permutation matrices.For this purpose, the following definition is proved to be useful.Definition 4. Let Q be an orthogonal matrix with rational entries.The level of Q, denoted by (Q) or simply , is the smallest positive integer l such that lQ is an integral matrix.
Clearly, is the least common denominator of all the entries of the matrix Q.If = 1, then Q is a permutation matrix.
When dealing with integral and rational matrices, the Smith Normal Form (SNF for short) is a useful tool.An integral matrix V of order n is called unimodular if det V = ±1.The following theorem is well-known.

Theorem 5 ([3]
).For an integral matrix M , there exists unimodular matrices Note that the SNF of a matrix can be computed efficiently (see e.g.page 50 in [12]).
Theorem 6 (Wang and Xu [13]).Let W (G) be the walk-matrix of a graph G ∈ G n , and Q ∈ Q(G) with the level .Then | d n , where d n is the n-th elementary divisor of the walk-matrix W (G).
By the above theorem, is always a divisor of d n , and hence is a divisor of det W .However, the following theorem shows that not every divisor of det W can be a divisor of .
Theorem 7 (Wang [15]).Let G ∈ G n and Q ∈ Q(G) with level .Let p be an odd prime.If p | det W and p 2 det W , then p cannot be a divisor of .
By the above theorem, if det W = ±2 m b with b being odd and square-free, then can only be a power of 2. This fact will be heavily used in the sequel.Now, we introduce the following family of Eulerian graphs which are the main focus of this paper: is odd and square-free}.
We remark that det W (G)/2 3n−3 2 is always an integer (see Corollary 17 in Section 3).By the definition of Σ n , the exponent of 2 is much higher in the prime factorization of det W (G) than that in [15,16], and unfortunately, all the previous methods cannot be applied to this situation, therefore we need to deal with the case from a new perspective.

Notations:
We shall use the finite field notation F p and mod p (for a prime p) interchangeably, and shall use rank p (M ) to denote the rank of an integral M over F p .

The level of
By the previous discussions, we know that for a given graph G, in order to tell whether G is DGS or not, it is crucial to determine the level of every regular rational orthogonal matrix Q ∈ Q(G).In this section, we show for any Eulerian graph G ∈ Σ n , the level of Q ∈ Q(G) is very restrictive.The main result of this section is the following The proof of Theorem 8 is based on the following two theorems: where b is an odd and square-free integer.
We postpone the proofs of the above theorems to the end of this section.

A simple arithmetic property of Eulerian graphs
In this subsection, we will present an arithmetic property about Eulerian graphs, which plays an important role in the sequel.
Lemma 11.Let Q ∈ Q(G) with level .Suppose is even.Then there exists a (0,1)vector v ≡ 0 (mod 2) such that for any k 0.
Proof.By definition, Q ∈ Q(G) implies that Q T AQ = B for some (0,1)-matrix B. Let v be the i-th column of Q such that v ≡ 0 (mod 2) (such a v always exists by the definition of ).It follows from be a (0,1)-vector for some integral vector β.Then The last assertion follows from the fact that Q T A k Q = B k and Qe = e imply that W T Q is an integral matrix.Thus W T v ≡ 0 (mod 2) holds.
Suppose m is the number of odd entries of v. Then m ≡ 0 (mod 4).
the electronic journal of combinatorics 26(1) (2019), #P1.9 Proof.Since ≡ 0 (mod 4), we have = 4t for some integer t.Moreover, v/ is a column of Q, it follows from the orthogonality of . For an odd entry v i , we have v i = 2t i + 1 for some integer t i .It follows that v The following lemma plays a significant role in establishing the main result in the paper.
Proof.Let v be any column of Q with v ≡ 0 (mod 4).Then we have W T v ≡ 0 (mod 4).Let v = v + 2u, where v is a (0, 1)-vector and u is an integral vector.Then W T v = W T (v + 2u) ≡ 0 (mod 4).i.e., Since G is an Eulerian graph, we have Ae T with every d i being even, for i = 1, 2, • • • , n. Therefore u T A i e ≡ 0 (mod 2), and hence v T A i e 2 ≡ 0 (mod 2), for i = 1, 2, • • • , n − 1.In addition, the parity of v is the same as that of v.By Lemma 12, we have v T e = m ≡ 0 (mod 4), where m is the number of odd entries of v.It follows that According to Lemma 11, v satisfies v T A k v ≡ 0(mod 4) for any k = 0, 1, • • • .The proof is complete.
By Lemma 13, we introduce a new matrix which is clearly an integral matrix and satisfies W T v ≡ 0 (mod 2).The matrix W plays a similar role as that of W (G) in [16].

Some auxiliary lemmas
In this subsection, we will present some lemmas, which are needed in the proof of the theorems in this section.
Lemma 14.Let G be an Eulerian graph with adjacency matrix A. Then e T A 2 e ≡ 0 (mod 4), and e T A k e ≡ 0 (mod 8), for any integer k 3, Proof.First, we show e T A 2 e ≡ 0 (mod 4) holds.
Then it is clear that d is an integral vector since every vertex degree is even in an Eulerian graph.It follows that e T A 2 e = (Ae) T (Ae) = (2 d) T (2 d) = 4 dT d ≡ 0 (mod 4).
Next, we show e T A k e ≡ 0 (mod 8) for any k 3.
Note that e T A k e = (Ae) , then e T A k e = 4 dA l d.So it suffices to show dT A l d ≡ 0 (mod 2) for any l 1. Next, we distinguish the following two cases: Combining Cases 1 and 2, the proof is complete.
the electronic journal of combinatorics 26(1) (2019), #P1.9 Proof.We shall prove the lemma in two cases: n is even and n is odd.First suppose n is even.Let W 1 be the matrix obtained from W by doubling the first two columns.Then it follows from Lemma 14 that . . .
It follows that rank . Now suppose n is odd.We further distinguish the following two cases: Case 1.If e T Ae ≡ 0 (mod 4), then from Lemma 15, we can get e T A 2 e ≡ 0 (mod 8).So it follows from Lemma 14 that Note that 2rank . . .
. This completes the proof.

As a corollary, we have
Corollary 17.Let G be an Eulerian graph.Then the exponent of 2 of in the prime factorization of det W is at , the number of even numbers in the diagonal entries of W is at least The following lemma gives the SNF of W , the proof of which follows the same idea as Lemma 3.5 in [16].
2 and the SNF of W is as follows: where b is an odd square-free integer. Proof.
is odd and square-free, we have det , where p i 's are distinct odd primes for each i.Thus the SNF of W can be written as . Thus, we have . The proof is complete.
The following corollary says that if G ∈ Σ n is an Eulerian graph with an even number of vertices, then the number of edges of G must be odd.
For convenience, next, we fix some notations.Let Ŵ be the matrix defined as follows: if n is odd and e T Ae ≡ 0 (mod 4). [ if n is odd and e T Ae ≡ 2 (mod 4).
It follows from Eq. ( 11) that the electronic journal of combinatorics 26(1) (2019), #P1.9 for some integral vector β.Left-multiplying A on both sides of Eq. ( 12) gives that Similarly, we have for any t 0. Thus we have rank 2 ( W ) m k − 1; a contradiction.This completes the proof.
Lemma 21.Let G be an Eulerian graph and Then det( W ) = ± det( W ).
Proof.By Cayley-Hamilton's Theorem, we have Right-multiplying the vector e on both sides of Eq. ( 15) and using the fact A i e ≡ 0 (mod 2) for i = 1, 2, • • • , n gives that c n is even.Thus we have i.e., , By performing the elementary column operations to W , it is easy to see that det . Then the matrix W can be obtained by switching the columns of W .It is obvious that det( W ) = ±det( W ).So the lemma follows and the proof is complete.
Let W 1 be the matrix obtained from the matrix W and defined as follows: if n is odd and e T Ae ≡ 0 (mod 4); if n is odd and e T Ae ≡ 2 (mod 4).
if n is odd and e T Ae ≡ 0 (mod 4). [ if n is odd and e T Ae ≡ 2 (mod 4).
, where W is defined as in Eq. ( 14).If n is odd, then rank 2 ( Proof.Let n be even.Let W 1 be defined as the above.It is easy to verify that is an integral matrix.By Lemma 18 and Lemma 21, it can be computed that Therefore, the column vectors of matrix are linearly independent, over F 2 .It follows that rank 2 ( ) equals the number of columns of Ŵ1 , which is n−2 2 = n−1 2 .Let n be odd.By the definitions of W and W 1 , it is easy to see that is an integral matrix.By Lemma 18, it can be computed that Therefore, the column vectors of matrix are linearly independent, over F 2 .Thus, rank 2 ( ) equals the number of columns of Ŵ1 , which is n−1 2 = n−1 2 .

Proof of Theorem 10
Now we are ready to present the proof of Theorem 10: Proof.We prove the theorem by contradiction.Suppose on the contrary that 4 | .It follows from Lemma 11 and Lemma 13 that there exists a vector v ≡ 0 (mod 2) such that v T A k v ≡ 0 (mod 4) for any k 0 and W T v ≡ 0 (mod 2).Note that v is a solution to the system of linear equations W T v ≡ 0 (mod 2).Note that G ∈ Σ n .It follows from Lemma 18 that rank 2 ( W) = n+1

2
. Moreover, it is not difficult to verify that W T Ŵ ≡ 0 (mod 2).Since rank 2 ( W) = n+1 2 , the dimension of the solution space of the system of linear equations , which is equal to rank 2 ( Ŵ ).Therefore, the column vectors of Ŵ can be chosen as a basis of the solution space of W T x ≡ 0 (mod 2).It follows the electronic journal of combinatorics 26(1) (2019), #P1.9 that v can be written as the linear combination of the column vectors of Ŵ over F 2 , i.e., v = Ŵ u + 2β, where u and β are integral vectors and u ≡ 0 (mod 2).Thus we have First, we prove the case that n is even.Note that . . .
Next, we prove the case that n is odd.According to the different definition of Ŵ (resp.Ŵ1 ) when n is odd, we further distinguish the following two cases.
Case 2. e T Ae ≡ 2 (mod 4).By Lemma 15, we have e T A 2 e ≡ 4 (mod 8).Note that Let . Using the same arguments as the case n is even, it follows that  where in deriving the congruence relations, we have used the facts that e T Ae ≡ 2 (mod 4), e T A 2 e ≡ 4 (mod 8) and e T A k e ≡ 0 (mod 8) for k 3.
This completes the proof.

Proof of Theorem 8 and Theorem 9
First we present the proof of Theorem 9: Assume without loss of generality that the SNF of , where b is an odd square-free integer.Then there exist two unimodular matrices U and V such that (a is an integer, α and β are column vectors of dimension n − 1, and V 1 is a square matrix of order n − 1).

It follows from [U
where the electronic journal of combinatorics 26(1) (2019), #P1.9 Taking determinant on both sides of Eq. ( 16) generates Note det V is odd.It is easy to see that Eq. ( 17) holds only if n−1 2 = n−1 i=1 (l i − 1) and det V = ±1.That is, V is a unimodular matrix.Thus, it follows from Eq. ( 16) that 1 0 0 Λ is the SNF of W .According to Lemma 18 we have ).
From the above equation we get immediately that the SNF of W is the one given as in Eq (2), as desired.This completes the proof.
Finally, we present the proof of Theorem 8: Proof.Note that G ∈ Σ n .It follows from Lemma 9 that the SNF of W is of the form given as in Eq. ( 2), where b is an odd and square-free integer.

Eliminating = 2
Let G ∈ Σ n and Q ∈ Q(G) with level .In the previous section, we have established that = 1 or 2. If the possibility that = 2 can be further eliminated, then = 1 and hence G is DGS.Unfortunately, however, this is not always the case (see the example below in this section).In this section, we shall give some simple sufficient conditions for excluding the case that = 2.
First, we need the following definitions.
where A is the adjacency matrix of G.
We have the following theorem.
with the level of .According to Theorem 8, we have = 2 or = 1.Suppose there exists a Q ∈ Q(G) with level = 2, then there must exist a column v of 2Q having exactly four "1" over F 2 and hence v ∈ S .However, this is a contradiction since S = ∅.Therefore = 1 for every Q ∈ Q(G), which means that Q(G) contains only permutation matrices.By Theorem 3, G is DGS.The proof is complete.
We remark that the condition of Theorem 25 can be efficiently verified, since we can check all (0,1)-vectors with exactly four "1" to see whether the condition v T A k v ≡ 0 (mod 4) holds for k = 0, 1, • • • , n − 1.This can be done in O(n n 4 ) = O(n 5 ) times.The following lemma says that S is closed under left-multiplication by A modulo 2.
be the characteristic polynomial of graph G.By Cayley-Hamilton's Theorem, we have So we can get v T A n v ≡ 0 (mod 4).Similarly we have v T A n+1 v ≡ 0 (mod 4).This completes the proof.
Next we assume that S = ∅ (and hence S = ∅).In order to determine whether an Eulerian graph G ∈ Σ n is DGS, we construct a new directed graph Γ = (V (Γ), E(Γ)) associated with G, as follows.
• The vertex set of Γ consists of the vectors in S.
• There is a directed edge from vertex v i to vertex v j if and only if Av i ≡ v j (mod 2), where v i , v j ∈ S.
We remark that the above definition is motivated by Lemma 26, which says that the adjacency matrix A acts on the set S, and the dynamics of its behavior can be fully captured by Γ.Also, the structural property of Γ is closely related to the existence of a regular rational orthogonal matrix Q ∈ Q(G) with = 2, as we shall see later.
By Lemma 26, for every vector in S, there is exactly one outgoing edge in Γ.Note that Γ may have loops, since we may have Av i ≡ v i (mod 2) for some i.In particular, we always have A 0 = 0, where 0 ∈ S is the zero vector.
Moreover, the directed graph Γ constructed above exhibits simple structural characteristics.
; it is a loop as l = 1 and a bidirectional edge as l = 2.It is easy to see that for any two directed cycles (including bidirectional edges and loops) C 1 and C 2 of Γ, the node sets V (C 1 ) and V (C 2 ) are disjoint.For any node v ∈ S, consider the trajectory, {v, Av, A 2 v, • • • } ⊂ S, of v. Since S is finite, there exist an n 0 > 0 and a directed cycle C such that A i v belongs to the cycle for i n 0 .Thus, Γ consists of one or more components, each of which contains exactly one directed cycle.Let us give an example for illustrations.
Example.Let A be the adjacency matrix of an Eulerian graph G.It can be easily computed that det W = 2 15 × 7. Thus we have G ∈ Σ n .By Definition 23, we obtain the set S which consists of the columns of the following matrix S (to save space, we write the vectors in S in a compact form).Then the directed graph Γ = (V (Γ), E(Γ)) can be easily constructed according to the above rules, as shown in Fig. 1.As it can be seen from Fig. 1, Γ consists of two components.The left component contains a directed cycle on vertices 5, 12 and 16, and the right component contains a loop.
Besides, we remark in this example, although G ∈ Σ n , there is a regular rational orthogonal matrix Q ∈ Q(G) with level 2 and hence, = 2 cannot be excluded.Actually, let .
Next, we show the structural property of Γ can help us in determining whether a graph G ∈ Σ n is DGS.
First we need the following Lemma 27.Let G ∈ Σ n .Suppose there is a regular rational orthogonal matrix S is an A-invariant subspace; 2. S ⊆ S.
Thus Av j ∈ S .So the first statement is true.
Next, in order to prove the second assertion, we only need to show v i + v j ∈ S for any v i ∈ S and any v j ∈ S .Since v i ∈ S , then v i ∈ S, we get v T i A k v i ≡ 0 (mod 4).Similarly we get v T j A k v j ≡ 0 (mod 4).It follows from

Thus we have
It follows that mod 4).So the second assertion follows.This completes the proof.
Let C 1 , C 2 , • • • , C s denote all the directed cycles of Γ such that the corresponding component containing C i has at least one vector from S , for i = 1, 2, • • • , s.Let V i := span(C i ) be the vector space spanned by the nodes in C i , for i = 1, 2, • • • , s. Next, we give a simple condition for an Eulerian graph G ∈ Σ n to be DGS, by using a simple dimension argument.
Then the graph G is DGS.
Proof.Suppose G is not DGS.Then there exists a Q ∈ Q(G) with level 2. Thus there is a column v of 2Q (mod 2) which belongs to S .Suppose that A k v is in some C i for sufficiently large k.Then V i is a subspace of S spanned by the column vectors of 2Q, according to Lemma 27.Thus dim F 2 (V i ) dim F 2 (S ) dim F 2 (span(S )).This contradicts the assumption of the theorem.So for any Q ∈ Q(G), the level = 1.Thus G is DGS.The proof is complete.

Numerical results
In this section, we shall give some numerical results for illustrations.All the Eulerian graphs are randomly generated using Mathematica 11.0.
First, we shall give several examples to illustrate Theorem 28.Example 1.Let the adjacency matrix of the Eulerian graph G be given as follows: Example 2. Let the adjacency matrix of the Eulerian graph G be gives as follows: Example 3. Let the adjacency matrix of the Eulerian graph G be gives as follows: Example 4. Let the adjacency matrix of the Eulerian graph G be gives as follows: Example 5. Let the adjacency matrix of the Eulerian graph G be gives as follows: the electronic journal of combinatorics 26(1) (2019), #P1.9Finally, we have conducted a series of numerical experiments to estimate the fraction of graphs satisfying Theorem 8 and Theorem 25.The main idea is as follows: Randomly generate m Eulerian graphs of order n (9 n 20) independently.Among these graphs, if there are m graphs that belongs to Σ n , then the fraction of graphs in Σ n among Eulerian graphs is roughly m m ; if there are m graph in Σ n which satisfies S = ∅, then the probability that a graph is DGS in Σ n is roughly m m .Set m = 10000, Table 1 displays the result of one of such numerical experiments.It can be observed from Table 1, there are around 21% Eulerian graphs belong to Σ n when n is odd, and the ratio is about 11% when n is even.That is, the ratio for even n is almost half of that for odd n; it would be an interesting future work to give an explanation of this phenomena.
In addition, the proportion of graphs in Σ n satisfying Theorem 25 (and hence are DGS) approaches 1 as n increases, which suggests the following Conjecture 29.Almost all graphs in Σ n are DGS.

Conclusions and future work
In this paper, we have investigated the generalized spectral characterizations of a large family of Eulerian graphs Σ n , in which for every graph G, the power of 2 in the prime Besides Eulerian graphs, we mention that there are many other families of graphs for which Theorem 1.1 does not work, e.g., • For graphs with the degrees of all vertices being odd (of course the order of these graphs are even), Theorem 1.1 fails.This is because, for such graphs G, all the entries of W (G) are odd and hence det W (G) is divisible by 2 n−1 ; • For all regular graphs G, Theorem 1.1 fails, since det W (G) always vanishes; • For graphs G obtained through some graph operations (such as Cartesian product, tensor product), the power of 2 in the prime factorization of det W (G) is usually high, although this is less obvious.
As a future work, we would like to investigate the extent to which Theorem 1.1 can be generalized, and find more families of graphs that are determined by their generalized spectra.

Ae 2 e T A 2 e 4 e T A 3 e 4 . . . e T A n e 4 e T A 2 e 2 e T A 3 e 4 e T A 4 e 4 . . . e T A n+1 e 4 .e T A n−1 e 2 e T A n e 4 e T A n+1 e 4 .
. .e T A 2n−2 e

2 . 2 . 9 it 2 e T A 2 e 2 . . . e T A n−1 e 2 e T Ae 2 e T A 2 e 4 e T A 3 e 4 . . . e T A n e 4 e T A 2 e 2 e T A 3 e 4 e T A 4 e 4 . . . e T A n+1 e 4 .e T A n−1 e 2 e T A n e 4 e T A n+1 e 4
Case If e T Ae ≡ 2 (mod 4), then from Lemma 15, we can get e T A 2 e ≡ 4 (mod 8).So the electronic journal of combinatorics 26(1) (2019), #P1.

e T Ae 2 e T A 2 e 2 .T Ae 2 e T A 2 e 4 e T A 3 e 4 . 2 e T A 3 e 4 e T A 4 e 4 . . . e T A n e 4 .e T A n−1 e 2 e T A n e 4 e T A n+1 e 4 .
. .e T A n−1 e 2 e . .e T A n e 4 e T A 2 e . .e T A 2n−2 e

2 =
: k according to Lemma 18.It suffices to show that the first k columns of W are linearly independent over F 2 .For contradiction, suppose e, Ae 2 , • • • , A k−1 e 2 are linearly dependent, i.e., there exist c 0

e 4 − 4 − e T A 2+k e 2 .A 3+k e 4 − e T A 2+k e 2 e T A 4+k e 4 .
e T A 1+k e + e T A k e e T A 3+k e

M 4 e T A 2 e−4e T Ae+4e T e 8 e T A 4 e 8 e T A 6 e 8 . 8 e T A 5 e 8 e T A 7 e 8 . 8 e T A 6 e 8 e T A 8 e 8 . . . e T A n+1 e 8 .e 8 e T A n+3 e 8 e T A n+5 e 8 .Then M 4 e 8 e T A 5 e 8 e T A 7 e 8 . 8 e T A 6 e 8 e T A 8 e 8 . . . e T A n+1 e 8 .e 8 e T A n+3 e 8 e T A n+5 e 8 . . . e T A 2n−2 e 8 e T A 2 e−4e T Ae+4e T e 4 e T A 4 e 4 e T A 6 e 4 . 8 e T A 5 e 8 e T A 7 e 8 . . . e T A n e 8 e 8 e T A 6 e 8 e T A 8 e 8 . . . e T A n+1 e 8 .e 8 e T A n+3 e 8 e T A n+5 e 8 .
. .e T A n−1 e 8 e T A 3 e−4e T A 2 e+4e T Ae . .e T A n e 8 e T A 4 e−4e T A 3 e+4e T A 2 e T A n+1 e−4e T A n e+4e T A n−1 e is an integral matrix and we have M 4 u ≡ 0 (mod 2T A 3 e−4e T A 2 e+4e T Ae . .e T A n e 8 e T A 4 e−4e T A 3 e+4e T A 2 e T A n+1 e−4e T A n e+4e T A n−1 e . .e T A n−1 e 4 e T A 3 e−4e T A 2 e+4e T Ae T A 4 e−4e T A 3 e+4e T A 2 e T A n+1 e−4e T A n e+4e T A n−1 e electronic journal of combinatorics 26(1) (2019), #P1.9

Figure 1 :
Figure 1: The directed graph Γ constructed by the set S.

Figure 2 :
Figure 2: The directed graph Γ constructed by the set S. It can be computed easily using Mathematica 11.0 that det W (G) = 2 16 × 27925453.Thus we have G ∈ Σ n .The set S consists of 2 3 vectors over F 2 .The directed graph Γ consists of exactly one directed 7-cycle and a loop, as shown in Fig. 2. Clearly the loop consisting of the zero vector can be ignored, since the set S does not have zero vector.Denote the seven circle by C 1 .Then dim F 2 (span(C 1 )) = 3. However dim F 2 (span(S )) = 2, where the set S = {v 1 , v 2 } consists of the first two columns of the matrix S. Note

Figure 3 :
Figure 3: The directed graph Γ constructed by the set S. It can be computed easily using Mathematica 11.0 that det W (G) = −2 18 × 7 × 13 × 17 × 1185703.Thus we have G ∈ Σ n .Furthermore, the set S consists of 2 4 vectors over F 2 .The directed graph Γ consists of exactly three directed 5-cycles and a loop, as shown in Fig. 3. Ignoring the loop consisting of the zero vector, let us denote the three directed cycles by C 1 , C 2 and C 3 .Then it is easy to compute that dim F 2 (span(C i )) = 4 for i = 1, 2, 3.However, dim F 2 (span(S )) = 3, where S = {v 1 , v 2 , v 3 , v 4 } consists of the first four columns of the matrix S. Since dim F 2 (span(S )) < dim F 2 (span(C i )) for i = 1, 2, 3, according to Theorem 28, the graph G is DGS.

Figure 4 :
Figure 4: The directed graph Γ constructed by the set S.

Figure 5 :
Figure 5: The directed graph Γ = (V (Γ), E(Γ)) constructed by the set S. It can be computed easily using Mathematica 11.0 that det W (G) = 2 21 × 753033383825423.Thus we have G ∈ Σ n .Furthermore, the set S consists of 2 5 vectors over F 2 (which are omitted to save space).The directed graph Γ consists of exactly four components, as shown in Fig. 5.The two components in the left-bottom and right-bottom of Fig. 5 can be ignored, since the vertex set of which do not have any entry from S .Denote the two components in the left-top and right-top of Fig. 5 by C i for i = 1, 2. Then we have dim F 2 (span(C i )) = 4 for i = 1, 2. However, dim F 2 (span(S )) = 3, where the set S = {v 1 , v 2 , v 3 , v 4 } consists of the first four vectors in S. Since dim F 2 (span(S )) < dim F 2 (span(C i )) for i = 1, 2, according to Theorem 28, the graph G is DGS.
e T A 2+k e e T A n+1 e 8 e T A n+3 e 8 e T A n+5 e 8 ...e T A 2n−2 e.Then M 2 is an integral matrix and M 2 u ≡ 0 (mod 2).
By Theorem 6 we have | 4b.Next we show p for any odd prime p.For otherwise, if p | , then p | b and p 2 b (since b is odd and square-free).According to Theorem 7 we get p ; a contradiction.Therefore, we have | 4. It follows from Theorem 10 that 4 , we get is equal to either 1 or 2.

Table 1 :
Fraction of DGS graphs in Σ n n (G) is high.Thus the existing method does not work.We first show that the level of every Q ∈ Q(G) is either 1 or 2. Then we present a simple method to eliminate the possibility that = 2, by the means of constructing a digraph associated with G and then using a simple dimension argument.Numerical experiments have also been presented to illustrate the proposed method.