Deza graphs with parameters $(n,k,k-1,a)$ and $\beta=1$

A Deza graph with parameters $(n,k,b,a)$ is a $k$-regular graph with $n$ vertices in which any two vertices have $a$ or $b$ ($a\leq b$) common neighbours. A Deza graph is strictly Deza if it has diameter $2$, and is not strongly regular. In an earlier paper, the two last authors et el. characterized the strictly Deza graphs with $b=k-1$ and $\beta>1$, where $\beta$ is the number of vertices with $b$ common neighbours with a given vertex. Here we deal with the case $\beta=1$, thus we complete the characterization of strictly Deza graphs with $b=k-1$. It follows that all Deza graphs with $b=k-1$ and $\beta=1$ can be made from special strongly regular graphs, and we present several examples of such strongly regular graphs. A divisible design graph is a special Deza graph, and a Deza graph with $\beta=1$ is a divisible design graph. The present characterization reveals an error in a paper on divisible design graphs by the second author et al. We discuss the cause and the consequences of this mistake and give the required errata.


Introduction
A k-regular graph Γ on n vertices is called a Deza graph with parameters (n, k, b, a) if the number of common neighbours of two distinct vertices takes on only two values a or b (a ≤ b). If the number of common neighbours of two vertices only depend on whether the vertices are adjacent or not, then Γ is a strongly regular graph with parameters (n, k, λ, µ), where λ (µ) is the number of common neighbours of two adjacent (non-adjacent) vertices; so {a, b} = {λ, µ}. A Deza graph is called a strictly Deza graph if it has diameter 2 and is not strongly regular. Note that the complete graph K n (which is normally excluded from being strongly regular) is a Deza graph which is not strictly Deza because it has diameter 1.
Let Γ be a Deza graph with parameters (n, k, b, a), and let v be a vertex of Γ. Denote by N (v) the set of neighbours of a vertex v, and let β(v) be the number of vertices u ∈ V (Γ) such that |N (v) ∩ N (u)| = b.
Theorem 1 ( [9]). Let Γ be a strictly Deza graph with parameters (n, k, b, a) and β > 1. The parameters k and b of Γ satisfy the condition k = b + 1 if and only if Γ is isomorphic to the strong product of K 2 with the complete multipartite graph with parts of size (n − k + 1)/2.
In this paper we characterize strictly Deza graphs with parameters (n, k, b, a), where k = b + 1 and β = 1.
Note that the adverb 'strictly' in Theorem 1 can not be removed, as is shown by the n-cycle with n ≥ 5. However, in the present characterization, we will see that there are no examples which are not strictly Deza, except for K 2 .

Construction 1.
Let Γ 1 be the strong product of K 2 and ∆. The graph Γ 1 is a strictly Deza graph with parameters (n, k, k − 1, a) and β = 1, where n = 2ℓ, k = 2ℓ + 1, a = 2µ. So, if B and A 1 are the adjacency matrices of ∆ and Γ 1 , respectively, then A 1 = B ⊗ J 2 − I n (J m is the m × m all-ones matrix, and I m is the identity matrix of order m).
Suppose that ∆ has an involution that interchanges only non-adjacent vertices. Let P be the corresponding permutation matrix, then B ′ = P B is a symmetric matrix (because P = P ⊤ and P BP = B) with zero diagonal (because P interchanges only nonadjacent vertices). So B ′ is the adjacency matrix of a graph ∆ ′ (say), which is a Deza graph because B ′2 = P BP B = B 2 . This construction was given in [3] and the method has been called dual Seidel switching; see [4].
Next, let Γ ′ 2 be the strong product of K 2 and ∆ ′ . Modify Γ ′ 2 as follows: For any transposition (x y) of the involution, take the corresponding two pairs of vertices x ′ , x ′′ and y ′ , y ′′ in Γ ′ 2 , delete the edges {x ′ , x ′′ } and {y ′ , y ′′ }, and insert the edges {x ′ , y ′′ } and {x ′′ , y ′ }. Define Γ 2 to be the resulting graph. If A 2 is the adjacency matrix of Γ 2 , then we can also construct Γ 2 from Γ 1 using dual Seidel switching in the following way: We easily have that A 2 2 = A 2 1 , which shows that Γ 2 is a Deza graph with the same parameters as Γ 1 .
Note that in Γ 1 any two vertices with b common neighbors are adjacent. For Γ 2 this is not true, therefore Γ 1 and Γ 2 are non-isomorphic.

Proof of the characterization
Let Γ be a Deza graph with parameters (n, k, k − 1, a) with k > 1 and β = 1. For a vertex x of Γ, denote by x b the vertex of Γ that has b common neighbours with x. Note that ( fact that x, x b are A-vertices implies that y b is adjacent to both the vertices x and x b . Let x be a N A-vertex. Then N (x) contains precisely one vertex which is not adjacent to x b . Denote this vertex by x ′ .
Lemma 6. The following statements hold.  Proof. Since the vertices x, x ′ , x b , (x b ) ′ are neighbours of y, and |N (y, y b )| = k − 1, the vertex y b is adjacent to at least three of them. By Lemma 6(6), the vertex y b is adjacent to all of them. Lemma 8. The parameter a is even.
Proof. It follows from Lemma 7 and the fact that |W (x)| = a.
Lemma 9. For any N A-vertex x, the vertex x ′ is a N A-vertex.
Proof. Suppose that (x ′ ) b and x ′ are adjacent. Since x is a neighbour of x ′ , then, in view of Lemmas 3 and 4, the vertices x, x b belong to N (x ′ , (x ′ ) b ). In particular, we obtain that x ′ is adjacent to x b , which is a contradiction because x ′ is not adjacent to x b by definition.
Let us count the number of common neighbours of the vertices ( Lemma 11. For a N A-vertex x, the equality x ′′ = x holds. Proof. The vertex x ′ is adjacent to x be definition. By Lemma 10, the vertex (x ′ ) b is not adjacent to x. This proves the lemma.
Proof. It is enough to show that (x ′ ) b and (x b ) ′ have at least a + 1 common neighbours.
By Lemma 6(5), the vertex ( The vertex x ′ has k − a − 1 neighbours in N 2 (x), which are neighbours of (x ′ ) b too. Let us take a vertex y ∈ N (x ′ , (x ′ ) b ) ∩ N 2 (x). Since y is adjacent to x ′ and x b is not adjacent to x ′ , the vertices y and x b have precisely a − 1 common neighbours in N (x). This implies that the vertex (x b ) ′ , which is a unique neighbour of x b in N 2 (x), is a common neighbour of x b and y, and, in particular, y is adjacent to (x b ) ′ . Thus, the vertices (x ′ ) b and (x b ) ′ have at least a + 1 common neighbours, which proves the lemma.
Further, in view of Lemma 12, we use the simplified notation Proof. It follows from the equality (x b ) b = x, Lemma 11 and Lemma 12.
Proof. It follows from Lemma 13.
Lemma 15. Let x be a N A-vertex, and let y be a vertex, y does not belong Then there are either all possible edges between {x, x b } and {y, y b } or no such edges.
Proof. If y is an A-vertex, then the result follows from Lemma 4. Let us assume without loosing of generality that y is a N A-vertex. In view of Lemma 13, it is enough to show that if x and y are adjacent, then x is adjacent to y b , x b is adjacent to y, and x b is adjacent to y b . Suppose x is not adjacent to y b . Then x = y ′ holds, which is a contradiction since, by Lemma 14, the condition {x, Using the similar arguments, we can show that x b is adjacent to y and x b is adjacent to y b . The lemma is proved.
Lemma 16. For any two N A-vertices x, y such that C(x) = C(y), the following statements hold.
(1) The vertices x ′ and y are adjacent if and only if the vertices x and y ′ are adjacent.
(2) The vertices x and y are adjacent if and only if the vertices x ′ and y ′ are adjacent.
such that x ′ is adjacent to y and x is not adjacent y ′ . For any vertex z ∈ N (x, y), z = x ′ , y ′ , in view of Lemma 15, there are all possible edges between {z, z b } and {x, x b }, {z, z b } and {y, y b }. In particular, z b belongs to N (x, y). Moreover, x ′ belongs to N (x, y) and y ′ does not belong to N (x, y). This means that a = |N (x, y)| is odd, which contradicts to Lemma 8.
Lemma 17. Let x, y be two N A-vertices, C(x) = C(y). Then the following statements hold.
The proof is similar to the one from item (1).
Lemma 18. Let x be a N A-vertex and y be an A-vertex. The following statements hold.
(1) If x and y are adjacent, then the vertices y and y b lie in W (x).
(2) There are either all possible edges between the sets {x, x ′ , x b , x ′ b } and {y, y b }, or no such edges.
Proof. (1) Let z be a vertex in N (x, y), z = x ′ , y b . We prove that z b belongs to N (x, y). If z is an A-vertex, then it follows from Lemma 4. Suppose z is an N A-vertex. Then, by Lemma 4, z b belongs to N (y), and by Lemma 15, z b belongs to N (x), which implies that z b belongs to N (x, y). Moreover, y b belongs to N (x, y). In view of Lemma 8, |N (x, y)| is even, which implies that x ′ belongs to N (x, y), and, in particular, y is adjacent to x ′ . This means that y belongs to N (x, x ′ ), and, consequently, y b belongs to N (x, x ′ ).
Let Γ ′ be the graph obtained from Γ by removing all edges {x, Lemma 19. The following statements hold.
(2) For any vertices x, y in Γ ′ such that y = x b , there are either all possible edges between {x, x b } and {y, y b } or no such edges.
(3) For any vertices x, y in Γ ′ such that y = x b , the equalities Proof. (1) It follows from the fact that precisely one edge was removed for each vertex.
(2) If x, y are A-vertices, then it follows from Lemma 3. If one of the vertices x, y is an A-vertex and the other is a N A-vertex, then it follows from Lemma 18. If x, y are N A-vertices, then it follows from Lemma 17.  Lemma 21. The following statements hold.
(1) The adjacency matrix of Γ ′′′ can be obtained from the adjacency matrix of Γ ′′ by permuting rows in pairs corresponding to the vertices {x, In particular, the adjacency matrices of Γ ′′′ and Γ ′′ coincide in the case when Γ has no N A-vertices.
In view of Lemma 21, if Γ has no N A-vertices, then it comes from Construction 1. Let us prove that, if Γ has N A-vertices, then it comes from Construction 2. Lemmas 20 and 21 imply that the adjacency matrix of the Deza graph Γ ′′ can be obtained from the adjacency matrix of the strongly regular Γ ′′′ by swapping rows in pairs corresponding to the vertices {x, It follows from [3, Theorem 3.1], that this permutation is an order 2 automorphism of Γ ′′′ that interchanges only non-adjacent vertices. Thus, Γ ′′ can be obtained from Γ ′′′ by dual Seidel switching, which implies that Γ comes from Construction 2 by definition. The theorem is proved.

Strongly regular graphs with
Note that, if Γ is a strongly regular graph with λ = µ − 1, then so is its complement. So both Γ and its complement satisfy the condition for Construction 1. For Construction 2, Γ needs an involution that interchanges only nonadjacent vertices. In this section we survey strongly regular graphs with λ = µ − 1, and look for the desired involutive automorphism.

Paley graphs of square order
The Paley graph P (r) is a graph with vertex set F r , where r is a prime power such that r ≡ 1 mod 4. Two vertices x and y of P (r) are adjacent whenever x − y is a non-zero square in F r . See [8] for an excellent survey of Paley graphs. The Paley graph is a strongly regular graph with parameters (r, r−1 2 , r−5 4 , r−1 4 ), so it satisfies the conditions of Construction 1, which leads to strictly Deza graphs with parameters (2r, r, r − 1, (r − 1)/2). The complement of P (r) is isomorphic to P (r), since for any non-square a in F r the map x → ax interchanges edges and non-edges in P (r).
If r = q 2 is a square, then the Paley graph P (q 2 ) satisfies the conditions of Construction 2. To explain this we need some properties of the field F q 2 . Let d be a non-square in F * q . The elements of the finite field of order q 2 can be considered as where α is a root of the polynomial f (t) = t 2 − d.
Let β be a primitive element of the finite field F q 2 . Then we have F * q = {β i(q+1) | i ∈ {0, . . . , q − 2}}. Since q + 1 is even, each element of F * q is a square in F * q 2 . It also follows that x q−1 = β q 2 −1 = 1 for every x ∈ F * q .
Lemma 22. For any γ = x + yα from F q 2 , the following equalities hold.
The norm mapping is a homomorphism from F * q 2 to F * q with Im(N ) = F * q . Thus, the kernel Ker(N ) is the subgroup of order q + 1 in F * q 2 . Now we make some remarks on squares in finite fields.
Lemma 23. (1) The element −1 is a square in F * q iff q ≡ 1(4); (2) For any non-square d in F * q the element −d is a square in F * q iff q ≡ 3(4). The following lemma can be used to test whether an element γ = x+yα ∈ F * q 2 is a square.

Lemma 24 ([1], Lemma 2). An element
Lemma 25 immediately follows from Lemma 24, Lemma 23 and the fact that Lemma 25. The element α is a square in F * q 2 iff q ≡ 3(4). Denote by ϕ the automorphism of P (q 2 ) that sends γ to γ q . Note that ϕ fixes the elements from F q .
Lemma 26. The following statements hold.

Symmetric conference matrices
An m × m matrix C with zero's on the diagonal, and ±1 elsewhere, is a conference matrix if CC ⊤ = (m−1)I. If a conference matric C is symmetric with constant row (and column) sum r, then r = ± √ m − 1, and B = 1 2 (J m − I m − C) is the adjacency matrix of a strongly regular graph with parameter set Note that P(−r) is the complementary parameter set of P(r). Symmetric conference matrices with constant row sum have been constructed by Seidel (see [10], Thm. 13.9). If q is an odd prime power and r = ±q, then such a conference matrix can be obtained from the Paley graph of order q 2 . Let B ′ be the adjacency matrix of P (q 2 ), and put S = J q 2 − I q 2 − 2B ′ (S is the so-called Seidel matrix of P (q 2 )). Define (1 is the all-ones vector). Then C ′ is a symmetric conference matrix of order m = q 2 + 1. However, C ′ doesn't have constant row sum. Next we shall make the row and column sum constant by multiplying some rows and the corresponding columns of C ′ by −1. This operation is called Seidel switching, and it is easily seen that Seidel switching doesn't change the conference matrix property. To describe the required rows and columns, we use the notation and description of P (q 2 ) given in the previous subsection. If q ≡ 3 mod 4 we take the complement of the described Paley graph. Then the involution ϕ given in Lemma 26 interchanges only non-adjacent vertices in all cases. For x ∈ F q define V x = {x + yα | y ∈ F q }. Then the sets V x partition the vertex set of P (q 2 ), and each class is a coclique. Moreover, the partition is fixed by the involution ϕ. Let V be the union of 1 2 (q − 1) classes V x . Then V induces a regular subgraph of P (q 2 ) of degree 1 4 (q − 1) 2 − 1 with 1 2 q(q − 1) vertices. Now make the matrix C by Seidel switching in C ′ with respect to the rows and columns that correspond with V . Then C is a regular symmetric conference matrix, and B = 1 2 (J − I − C) is the adjacency matrix of a strongly regular graph Γ with parameter set P(q), and ϕ remains an involution that interchanges only nonadjacent vertices. So Γ satisfies the conditions for Construction 1 and 2. For the complement of Γ we found no involutions that interchanges only nonadjacent vertices. Thus we find Deza graphs with parameters (q 2 + 1, 1 2 (q 2 − q), 1 4 (q − 1) 2 , 1 4 (q − 1) 2 − 1 ) (by dual Seidel switching in Γ), (2q 2 +2, q 2 −q +1), q 2 −q, 1 2 (q −1) 2 ) (by Construction 1 and 2 applied to Γ), and (2q 2 +2, q 2 +q+1), q 2 +q, 1 2 (q+1) 2 ) (by Construction 1 applied to the complement of Γ). If q = 3, Γ is the Petersen graph. It has a unique involutive automorphism that interchanges only non-adjacent vertices. This automorphism has 4 fixed and 6 moved vertices. The Deza graph obtained from the Petersen graph with dual Seidel switching has diameter 3. However, in all other cases the obtained Deza graphs are strictly Deza. The complement of the Petersen graph has no involutive automorphisms that interchanges only non-adjacent vertices. We expect that this is the case for all strongly regular graphs with parameter set P(r) and r < 0.
The 50 vertices of Hoffman-Singleton graph are grouped into 5 pentagons P 0 , . . . , P 4 and 5 pentagrams Q 0 , . . . , Q 4 labeled in such a way that the pentagrams are the complements of the pentagons; there are no edges between any two distinct pentagons, nor between any two distinct pentagrams. Edges between pentagon and pentagram vertices are defined by the rule: each vertex i ∈ {0, 1, 2, 3, 4} of a pentagon P j , j ∈ {0, 1, 2, 3, 4} is adjacent to the vertex (i + jk) mod 5 of a pentagram Q k , for any k ∈ {0, 1, 2, 3, 4}.
Let ϕ be the permutation of vertices of Hoffman-Singleton graph that fixes each vertex of P 0 , Q 0 , and interchanges P 1 , Q 1 with P 4 , Q 4 and P 2 , Q 2 with P 3 , Q 3 . The permutation ϕ is a unique involutive automorphism of Hoffman-Singleton graph that interchanges only non-adjacent vertices. The Deza graph obtained from Hoffman-Singleton graph with dual Seidel switching has diameter 3. However, each of Constructions 1 and 2 produces a strictly Deza graph with parameters (100, 15, 14, 2). Construction 1 applied to the complement gives a strictly Deza graph with parameters (100, 85, 84, 72).

Divisible design graphs
For a Deza graph Γ with parameters (n, k, b, a) and a = b, we define two graphs Γ a and Γ b on the vertex set of Γ, where two vertices x and y are adjacent in Γ a (Γ b ) if x and y have a (b) common neighbors. Clearly Γ a and Γ b are each others complement, and regular of degree α and β, respectively (Γ a and Γ b have been called the children of Γ).
If Γ a or Γ b is the disjoint union of complete graphs, then Γ is called a divisible design graph (DDG for short). DDGs are interesting structures on their own, and have been studied in [6] and [2]. If A is the adjacency matrix of a DDG, then A also satisfies the conditions for the incidence matrix of a divisible design, which explains the name.