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A CHARACTERIZATION OF TETRAHEDRAL GRAPHS
by
R. C. BOSE and RENU IASKAR
March 1967
Abstract
A tetrahedral graph may be defined as a graph G,whose vertices may be
identified with the n(n-l)(n-2)/6 unorde~~d triplets on n symbols, such
that two vertices are adjacent if and only if the corresponding triplets have
two symbols in common. If d(x,y) denotes the distance between two vertices
x and y and 6(x,y) denotes the number of vertices adjacent to both x and y,
then a tetrahedral graph G has the following properties: (b l ) The number of
vertices is n(n-l) (n-2)/6. (b) G is connected and regular of valence 3(n-3).
(b ) For any two adjacent vertiges x and y, 6(x,y)=n-2. (b4 ) 6(x,y)=4 if d(x,y)=
2. 3 We show that if n > 16, then any graph G (without loops
and with utmost
one edge connecting two vertices) having the properties (b l )-(b 4) must be a
tetrahedral i graph.
"
This research was supported by the National Science
Foundation Grant No. GP-3792 and the Air Force Office of
Scientific Research Grant No. AF-AFOSR-760-65.
DEPARTMENT OF STATISTICS
UNIVERSITY OF NORTH CAROLINA
Cha.pel Hill, N. C.
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A CHARACTERIZATION OF TETRAHEDRAL GRAPHS
R. C. BOSE and RENU LASnR
Department of Statistics, University of North Carolina
I
Chapel Hill, North Carolina
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I. Introduction
1.
We shall consider only finite undirected graphs, with at most one
edge joining a pair of vertices and no edge joining a vertex to itself.
The valence d(u) of the vertex u of a graph G, is defined to be the number
of vertices adjacent to
u.
If all vertices of G have the same valence n ,
l
the graph G is said to be a regular graph of valence n •
l
A chain xl' x ' "', x is a sequence of vertices of G, not necessarily
n
2
all different, such that any two consecutive vertices in the chain are adjacent.
Thus the pairs (xl,x2)(x2'~)' ••• (x _l , x ) are edges of G. The number of
n
n
edges n-l is said to be the length of the chain. The chain i.s said to begin
'
n
The graph G is said to be connected if for every pair of distinct vertices
at xl and terminate at xn '
and is said to join xl and
X
x and y, there isa chain beginning at x and terminating at y.
For a connected
graph the distance d(x,y) between two vertices x and y is defined to be the
length of the shortest chain joining x and y.
For any two vertices u and v,
adjacent to both u and v.
6. (u, v) denotes the number of vertices w,
If u and v are adjacent, Le. d(x,y)=l, 6. (u, v) is
This research was supported by the National Science Foundation Grant No. GP3792 and the Air Force Office of Scientific Research Grant No. AF-AFOSR-76o-65.
\
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called the edge degree of the edge (u, v).
A regular graph G for which all
edges have the same edge-degree 1::., is said to be edge-regular, with edgedegree-A •
2.
A graph G is said to be triangular if the vertices of G can be
identified with unordered pairs on n symbols, so that two pairs are adjacent
if ar..d only if, the correspondiag pairs have one symbol in common.
A triangular
graph G obviously possesses the following properties:
(a )
l
The number of vertices in G is n(n-l)/2.
(a )
2
G is regular of valence 2(n-2).
(a, )
G is edge regular with edge degree n-2, i. e.
I::. (u, v) = n-2
-,
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if u and v are adjacent.
(a4)
I::.(u, v) = 4, if u and v are non-adjacent.
Connor [4] showed (with a somewhat different terminology) that for n > 8,
the properties (a )-(a ), characterize a triangular graph, i.e. if G has the
l
4
properties (&1)-(a4)' then G must be triangular.
Shrikhande [9], Li-chien
(7,8] and Hoffman (5,6] completed Conner's work by demonstrating that the
same result holds for n < 8, but if n=8, then there exist other non-isomorphic
graphs Which are not triangular.
;.
graphs.
In this paper we consider the problem of characterization of tetrahedral
A tetrahedral graph may be
defined as a graph G whose vertices can
be identified with unordered triplets on n s,ymbols, such that two vertices
are adjacent if and only if the
corres~onding
triplets have two common symbols.
It is readily seen that G has the following properties:
(b ) The number of vertices in G is n(n-l)(n-2)/6.
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(b ) G is connected and regular of valence 3 (n-3).
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(b ) G is edge-regular, with edge degree n-2, i.e.
3
if d(x,y)=1.
(b ) .6(x,y)==4
4
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••
;
a(x,y)=2.
In Section III we prove that for n > 16, the
propert..:f~6 ,(b
1 )-(b4)
characterize a tetrahedral graph, i.e., if G posse~es properties (b l )-(b4)
and n > 16, then it is possible to establish a
(l~)
correspondence between the
vertices of G, and the unordered triplets on n symbols, such that two vertices
of G are adjacent if and only if the corresponding triplets have a pair of
symbols in common.
The proof is based on certain theorems regarding the existence or nonexistence of cliques and claws in edge-regular graphs, which are proved in
Section II.
These theorems generalize the previous work of Bruck [2J and
(one of the authors) Bose [lJ.
Other applications of these theorems will be
given in subsequent communications.
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if
.6 (x,y)=n-2
II.
1.
Claws and Cliques in Edge-regular gra.phs.
In this section we shall consider a graph G which has the following
properties:
(c ) G is connected and regular of valence r(k-l) •
l
(c ) G is edge-regular with edge-degree (k~2) + 0: •
2
(c )
3
.6 (x,y) ~ 1 + ~, for all pairs of non-adjacent vertices,
x and y of G.
3
In the above r, k,
k
~
2, a
~
0,
~ ~
a,~,
° and
are fixed positive integers, such that
r~-2a ~
0.
r
~
1,
All the leIllJl.8,s and theorems in paragraphs
2 and 3 of this section are about the edge regular graph G, with properties
(c ),(c ) , (c )·
l
2
3
We define here some functions of the parameters r, k, a,
and
~,which
plS\V an important role in subsequent developments.
(2.1.1)
7(r,a)
=
1 + (r-l)a •
(2.1.2)
~(r,a)
=
1 + (2r-l)l.
(2.1.3)
p(r,a,~)
=
1 + ~ + (2r-l)l.
(2.1.4)
p(r,a,~)
=
1 +
i
(r+l)(r~ -2a).
We shall denote as usual the cardinality of a set S by
A clique
I:
Is I.
of· a graph is a set of vertices adjacent to each other.
A
clique I: will be called complete if we cannot find a vertex x, not contained
in K such that
xUJ( is a clique.
Thus a complete clique cannot be extended
to a larger clique by the adjunction of a new vertex belonging to the graph.
Now consider the graph G with the properties, (c ) , (c ) , (c ) • A clique :K
2
l
3
of G Will be called a major clique if
(2.1.5)
IKI
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~ 1 + k -7 (r,a) = k - (r-l)a.
A clique I: of. G will be called a stand clique if it is both major and
I:
complete.
A claw (p, S] of G , consists of a vertex p , the vertex of the claw and
a non-empty set S of vertices of G, not containing p, such that p is adjacent
to every vertex in S, but any two verticel in Sare 'non-adjacent.
of the claw i8 defined to be the number
8
The order
= Is I.
In the next two paragraphs of this section, we obtain a number of theorems
about ,claws and cliques in a graph G having the properties (c ), (c ), (c ).
l
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3
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These theorems are very similar to those obtained in an earlier paper [ 1],
but are now proved under less restrictive conditions, thereby substantially
increasing their range of applicability.
III,
This will be illustrated in Section
where they will be used to obtain a geometric characterization of
tetrahedral graphs (defined in Section I).
Further applications will be
given in sUbsequent communciations.
]
]
')
)
Je
2.
order
r+l
1
J
Je
1
J
,~
), there cannot exist a claw of
Suppose there exists in G a claw [P, S] of order s.
of vertices of G, not belonging to
Let T be the set
(P, S], and adjacent to p.
Let f(x)
denote the number of vertices q in T, such that q is adjacent to exactly x
vertices in S.
Counting the number of vertices in T we have from (c ),
l
s
x~
f(x)
= r(k-l)-s = rk-r-s,
Counting the number of ordered pairs (b, q) where b and q are adjacent,
b belongs to S, and q belongs to T, we have from (c )
2
)
]
> p(r,a
in G.
]
1
If k
Theorem (2.2.1).
s
x~
x f(x)
= s(k-2~).
Again counting the triplets (b , b , q), where b , b is an ordered
2
l
l
2
pair of vertices in S, q is a vertex in T, and b , b are both adjacent to
l
2
q, we have from (c )
3
s
Z
x=O
x(x-l) f(x)
5
<
s(s-l)~.
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If a claw of order r+l exists, putting
(2.2.4)
f(O) +
i
r+l
~
x=l
s=r+l, we have fram (2.2.1),
(x-l)(x-2) f(x) ~.k + 1 +
i
(r+l)(~-2a) = -k+p(r,a,~)
Since the left hand side is essentially non-negative whereas
by hypothesis, we have a contradiction.
Theorem (2.2.2).
If k >
k >
p(r,a,~)
This proves our theorem.
r(r,a), then any claw of G of order s < r,
can be extended to a claw of order r.
Suppose there exists in G a claw [p,S] of order s.
From (2.2.1) and
(2.2.2)
s
f(O)
Z (x-l) f(x) = (k-l)(r-s)-as.
-,
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x=l
,
If k >
Ii
r(r,a) = l+(r-l)a, and s < r
f(O) > a r(r-s-l) ~ 0
I:
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Hence f(O) > 0,
which shows that there exists a vertex
is not adjacent to any vertex of S.
to the claw [p,S*] of order s+1.
If
q
in T,
which
If S* =SUq. we can extend the claw [p, S]
s+l < r
we can continue the process till
we arrive at a claw of order r.
Theorem (2.2.3).
least
k-r(~a)
I.:
j
I!
1
Given a claw (p,S] of G of order r-l there exist at
distinct vertices q of G such that [p,SUqJ is a claw of order r.
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Putting s=r-l, in (2.2.5) we have
f(O)
> k-l-a(r-l)
=
k-r(r,a) •
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ell,
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Hence there exist at least k-Y(r,a) vertices q in T such that [p,SUq]
is a claw of order r.
3.
Lemma (2.3.1).
If k >
r (r;a) and if G has no claw of order r+l,
then any pair of adjacent vertices p and q is contained in at least one
major clique.
From theorem (2.2.2) we can extend the
cl~
[p,q] to a claw [p,S] of
order r.
Let b , b , ••• b be vertices in S other than q. Let n be the
l
2
r
set of vertices w, which when adjoined to S-q give a claw [p,S*] of order
r, i'There S*
= (S-q)Uw.
Of course q is contained in n and from theorem
Inl > k
-
r(r,a).
The vertices in n are all adjacent to one another.
If any two were
not adjacent they could be added to b , b , •.. b to give a claw of order
l
2
r
r+l. Let K = pun.
Then K is a major clique since
IKI
~ 1 + k -
r(r,a).
In Lennna (2.3-l), the hypothesis may be replaced by
Corollary (2.3.1).
k
> max
~(r,a),
p(r,a,~)].
This follows at once from theorem (2.2.1).
Corollary (2.3.2).
When the conditions of Lennna (2.3.1), or corollary
(2.3.1) are satisfied, then any pair of adjacent vertices p and q, is
conta~ned
in at least one grand clique.
We can extend the major clique K by adding new vertices till it is complete
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and therefore a grand clique.
Lemma (2.3 .2).
I KnL l:s
then
Since
If K and L are cliques of G, and KUL is not a clique,
l~.
XUL is not a clique, there exists in KUL a pair of vertices c
and d 1>Thich are non-adjacent, such that c belongs to K and d to L.
Any vertex
belonging to mL must be adjacent to both c and d.
2: ImLI.
Hence 6(c,d)
The lemma follows from (c ).
3
Lemma. (2.3.3).
If K and L are cliques of G andKf) L
least two vertices
a'
. and b, then IlULI
-<
k +
contains at
o.
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Every vertex in XUL, other than a and b is adjacent to both a and b.
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2: /KULI
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Hence
A (a,b)
It follows from (c ), that
l
-2.
_I:
IKU LI:s k + ex.
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Lemma (2.3.4).
If K and L are cliques of G, KULis not a clique and
KnL contains at least two vertices, then
he I ...
IL l:s 1 + k +
0 ...
f3.
This follows at once from Lemmas (2.3.2) and (2.3.3) by noting that
I~
Ii
I;
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Theorem (2.3.1A).
If k > p(r,O ,(3) and G has no claw of order r+l,
then any pair of adjacent vertices p and q is contained in one and only one
grand clique.
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The existence of at least one grand clique containing p and q follows at
onc~
from corollary (2.3.2) by noting that
8
per, 0, (3)
2:
r(r, ex).
el'!
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Suppose there exist at least two distinct grand cliques Ie and L both
containing the adjacent vertices p and q.
is not a clique.
IK I
+ IL I
=s 1
But K and L are both major cliques.
IK I +
+ k + 0: +
13 •
Hence
IL I 2= 2 {k-(r-l)O:}.
vlhich shows that
k <1 +
13
+ (2r-l)0:
= per, 0:, 13),
contrary to the hypothesis.
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The previous theorem can be written in the follovTing alternative form:
Theorem (2.3.lB).
tvlO
If k > max [ per, 0:,
13) per, 0:, 13)], then any
adjacent vertices p and q of G are contained in exactly one grand clique.
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This follows at once from Theorems (2.2.1) and (2.3.lA).
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Theorem (2.3.2A).
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KlJL
Hence from Lemma (2.3.4)
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Since Ie and L are complete,
If k > q(r,o:) , there exists no claw of order
r+l
in G, and every pair of adjacent vertices of G is contained in utmost one grand
clique of G, then each vertex of G is contained in exactly r grand cliques.
If we note that
q(r,O:) 2= r(r,o:) it follows from corollary (2.3.2),
that any pair of adjacent vertices is contained in exactly one grand clique.
Again from Theorem (2.2.2),
r.
p is the vertex of at least one claw of order
Let [p,S] be a claw of order r, where S
= (b1 ,
b2 , ••• ,br J.
As in Theorem
(2.2.1), let T be the set of vertices not belonging to S, which are adjacent
to p.
9
Let Hj be the set consisting of p, b j and q belonging to T, such that
q is adjacent to b but not adjacent to b , i F j . As in Theorem (2.2.1)
j
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let rex) denote the number or vertices in T, 'Which are adjacent to exactly
x vertices in S.
r+l.
Then reO)
= 0,
otherwise there 'WOuld exist a claw of ord .
Putting s=r in (2.2.1) and (2.2.2), we have
r
E
=
rex)
r(k-2).
x:l
r
= r(k-2~),
E x rex)
x:l
Hence
r
E
=
(X-l) rex)
ra •
x=2
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Since
r
-f(l) + E rex)
x=l
=
r
x~f(X)
<
-
r
E (x-l) f(x)
x=2
=
rex
it rollows that
r(l)
~
r(k-2-a).
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Any two vertices of H. are adjacent to one another, otherwise there would
J
exist a claw of order r+l.
Put Hj
= Hj
Thus H is a clique.
j
- (b j Up).
T vlhich are adjacent to b
j
Then
Hj consists of exactly those vertices of
but to no other vertex of S.
Hence
H!,
Now there is a unique grand clique Kj
containing
b j and p.
of vertices in K cannot be less than the number of vertices in H •
j
j
H~
The number
If possible
let IKjl< IHjl. Since K is a grand clique it follows that H is a major clique
j
j
and contained in some grand clique
Kj'
they must coincide.
Kj.
Since b j and p are contained in K and
j
Hence K contains H , Which 'contradictsIKjl<IH
j
j
j
I.
Now consider the .r grand cliques, lC , 1(2' ..• , K • Then lCi]?' K -p, ••• ,
l
r
2
K -p are disjoint. For if K. -p and K.-P, i :f j, have a common vertex
r
J
~
q, then K and K would coincide, and would contain both b and b which is
j
i
j
i
impossible since b
(2.3.4)
i
is not adjacent to b . Remembering (2.3.3), we have
j
r
j~l IKj;p1 > E IHj-p I
r
r + E IH*I
j
j=l
]
=
r + f(l)
J
>
r(k-l-O:) •
If possible, suppose there is another grand clique K + containing p.
r l
]
The vertices in Kr+l-P must be disjoint from the vertices in lCl-P,
K -:po
r
~-P,
••• ,
Since K + is a grand and therefore a major clique, IKr+l-P I ~ k-l-(r-l)ct.
r l
]e
]
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1
.••
are disjoint sets, and the total number of vertices in these sets in f(l).
=
1
H~,
I
I
But from (c ), the number of vertices adjacent to p is exactly r(k-1).
1
from (2.3.4)
r+l
r(k-l) > E
j=l
IKj"-p I
~
which is a contradiction.
Theorem (2.3.2B).
order
r+l
2:
Hence
r(k-l) + k - 1 - (2r-l) 0:.
k < 1 + (2r-l)0:
=
q(r,o:)
Thus p is contained in exactly r grand cliques.
If k > p (r, 0:, 13), and there exists no claw of
in G, then each vertex of G is contained in exactly r grand
cliClues.
This follows at once from the previous theorem and Theorem (2.3.lA),
remembering
p(r, 0:, 13)
Theorem (2.3.2C).
> q(r, 0:) •
If k > max
[p(r, 0:,13),
p(r, ex, 13)], then each
vertex in G is contained in exactly r grand cliques.
This follows :rrom the previous theorem and theorem (2.2.1).
-.
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~II.
Characterization of Tetrahedral Graphs
1.
As mentioned earlier in the introducation a tetrahedral graph G is a
gra)!1
~:hose
vertices can be identified with the
trinlets on
n
symbols, such that any two vertices are adjacent if and only
if the correspondingiriplets have a pair of common symbols.
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Then G clearly pos-
sesses tl1e rroperties (bl)-(bl~) given in Section I, paragraph 3.
We shall
here 2lrove that if n > 16, the converse also holds.
In the following lemmas G is a graph satisfying the conditions (b )-(b ),
4
l
and such that
n > 16.
If we set
7
= 3,
K = n-2,a
= 2,
= 3,
~
then the conditions (b ), (b ),
2
3
(b ) are the same as (c ),(c ), (c ) of Section II. Also from (2.1.1), (2.1.3),
4
l
2
3
and (2.1.4)
Ie
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n(n-l)(n-2)/6 unordered
p(r,a,~) =
y(r ,a) = 5,
Hence a clique K of G is a major clique if
it is a grand clique.
p(r,a,~)]
Since
is satisfied.
Lemma (3.1.1).
grand clique.
n
IKI
14, p(r,a,~)
= 11.
~ n-6, and if it is complete
> 16, the condition K >.max (p(r ,a,~)
,
Hence from theorems (2.3.18) and (2.3.2C) we have:
Any two adjacent vertices of G are contained in exactly one
Each vertex of G is contained in exactly 3 grand cliques,.
The unique grand clique containing any two given adjacent vertices x and
y, may be denoted by K(x,y).
The null set will be denoted by
¢.
The following six lemmas are directed towards proving that
any grand clique
K in G.
Lemraa (3.1.2).
If
K is a grand clique in G, then
n-4
<
IK I
.:s n.
13
IKI
= n-2,
for
x and y be any t~'lO vertices in K.
Let
There are
IK 1-2 vertices in
K other than x and y, and by definition each of these is adjacent to both
x
and y.
If
IKI > n, then
IKI-2 > n-2
which would contradict (b ).
3
Hence
IKI::: n
Let A be the set of all vertices adjacent to both x and y but not
contained in K.
Then from (b )
4
IKI-2 + IAI= n-2.
i.e.
If 8 1 and S2
are the t"TO other cliques containing x and T , T be
l
2
those containing y, then any vertex in A must be of the for.m
Zij = Si
if it exists at all.
n Tj
,
i,j = 1,2,
Since two distinct cliques can have at most one vertex
in common, we have
IA I::: 4.
Hence from (3.1.1)
IK I 2: n-4.
Lemma (3.1.3).
If K is a grand clique in G, then
IK I
f
n-4.
Suppose
(3.1.2).
y,
IKI = n-4, and let x, y, A, Si' Tj (i,j = 1,2) be as in lemma
Since there are only n-6 vertices in K adjacent to both x and
1fe must have
IAI
=4.
Hence,
Si
n Tj f ¢
for
i,j = 1,2.
from (b 2 ) that
(3.1.2)
Since
IK-x I + Isl-x I +
IK-xl = n-5,
(3.1.3)
Is2 - x l
we have
Isl-x I + Is2 -x I = 2n-4.
14
= 3n-9·
It follows
I
I
e
I
I
I
I
I
I
I
Ie
I
I
I
I
I
I
Ie
I
I
Thus, at least one
Si-x, say
Is l
(3.1.4)
Consider the tvl0 vertices
18 1 1-2
other vertices in
(3.1.4) it follows that
Lemma (3.1.4).
Suppose
Sl-x, has at least
Sl
If
y
zll
in
Sl'
as well as to
~ n-l,
y
They are both adjacent to
and
IAI = 3,
x
not in 8 ,
1
(b ).
3
and one of the grand
S2' containing
x, must
T , T , say T •
l
2
2
from (3.1.2) we have
It follows then, that one of Sl-x and S2-x has at least n~2 vertices.
. /31- x
I>- n-2,
From
must intersect both of the other two grand
and the other grand clique
= n-4,
z2l
which contradicts
Then from (3.1.1),
intersect exactly one of
Since IK-xl
and.
K is a grand clique in G, then IKlf n-3.
cliques, say Sl' containing
cliques containing
> n-1.
,
6(x, Zll)
= n-3.
IKI
x
n-2 vertices, and therefore
then I Sl' > n-1 and considering the vertices x and z
-
at most one vertex not in Sl adjacent to both.
z22 and y are adjacent to both.
~
If
we have
This is contradicted since
Ifl S2 1 ~ n-1, the same argument can be
applied to x and z22.
Lermna (3.1.5).
Suppose IKI
of
8 , S2
1
If
= n-1,
K is a grand clique in G,
then from
(3.1.1),
intersects exactly ore of
and
Since IK-xl
= n-2,
then IK If n-1.
= 1.
IAI
T , T •
l
2
8
2
n T2
=
¢.
it follows from (3.1.2) that
lSI-xi
+
15
Is2 -xl =
2n-7.
Hence, exactly one
This follows i.mmediate1y from lemmas (3.1.2)--(3.1.6).
I
I
it
I
I
I
I
I
I
I
Lemma (3 .1. 8).
el
Hence for one
lSi-xi < n-4
i,
(3.1.3), (3.1.4), we have
IKI> n-2
If
Lemma (3.1.6).
K
lKI
(j
= 1,
2,).
= n,
then
Since IK-xl
\Sil < n-3.
for every
But by lemmas (3.1.2),
G.
K in
is a grand clique in G, then
IKI
Suppose
and then
IAI
= n-1,
f
=0
n •
and hence
Si
n Tj = ¢,
(i
= 1,
2,)
from (3.1.2) we have
=
2n-8.
Hence at least one of Sl-x andS -x, say Sl-x, has at most n-4 vertices.
2
Thus Is 1 , < n-3 which contradicts at least one of the lemmas (3.1.2), (3.1.3),
(3.1.4).
Lemma (3.1.7).
If K is a grand clique in G, then
Let
IKI
= n-2.
x be a vertex in G and let L be a grand clique
K1, K2, K3 all meet L, and Let· Yi = Ki n L, i = 1, 2, 3.
From Lemma (3.1.1) the vertices Yi , i = 1, 2, 3 are all distinct. Let
Suppose
= 1, 2, 3 be the third grand clique containing y. in addition to Ki
S., i
~
~
and L.
Suppose
Si
n Kj f ¢,
Then, the vertices
for some pair
i,j, i
~
(i, j, K are different).
Yj
and
let
z
= S1 n Kj~
. J
~
Z,
j,
z, y. and Y are such that, each of these is adjacent to both
k
x and y. but none contained in the grand clique K.
vertices in
f
From lemma (3.1.7) IKil
containing x and Yi
.
= n-2,
and hence the n-4
K. , othen than x and y. , together ,nth the three vertices
~
~
and Y constitute a set of n-1
k
16
vertices which are adjacent to both
I
I
I·
I
I
I
ell
I
I
I
I
re
I
I
I
I
I
I
I
Ie
I
I
I
I
I
I
1(I
I
~
and Y '
i
This contradicts
(b ).
3
s.3. n K.J = ¢ for i f j, i, j=l, 2, 3
(3.1.6)
By lemma (3.1.7) there are
Y"j'
vertices in L other than Yl , Y and
2
Each of these vertices must be non-adjacent to x, for otherwise we 1'TOuld
have
n-5
4 grand cliques containing x contradicting lemma (3.1.1).
Let
z
vertices in L-yl -y2 -y • Then, d(z,x) ~ 2 and by
3
there are exactly 4 vertices adjacent to both z and x. Clearly three
be one of these
(b .~l )
Hence
n-5
(1 < i < 3)
Yl ' Y2 , Y and the fourth vertex must be on some K.,
3.
-3
and be distinot from x and y3..• Thus for each of the n-5 vertices in
of these are
L-Y l - Y2-Y3 ' there is exactly one vertex in the set
T
=
3
U
i=l
( K
i
-X-Yi )
vn1ich is adjacent to it.
Let us define three sets Al , A , A where Ai
2 3
consists of all those vertices in K.-x-y.
which are adjacent ot at least one
J.
J.
vertex of L other than Yi ' Since from (3.1.6) no vertex z' € K -x-Y can
i
i
be ad.jacent to Yj ' j f i, it is clear that Ai consists of all those vertices
in Ki-X-Yi Which are adjacent to a vertex of
Now, since each vertex in L-Y l - Y2-Y3 is
T, and since
IL-y1-y2-y
3I
vle
=
~1-Y2-Y3
•
adjacent to exactly one vertex
n-5,
have
(3.1. 7)
3
E
i=l
IA i
I<
Again, since IKi-X-Yi ' = n-4, and the
we have
17
n-5.
Ki-K-Yi )
i
= 1, 2,3
are disjoint,
ITI =
If 'I'Te define
.3
L:
i=l
I
I
I K. -x-y.J. I = 3n-12.
~
Bi
to be tl1e set of all those vertices in K -x-yi Which
i
are adjacent to no vertex of L other than y., then clearly A
is disjoint
i
~
from B.
J.
and
A.~ U B.~ = Ki-x-y.J.
Hence, from (3.1. 8),
Combining (3.1.7)
and
(3.1.9),
we have
3
• L:
IBi
i=l
I >
(3.1.11)
NOi'T, let
>
b €
B.
= i O'
i
we have
2n-7
3·
and consider the vertices
b
and y. •
~O
J. O
Since
IK.~OI
= n-2
from lemma (3.1.7), it fol101'18 from (b ) that there are exactly two vertices
3
adjacent to both b and y. and not in K • From the definition of B. ,
i0
~O
~O
neither of these two vertices can be in L. Hence they nmst both lie in
Let
b
and b t
be
t'l-TO
distinct vertices in
the two vertices of Si -Yo
adjacent to
of S - Yi adjacent to b t
iO
0
Suppose
o
~O
•
18
I
I
I
I
I
I
_I
2n-7 •
It follows from (3.1.10) that for at least one
••
B.
:1,0
and let s l' s2
be
b andsi, s2 be the two.vertices
I
I
I
I
I
I
_I
I
I
I
I
••
I
I
I
I
I
I
8
Ie
I
I
=
for some pair
Sj'
Without loss of generality, suppose
containing the adjacent vertices
in
i,j,
= sl
sl
band.
1
Let
sl'
~
M
~
i, j
2 •
be the grand clique
Then, there are
n-4
,.vertices
M adjacent to both b and sl as well as 3 others not in M, namely,
s2' b' and y.
It follows that, there are at least
~O
n-1
vertices adjacent to
both b and s 1 which contradicts (b ). Hence the 4 vertices s
s
s'
s'
l' 2' l ' 2
3
nmst all be distinct. Hence the number of distinct vertices in s. -yo
adjacent to vertices in B.
~O
or from (3.1.11)
a IBi I •
is exactly
0
~O
~O
Consequently
and 1ermna (3.1.7) ,
Ie
I
I
I
I
I
I
i
n-3
2n-7 ) ,
2( ~
2::
vlhich contradicts the assumption
Lemma (3.1. 9).
Let
x
n > 16.
and y
This completes the proof.
be two adjacent vertices in G and let
K be. the grand clique containing both
x
and y.
Let
Sl' S2 be the other
tyro grand cliques containing x
containing y.
and T , T be the other two grand cliques
l
2
Then the grand cliques S. may be put in (1,1)' correspondence
J.
with grand cliques
Since
T
j
Ixl = n-2,
so that only corresponding cliques intersect,
it is clear that exactly two of the
i, j
= 1,
19
2
4 intersections
I
I
are non-empty.
If Sl and S2 both intersect one of the grand cliques containing
y , other than K, say
T
l
and
x
cliques
Si
T , then all three grand cliques containing x
l
This contradicts lemma (3.1.8).
'T •
1
intersect·
Hence each of the grand
intersects one and only one of the grand cliques and vice versa.
Lemma (3.l.l0).
Let
x
and
Then there is one grand clique
grand clique containing y,
y
S3
be two vertices in G such that,
d{x,y)
caDtaining x which does not intersect any
and one grand clique
= 2.
-.
I
I
I
I
I
I
el,
I'
I'
I
I
I;
T containing y which
3
does not intersect any grand clique containing x. The other two grand cliques S1'8
2
containing x and the other two grand cliques Tl'T containing y mutually intersect.
2
From (b ),
there are exactly 4
4
Let
zll be one of these and let
zll
and Tl
since
clique
S2
and y.
Sl be the grand clique containing x
be the grand clique containing y
d{x,y)
Since
vertices adjacent to both x
and
zll'
Clearly Sl
and
f
Tl ,
= 2.
d{X'Zll) = 1 by lemma (3.l.9) there exists exactly one grand
containing
x,
other than
Sl' which intersects
T •
Let
1
z2l
=
l\
Tl •
Clearly zll f z2l' Y f z2l· Similarly since
d{zll'Y) = 1, .
there exists exactly one grand clique T2 containing y, which intersects 8 1 in
82
z12 say.
Clearly x '/: z12'
zll '/: z12'
Suppose
T2
n S2
d{z12' x) = 1, from the previous lemma the thir.d grand clique
x
must intersect
containing y
z32
E
T2
T
2
in
z32' say.
must intersect
and
S2
n
T2
= ¢,
vertices adjacent to both x
This contradicts
then have
S2
(b ) 'and hence'
4
in
z23 say.
z23 '/: z32.
Y, namely,
n T2
2
8
Since
(b )
4
S3
z23 E"8 2
containing
that
T
3
Fd
But then we have
,
zll' z12' z2l' z23' z32'
'/:
¢.
Let
z22
4 vertices adjacent to both x and y , namely,
It follows from
¢. Then. since
Similarly, the third grand clique
we must have
and
•
= 82 n .T2 , we
zij' i, j = 1,2.
I:
el~
I:
20
I
I
I
.I
I
I
I
I
I
Ie
I
I
I
I
I
I
Ie
S3 (\ Tj
II
Si
Lemma (3.1.11).
T
3
j
= ¢,
i = 1,2,3.
K and K of G,
l
2
vdth a common vertex x, there is a (1,1) correspondence between the vertices
of
K and K
1
2
Given tvro distinct grand cliques
such that the corresponding vertices are contained in a grand
clique.
Let
Y
l
d(yl,z) .:s 2.
F
F
K, Y
x. If Z
x is a vertex of K , then
l
l
2
It follows as in the previous lemma that every vertex of ~ is
be a vertex of
Hence there exists a vertex z in K such that d(Y , z)
2
l
not adjacent to Y '
1
From lemma (3.i.10) there is another clique
Yl , and intersects
~.
Let
Y
2
K*,
besides
K,
1
= 2.
Which contains
= K*
Also the third clique containing
f\ K • Then Y1 is adjacent to Y '
2
2
Y does not intersect ~. Thus there is
l
exactly one vertex Y in K which is adjacent to Y ' In the same way we
2
2
l
can start from a vertex Y in K and show that there is exactly one vertex
2
2
in K l
adjacent to
Lemma (3.1.12).
Yl '
There are exactly n(n-l)/2
Consider ordered pairs
(x,K)
clique of G containing x.
we get
G.
3v such pairs,
where
Since
where
v
x
is a vertex and K is a grand
= n(n-l)(n-2)/6
But each grand clique accounts for n-2 pairs.
cliques is 3v/(n-2)
Lemma (3.1.13).
2(n-2)
grand cliques in G.
each vertex is contained in 3
grand cliques
is the number of vertices in
Hence the number of grand
= n(n-l)/2.
Each grand clique in Gis intersected by exactly
other grand cli,ques.
This follows at once by noting that each vertex of a grand clique
contained in exactly two other grand cliques.
I
~
= 1,2,3
= ¢
21
K is
Lemma (3.1.14).
G,
If
there exist exactly
Let
x
=~ n
K •
2
K
l
and
n-2
K
2
are two intersecting grand cliques in
grand cliques which intersect both
Then through
x
K
l
~.
and
there passes another grand clique
K
(intersecting K and K in x). Again if y 1 is anyone of the other
l
2
3
n-3 points on K , then from Lemma (3.1.11) there exists a corresponding point
l
Y2 on K such that
2
intersecting both
Y
l
K
l
and Y are adjacEnt, K(y , Y ) being a grand clique
l
2
2
and ~. Each pair of corresponding points gives one such
clique and vice versa.
The following three lemmas are directed towards proving that there exist
exactly
4
grand cliques, vmich intersect each of two given non-intersecting
grand cliques
K and K •
2
l
Lemma (3.1.15). If K and
l
G,
K are two non-intersecting grand cliques in
2
Xl and x are two adjacent vertices belonging to ~ and K respectively,
2
2
then there exist vertices
that
xl' x2 ' Yl ' Y2
Since
K
l
and
exists a vertex
Y , Y belonging to
2
l
are nru.tually adjacent (Yl
~
f
and
xl'
~
respectively, such
Y2
f
x2 )·
K(x ,x ) are intersecting cliques, by lemma (3.1.11), there
l 2
in K i'Thich corresponds to x
and is therefore adjacent
2
l
to it. Similarly there exists a vertex Y2 inK , which is adjacent to xl'
2
Clearly d(y ,y ).:s 2. We want to show that Y and Y are adjacent.
l 2
l
2
If not, then
Y
l
d(y ,y )
l 2
= 2.
containing Y , and a clique
l
any clique containing
Hence
Y,
2
Then by lemma. (3.1.l0) there exists a clique
~
~
containing Y
2
K! does not intersect
does not intersect any clique containing Y •
l
~ nru.st be distinct from
from K
2
such that
K!
K
l
and
and
K(Y ,x ).
l 2
Also
K~
nru.st be distinct
K(y ,x ). The cliques K and k(y ,x ) nru.st therefore
l
l 2
2 l
intersect the cliques K and K(Y ,x ). Since by hypothesis K does not
l
2 l
2
intersect K , this is a contradiction. Hence d(Y ,y ) = 1.
l 2
2
22
I
I
-.
I
I
I
I
I
I
_I
I
I
I
I
I
I
_I
I
,
I
I
.e
I
I
I
I
I
I
Ie
I
I
I
I
I
I
Ie
I
~\
Lemma (3.1.16).
JS.
and K are two non-intersecting grand cliques in
2
G, such that there exist vertices xl and x belonging to K and K
l
2
2
respectively with d(x ,x ) = 2, then there exists at least one pair of adjacent
l 2
vertices one of which belongs to K and the other to ~.
l
From lemma (3.1.10), there exist two cliques (different from K ) conl
taining xl' which intersect each of the two cliques (other than K ) con2
taining x • Let z be the vertex of intersection of one of the cliques con2
taining xl wi th one of the cliques containing x • Then z is different
2
from xl and x • Since Kl and K(x , z) are intersecting cliques, from
l
2
lemma (3.1.11), there exists a vertex Y in Kl (Y l f xl)' such that Yl
l
is adjacent to z. Similarly there exists a vertex Y in K adjacent to
2
2
z. xl and x are non.-adjacent by hypothesis. If the lemma is false then the
2
pairs (xl ,y ), (Yl'~) and (Yl ,Y ) must also be non-adjacent. Hence K(z,x l ),
2
2
K(Z,x ), K(z,y ) and K(z,y ) must all be different grand cliques, contradicting
l
2
2
the lemma (3.1.1).
Lemma (3.1.7).
If
If
K1
and K are two non-intersecting grand cliques in
2
G, there must exist a pair of adjacent vertices one of which belongs to K
l
and the other to
K2.
xl be a vertex in K and x a vertex in K • If d(x l ,x ) = 1,
l
2
2
2
the lemma is true. If d(x ,x ) = 2, the required result follows from the
l 2
previous lemma. Suppose d(Xl'~) = r > 2. Then, there exists a chain xl'
Let
zl' z2' ••• zr_l' x2 joining xl and x2 ' Clearly d(zr_2'x2 ) = 2. Consider
the grand. cliques K(z r-3'z r-c:.~) and K_.
These must be non-intersecting, other-~
wise
d(zr_3' xa' =2, and X and Y would beat distance
r-l.
Hence from
the previous lemma we can find a vertex Y2 in K(zr_3' zr_2) and z~_2 in
K such that Y and z~_2 are adjacent. Then d(Xl'Y2 ) = r-1. By successive
2
2
23
I
repetition of the same process we can find a vertex w in
2
the required result then follow·s from the previous lemma.·
K2
such that
i
Lemma (3.1.18).
If K and K are two non-intersecting grand cliques
1
2
in G, then there exist exactly 4 grand cliques, which intersect each of
K and K •
2
l
From the previous lemma we can find a vertex xl in K and a vertex x
l
2
in K such that xl is adjacent to x ' Now from lemna (3.1.15) we can find
2
2
in Kl a vertex y1
f
xl' and in
1<2
a vertex Y2
f
x2' such that
xl' x2 '
Y1 , Y2 are .mutually t'l.djacent. Hence K(xl ,x2 ), K(~'Y2)' K(Yl'x ) and K(y ,y )
2
1 2
are 4 grand cliques each of which intersects K and
l
If possible let there be another grand clique Which intersects K in zl
l
and K2 in z2' Then zl is distinct from xl and Yl and z2 is distinct from
K2'
x and Y ' Otherwise there would be 4 grand cliques containing some vertex.
2
2
Again zl is non-adjacent to x • otherwise there would be 4 grand cliques
2
containing x ' Hence d(zr' x ) = 2, The grand cliques K and K containing
l
2
2
2
zl and x respectively are non-intersecting. Since the grand cliques K(x ,x l )
2
2
and K(x ,y ) intersect the grand clique K containing zl' it follows
l
2 l
from lemma (3.1.10), that K the third grand clique containing x ' cannot
2
2
intersect any grand clique containing zl' This is a contradiction since
K(zl,z2)
2.
intersects
(b ) hold.
4
K2 in z2'
Theorem (3.2,1).
For a tetrahedral graph G,
(b )l
Conversely if n > 16 and conditions (b )-(b4 ) are satisfied then
l
the conditions
G is tetrahedral.
Given n
symbols,
1,2, ••. n, a tetrahedral graph G is the graph whose
vertices are the unordered triplets on these symbols, two triplets being
adjacent when the corresponding triplets have two symbols in common.
24
2 )=2.
d(X , w
.
eI
I
I
I
I
I
I
I
el
I
I
I
I
I
I
_I
The
I
i
I
I
.I
I
I
I
I
I
Ie
conditions
(i)
(b )-(b ) are easily checked.
4
l
The number of unordered triplets on n
symbols is clearly n(n-l)
(n-2)/6 which is the number of vertices in G.
(ii)
Let
x be the vertex corresponding to the symbol (i,j,k), for a
vertex adjacent to x,
symbols i,j,k
the corresponding triplet must contain 2 of the
and one of n-3 symbols other than i, j,k.
number of vertices adjacent to
x is
3(n-3).
Hence the
Hence G is regular of
valence 3(n-3).
Given any two triplets we can readily construct a chain of triplets
beginning with the first and ending with the second so that two consecutive
triplets have two symbols in common.
(iii)
Let
x
Hence G is connected.
and y be two adjacent vertices of G.
Let
x and y
correspond to (i, j,k1 ) and (i, j,k ). Then the triplets which have two
2
symbols in common with each of the two triplets are the triplets (i,k l ,k ),
2
(j,k l ,k ) and the n-4 triplets (i,j,s) where s is any of the n symbols
2
other than i!j,k ,k • Hence the number of' vertices adjacent to both x
l 2
w.d Y is, n-2, i.e.
6(x,y) = n-2.
~
•
(iv)
Let x
and y be two vertices such that
d(x,y)
= 2.
Then we may
take x to correspond to
(i,jl,k ) and y to correspond to (i,j2,k ).
l
2
Then the only triplets which have two symbols inccommon with both these
triplets are (i,jl,j2)' (i,jl,k2 ), (i,j2,k l ) and (i,k l ,k2 ).
6(x,y) = 4 if d(x,y) = 2.
for
Thus
Conversely, suppose n > 16 and the conditions (b )-(b4 ) are satisfied
l
G. Then lennnas (3.1.1) - (3.1.18) hold. Let H be a graph whose
vertices are the grand cliques of' G, two vertices of H being adjacent if'
and only if the corresponding grand cliques of G have a non-empty intersection.
25
\
..
Then
I
I
-.
H satisfies the following conditions:
H is n(n-l)/2.
(a l )
The number of vertices in
(a )
2
H
(a )
3
H is edge regular with edge degree
(a4)
is regular and of valence
if u
and
v are adjacent.
~(u,v)
=4
if u
and
v
2(n-2).
n-2, ie.
~(u,v) =
n-2
are non-adjacent.
Conditions (a ),(a ),(a ),(a4) are satisfied in virtue of lemmas
2
l
3
(3 ~.12), (3~~ 13), (3~.14) and (3.1.18).
It now follows from Connor's theorem
(see section I), that for n > 8, H is triangular.
Hence to each vertex of H
we can associate an unordered pair (i,j) of two distinct symbols chosen out
of n symbols 1,2, ••• ,n so that two vertices of H are adjacent if and only
if the corresponding pair have a common symbol.
From the correspondence
between Hand G it follows that to each grand clique in G we can associate
an unordered pair of symbols chosen out of n symbols, such that the pairs
corresponding to two grand cliques have a common symbol if and only if the
two grand cliques intersect.
Let K and K be any two intersecting grand cliques of G, having the
2
l
vertex x in common and let (i,j) and (i,k) be the corresponding pairs. To the
vertex x we associate the unordered triplet (i,j,k).
K3 containing x, intersects both
pair (j,k).
The third grand clique
and K and hence must correspond to the
2
Note that the triplet assigned to x is unambiguously determined
~
by any two of the three cliques intersecting in x.
Thus to each of the
n(n-l)(n-2)/6 vertices of G we can associate a unique triplet.
If two vertices x and y of G are adjacent then there is a unique grand
clique K containing x and y.
If (i,j) is the pair corresponding to K, then
26
I
I
I
I
I
I
el
I
I
I
I
I
I
_I
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~
I
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I
I
I
I
the triplets corresponding to x and y must contain the symbols i and j •
Conversely if the triplets corresponding to the vertices x and y contain the
symbols i and j, then x and yare contained in the grand clique K corresponding
to the pair (i,j), and must therefore be adjacent.
Thus two vertices of G
are adjacent if and only if the corresponding triplets have a common pair
of symbols.
Hence G must be tetrahedral.
Acknowledgement.
Our thanks are due to Thomas A. Dowling for suggestions
which resulted in a proof of the main theorem (;.2.1) ~r assumptions less
restrictive than in our original version [2].
Ie
I
I
I
•
27
References
1.
2.
R. C. Bose, Strongly regular graphs, partial geometries, and partially
balanced designs, Pacific ~. Math., 13 (1963), 389-419.
R. C. Bose and Renu Laskar, A characterization of tetrahedral graphs,
Noti~es Ann. Math. Soc., 13 (1966), 505.
R. H. Bruck, Finite nets II, Uniqueness and imbedding, Pacific J.
Math., 13 (1963), 421-457.
4.
w.
5.
A. J. Hoffman, On the uniqueness of the triangular association scheme,
.Ann. Math. Stat., 31 (1960), 492-497.
6.
A. J. Hoffman, On the exceptional case in a characterization of the arcs
of a complete graph, IBM J., 4. (1960), 487-496.
S. Conner, The uniqueness of the triangular association scheme,.
Ann. Math. Stat.,. 29 (1958), 262-266.
.-
I
I
I
I
I
I,
I'.
~
I,
I .j
I
l
7.
8.
9·
Chang Li-chien, The uniqueness and non-uniqueness of the triangular
association schemes, Science Record, Math. New Ser., 3 (1959),
604-613.
Chang Li-chien, Association schemes of partia1 y balanced designs with
2
parameters v = 28, nl = 12, n2 = 15 and Pll = 4. Science Record,
Math. New Ser., 4 (1960), 12-18.
S. S. Shrikhande, On a characterization of the triangular association
scheme, Ann. Math. Stat., 30 (1959), 39-47.
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