0 7 8 9
6 1 7 8
5 0 2 7
4 6 1 3
0 6 5 4
7 1 0 6
8 7 2 1
9 8 7 3
1 9 8 7
3 2 9 8
5 4 3 9
2 3 4 5
4 5 6 0
6 0 1 2
9
5
0
2
4
7
8
6
1
3
1 3 5 2
9 2 4 3
8 9 3 4
7 8 9 5
8 6
8 7 1 2 3
7 0
9 8 2 3 4
6 1
6 9 3 4 5
0 7
1 0 4 5 6
1 9
3 2 5 6 0
2 8
5 4 6 0 1
7 6 0 1 2
0 1 7 8 9
2 3 8 9 7
4 5 9 7 8
4 6
5 0
6 1
0 2
1 3
2 4
3 5
8 9
7 8
9 7
COM BIN ATOR I A L
MAT HEM A TIC S
YEA R
February 1969 - June 1970
A GEOMETRIC CHARACTERIZATION OF THE LINE GRAPH OF
A SYMMETRIC BALANCED INCOMPLETE BLOCK DESIGN
by
Martin Aigner
Thomas A. Dowling
Department of Statistics
University of North Carolina
Chapel Hill, N.C.
Institute of Statistics Mimeo Series No. 600.5
FEBRUARY 1969
Research partially sponsored by the United States Air Force under
AFOSR Grant No. 68-1406, under AFOSR Grant No. 68-1415; and the
National Science Foundation, under Grant No. 68-8624.
A GEOMETRIC CHARACTERIZATION OF THE LINE GRAPH OF
A SYMMETRIC BALANCED INCOMPLETE BLOCK DESIGN
by
Martin Aigner
Thomas' A. Dowling
I.
INTRODUCTION
A symmetric balanced incomplete block (SBIB) design
is an arrangement of
v
objects, called treatments, into
called blocks, such that each block consists of
treatments, each treatment appears in
distinct treatments appears in
A
<
k
and
= k(k-l).
A(v-1)
the bipartite graph
and blocks of
H(rr)
n(v,k,A)
k
A blocks.
The graph of
k
v
distinct
blocks, and each pair of
The parameters satisfy
n(v,k,A)
whose vertices are the
2v
is defined as
treatments
rr, with two vertices adjacent if and only if one is
a block and the other is a treatment contained in the block.
line graph
G(rr)
of
rr(v,k,A)
is the line graph of
the graph whose vertices are the edges of
in
G(rr)
sets,
The
H(rr), i.e.
Herr), with two vertices
adjacent if and only if the corresponding edges in
H(rr)
have a common end point.
Let
{G(rr)}
denote the set consisting of the line graphs of
all SBIB designs with parameters
and Ray-Chaudhuri [4] proved that
a regular connected graph on
vk
V,k,A.
G
E
In a recent paper, Hoffman
{G(rr)}
if and only if
G is
vertices and the distinct
Research partially sponsored by the United States Air Force under
AFOSR Grant No. 68-1406, under AFOSR Grant No. 68-1415; and the
National Science Foundation, under Grant No. 68-8624.
2
eigenvalues of the adjacency matrix of
unless
G are
-2, 2k-2, and
k_2±(k_A)1/2,
3, A = 2, when the sufficiency of these conditions
v = 4, k
fails due to the existence of a single exceptional graph.
a characterization of
We give here
in terms of several geometric
{G{1T)}
properties of these graphs, with again one exceptional case (Figure 1)
=
for v
7, k
=
2.
4, A
This result generalizes an earlier charac-
terization [2] of the line graph of a finite projective plane
but the conditions for
those of [2].
A=l
given here are slighly different from
(See Remarks below.)
II.
By a graph
DEFINITIONS
G we mean a finite undirected graph without loops
or multiple edges.
V(G), E(G)
denote, respectively, the vertex set
and the edge set of
G.
vertices
and define further
x
and
y
{y
D. (x)
1.
The degree of
x
x
E
We denote by
E
V(G)
= IDl(X) n Dl(y) I
~(x,y)
and
y.
E
E(G).
is written deg x
are adjacent.
the distance between
i
0,1, ...
(= IDl(x) I ).
is the number of vertices adjacent to both
G is regular if
A clique
d(x,y)
V(G) : d(x,y) = i},
and edge-regular if, further,
(x,y)
(A=l) ,
~
x
~(x,y)
is constant for all
x E V(G),
is constant for all
K is a set of vertices, any two of which
3
III.
THEOREM.
Let
V,k,A
be positive integers with
= k(k-J.) , and let
A(v-l)
THE THEOREM
{G(n)}
A
<
k
and
be the set of line graphs of all
symmetric balanced incomplete block designs with parameters
If
G
{G(rr) } , then
to
V,k,A.
G is connected and has the following
properties:
IV(G)I
(P2)
deg x
(P3)
L::.(x,y)
(P4)
ll(x,y)
(PS)
ID (x) n Dl(y) I
3
(P6)
d(x,y)
Conversely, if
then
= vk,
(Pl)
G
E
2k - 2
~
<;
x
for all
E
V(G),
k - 2
if
(x,y)
E
E(G) ,
2
if
(x,y)
¢
E(G),
3
k - A if
for all
x,y
E
d(x,y)
or else
v
=
=
7, k
1,
V(G).
G is a connected graph satisfying
{G(n)}
2, ll(x,y)
4, A = 2, and
(Pl) - (P6),
G is the
graph shown in Figure 1.
Remarks.
Properties
(Pl) - (PS)
are the counterparts of the
characterizing properties of the line graph of a finite projective
plane
(A
= 1)
given in [2], except that the analogue of
in that paper is
bound for
(P6)
here.
L::.(x,y)
L::.(x,y) ~ 1
if
required when
That neither
(PS)
(x,y) ~ E(G).
A
nor
>
1
(P4)
The increased upper
necessitates the addition of
(P6)
is redundant in all cases
is demonstrated by the example given in [2] and the graph of Figure 2.
On comparing our characterization with that of Hoffman and
Ray-Chaudhuri [4], we meet with the intriguing problem of the
4
relationship of the eigenvalues of the graph to its geometric
properties.
Apart from the fact that in the case of regular graphs
the dominant eigenvalue is the degree of regularity, very little
is known.
An important tool is the polynomial of a graph [3], which
for regular graphs gives an upper bound to the diameter.
the eigenvalues of
G(TI)
other than the degree are
For example,
-2, k_2±(k_A)1/2,
and the fact that there are just three immediately yields property
(P6) of the theorem.
the fact that
(P4)
-2
Using certain "impossible subgraphs" of [4],
is the minimal eigenvalue is easily shown to imply
except for one possible configuration.
directly prove edge-regularity
(P3)
for
of the relationship between eigenvalues
(P5)
Hoffman and Ray-Chaudhuri
k>4.
The exact nature
k_2±(k_A)1/2
and condition
is unknown.
IV.
The necessity of
PROOF OF THE THEOREM
(PI) - (P6)
when
verified and the proof is omitted here.
G is a connected graph and satisfies
By
K in
(P3)
G.
IKI=
sense.
it is evident that
G
E
{G(A)}
is easily
We therefore assume that
(PI) - (P6).
IKj;
k
for any clique
Since we shall only consider cliques
K in
G such that
k, let us agree to use the term "clique" in this restricted
We then have
LEMMA 1.
Each vertex of
and each edge of
G is contained in exactly two cliques,
G is contained in exactly one clique.
5
Proof.
If
k
4,
>
the lemma follows from
(P2) -
theorem of Bose and Laskar [1] on edge-regular graphs.
k
=
2, 3, 4
that
(P4)
by a
The cases
must be examined separately, and it is easily verified
(P2) - (P4)
imply the lemma except for
k
=4
, when these
three conditions alone do not rule out the possibility that a vertex
x ~ V(G)
exists such that the subgraph generated by
6-cycle.
If this is the case, connectivity implies that the
subgraph generated by
Dl(y)
Dl(x)
is a
y ~ V(G).
is a 6-cycle for all
We can then show with little difficulty that no such graph satisfies
all the conditions
satisfies
(PI) -
(PI) - (P6)
if
(P6).
(We do encounter one graph that
A = 4 ( = k).
This graph is the
exceptional case in Shrikhande's characterization [5] of the
association scheme (square lattice graph).
arguments
include this additional exception.)
A
A=k = v
k.
<
provided we
As the proof involves a case
by case investigation and bears little connection to the main
argument, it has been omitted.
COROLLARY 1.1.
Let
C(G)
The number of cliques in
denote the set of cliques in
set of unordered pairs
K n L
+
~.
IK n LI
x
=
(K,L) .
1.
(K,L)
Then by Letuma. 1,
2
To avoid some special
in the proof of the theorem, we have assumed
However, the theorem remains valid when
L
G
is
G, and
2v.
I(G)
the
of distinct cliques such that
(K,L)
E
I(G)
We denote the single vertex
i f and only if
x E K n L
by
6
C = (K ' K , " ' , Km)
O l
A clique chain
(Ki,K + )
i l
cliques such that
of
is the number
C
E
I(G)
Proof.
G.
Let
Suppose
Dl(x )
l
contradicting
(P3).
For
x
V(G),
and
X
1,2.
o 4 K2 ,
Then
~(xl,x2) >
which implies
k-l,
we define
{y : d(x,y) = 2,
2
j
KO = Km
is a clique cycle in
(subscripts modulo 3).
D (j) (x)
for
If
C is a clique cycle.
= (K ' K , K , K )
2
O l
O
C
(Ki,K i +l ),
xi
D (x )
l 2
oE
The length
G contains no clique cycle of length three.
n
X
i=O,l, ... m-l.
xi = (Ki,K i +l >·
m of vertices
and no other clique appears twice,
LEMMA 2.
for
is a sequence of distinct
(P4), the sets
Then by
~(x,y)
= j}
D (l)(x), D (2)(x)
2
2
Then by properly labeling the four cliques
K. ,L. ,
~
J
i,j
partition
=
0,1,
E
I(G).
we have
(1)
Y
E
DO(x)
iff
K
O
LO'
K = L ·
1
1
(ii)
y
E:
Dl(x)
iff
K
O
LO'
K
1
(iii)
y
E:
(iv)
Y
E
D (2)(x)
2
D (1) (x)
2
Proof.
E
Ll •
iff
(KO,L O)
E
I (G) ,
(K ,L )
1 1
iff
(KO,L O)
E
I(G),
(K ,L ) 4 I(G) .
1 l
Follows immediately from Lemmas 1 and 2.
LEMMA 4.
(KO,L )
O
=1=
I(G),
If
x = (K ,K ),
O 1
(K ,L )
1 l
4
y = (L ,L ), where
O 1
I(G), then y E D (l)(x)
2
and there are
7
exactly
A cliques, including
which intersect
L
l
and
KO,K .
l
one of
Proof.
there are
in
L .
l
If
z
k-Z
vertices
Hence the
= <Ll,K)
Proof.
cycle in
define
y E D (l)(x)
2
Clearly
LEMMA 5.
and
La,
r~maining
Lemma 3 - (iv).
G
By
A vertices of
L
l
are in
(K,K )
1.
= (KO' Kl , K2 , K3 , K4 , KO) is a clique
Let xi = (Ki,K + > (subscripts modulo 5) and
i l
Then
K .•
1.
xi
5,
DZ
E
(1)
we infer that all
D (1) (x )
2
o
,
a
i
so by the
(x i + 2 ),
Since
K.1.- 1
(i+l)- (i-I)
= a ,say,
=Z
A = Za.
and
= ( K,L ) ,where K intersects neither
If
Y
E
K nor
O
K
, and L intersects exactly one of KO,K l ·
D (l)(x ) I
A = 2a. Since Zk cliques intersect
Z
O
IK n
Then
K or
O
K
(including
1
y
K ,K )
O 1
and
G
contains
<
To show
(1)
intersecting
xl
E
D
Z
meeting
(1)
or
C
preceeding lemma,
I
(x).
contains no clique cycle of length five.
K + , including
i 1
is prime to
D
2
I(G)
E
O
as the number of cliques intersecting both
a.
(P5)
D (x) n Dl(y) , and these must be
3
(1)
is one of these, then either
Suppose
G.
in
by
is impossible, let
KO' but not
K '
Z
La
Then if
yO
such cliques meeting
K or
O
K and
Z
La.
cliques in all,
Zv.
be one of the
(yO)' and so by Lemma 4, there are
K which meet
Z
Zv
(1)
k-a
= (KO,L O)'
Za
cliques
we find
cliques
But there are exactly
K ' so also there are
O
a
a
1
cliques
=a
8
meeting
a
K
2
and
L .
O
L
cliques meeting
k-a
exactly
O
and
K .
l
(1)
DZ
vertices of
k-a
any of the
Then by the same argument, there must be
cliques
L
(x ).
O
contains
O
We can repeat the argument for
meeting
1
L
It follows that
K , obtaining
K , but not
1
4
finally
ID Z(l)(xo)! ~
which is easily seen to contradict
COROLLARY 5.1.
and
L
If
K ,K
O 1
(1).
are two disjoint cliques in
is a clique intersecting both
number of cliques intersecting both
A or
L • is either
Proof.
k
~
X
o
E
K
O
K
O
and
and
G
K • then the
1
K • including
1
k •
xi = {Ki,L).
cliques meeting
such that
are
Let
Z(k-a)2.
K
O
and
K
i
(1)
DZ
(Y1).
O
If there are fewer than
' then there exists
Now if
K
cliques meeting
i=O,l.
I
K
O
E
K
1
\
Y2 = \Kl,L l/ , there
K ,L .
l 1
which meet one of
if such a clique were to meet
Yl
and
L
l
,
G
But
would contain
a clique cycle of length five.
H , the clique graph of
Consider now a graph
by
V(H)
=
C(G) ,
E(H)
=
I(G)
is a one-to-one function of
adjacent edges of
H
if and only if
G
onto
V(G)
such that
into adjacent vertices of
maps adjacent vertices of
follows that
~ : (K,L) ~ <K,L)
The mapping
E(H)
G
G
of
H.
Thus
G
H.
€
~
~-l
and
into adjacent edges of
is the line graph
H E {H(rr)}.
G, defined
It
{G(rr)}
We shall restrict our attention
maps
9
henceforth to
R.
We first summarize some properties of
R is connected and has the following properties:
LEMMA 6.
2v,
\V(R)I
(Ql)
deg K
(Q2)
K
V(R) ,
= 2,
(Q4)
R contains no cycles of length three or five.
A or
(Ql)
k
is Corollary 5.1, and
LEMMA 7.
d(K,L)
~(K,L) <
If
VI = {K}
(Q4)
k
(Q3)
= 2.
Then
VI
If two vertices of
VI
edges of
V
2
R by
V = VCR) - VI
2
vertex sets
(Ql)
R .
VCR)
K,L
we have
d(K,L)
Let
K
VCR)
E
E
such that
=A
for all
and define
(Q2) - (Q4) imply
=
1 + k(k-l)/A
are adjacent, then
are all distinct.
and
=
v.
R contains as-cycle.
vk
edges incident
Since these are all the
(Q2) , the complementary set
is also independent.
Thus
V is bipartite with
V 'V 2 .
l
Finally let
VI
follows from Lemmas 2 and 5, since
is an independent set, and so the
with vertices of
follows from Lemma 1,
for all
Ivll
Thus
(Q2)
{R(TI)}.
d(K,L)
D (K).
2
u
E
In view of
such that
d(K,L)
G corresponds to a cycle in
= 2, then R
Proof.
if
is Corollary 1.1,
a clique cycle in
in
E
~(K,L)
(Q3)
K,L
for all
k
(Q3)
Proof.
R in
such that
n
be the number of unordered pairs
~(KO,Kl)
= A.
Then
KO,K
l
nA = vk(k-l)/2, since
10
each of the
k(k-1)/2
v
vertices in
such
pairs
V
2
K ,K .
O 1
for all KO,K 1 € Vl,KO
+ K1 .
is in
D (K ) n D (K )
1 O
1 1
for exactly
i1(K ,K )
O 1
n = v(v-1)/2, Le.
Thus
If we identify the vertices of
with treatments and those of
V
2
>..
V
1
with blocks, and define the
treatment to be contained in a block if and only if the corresponding
vertices are adjacent in
H , then it is clear that
graph of an SBIB qesign
rr(v,k,>") ,i.e.
H
E
H is the bipartite
{H(n)}. 1)
We consider now the case where there exists two vertices,
K,L
in
H such that
vertices
K ,K
O l
d(K,L)
to be
equiva1en~
L1(K ,K ) = k.
O 1
d(KO,K ) = 2,
1
if and only if
= 2,
By
class containing
K
€
V(H)
and write
(Q2)
D (K ) = Dl(K ).
1 O
l
= k.
i1(K,L)
Let
Let us define two
K :: K , if KO=K
l
O
1
we then have
K :: K
O
1
K denote the equivalence
(:: is readily seen to be, in fact,
an equivalence relation.)
LEMMA 8.
same number
to
~
t
Proof.
first that
Any two equivalence classes in
Let
2
contain the
of vertices.
K ,K
O 1
(KO,K )
l
V(H)
E
€
V(H) ,
E(H).
t.
1
= IK.1 I ,
i =0 , 1.
Suppose
If we can show that in this case
= t 1 = t , then the lemma will follow by the connectedness of
H.
Consider then the number
(KO,L )
O
that
E
E(H), (K ,L )
1 1
n
E
E(H),
L
O
L
can be chosen in
O
~
(L ,L )
O 1
of edges
~
E
E(H)
+ KO'
K ,
1
L
1
k-1
ways, while
K - {K } ;:',:L
can be chosen. in",·",>..-l
O
O
O
Hence
if
L
l
1)
It is interesting to note that so far we have not made
use of
(P6).
such
wa~s2.
or
11
n
= (to-l)(k-l) + (k-tO)(A-l).
choose
k-2
L , we obtain
l
0 , this implies
>
t ~ 2
(Note
Suppose
L
l
n
to
(tl-l)(k-l) + (k-t ) (A-l).
l
=
(K,L)
E
(K,L )
l
E(H).
+L
, and if
t + t
V(H)
K
l
(Kl,L )
l
K u L
vertices.
E
E
K
E(H).
is the
The equivalence
induces a homomorphism
H+ H ,
H is the graph defined by
where
v
VCR)
{K : K
E(H)
{(K, i)
V(H)},
E
(K, L)
E
E(H)}.
H is connected and has the following properties,
= v/t,
k
= kit,
(Ql)
l"v(R) I
(Q2)
deg K
A = A/t:
2v,
=
k
for all
A
if
K
V(H),
= 2,
t,(K,L)
(Q4)
H contains no cycles of length three or five.
(Ql)
and
Lemma 6 , since now
= L.
(Q2)
d(K,L)
E
(Q3)
Proof.
K
K
H on the vertices of
defined on
LEMMA 9.
of
Since
' and the proof is complete.
and therefore
complete bipartite graph on
where
l
Then
E(H)
E
Hence the sub graph of
relation
t
L
and
O
by hypothesis.)
L ,then
E
=
Similarly if we first fix
are obvious.
~(K,L)
(Q3)
follows from
= k would imply K = L
We finally observe that if
length three or five, then so would
(Q3)
and hence
H were to contain a cycle of
H , contradicting (Q4) of Lemma 6.
lZ
v, k, A reads
The equation relating
A(vt-l)
(Note that
v
need not be an integer.)
Consider now a fixed edge
A
a
B
l
A
Z
(Z)
= k(kt-l).
(Ka,La)
{K } Da(K ),
a
a
D (K )
Da(La),
l a
D (K ) - D (La)'
Z a
l
E(H)
E
and define
Al
{La} = Da(La),
D (La) - D (Ka),
a
l
B
Z
D (La) - Dl(Ka )·
Z
B
a
Then
Da(Ka) = Aa ,
D (K ) = B U
l a
a
D (K ) = A U
Z a
1
It follows from
(Q4)
Da(La) = Ba,
Dl (La) = Aa U AI'
DZ(La) = B1 U BZ·
B ,
l
A .
Z
that the six sets
are pairwise disjoint, and thus the two sets
disjoint for each
vertices of
Using
(QZ)
i=a,l,Z.
Di(Ka )
and
or of
(Q3)
Also by
(Q4)
A.,B.
1
1
(i=a,l,Z)
Di(K ),
a
are
we see that no two
can be adjacent for
Di(La )
Di(L )
a
i=a,l,Z.
we can then easily determine the number of
vertices in each set, obtaining
Let
using
IAal
IBal
1,
IBII
JAIl
k"-l,
IAZI
IBZI
(k-l) (k-A) / A.
C denote the set of vertices not in any of these sets.
(Ql),
(Z), and
(3)
(3)
Then
we have
(4)
13
Thus
t
2(k-~),
divides
and so
2
LEMMA 10.
Proof.
K
'0
Let
C.
x
Clearly we have
C and
E
Then by
=
K,L, say
K
E:
C
.
If
Proof.
Let
Let
<
3
(K,L) ,
in
G.
n
and let
If
D (K)
l
.
, we have
K be a fixed vertex of
ID (K) n BZI
l
H ,
is even simpler.
~
A u B
Z
Z
s
for all
= ID I (K) n DZ(L) I
DZ(K) n D (1)
l
n
= rA = sA ,
=s .
b
KO,L O
= 3 , then
(Q3)
= IDl (K) nAZi
E(H) .
a contra-
of edges joining
a
for all
3
E
meeting one of
l
r = IDZ(K) n D (1) I
l
Using
r
=
=
Z in
Lemma 10 that
d(K,1)
Consider the number
x
K
C ~ D (K ) u D (1 )
3 O
3 0
If
L
(K,L)
then
d(xO'x)
o = K or
f Z , the result
follows from
LEMMA 11.
i.e.
(P6) ,
d(LO,K)
K
d(xO'x)
Hence
3
defined by
K and
O
which implies either
It
G
=
E(H) ,
E
(5)
C are adjacent.
3 , there exists a clique
and one of
diction.
z(iZ-i) .
d(K ,K)
O
(K,L)
be the vertex of
d(xO'x)
<
t
No two vertices of
K,L
If
f
C and define
and
14
a + b
Then
=
0.
k
l
IDZ(K) n Bli
bl
IDZ(K) nAzi
a
If we define
= a , since clearly
then it follows from Lemma 11 that
a
DI(K) n DZ(K O)
DZ(K) n DI(KO)
b = b
Similarly
b
~
I
a ,;, I
D (K) n A
I
Z
.
1
Since
Hence
implies
.
C
= D/KO) n D3 (LO).
Again,
and hence there exists
=A
with
be in
DZ(K) n BI
<
a = I DZ(K) n Bli =
IB11 = k-l , we have
>
I
a =
Similarly
and
l
Dl(K) n A ' and therefore
Z
Each of these vertices must
> =
A • Similarly b
a
~ ~
and we have
<
a,b
where
a + b
k-A
(6)
k
We wish to establish an upper bound on the
number of vertices
Dl(K)
L
= (DICK) n AZ)
E
U
DZ(K)
(DI(K) n B ) ,
Z
DI(K) n A
Z
some vertex of
which are not in
such an
C
L
or to some vertex of
Since
is adjacent to
DI(K) n B .
Z
Assume without loss of generality the former and call the vertex
K ·
Z
It then follows that
for some vertex
Kl
E
L
Dl(L ) .
O
tains a cycle of length five.
set of
a
vertices.
IDZ(K ) n D (LO) I
Z
I
Thus at most
a
<
E
DZ(L ) ,
O
If
Hence
K
l
E
By Lemma
of the
k
L
E
Dl(K )
l
DZ(K) n AI' then
K E A U (Al-DZ(K»
I
O
Thus we have, since
a
and hence
11,
K
I
E
D (K )
Z Z
H
, a
,
ID (K ) n DZ(L ) I
I Z
O
vertices adjacent to
KZ E
con-
D (K)
I
<
=
a
n
A
Z
15
are in
DZ(K) n DZ(L )
O
and since any vertex
of
DlCK)
0
D (L )
2
we have at most
DI(K )
Z
D (L
1
D (K) n D (L )
2
2
)
A
K
€
2
I
[2
lei
>
€
2
<'K)
2
which are also in
2 (K)
~ E D
vertices
E
D
Dl(K) n B ) ,
A similar argument shows that
2
b2 /
for some
La
Dl(K) n A "
and thus together with
D (K) n B
which are also
This now
2
1
gives us
<
(3)
1
+ I -1
[k(k-1) -
(a 2
+ b 2 )]
"
(7)
(4)
z(k-I) (1-l/t)
>
The inequality remains valid on replacing
2 (k-:)
(by (Q4),
of them
I
cl
Then by
adjacent to one vertex
O
a2 /
vertices
DI(K) n A '
Z
vertices in
cannot be adjacent to any vertex of
for some
2
€
a
O
there are at least
in
La
AZ is adjacent to exactly
n
La ~ DZ(K)
Since there are
of
by the upper bound
(5) , and we then have
k2
Substituting
t
k-a
for
b
3k + 3). + 1 "
and simplifying,
a 2 - ka + 1/2[3(k-I) --1]
>
o.
( 8)
16
Assuming with no loss of generality that
verified that the only values of
(6)
and
(8)
k
~ 2,
(ii)
k
=
from
(3)
sets
A. ,B.
1
4,
2,
k
and
1
a,b,k,A
satisfying both
are
(i)
If
<
= b , it is easily
a
(i
A
=
A = 1, a
= 1, b
A
=
=
1,
1, a
a
=
b
b
=
=
k-l,
2.
1 ,then
t
=
2
by
(5), and
(4) , we have exactly one vertex in each of the
= 0,1,2)
and
C.
In this case
H is a cycle
of length seven,
v = 7, k = 4, A = 2,
Figure 1
The lines in the figure represent cliques, two
lbel~.
vertices being adjacent in
Figure 1.
perties
and
G is the graph
01
G if and only if they are collinear.
A graph
(PI) - (P6)
G
4
{G(n)}
with
v
satisfying the pro-
= 7, k = 4, A = 2 .
17
->
Consider next the case
is the vertex of
A
Z
b
=
k-1
or
k-1
of
Co
Co = Dl (12 )
a
= k-l,
b
K
for some
K
Z
C ,C
O 1
(9)
b.
1.
a.
1
C.
=
= k-1
A
Z
and
For
=
k
is adjacent to
Z
Since
C
is contained in at least one
E
B
Z
<
Ici
.
By
K
c
such that either
(4)
~
C
(KZ)
Co
D
l
and
(5)
or
we have
(9)
Zk-3
(Q3)
A = 1.
,since
CO,C
l
~ C
K
k-1
~ 3
ICo n Cli
with
, while in the latter
C
Hence it follows
o~
C
.
=
1
C
'
,
and let
Then since
a
Ici = Zk-3 .
(Q3) , that the number of edges
=
i
!Dl(K ) nAZi
i
be the number of
a
or else
In the former
is equal to the number of edges joining
E C ,set
i
ta i
=
\
Lb.1.
,
B
Z
and
K.1. 's
flor whi{!h
we obtain
;:(k-l) + (I C 1-;:)
+ (I,c 1-;:) (k-l)
a
and since
K
If
ID (K ) n BZI
2 i
1, b
since
k-l
for every choice of
We can easily verify, using
and
there are exactly
that either there is only one such set
ICI
joining
K
Then if
k-l
are two such sets, they must have a non-empty
there are two sets
case,
E
=
1
C
<
6(K,L) ~ Z ,contradicting
from
C
A , L
Z
2
E
k-l
intersection.
E
vertices in
,~.
so that if
K
b
by our earlier argument.
DZ(L )
O
it follows that each vertex
set
= 1,
adjacent to
exactly one vertex of
= 1,
3, A
D (K ) n C , including
l
2
vertices in
a
-
=
k
,
Ici
=
Za.
(10)
18
Ici = 2k-3
This rules out the possibility
=
ICI
k-l , it follows from
K ,K c C
2
I
LI
E
B
2
with, say
a
=
l
(10)
1
that we must have two vertices
and
b
l.
2
1 ,12
1
Let
Kl ,K2
be the vertices adjacent to
by our earlier argument
As to the case
L
A
2
E
2
and
' respectively, then
are both adjacent to all of
C,
~(Kl,K2) ~ 2 , a contradiction.
which implies
k - 4, A
The remaining case
=
2, a
=b =
2
can be shown
to be impossible by a rather involved argument which demonstrates
that
H
must cpntain a cycle of length five.
We have omitted
the proof here.
Concluding Remarks.
(PI) - (P6)
The question whether the conditions
are redundant is difficult to answer in general.
The example given in
that
(PS)
[2]
with
7, k
=
3, A
= 6, A =
2
Figure 2
H
shows
with
v
demonstrates that the same holds true for
t
=
The two vertices of
in
1
below with
For ease of exposition we have drawn the graph
Figure 2
=
cannot be dropped without admitting additional
exceptions, while the graph in
k
=
v
H
in
3, A
1
corresponding to the edges
e
2 , and thus
G
v
=
8, k
are readily seen to be at distance
=
4
=
16,
(P6).
and
f
from each other.
19
e
Figure 2.
G ~ {G(n)}
(PI) - (PS)
A graph
H whose corresponding graph
and satisfies the properties
with
v
= 16,
k
= 6,
A
=2
.
20
REFERENCES
[1]
Bose, R. C.
and
Laskar, R.:
Tetrahedral Graphs,
3
[2]
A Characterization of
J. Combinatorial Theory
(1967), 366-385
Dowling, T. A.
and
Laskar, R.:
A Geometric Characteri-
zation of the Line Graph of a Projective Plane,
J. Combinatorial Theory
[3)
Hoffman, A. J.:
3, (1967), 402-410
On the Polynomial of a Graph,
Amer. Math. Monthly 1Q (1963), 30-36
(4]
Ho ffman, A. J.
and
Ray-Chaudhuri, D. K.:
On the Line
Graph of a Symmetric Balanced Incomplete Block
Des~gn,
[5)
Trans. Amer. Math. Soc.
Shrikhande, S. S.:
Scheme,
11 (1965), 238-252.
The Uniqueness of the
Ann. Math. Stat.
L
Association
2
30 (1959), 39-47.
© Copyright 2024 Paperzz
