STRONGLY REGULAR
GRAPHS
By:
BAHMAN AHMADI
A Project Report for The Course
ALGEBRAIC COMBINATORICS
Instructor:
Dr. Karen Meagher
Department of Mathematics and Statistics
University of Regina
April, 2009.
In The Name of GOD,
The Most Beneficent,
The Most Merciful.
1
Introduction
One of the most important family of regular graphs is the family of Strongly
Regular Graphs (abbreviated SRG) which has so many beautiful properties. One of
these well-known properties is the fact that the eigenvalues of all SRGs depend only
the parameters of these graphs. There are many SRGs arising from combinatorial
concepts such as orthogonal arrays, Latin squares, conference matrices, designs and
geometric graphs. In this report, we try to familiarize the reader with the concept
of strongly regular graphs and their nice properties and connections with these
combinatorial objects and to describe an interesting open problem related to the
cores of SGRs. The main reference for this report is [1].
2
1. Definition and basic properties
Definition 1.1. Let X be a graph which is neither empty nor complete. Then
X is said to be a strongly regular graph (SRG) with parameters (n, k, a, c) if X
is a k−regular graph on n vertices in which every pair of adjacent vertices has a
common neighbors and every pair of non-adjacent vertices has c common neighbors.
Throughout the text, when X is supposed to be a SRG, unless otherwise stated,
we assume that the parameters of X are (n, k, a, c). We will denote the adjacency
matrix of any graph X with A(X).
It easily follows from the definition that if X is a bipartite SRG, then a = 0; but
as in the following example, the converse is not true.
Example 1.2. i) The cycle C4 is a (bipartite) SRG with parameters (4, 2, 0, 2).
The cycle C5 is a (non-bipartite) SRG with parameters (5, 2, 1, 0). It is easily seen
that for any n ≥ 6 the cycle Cn is not a SRG.
ii) The Petersen graph P10 is a SRG with parameters (10, 3, 0, 1); indeed since the
diameter is 2, every nonadjacent pair of vertices has at least a common neighbor,
and since the girth is 5, this common neighbor is unique. Also a = 0 follows from
the fact that the girth is 5 and so P10 has no triangle.
iii) A connected graph X is said to be a Moore graph if diam(X) = d and
girth(X) = 2d + 1. It can be seen ([1], p.90) that X is regular. Now if X is a
Moore graph of diameter 2 on n vertices with valency k, then by similar argument
as in part (ii), we can see that X is a SRG with parameters (n, k, 0, 1).
iv) Let X be L(Kn ), the line graph of Kn . Then the number of vertices of X,
is the number of edges of Kn , namely
n(n−1)
.
2
Let x, y ∈ V (Kn ). Then the edge
e = xy is ‘adjacent’ to n − 2 edges through x and to n − 2 other edges through y.
Therefor X is (2n − 4)−regular. Let e = xy and e0 = xz be two ‘adjacent’ edges in
3
Kn (i.e. e ∼ e0 as vertices in X). Then e and e0 are both adjacent to n − 3 other
edges through x, and both are adjacent to the edge yz. Thus every pair of adjacent
vertices in X has n − 2 common neighbors. Finally if e = xy and e0 = wz are two
‘nonadjacent’ edges in Kn ; then both of them are adjacent to the edges xw, xz,
yz and yw. Consequently every pair of nonadjacent vertices in X, has 4 common
neighbors. The result is that X is a SRG with parameters ( n(n−1)
, 2n − 4, n − 2, 4).
2
v) By a similar reasoning as in part (iv), we can see that L(Kn,n ) is a SRG with
parameters (n2 , 2n − 2, n − 2, 2).
Proposition 1.3. If X is a SRG, then X̄, the complement of X, is also a SRG
with parameters (n, k̄, ā, c̄), where
k̄ = n − k − 1,
ā = n − 2 − 2k + c,
c̄ = n − 2k + a.
Proof. It is obvious that X̄ is (n − k − 1)−regular and so k̄ = n − k − 1. Consider
two vertices x, y ∈ V (X) which are adjacent in X̄ and so nonadjacent in X. Then
x and y share c neighbors; each of them has k − c neighbors not in common with
the other. Therefore the number of all vertices in X \ {x, y} which are adjacent to
non of x and y is n − c − 2(k − c) − 2, and so ā = n − 2 − 2k + c.
Now if x and y are nonadjacent in X̄, then they are adjacent in X, and so have a
common neighbors. Therefore, by a similar argument, x and y will have n − 2k + a
common neighbors in X̄; that is, c̄ = n − 2k + a.
Definition 1.4. Let X be a SRG. Then X is said to be primitive if both X and
X̄ are connected. We say X is imprimitive if it is not primitive.
Lemma 1.5. Let X be a SRG. Then the following conditions are equivalent:
4
i) X is not connected;
ii) c = 0;
iii) a = k − 1;
iv) X is isomorphic to mKk+1 , for some m > 1.
Proof. Suppose X is not connected and let x, y ∈ V (X) be such that there is no
path joining them to each other. So x is not adjacent to y, and they cannot have
a common neighbor; therefore c = 0.
Let c = 0, and let x, y ∈ V (X) be adjacent. If z ∈ V (X) and z ∼ y, then x must
be adjacent to z; indeed, if x is not adjacent to z, since they have y as a common
neighbor, then c 6= 0. Thus x is adjacent to all the neighbors of y and so a = k − 1.
Obviously a = k − 1 implies that each component of X has to be a complete
graph, and since the degree is k, each component has to be isomorphic to Kk+1 ,
and the result follows.
The part (iv) ⇒ (i) is trivial.
Corollary 1.6. Suppose X is an imprimitive SRG. Then either X ∼
= mKk+1 or
X∼
= mKk+1 , for some m > 1.
It is reasonable to ask: if n, k, a, c are some positive integers such that k < n and
a, c ≤ k, whether they are parameters of some SRG or not. The following lemma
provides us with a simple feasibility condition.
Lemma 1.7. If X is a SRG with parameters (n, k, a, c), then
k(n − a − 1) = (n − k − 1)c.
Proof. Fix a vertex u ∈ V (X), and let A be the set of all neighbors of u, and
B = V (X) \ A. Note that |A| = k and |B| = n − k − 1. We count t, the number of
all edges between A and B, in two ways.
5
First, each vertex in A is adjacent to a vertices in A and so is adjacent to k −1−a
vertices in B. Thus
t = |A|(k − a − 1) = k(k − a − 1).
Second, any vertex x ∈ B, is adjacent to c vertices in A, since x is not adjacent
to u. Therefor
t = |B|c = (n − k − 1)c,
and the result follows.
2. Eigenvalues
In this section we try to determine all the eigenvalues of the strongly regular
graphs using their parameters. We shall show that all SRGs have at most three
distinct eigenvalues and evaluate their multiplicities.
Let X be a SRG and let A be the adjacency matrix of X. Note that the ij−th
entry of the matrix A2 is the number of walks of length 2 joining i to j. Consider
the following cases:
If i = j, then (A2 )ij is the degree of i. So all the diagonal entries of A2 are equal
to k.
If i 6= j and they are adjacent, then (A2 )ij is the number of common neighbors
of i and j. Thus (A2 )ij = a.
If i 6= j and they are nonadjacent, then (A2 )ij is the number of common neighbors
of i and j. Thus (A2 )ij = c.
Consequently we have the following identity:
A2 = kI + aA + c(J − I − A),
where J is the n × n all 1’s matrix. We may rewrite it as
A2 − (a − c)A − (k − c)I = cJ.
6
Since X is k−regular, k is an eigenvalue of A with the eigenvector 1, which is the all
1’s vector of length n. Furthermore, if z is another eigenvector of A with eigenvalue
λ, then z has to be orthogonal to 1. Therefore
A2 z − (a − c)Az − (k − c)Iz = cJz = 0,
so
λ2 − (a − c)λ − (k − c) = 0.
This means that the other eigenvalues of A are the roots of the monic quadratic
equation
x2 − (a − c)x − (k − c) = 0,
(2.1)
Set ∆ = (a − c)2 + 4(k − c). Then the roots of (2.1), i.e. the eigenvalues of A other
than k, are
(2.2)
θ=
(a − c) +
2
√
∆
,
τ=
(a − c) −
2
√
∆
.
Now θτ = (c − k) and therefore, for the cases in which c < k, θ and τ have opposite
signs. We always assume that τ < 0. Note that c = k if and only if X ∼
= mKk+1 ,
for some m > 1. Thus for a strongly regular graph X, which is not isomorphic to
mKk+1 , (m > 1), we have τ < 0 and θ > 0.
Next we try to find the multiplicities of the eigenvalues. We do the work for
connected SRGs and the result for all SRGs will follow easily.
Let X be a connected SRG with eigenvalues k, θ and τ . Since X is connected,
the multiplicity of k is 1. Denote by mθ and mτ the multiplicities of θ and τ ,
respectively. Because tr(A) = 0, we have that mθ +mτ = n−1 and mθ θ+mτ τ = −k.
Thus following equalities hold:
mθ = −
(n − 1)τ + k
,
θ−τ
mτ =
(n − 1)θ + k
.
θ−τ
7
On the other hand, (θ − τ )2 = (θ + τ )2 − 4θτ = (a − c)2 + 4(k − c) = ∆. We can,
therefore, write mθ and mτ as
(2.3)
mθ =
1
2k + (n − 1)(a − c)
√
[(n − 1) −
],
2
∆
mτ =
1
2k + (n − 1)(a − c)
√
].
[(n − 1) +
2
∆
Note that in equations (2.3), right hand sides of the equations have to be integers.
Therefore, (2.3) also can be regarded as other feasibility conditions on a parameter
set (to be parameters of some SRGs).
Following proposition shows that having three distinct eigenvalues, is indeed a
sufficient condition for connected regular graphs to be strongly regular.
Proposition 2.1. If X is a connected k−regular graph with exactly three distinct
eigenvalues k, θ and τ , then X is strongly regular.
Proof. Consider the matrix
(2.4)
M=
1
(A − θI)(A − τ I),
(k − θ)(k − τ )
where A is the adjacency matrix of X and n is the number of vertices of X. Since
M is a polynomial in A, we can see that all the eigenvalues of M are 0 or 1. Let z
be an eigenvector of A with eigenvalue θ. Then
Mz =
1
(k−θ)(k−τ ) (A
− θI)(A − τ I)z
=
1
(k−θ)(k−τ ) (A
− θI)(Az − τ z)
=
1
(k−θ)(k−τ ) (A
− θI)(θz − τ z)
=
1
(k−θ)(k−τ ) (θAz
=
1
2
(k−θ)(k−τ ) (θ z
− τ Az − θ2 z + θτ z)
− τ θz − θ2 z + θτ z) = 0.
Therefore all eigenvectors of A corrosponding to the eigenvalue θ are in the kernel
of M . The same argument shows that this is also the case for all the eigenvectors
of A corresponding to the eigenvalue τ . Since the set of all eigenvectors of A is
8
a basis for Rn , this implies that the rank of M is equal to the multiplicity of the
eigenvalue k which is 1 (since X is connected). That is, all the rows of M are
linear combinations of one row. On the other hand, 1 is an eigenvalue of M with
eigenvector 1, and so M 1 = 1. Therefor M =
A2 = θτ I − (θ + τ )A −
1
n J.
Substituting this in (2.4) yields
(k − θ)(k − τ )
J.
n
Since we could write A2 as a linear combination of I, A and J, this shows that A
is a strongly regular graph (i.e. the parameters are well-defined).
Definition 2.2. Let X be a SRG with mθ = mτ . Then X is said to be a conference
graph.
This name arises from the fact that these graphs are associated to so-called
symmetric conference matrices. A conference matrix C of order n is an n × n,
(0, 1, −1) matrix whose diagonal entries are all 0 and CC T = (n − 1)I. If C is a
symmetric conference matrix, then by permuting rows and columns (if necessary)
we may assume that the first row and the first column of C are all 1’s except the
element C11 = 0. Now delete the first row and the first column and switch all +1’s
to 0 and all −1’s to 1 to obtain the (n − 1) × (n − 1) matrix AC . Let XC be a graph
whose adjacency matrix is AC . It is not hard to show that XC is a conference
graph. It has been shown that if n ≡ 2 (mod2), then a conference matrix of order
n exists if and only if n − 1 is the sum of two squares. (For a proof of this, you may
refer to [2] or [3]). Therefore we have the following result.
Corollary 2.3. A conference graph with n vertices exists if and only if n is the
sum of two squares.
Example 2.4. Dose a SRG with parameters (21, 10, 4, 5) exists?
9
Answer: If X is a SRG with parameters (21, 10, 4, 5), then using (2.3), we see that
mθ = mτ = 10, and so X would be a conference graph; but this is contradiction,
since 21 is not sum of two squares.
Recall that a graph X is said to be vertex-transitive if for any pair {x, y} ∈ V (X),
there is an automorphism g on X such that g(x) = y. One of the obvious properties
of vertex-transitive graphs is that they are regular.
Example 2.5. In this example we introduce an important (infinite) family of
conference graphs, namely the Paley graphs. Let q be a prime power such that
q ≡ 1 (mod4), and let GF (q) be a filed with q elements. Then the Paley graph
P (q) is a graph with vertex set GF (q), in which two vertices are adjacent if and
only if their difference is a square of an element of GF (q). To show that P (q) is
not directed, we should show that −1 is a square. Since GF (q)∗ = GF (q) \ {0} is
a cyclic (multiplicative) group, there is an α ∈ GF (q)∗ such that
GF (q)∗ =< α >= {α, α2 , . . . , αq−1 = 1}.
Let C = {(α1 )2 , (α2 )2 , (α3 )2 , . . . , (α(q−1)/2 )2 } be the set of all squares in GF (q)∗ .
Obviously C is a subgroup of GF (q)∗ of order (q − 1)/2. We claim that −1 ∈ C.
Note that (α(q−1 )/4) ∈ GF (q)∗ , (since q ≡ 1 (mod4)), and
((α(q−1)/4 )2 )2 = αq−1 = 1 ⇒ (α(q−1)/4 )2 = ±1;
and since GF (q)∗ =< α >, we must have −1 = (α(q−1)/4 )2 ∈ C (since otherwise
we would have | < α > | < |GF (q)∗ |).
Now for any β ∈ GF (q) consider the homomorphism x 7→ x + β. It is easily
seen that this homomorphism is an automorphism, and therefore P (q) is vertextransitive. Consequently P (q) is regular. On the other hand the vertex 0 is adjacent
to all the elements of C. Thus P (q) is ( q−1
2 )−regular.
10
We claim that P (q) is self-complementary; i.e. P (q) ≡ P (q). Let 0 6= β ∈
/ C and
define φ : P (q) −→ P (q) by x 7→ βx. This map is clearly well-defined and bijection.
We claim that φ is an isomorphism. To show that, first note that for some odd
number s, β = αs . If x, y ∈ V (P (q)) are adjacent in P (q), then x − y = (αi )2 , for
some 1 ≤ i ≤ (q − 1)/2. Therefore φ(x) − φ(y) = βx − βy = βα2i = α2i+s ∈
/ C;
that is, φ(x) and φ(y) are not adjacent in P (q), and so are adjacent in P (q). Also
if x and y are not adjacent in P (q), then x − y = αt for some odd number t. So
φ(x) − φ(y) = βx − βy = βαt = αs+t ∈ C; that is, φ(x) and φ(y) are adjacent in
P (q), and so are not adjacent in P (q). Consequently φ is an isomorphism and P (q)
is self-complementary.
Now let γ ∈ C, and consider the map x 7→ γx. It is easily seen that this is an
automorphism of P (q), which fixes 0 and maps neighbor 1 of 0 to γ. We can deduce
that the number of common vertices of 0 and γ is independent from the choice of
γ ∈ C. Thus the number of triangles on an edge containing 0 is independent of the
edge. Since P (q) is vertex-transitive, it follows that any pair of adjacent vertices in
P (q) has the same number of common vertices. On the other hand since P (q) is selfcomplementary, every pair of nonadjacent vertices of P (q) has the same common
neighbors as well. Therefor P (q) is a strongly regular graph with parameters
(n = q, k =
q−1
, a , c).
2
Finally we determine a and c. Since a = ā, from Proposition 1.3, we have
a − c = n − 2k − 2 = q − 2 q−1
2 − 2 = −1, and using the fact that k(k − a − 1) = k̄c,
we have a + c = k − 1. Together these imply that c = k/2 and a = c − 1; thus
a = (q − 5)/4 and c = (q − 1)/4, and so P (q) is a SRG with parameters
(q,
q−1 q−5 q−1
,
,
).
2
4
4
11
Using these parameters we see that θ = (−1 +
√
q)/2 and τ = (−1 −
√
q)/2.
Furthermore the multiplicities are mθ = mτ = (q − 1)/2. Consequently P (q) is a
conference graph.
As we saw in the example above, all the parameters of the Paley graph can be
written in terms of q, the number of vertices. We show that this is the case for all
conference graphs. We need the following lemma.
Lemma 2.6. Let X be a SRG with distinct eigenvalues k, θ and τ . Then
nk k̄
.
(θ − τ )2
mθ mτ =
Proof. Let

k̄
−1 − θ 
−1 − τ

1 k
P = 1 θ
1 τ
in which first, second and third column are the eigenvalues of I, X and X̄. We
show that

1
P T 0
0
(2.5)
0
mθ
0


0
1
0  P = n 0
mτ
0
0
k
0

0
0 .
k̄
Denote the left hand side of (2.5) by M . Then we have
M12 = k + mθ θ + mτ τ = 0,
M13 = k̄ + mθ (1 − θ) + mτ (−1 − τ ) = tr(A(X̄)) = 0,
M23 = k k̄ + mθ θ(−1 − θ) + mτ τ (−1 − τ ) = tr(A(X)A(X̄)) = 0,
where A(X) is the adjacency matrix of X. To see that tr(A(X)A(X̄)) = 0, it may
help to note that
tr(BC) =
X
Bij Cji .
ij
Given that M is symmetric, these equations show that all the off-diagonal entries
of M are zero. Similar calculations show that the diagonal entries of M are n,
tr(A(X)2 ) and tr(A(X̄)2 ). But tr(A(X)2 ) is the number of closed walks of length
12
two in X, namely twice the number of edges. Therefore tr(A(X)2 ) = nk. Similarly
tr(A(X̄)2 ) = nk̄ and the claim is proved.
Now if we take determinant of both sides of (2.5), we will find
(det(P ))2 mθ mτ = n3 k k̄ ⇒ (n(τ − θ))2 mθ mτ = n3 k k̄;
and the lemma follows.
Proposition 2.7. Let X be a conference graph with parameters (n, k, a, c). Then
k = (n − 1)/2, a = (n − 5)/4 and c = (n − 1)/4.
Proof. Since mθ + mτ = n − 1 and mθ = mτ , we have m := mθ = mτ = (n − 1)/2.
Obviously gcd(m, n) = 1 (i.e. m and n are relatively prime). By Lemma 2.6, we
have
m2 |nk k̄ ⇒ m2 |k k̄.
On the other hand k + k̄ = n − 1, and so k k̄ ≤ (n − 1)2 /4 = m2 and equality holds if
and only if k = k̄. Therefore k = k̄ = m. Since m(θ + τ ) = mθ + mτ = −k, we see
that a−c = θ +τ = −k/m = −1 and so a = c−1. Finally because k(k −a−1) = k̄c,
we find c = k − a − 1, and hence c = k/2. Putting these all together shows that X
has the stated parameters.
If a SRG, X is not conference graph, then mτ − mθ will be a non-zero integer.
But using (2.3), we have
mτ − mθ =
2k + (n − 1)(a − c)
√
.
∆
Therefore ∆ must be a perfect square, and so θ and τ are rational numbers. On
the other hand θ and τ are the roots of the monic quadratic equation (2.1). Consequently θ and τ are integers. Therefore we have the following interesting result.
Corollary 2.8. Let X be a SRG with eigenvalues k, θ and τ . Then either
13
i) X is a conference graph, or
ii) (θ − τ )2 is a perfect square and θ and τ are integers.
When a SRG has p or 2p vertices, where p is a prime number, using the facts
above, we would have very interesting information about the parameter set; in what
follows we prove these results.
Proposition 2.9. If X is a SRG on p vertices, where p is a prime number, then
X is a conference graph.
Proof. By Lemma 2.6, we have
(θ − τ )2 =
(2.6)
pk k̄
.
mθ mτ
By Corollary 2.8, if X is not a conference graph, then (θ − τ )2 will be a perfect
square. Thus
pkk̄
mθ mτ
pkk̄
mθ mτ
is a perfect square having p as a factor. So p2 must divide
; that is, p must divide
kk̄
mθ mτ
. But this cannot be the case, since all of the
numbers k, k̄, mθ and mτ are non-zero numbers less than p. This contradiction
shows that X has to be a conference graph.
Lemma 2.10. Let X be a primitive SRG with an eigenvalue λ of multiplicity n/2.
If k < n/2, then the parameters of X are
((2λ + 1)2 + 1, λ(2λ + 1), λ2 − 1, λ2 ).
Proof. Since X is connected, λ 6= k, and so λ is either θ or τ . Since mθ = n−1−mτ ,
we see that mθ 6= mτ and hence θ and τ are integers.
We will show by contradiction that λ = θ. Suppose λ = τ and so mτ = n/2.
Note that
mτ =
nθ + k − θ
(n − 1)θ + k
=
⇒ mτ |(nθ + k − θ).
θ−τ
θ−τ
14
On the other hand mτ |n. Therefore mτ must divide k − θ. Now we claim that
0 < θ < k. Indeed if θ = 0, then using the fact that θτ = (c − k), we will have
c = k. Therefore by Proposition 1.3, ā = n − k − 2 = k̄ − 1, and so by Lemma 1.5,
X̄ would be disconnected which is a contradiction. Consequently θ > 0. Also if
θ = k, then using the fact that mθ = n − 1 − mτ , we will have that the multiplicity
of k as an eigenvalue of X is
n
2
− 1 which contradicts with the connectivity of X.
So the claim is proved; that is 0 < k − θ < n/2 and so mτ (= n/2) cannot divide
k − θ. This contradiction shows that we must have λ = θ and so mθ = n/2.
Next, note that
mθ =
nτ + k − τ
(n − 1)τ + k
=
θ−τ
θ−τ
⇒ mθ |(nτ + k − τ ),
and since mθ |n, mθ must divide k − τ . But since −k ≤ τ , k − τ ≤ 2k < n, and so
we must have mθ = k − τ .
Set m = mθ = n/2, so mτ = m − 1. Since tr(A(X)) = 0, we have
k + mθ + (m − 1)τ = 0
⇒ k + mθ + mτ − τ = 0
⇒ k + mθ + mτ − k + m = 0
⇒ 1 + θ + τ = 0,
and since tr(A(X)2 ) = nk, we will have
k 2 + mθ2 + (m − 1)τ 2 = 2mk
⇒ k 2 + mθ2 + mτ 2 − τ 2 = 2mk
⇒ k 2 + mθ2 + mτ 2 − k 2 + 2mk − m2 = 2mk
⇒ θ2 + τ 2 = m.
Combining these we will get m = θ2 + (θ + 1)2 and hence k = θ(2θ + 1). Also
since c − k = θτ = −(θ2 + θ), we have c = θ2 . Finally using a − c = θ + τ = −1,
we get a = θ2 − 1 and we are done.
15
Lemma 2.11. If X is a SRG with k and k̄ coprime, then X is imprimitive.
Proof. Note that k(k − a − 1) = k̄c. Thus k divides k̄c. Since gcd(k, k̄) = 1, k must
divide c. Since c ≤ k, either c = 0 or k = c. If c = 0, then Lemma 1.5 implies that
X is not connected, and if k = c, using Proposition 1.3, we will find ā = k̄ − 1, and
again by Lemma 1.5, we deduce that X̄ would not be connected. Therefore, in any
case X is imprimitive.
It can be inferred from the last lemma that if p is a prime number, then all SRGs
on p + 1 vertices are imprimitive.
Proposition 2.12. Let X be a primitive SRG on 2p vertices. Then the parameters
of X or its complement are
((2θ + 1)2 + 1, θ(2θ + 1), θ2 − 1, θ2 ).
Proof. By taking complement if necessary, we may assume that k ≤ (2p−1)/2. Note
that X cannot be a conference graph, since otherwise we would have 2mθ + 1 = 2p
which is contradiction. Therefor θ and τ are integers. By Lemma 2.6 we have
(θ − τ )2 mθ mτ = 2pk k̄, thus p divides (θ − τ )2 mθ mτ . Hence p must divide either
(θ − τ )2 or mθ mτ . In the former case, (θ − τ )2 must be divisible by p2 as well;
that is p2 must divide 2pk k̄. So p must divide k k̄. By assumption, k ≤ (2p − 1)/2,
thus p would divide k̄, and hence k̄ = p and k = p − 1; but by Lemma 2.11, this is
impossible, since X is primitive. Consequently p must divide mθ mτ and so either
mθ = p or mτ = p and the result follows form Lemma 2.10.
3. Latin Square Graphs
In this section we will discuss about the graphs arising from Latin squares, orthogonal arrays, generalized quadrangles and etc. and will try to investigate their
properties. We will, also, discuss the problem of cores of SRGs.
16
Definition 3.1. A Latin square L = [lij ] of order n is an n × n matrix with entries
from a set of size n (called symbols) such that each row and each column contains
each symbol exactly once.
Example 3.2. The matrix

1
2
L=
3
4
2
1
4
3
3
4
1
2

4
3

2
1
is a Latin square of size 4.
If G is a group of order n, the the multiplication table of G is a Latin square of
size n.
Given a Latin square L = [lij ] of size n, define the graph X to be a graph whose
vertex set is the set of all ordered triples (i, j, lij ) and two vertices are adjacent
if and only if they agree exactly in one position. The graphs associated to Latin
squares are special cases of the graphs associated to orthogonal arrays.
Definition 3.3. An orthogonal array with parameters k and n, OA(k, n), is a
k × n2 matrix with entries from a set of size n, in which n2 ordered pairs defined
by any pair of rows of the array are all distinct.
Example 3.4. The matrix
1
1
1
1
2
2
1
3
3
1
4
4
2
1
2
2
2
1
2
3
4
2
4
3
3
1
3
3
2
4
3
3
1
3
4
2
4
1
4
4
2
3
4
3
2
4
4
1
is an OA(3, 4).
Given a Latin square L of order n, we can build an OA(3, n) by taking the
three rows of the array to be ‘row numbers’, ‘column numbers’ and ‘symbols’. By
reversing this process, we can build a Latin square of order n from an OA(3, n).
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Given an orthogonal array OA(k, n), define the graph X to be a graph whose
vertices are the column vectors of OA(k, n) and two vertices are adjacent if and
only if their corresponding vectors have the same entry exactly in one position.
Theorem 3.5. The graph X defined by an OA(k, n) is a SRG with parameters
(n2 , (n − 1)k, n − 2 + (k − 1)(k − 2), k(k − 1)).
Proof. Let x = (x1 , . . . , xk ) ∈ V (X). Since for each 0 ≤ i ≤ k, the element xi
happens n times in the i−th row, x is adjacent to n − 1 other vertices (i.e. vectors)
which have the same entry in their i−th position. Therefore x is adjacent to k(n−1)
vertices and so X is regular of valency k(n − 1).
Let x = (x1 , . . . , xk ), y = (y1 , . . . , yk ) ∈ V (X) be adjacent and assume they agree
in i−th position. For an arbitrary vertex z = (z1 , . . . , zk ) ∈ V (X) to be adjacent
to both x and y, we must have either zi = xi = yi or zs = xs and zt = yt , for
some s 6= t ∈ {1, . . . , k} \ {i}. For the first case, there are n − 2 possibilities, since
the entry xi = yi occurs n − 2 times in the i−th row and columns other than x
and y. Also in the second case, there are (k − 1)(k − 2) possible choices for z,
since (s, t) can be chosen in (k − 1)(k − 2) ways and because of the property of
OA(k, n), z is identified by choosing (s, t). Therefor every two adjacent vertices
have a = n − 2 + (k − 1)(k − 2) common neighbors.
Let x = (x1 , . . . , xk ), y = (y1 , . . . , yk ) ∈ V (X) are not adjacent. For an arbitrary
vertex z = (z1 , . . . , zk ) ∈ V (X) to be adjacent to both x and y, we must have zi = xi
and zj = yj , for some i 6= j ∈ {1, . . . , k}. We can choose such i and j in k(k − 1)
ways, and because of the property of OA(k, n), z is identified by choosing (i, j).
Therefore every pair of nonadjacent vertices has c = k(k − 1) common neighbors
and the theorem is proved.
18
Note that the graph defined by an OA(2, n) is isomorphic to L(Kn,n ). Now
we study the chromatic numbers of the graphs associated to Latin squares and
orthogonal arrays.
Proposition 3.6. Let L be a Latin square of order n, X be the graph corresponding
OA(3, n), the orthogonal array defined by L. Then α(X) ≤ n ≤ χ(X).
Proof. Identify the n2 vertices of X with the cells of L. Since all the entries in a
row (or in a column) are mutually adjacent, any independent set of vertices of X,
can contain at most one element form each row. Therefore α(X) ≤ n. On the other
hand, if there is a coloring of X with fewer than n classes, then at least one of color
classes will contain more than n vertices, and since a color class is an independent
set, this is a contradiction and so χ(X) ≥ n.
An OA(k, n) is said to be extendible if we can extend it to an OA(k + 1, n) by
adding a new row to it. The following result states when the graph corresponding
to an OA(k, n) is n−colorable.
Theorem 3.7. Let X be the graph corresponding to an OA(k, n). Then χ(X) = n
if and only if OA(k, n) is extendible.
Proof. Assume first that χ(X) = n and that V (X) = V1 ∪ · · · ∪ Vn is the partition
of V (X) to the color classes of a proper n−coloring. Obviously |Vi | = n, for each
1 ≤ i ≤ n. Define the (k + 1)st row Rk+1 , such that its entry in j−th column is
i, where j ∈ Vi . To show that Rk+1 extends OA(k, n) to OA(k + 1, n), let Rl be
r
the l−th row of the OA(k, n), and suppose the ordered pair
appears in two
s
columns j1 and j2 by two rows Rl and Rk+1 . Then two vertices of X corresponding
to the j1 −th and j2 −th columns cannot be in the same color class, since they are
adjacent. Also since Rk+1 has element s in both j1 −th and j2 −th columns, by
19
definition, these two columns are in the same color class. This contradiction shows
that Rk+1 extends OA(k, n) to a OA(k + 1, n).
Now assume that we can extend OA(k, n) ro an OA(k + 1, n) and let Rk+1 be
the new added row. Then color each vertex j in X by the color i ∈ {1, . . . , n} where
the entry of column j in the row Rk+1 is i. It is easily seen that this coloring is
proper and each color class contains n vertices.
In the rest of this section we study SRGs arising from other combinatorial objects,
namely generalized quadrangles.
Definition 3.8. An incidence structure I = (P, L, I) is a set P of points, along
with a set L of lines (disjoint from P), and a relation
I ⊆P ×L
called incidence. If (p, L) ∈ I, then we say that the point p and the line L are
incident. A partial linear space is an incidence structure in which any two points
are incident with at most one line.
The definition of partial linear space implies that also any two lines are incident
with at most one point. From these we can see that the girth of any partial linear
space is at least six.
Given an incidence structure I, we define the incidence graph, X(I), of I, to be
a graph with vertex set P ∪ L, where two vertices are adjacent if and only if they
are incident. From the definition it can be easily inferred that X(I) is bipartite
with bipartition (P, L). Conversely, we can build an incidence structure from a
given bipartite graph: define P to be one part, and L to be the other part.
Terminologies.
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In an incidence structure, when a point p and a line L are incident, we usually
say that L contains p and p lies on L. Also when two points are incident to a
common line, we say that these two points are joined by a line or are collinear;
similarly, when two lines are incident to a common point, we say that these two
lines meet at a point or are concurrent.
Definition 3.9. A generalized quadrangle of order (s, t), GQ(s, t), is a partial
linear space in which every line contains s + 1 points and every point lies on t + 1
lines; in addition if p is a point not in a line L, then there is a unique point on L
collinear with p.
Example 3.10. Let P be the set of all edges of K6 and L be the set of all 1-factors
of K6 . We define the incidence of an e ∈ P and an F ∈ L to be simply e ∈ F .
Since any 1-factor of K6 is identified by two of its elements, we conclude that every
two points can be joined at most by one line. Therefore this incidence structure is
a partial linear space. Furthermore, if an edge e is not in a 1-factor F , then e has
no common vertices with exactly one element e0 ∈ F ; thus e and e0 belong to one
other 1-factor. The translation of this fact to the language of incident structures is
that, any point e not in a line L is collinear with exactly one point on L.
Finally, we see that every line contains 3 points and every points lies on 3 lines.
The conclusion is that this incident structure is a GQ(2, 2).
The generalized quadrangles are special cases of some other incidence structures
called partial geometries in which given a line L and a point p not in L, there are
exactly α points on L which are collinear with p.
Definition 3.11. The point graph of a generalized quadrangle GQ(s, t) is a graph
whose vertex set is P, the set of points of the GQ(s, t), and two vertices are adjacent
if and only if they are collinear.
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The number of vertices of the point graph of a GQ(s, t) is (s + 1)(st + 1). To see
that, let L be a fixed line in the GQ(s, t). It has s + 1 points. Each of these points
lies on (t+1)−1 lines other than L, each of which has (s+1)−1 points other than the
point which lies on L. Therefore we have counted (s+1)+(s+1)(ts) = (s+1)(st+1)
points. Now note that since each line meets L, we have considered all lines and that
no two of the mentioned lines meet out of L, since otherwise there would be a point
not in L, which is collinear with two distinct points of L, which is a contradiction,
and so the claim is proven.
The point graph of the GQ(2, 2) defined in Example 3.10 is L(K6 ) which is a
SRG. We show that this is the case for the point graphs of all nontrivial generalized
quadrangles.
Proposition 3.12. Let X be a point graph of a GQ(s, t), with s, t > 1. Then X
is a SRG with parameters
((s + 1)(st + 1), s(t + 1), s − 1, t + 1).
Proof. Let p be an arbitrary vertex of X, i.e. a point in GQ(s, t). Then p lies on
t + 1 lines, any two of which have exactly the point p in common. So p is collinear
with s(t + 1) points; thus X is regular of valency s(t + 1).
Now let p and p0 be two collinear points, and let both lie on the line L. there
are s − 1 other points on L which are so collinear with both p and p0 . It is clear
that no point out of L can be collinear with both of p and p0 . Therefore a = s − 1.
Finally let p and p0 be two non-collinear points. There are t + 1 lines through
p, and since p0 is not on any of them, p0 is collinear with exactly one point on each
of these lines. In other words, the vertices p and p0 have exactly t + 1 common
neighbors, and so c = t + 1, and the proposition is proved.
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Our final goal is to have a short discussion about the problem of the cores of
SRGs and to know what one can say about the core of a SRG. Recall that a graph
X is core if all the endomorphisms of X are automorphisms; and that a subgraph
Y of X is the core of X, denoted by X • , if Y is core and there is a homomorphism
X −→ Y . (To get more information about the cores see chapter 6 of [1].) For each
family of graphs it is always an interesting question to ask if the members are cores,
and if not, to find their cores. It is, therefore, reasonable to ask the same question
about the strongly regular graphs. Small examples show that we can not give the
same answer to all SRGs; in other words, C4 is a SRG with core K2 , while C5 is
a SRG which is core itself. As other examples, we see that the Petersen graph is
core as well, while the core of L(K5 ) is K3 . In all these examples, either the SRG
is core or has a complete graph as its core. Indeed it has been conjectured [4] that
this is the case for all SRGs. This problem is still open but in [5], along with some
other subfamilies of SRGs, it has been shown that the conjecture is true for point
graphs of generalized quadrangles.
Theorem 3.13 (5). Let X be the point graph of a generalized quadrangle GQ(s, t)
with s, t > 1. Then every endomorphism of X which is not an isomorphism, has a
clique of size s + 1 as its image.
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References
[1] C. Godsil, G. Royle, Algebraic Graph Theory, Graduate Texts in Math. 207 Springer-Verlag
(2001).
[2] Belevitch, V. (1950), Theorem of 2n-terminal networks with application to conference telephony. Electr. Commun., vol. 26, pp. 231-24.
[3] van Lint, J.H., and Seidel, J.J. (1966), Equilateral point sets in elliptic geometry. Indagationes
Mathematicae, vol. 28, pp. 335-348.
[4] P.J. Cameron, P.A. Kazanidis, Cores of symmetric graphs. Preprint, 2008.
[5] C. Godsil, G.F. Royle, Cores of geometric graphs. Preprint, 2008.