The Number of Columns (1,k-2,1) in the Intersection Array of a Distance-Regular Graph

The Number of Columns (1,k) 2,1) in the Intersection

Array of a Distance-Regular Graph

Akira Hiraki

Division of Mathematical Sciences, Osaka Kyoiku University, Kashiwara,Osaka 582-8582, Japan. e-mail: [email protected]

Abstract. In this paper we consider the number t of columns ð1; k � 2; 1Þ in the intersectionarray of a distance-regular graph. We prove that t is at most one if k � 58.

1. Introduction

One of the main open problem for distance-regular graphs is to prove that thereare only finitely many distance-regular graphs of fixed valency k with k � 3. Thisproblem is equivalent to construction of a diameter bound for distance-regulargraphs in terms of valency k.

Let C be a distance-regular graph with fixed valency k with k � 3. We denoteby ‘ðc; a; bÞ the number of columns tðc; a; bÞ in the intersection array of C. Thereare only finitely many kind of column vectors tðc; a; bÞ with cþ aþ b ¼ k. Tosolve the problem we have to bound the number ‘ðc; a; bÞ of columns tðc; a; bÞ inthe intersection array of C.

Bannai and Ito proved a lot of interesting results concerning this problem byusing eigenvalue technique. (See [2–5].) In [2] they showed ‘ðc; a; cÞ � 10k2k ifk � 3.

In [7] Biggs, Boshier and Shawe-Taylor showed ‘ð1; 1; 1Þ � 3 using circuitchasing technique. It was a key to their classification of distance-regular graphs ofvalency 3.

In [12] Higashitani and Suzuki showed that ‘ð1; k � 2; 1Þ � 46ffiffiffiffiffiffiffiffiffiffiffi

k � 3p

ifvalency k � 5.

In the previous paper [15] the author proved that ‘ð1; k � 2; 1Þ � 20.In [10] Brouwer and Koolen have classified distance-regular graphs of

valency 4, and from their classification it follows that ‘ð1; 2; 1Þ � 1.There are no known examples of a distance-regular graph of k � 5 with

‘ð1; k � 2; 1Þ � 2.Our purpose in this paper is to improve this upper bound for ‘ð1; k � 2; 1Þ and

to prove that it is at most 1 if valency k is enough large.The following are our results.

Graphs and Combinatorics (2003) 19:371–387Digital Object Identifier (DOI) 10.1007/s00373-002-0521-9 Graphs and

Combinatorics� Springer-Verlag 2003

Theorem 1. Let C be a distance-regular graph of valency k and t :¼ ‘ð1; k � 2; 1Þ.Then the following hold.

(1) If k ¼ 5, then t � 14.(2) If k � 6, then t � 7.(3) If k � 9, then t � 6.(4) If k � 12, then t � 4.(5) If k � 20, then t � 3.(6) If k � 30, then t � 2.(7) If k � 58, then t � 1.

Theorem 2. Let C be a distance-regular graph of valency k � 5. Let r :¼‘ð1; 0; k � 1Þ and ðcrþ1; arþ1; brþ1Þ ¼ � � �= ðcrþt; arþt; brþtÞ ¼ ð1; k � 2; 1Þ. Then thefollowing hold.

(1) t � 3.(2) If k � 6, then t � 2.(3) If k � 9, then t � 1.

In Section 2 we recall the definition and collect several known results. In Section 3and Section 4 we prove Theorem 1 and Theorem 2, respectively.

2. Preliminaries

All graphs considered are undirected finite graphs without loops or multipleedges. Let C be a connected graph with usual distance @C. We identify C with theset of vertices. Let d :¼ maxf@Cðu; vÞju; v 2 Cg which is called the diameter of C.For a vertex u in C, we denote by CjðuÞ the set of vertices which are at distance jfrom u. A graph C is called a regular graph of valency k, k-regular for short, ifjC1ðuÞj ¼ k for all u 2 C.

For two vertices x and y in C, let

pi;jðx; yÞ :¼ jCiðxÞ \ CjðyÞj:

A graph C is said to be distance-regular if pi;jðx; yÞ depend only on h ¼ @Cðx; yÞrather than individual vertices. In this case, we write ph

i;j for pi;jðx; yÞ with@Cðx; yÞ ¼ h which will be called the intersection numbers of C. Let

ci :¼ pii�1;1; ai :¼ pi

i;1; bi :¼ piiþ1;1 and ki ¼ p0

i;i

for all 0 � i � d. The array

iðCÞ ¼� c1 c2 � � � cj � � � cd�1 cd

a0 a1 a2 � � � aj � � � ad�1 ad

b0 b1 b2 � � � bj � � � bd�1 �

8

>

<

>

:

9

>

=

>

;

372 A. Hiraki

is called the intersection array of C. We denote by ‘ðc; a; bÞ the number of columnstðc; a; bÞ in the intersection array of C. It is clear from our definition a distance-regular graph is a regular graph of valency k ¼ k1 ¼ b0.

The following are basic properties of the intersection numbers which we useimplicitly in this paper.

ð1Þ ci þ ai þ bi ¼ k for all 1 � i � d � 1:

ð2Þ k ¼ b0 � b1 � � � � � bd�2 � bd�1 � 1:

ð3Þ 1 ¼ c1 � c2 � � � � � cd�1 � cd � k:

ð4Þ kibi ¼ kiþ1ciþ1 for all 0 � i � d � 1:

The rest of this paper C denotes a distance-regular graph of diameter d andvalency k � 3. Let A :¼ AðCÞ be the adjacency matrix of C. It is well known thatA has exactly d þ 1 distinct eigenvalues and the valency k is the largest eigenvalueof A. Let

k ¼ h0 > h1 > h2 > � � � > hd

be the eigenvalues of A, which are called the eigenvalues of C. We denote by mðhjÞthe multiplicity of hj in A.

The polynomials viðxÞ ð0 � i � dÞ are defined by the recurrence relation

xviðxÞ ¼ bi�1vi�1ðxÞ þ aiviðxÞ þ ciþ1viþ1ðxÞ for 1 � i � d � 1

with v0ðxÞ ¼ 1 and v1ðxÞ ¼ x. Then fviðxÞg0�i�d is a set of orthogonal polynomials.It is easy to see that viðkÞ ¼ ki for all 0 � i � d.

For all 0 � j � d we set

SjðxÞ ¼X

j

i¼0

viðxÞ2

ki:

Then it is well known that for any eigenvalue hj of C we have

mðhjÞ ¼jCj

SdðhjÞ:

The reader is referred to [1], [6] and [9] for general theory of distance-regulargraphs.

Next we collect several known results which we need in this paper.

Proposition 3 ([20]). Let r ¼ ‘ð1; 0; k � 1Þ � 1 and h be an eigenvalue of C withh 6¼ �k. Then

2ðk � 1Þr2 � mðhÞ:

The Number of Columns ð1; k � 2; 1Þ in the Intersection Array 373

Proposition 4 ([2, Proposition 2]). If cj < bj, then vjðxÞ has roots all less thank � cj � bj þ 2

ffiffiffiffiffiffiffiffi

cjbjp

.

Proposition 5. Let r :¼ ‘ð1; 0; k � 1Þ � 1 and crþ1 ¼ 1. Let a be a real number suchthat 2

ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

< a < k, and let w be the positive real number such that a ¼ 2ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

coshw. Then the following hold.

(1) For any 1 � j � r þ 1 we have

vjðaÞ ¼ffiffiffiffiffiffiffiffiffiffiffi

k � 1p j

sinhwsinhðjþ 1Þw� sinhðj� 1Þw

ðk � 1Þ

� �

:

In particular, vjðaÞ >ffiffiffiffiffiffiffiffiffiffiffi

k � 1p j

ejw.

(2) For any 1 � j � r þ 1 we have

SjðaÞ ¼ 1þ 2jk� ðk � 2Þ2j2kðk � 1Þ sinh2 w

þ sinh jw

2kðk � 1Þ sinh3 wM ;

where

M ¼ ðk � 1Þ2 coshðjþ 3Þw� 2ðk � 1Þ coshðjþ 1Þwþ coshðj� 1Þw:

Proof. (1) The first assertion is obtained from the recurrence relation withðci; ai; biÞ ¼ ð1; 0; k � 1Þ for all 1 � i � r and crþ1 ¼ 1. (See [3, Proposition 3].)Since w is positive, we have

sinhðjþ 1Þw� sinhðj� 1Þwðk � 1Þ

� �

> ejw sinhw:

The desired result is proved.

(2) This follows from (1). (See [3, Proposition 6].) (

Lemma 6. Let q be an integer with 2 � q � d. Let k be the largest root of vqðxÞ.Then for any real number a with k < a < k we have

vqðaÞkq

>a� kk � k

� �

vq�1ðaÞkq�1

:

Proof. First we remark that

a� ck � c

� �

>a� bk � b

� �

for any real numbers a;b; c with c < b < a < k.

374 A. Hiraki

Let k ¼ k1 > � � � > kq be the roots of vqðxÞ and l1 > � � � > lq�1 be the roots ofvq�1ðxÞ. Since fviðxÞg0�i�d is a set of orthogonal polynomials, the roots of vq�1ðxÞseparate those of vqðxÞ. It follows that kiþ1 < li < ki for all 1 � i � q� 1. Hencewe have

vqðaÞvqðkÞ

¼Y

q

i¼1

a� ki

k � ki

� �

>a� k1k � k1

� �

Y

q�1

i¼1

a� li

k � li

� �

¼ a� kk � k

� �

vq�1ðaÞvq�1ðkÞ

:

The desired result is proved. (

Lemma 7. Let r :¼ ‘ð1; 0; k � 1Þ � 2; t :¼ ‘ð1; k � 2; 1Þ and g :¼ maxfi j ci ¼ 1g.Then the following hold.

(1) g � 2r þ 2.(2) If c2rþ2 ¼ 1, then a2rþ1 > a2r.(3) If g ¼ 2r þ 1, then arþ1 ¼ 1.(4) If g ¼ 2r þ 2, then arþ1 ¼ arþ2 ¼ 1 and r � 0 ðmod 3Þ. In particular, t � 2.

Proof. (1) This is proved in [17, Theorem 2].(2) This follows by putting ðm; hÞ ¼ ðr; r þ 1Þ in [15, Corollary 2.5].(3) This is proved in [13, Corollary 1.2]. ( See also [18, Proposition 4.1]. )(4) The first assertion is proved in [18, Proposition 4.1] and [8]. Suppose t � 3.Then we have a2rþ1 ¼ a2r ¼ k � 2 which contradicts (2). The lemma is proved.

(

For any non-negative integer a with a � k let

d :¼ dðaÞ ¼ minfi j 1 � i � d; a � ci þ aig;

bi :¼ biðaÞ ¼ a� ci � ai for 0 � i � d;

ji :¼ jiðaÞ ¼b0 � � � bi�1

c1 � � � cifor 1 � i � d

and

NðaÞ :¼ 1þ j1 þ � � � þ jd:

In [11, §4] C-W. Weng studied regular subgraphs of a distance-regular graph andobtained lower bound for the number of vertices. He showed that jXj � NðaÞ forany a-regular subgraph X of C.

Using this result we have the following generalization of the Nomura’s result[19].

Proposition 8. Let C be a distance-regular graph of diameter d and valency k. Let hand j be positive integers with hþ j � d. Then the following hold.


(1) If ch ¼ chþj, then

NðahÞ �bj � � � bhþj�1

c1 � � � ch:

(2) If bh ¼ bhþj, then

NðahþjÞ �bj � � � bhþj�1

c1 � � � ch:

(3) If bh ¼ cj, then

NðajÞ �cjþ1 � � � chþj

c1 � � � ch:

Proof. Let ðx; yÞ be a pair of vertices at distance j and let D be the subgraphinduced by ChðxÞ \ ChþjðyÞ. If ch ¼ chþj, then D is an ah-regular graph with pj

h;hþjvertices. Thus

NðahÞ � jDj ¼ pjh;hþj ¼

bj � � � bhþj�1c1 � � � ch

:

We have (1). Similarly we have (2) as D is ahþj-regular if bh ¼ bhþj. (See [19].)Let ðu; vÞ be a pair of vertices at distance hþ j and look at the subgraph D0

induced by ChðuÞ \ CjðvÞ. Then we have (3). The proposition is proved. (

Corollary 9. Let r :¼ ‘ð1; 0; k � 1Þ and g :¼ maxfi j ci ¼ 1g. Let h be an integerwith r þ 1 � h � g� 1.Then NðahÞ � bg�hbg�hþ1 � � � bg�1.

Proof. Since ch ¼ cg ¼ 1, the assertion immediately follows by putting j ¼ g� hin Proposition 8 (1). (

Corollary 10. Let r :¼ ‘ð1; 0; k � 1Þ � 1 and ðcrþ1; arþ1; brþ1Þ ¼ � � � = ðcrþt; arþt;brþtÞ ¼ ð1; k � 2; 1Þ. Then the following hold.

(1) If t � 2, then ðk � 2Þ2ðk � 3Þr�1 < ðk � 1Þrþ2�t.(2) If k ¼ 5 and t � 4, then r � 7.(3) If k � 6 and t � 3, then r � 7.(4) If k � 9 and t � 2, then r � 8.

Proof. (1) Put h ¼ r þ 1 in Corollary 9. Then dðarþ1Þ ¼ r þ 1 and

ðk � 2Þ2ðk � 3Þr�1 ¼ jrðarþ1Þ þ jrþ1ðarþ1Þ < Nðarþ1Þ � ðk � 1Þrþ2�t:

(2)–(4) These follow from (1). (

Lemma 11. Let r :¼ ‘ð1; 0; k � 1Þ; t :¼ ‘ð1; k � 2; 1Þ and g :¼ maxfi j ci ¼ 1g.Suppose t � 2, Then a1 ¼ 0 and

k � 2þ 2 cospt

� �

< h1:

Moreover the following hold.

376 A. Hiraki

(1) If r � 2 and k � 5, then k ¼ 5 and t ¼ r ¼ 2.(2) If g ¼ 2r þ 1 and k � 10, then brþ1 ¼ k � 2 and brþ2 >


k � 1p

.(3) If g ¼ 2r þ 2 and k � 11, then brþ1 ¼ brþ2 ¼ k � 2 and brþ3 >


k � 1p

.

Proof. The first assertion is proved in [15, Lemma 3.2, Lemma 4.5]. Hence r � 1.

(1) Suppose ðcrþ1; arþ1; brþ1Þ ¼ ð1; k � 2; 1Þ. Then Corollary 10 (1) implies that

ðk � 2Þ2ðk � 3Þr�1 < ðk � 1Þr

which contradicts our assumption. Hence we have ðcrþ1; arþ1; brþ1Þ 6¼ ð1; k � 2; 1Þand thus g � ðr þ 1Þ þ t � r þ 3. It is known that there are no distance-regulargraph with k � 5; r ¼ 1 and g � 4. (See [14, Corollary 4.7].) Thus we have r ¼ 2.Hence g ¼ 5; t ¼ 2 and a3 ¼ 1 by Lemma 7. It follows by putting h ¼ 4 inCorollary 9, that

Nða4Þ � b1b2b3b4 ¼ ðk � 1Þ2ðk � 2Þ:

On the other hand we have dða4Þ ¼ 4 and

Nða4Þ > j3ða4Þ þ j4ða4Þ ¼ ðk � 2Þðk � 3Þ3:

Hence we have k ¼ 5. The assertion follows.

(2) Suppose brþ2 �ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

to derive a contradiction. We have arþ1 ¼ 1 andbrþ1 ¼ k � 2 from Lemma 7 (3). Let a :¼ arþ2. It follows, by putting h ¼ r þ 2 inCorollary 9, that

NðaÞ � br�1 � � � b2r < ðk � 1Þrþ42

since br�1 ¼ br ¼ k � 1 > brþ1;ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

� brþ2 � � � � � b2r�1 and b2r ¼ 1.On the other hand, dðaÞ ¼ r þ 2 and

NðaÞ > jrþ1ðaÞ þ jrþ2ðaÞ ¼ aða� 1Þrþ1 > ða� 1Þrþ2:

This implies

k �ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

� 2 � a� 1 < ðk � 1Þrþ42rþ4 � ðk � 1Þ

710

which contradicts our assumption k � 10. The desired result is proved.

(3) Suppose brþ3 �ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

to derive a contradiction. We have arþ1 ¼ arþ2 ¼ 1and brþ1 ¼ brþ2 ¼ k � 2 from Lemma 7 (4). Let a0 ¼ arþ3. Then we have

Nða0Þ � br�1 � � � b2rþ1 < ðk � 1Þrþ62

by putting h ¼ r þ 3 in Corollary 9.


On the other hand, dða0Þ ¼ r þ 3 and

Nða0Þ > jrþ1ða0Þ þ jrþ2ða0Þ þ jrþ3ða0Þ > ða0 � 1Þrþ3:

It follows that

k �ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

� 2 � a0 � 1 < ðk � 1Þrþ62rþ6 � ðk � 1Þ

34

which is a contradiction as k � 11. The lemma is proved. (

Remark. It is not hard to show that a distance-regular graph with k ¼ 5 andr ¼ t ¼ 2 does not exist. But we skip its proof as we do not need it in this paper.

3. Proof of Theorem 1

Throughout this section C denote a distance-regular graph of valency k � 5 andt ¼ ‘ð1; k � 2; 1Þ � 2. Let r :¼ ‘ð1; 0; k � 1Þ; s :¼ maxfi j bi > 1g and g :¼maxfi j ci ¼ 1g. Then C has the following intersection array.

iðCÞ ¼

� 1 � � � 1 1 � � � 1 1 � � � 1 csþtþ1 � � �

0 0 � � � 0 arþ1 � � � as k � 2 � � � k � 2 asþtþ1 � � �

k k � 1 � � � k � 1 brþ1 � � � bs 1 � � � 1 � � �

8

>

>

<

>

>

:

9

>

>

=

>

>

;

r t

First we recall the following result. (See [12, Lemma 3.1].)

Lemma 12.

jCj < ksþ1 t þ 1þ 1

bs � 1

� �

� ksþ1ðt þ 2Þ:

Proof. We have ksþ1 ¼ � � � ¼ ksþt from our basic properties of intersection num-bers. Let b :¼ bs and c :¼ csþtþ1. For any 0 � j � s we have

ksþ1 ¼ kjðbj � � � bsÞ � kjbs�jþ1:

Thus

X

s

j¼0kj � ksþ1

1

bsþ1 þ � � � þ1

b

� �

< ksþ11

b� 1:

378 A. Hiraki

For any sþ t þ 1 � j � d we have

ksþt ¼ kjðcj � � � csþtþ1Þ � kjc j�s�t:

Since c � 2 if d � sþ t þ 2, we have

X

d

j¼sþtþ1kj � ksþ1

1

cþ � � � þ 1

cd�s�t

� �

� ksþ1:

It follows that

jCj ¼X

d

j¼0kj ¼

X

s

j¼0kj þ

X

sþt

j¼sþ1kj þ

X

d

j¼sþtþ1kj < ksþ1

1

b� 1þ t þ 1

� �

:

Since b � 2, we have the desired result. (

Let h :¼ h1 be the second largest eigenvalue of C. We assume t � 3 when k � 6.Then we have r � 3 and

2ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

� k � 2þ 2 cospt

� �

< h

from Lemma 11. Let / be the positive real number such that h ¼ 2ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

cosh/:For any positive integers b;m with 2 � b � k � 1 we define

Dðb;mÞ :¼ðffiffiffi

bp� 1Þ2 � 2þ 2 cosðpmÞðffiffiffi

bp� 1Þ2

!

:

We remark that Dðb;mÞ < 1 and Dðb;mÞ � Dðb0;m0Þ if b � b0 and m � m0.

Lemma 13. Let q be an integer with 2 � q � s: If ðffiffiffiffiffi

bqp

� 1Þ2 � 2� 2 cospt

� �

; then

vqðhÞkq

>vq�1ðhÞ

kq�1Dðbq; tÞ:

In particular, for any 1 � h � q we have

vqðhÞkq� vhðhÞ

kh

Y

q

j¼hþ1Dðbj; tÞ:

Proof. Let b :¼ bq and k be the largest root of vqðxÞ. Then Proposition 4 impliesthat

k < k � ðffiffiffi

bp� 1Þ2 � k � 2þ 2 cos

pt

� �

< h:


Since

h� kk � k

� �

>h� ðk � ð

ffiffiffi

bp� 1Þ2Þ

k � ðk � ðffiffiffi

bp� 1Þ2Þ

!

> Dðb; tÞ;

the first assertion follows from Lemma 6.The second assertion follows from the first one and an easy induction. (

The method we use to prove Theorem 1 is essentially the same as the Bannai-Ito’s one. (See [2]. See also [12, 15].)

In [17] we showed that sþ t � g � 2r þ 2. Using this new upper bound for sand refinement computation we could obtain an improvement of bound for t.

The next lemma is our criterion inequality.

Lemma 14. Let b be a real number with 1 � b < k � 1 and q ¼ qðbÞ :¼ max

fi j bi > bg. Let cj :¼ Dðbj; tÞ. Suppose ðffiffiffiffiffi

bqp

� 1Þ2 � 2� 2 cospt

� �

. Then the fol-lowing hold.

(1) If q � r þ 1, then

Y

q

j¼rþ2c2j b

!

e2/

bffiffiffiffiffiffiffiffiffiffiffi

k � 1p

� �rþ1

<k2bqðt þ 2Þ

2ðk � 1Þ32b2rþ2�s

:

(2) If c2qb > 1 and bffiffiffiffiffiffiffiffiffiffiffi

k � 1p

< e2/, then

1 <k2bq

2ðk � 1Þ32

� ðt þ 2Þb2rþ2�s :

Proof. We have r � q � s from our definition.

(1) Since ksþ1 ¼ kqðbqbqþ1 � � � bsÞ � kqbqbs�q, Proposition 3 and Lemma 12 show

that

SdðhÞ ¼jCj

mðhÞ <ksþ1ðt þ 2Þ2ðk � 1Þ

r2� kqbqb

s�qðt þ 2Þ2ðk � 1Þ

r2

:

On the other hand, Lemma 13 implies that

SdðhÞ >vqðhÞ2

kq¼ kq

vqðhÞkq

� �2

� kqvrþ1ðhÞ

krþ1

� �2Y

q

j¼rþ2c2j :

Since krþ1 ¼ kðk � 1Þr and vrþ1ðhÞ > ðk � 1Þrþ12 eðrþ1Þ/ from Proposition 5 (1), the

desired result is proved.

(2) Suppose q � r þ 1. Then brþ1 � � � � � bq and thus crþ1 � � � � � cq > 0. Theassertion immediately follows from (1).

380 A. Hiraki

Suppose q ¼ r. Then ksþ1 ¼ krþ1ðbrþ1 � � � bsÞ � krþ1bs�r and

vrþ1ðhÞ2

krþ1< SdðhÞ ¼

jCjmðhÞ �

krþ1bs�rðt þ 2Þ

2ðk � 1Þr2

:

Hence we have

1 <e2/

bffiffiffiffiffiffiffiffiffiffiffi

k � 1p

� �rþ1

<k2bðt þ 2Þ

2ðk � 1Þ32b2rþ2�s

:

Since bq ¼ br > b, the desired result is proved. (

Lemma 15. Set q be as follows. Then e2/ > ðk � 1Þq.

q ¼

0:970 if t � 15 and k ¼ 5;

0:924 if t � 8 and k � 6;

0:943 if t � 7 and k � 9;

0:924 if t � 5 and k � 12;

0:935 if t � 4 and k � 20;

0:929 if t � 3 and k � 30;

0:929 if t � 2 and k � 58:

8

>

>

>

>

>

>

>

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>

<

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:

Proof. Let

f ðk; tÞ ¼ e2w

ðk � 1Þ where k � 2þ 2 cospt

� �

¼ 2ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

coshw; ðw > 0Þ:

Then it is not hard to verify that f ðk; tÞ � f ðk0; t0Þ if k � k0 and 2 � t � t0. Thedesired result follows. (

Proof of Theorem 1. (1) Suppose k ¼ 5 and t � 15 to derive a contradiction. Thensþ t � g � 2r þ 1 from Lemma 7 (1), (4).

Let q ¼ 0:97; b ¼ qffiffiffiffiffiffiffiffiffiffiffi

k � 1p

¼ 1:94 and q ¼ qðbÞ :¼ maxfi j bi > bg. Then bq � 2;cq ¼ Dðbq; tÞ � Dð2; 15Þ > 0:745 and c2qb > 1. We have

ðffiffiffiffiffi

bq

p

� 1Þ2 � ðffiffiffi

2p� 1Þ2 � 2� 2 cos

p15

� �

� 2� 2 cospt

� �

and bffiffiffiffiffiffiffiffiffiffiffi

k � 1p

¼ qðk � 1Þ < e2/ from Lemma 15. It follows, by Lemma 14 (2), that

1 <k2ðk � 1Þ2ðk � 1Þ

32

� ðt þ 2Þbtþ1 <

25

4� 17

b16< 1;

which is a contradiction. The desired result is proved.


(2) Suppose k � 6 and t � 8 to derive a contradiction. Then sþ t � g � 2r þ 1from Lemma 7 (1), (4).


k � 1p

and q ¼ qðbÞ :¼ maxfi j bi > bg. Then b �qffiffiffi

5p� 2:06 and bq � 3. Thus we have cq ¼ Dðbq; tÞ � Dð3; 8Þ > 0:715 and

c2qb > 1. We have

ðffiffiffiffiffi

bq

p

� 1Þ2 � ðffiffiffi

3p� 1Þ2 � 2� 2 cos

p8

� �

� 2� 2 cospt

� �

and bffiffiffiffiffiffiffiffiffiffiffi

k � 1p

¼ qðk � 1Þ < e2/ from Lemma 15. It follows, by Lemma 14 (2), that

1 <k2ðk � 1Þ2ðk � 1Þ

32

� ðt þ 2Þbtþ1 <

k2

2ðk � 1Þ12

� 10

b9¼ 5k2

q9ðk � 1Þ5:

This is a contradiction as k � 6. The desired result is proved.

(3)–(6) Using the same manner the assertions are proved by setting q as in Lemma15.

(7) Suppose k � 58 and t � 2 to derive a contradiction. Then t ¼ 2 andsþ t � g � 2r þ 2 from (6) and Lemma 7 (1).


k � 1p

and q :¼ qðbÞ ¼ maxfi j bi > bg: Then we haveb � q

ffiffiffiffiffi

57p

> 7:01 and cq ¼ Dðbq; tÞ � Dð8; 2Þ > 0:401. It follows, by Lemma 15,

that c2qb > 1:12 and bffiffiffiffiffiffiffiffiffiffiffi

k � 1p

< e2/.First we assume g � 2r. Then s � 2r � 2. It follows, by Lemma 14 (2), that

1 <k2bq

2ðk � 1Þ32

� ðt þ 2Þb2rþ2�s �

k2ðk � 1Þ2ðk � 1Þ

32

� 4

b4¼ 2k2

q4ðk � 1Þ52

which contradicts to k � 58.Next we assume g ¼ 2r þ 1. Then s � 2r � 1; brþ1 ¼ k � 2 and brþ2 >


k � 1p

from Lemma 11 (1). Hence q � r þ 2. It follows, by Lemma 14 (1), that

1 < ðc2qbÞ �4k2bq

2ðk � 1Þ32b3¼ 2k2bq

q3ðk � 1Þ3:

If bq � 22, then we have a contradiction as k � 58.If bq � 23, then cq ¼ Dðbq; 2Þ � Dð23; 2Þ > 0:86. Hence we have

5 < ðc2qbÞ �2k2bq

q3ðk � 1Þ3� 2k2

q3ðk � 1Þ2� 2 � 582

q3 � 572 < 3

which is a contradiction.Finally we assume g ¼ 2r þ 2: Then s � 2r; brþ1 ¼ brþ2 ¼ k � 2 and brþ3 >ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

from Lemma 11 (2). Hence q � r þ 3. Lemma 14 (1) implies that

382 A. Hiraki

ðc2rþ2bÞ � ðc2qbÞ <2k2bq

ðk � 1Þ32b2

:

Since crþ2 ¼ Dðk � 2; 2Þ � Dð56; 2Þ ¼ 0:95 and c2qb � 1:12, we have

1 < c2rþ2 � ðc2qbÞ <2k2bq

ðk � 1Þ32b3¼ 2k2bq

q3ðk � 1Þ3:

If bq � 22, then we have a contradiction as k � 58.If bq � 23, then cq � Dð23; 2Þ > 0:86. Hence we have

4 < c2rþ2 � ðc2qbÞ <2k2

q3ðk � 1Þ2� 2 � 582

q3 � 572 < 3

which is a contradiction. This completes the proof of the theorem. (

Remarks. (1) There are only two known examples of a distance-regular graph ofvalency k � 3 and t ¼ ‘ð1; k � 2; 1Þ � 2. They are the Biggs-Smith graph and thedodecahedron whose intersection arrays are

� 1 1 1 1 1 1 3

0 0 0 0 1 1 1 0

3 2 2 2 1 1 1 �

8

>

<

>

:

9

>

=

>

;

;

� 1 1 1 2 3

0 0 1 1 0 0

3 2 1 1 1 �

8

>

<

>

:

9

>

=

>

;

;

respectively.(2) If the upper bound for s can be improved, then we can improve the bound fort. In fact we can show that if k ¼ 5, then t � 7.

4. Proof of Theorem 2

In this section C denotes a distance-regular graph of valency k � 5 with r :¼‘ð1; 0; k � 1Þ and ðcrþ1; arþ1; brþ1Þ ¼ � � � ¼ ðcrþt; arþt; brþtÞ ¼ ð1; k � 2; 1Þ, wheret � 2. We assume t � 3 if k � 6. Let h ¼ h1 be the second largest eigenvalue of C.Then we have r � 3 and

2ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

� k � 2þ 2 cospt

� �

< h

from Lemma 11. Let / be the positive real number such thath ¼ 2


k � 1p

cosh/.


Lemma 16. Set q be as follows.

q ¼

0:616 if t � 4 and k ¼ 5;0:716 if t � 4 and k ¼ 6;0:523 if t ¼ 3 and k ¼ 6;0:622 if t � 3 and k � 7;0:483 if t ¼ 2 and k � 9:

8

>

>

>

<

>

>

>

:

Then the following hold.

(1) e2/ > ðk � 1Þq.(2) Let a :¼ k � 2þ 2 cos

pt

� �

; then

Srþ1ðaÞ < Srþ1ðhÞ <kðk � 1Þ

r2

2t þ k � 1

k � 2

� �

:

(3) We have

ðqffiffiffiffiffiffiffiffiffiffiffi

k � 1p

Þrþ1 < k2

2ðk � 1Þ32

t þ k � 1

k � 2

� �

:

Proof. (1) This is proved as same as the proof of Lemma 15.

(2) Proposition 5 (1) implies that the largest root of vjðxÞ is less than 2ffiffiffiffiffiffiffiffiffiffiffi

k � 1p

andthus 0 < vjðaÞ < vjðhÞ for all 1 � j � r þ 1. Hence we have Srþ1ðaÞ < Srþ1ðhÞ.

It follows, by Proposition 3 and Lemma 12, that

Srþ1ðhÞ < SdðhÞ ¼jCj

mðhÞ <kðk � 1Þr

2ðk � 1Þr2

t þ 1þ 1

k � 2

� �

:

The assertion follows.

(3) Proposition 5 (1) implies

Srþ1ðhÞ >vrþ1ðhÞ2

krþ1>ðk � 1Þrþ1e2ðrþ1Þ/

kðk � 1Þr :

Hence we have

ðqffiffiffiffiffiffiffiffiffiffiffi

k � 1p

Þrþ1 < e2/ffiffiffiffiffiffiffiffiffiffiffi

k � 1p� �rþ1

<k2

2ðk � 1Þ32

t þ k � 1

k � 2

� �

from (1), (2). The desired result is proved. (

Lemma 17. (1) 3ðt � 1Þ � r þ 2.(2) If k ¼ 5, then t � 3.(3) If k ¼ 6, then t � 2.

384 A. Hiraki

Proof. (1) This is proved in [16, Theorem 1].

(2) Suppose t � 4 to derive a contradiction. Then r � 7 from Corollary 10 (2).We have 3t � 4 � r þ 1 from (1). It follows, by Lemma 16 (3), that

ð2qÞ3t�4 � ð2qÞrþ1 < 25

16� t þ 4

3

� �

;

whereq ¼ 0:616.Hence t ¼ 4 and r � 9. Let a :¼ k � 2þ 2 cospt

� �

¼ 3þffiffiffi

2p

:Then

Srþ1ðaÞ <40

32r

from Lemma 16 (2). The left hand side can be precisely calculated by Proposition5 (2). Then we have a contradiction as 7 � r � 9.

(3) Suppose t � 3 to derive a contradiction. Then r � 7 from Corollary 10 (3).We have 3t � 4 � r þ 1 from (1). It follows, by Lemma 16 (3), that

ðqffiffiffi

5pÞ3t�4 � ðq

ffiffiffi

5pÞrþ1 < 36

10ffiffiffi

5p � t þ 5

4

� �

;

where q ¼ 0:523 if t ¼ 3, q ¼ 0:716 if t � 4. Hence we have t ¼ 3 and r � 11: Leta :¼ k � 2þ 2 cos

pt

� �

¼ 5: Then

Srþ1ðaÞ <51

4ðffiffiffi

5pÞr

from Lemma 16 (2). The left hand side can be precisely calculated by Proposition5 (2). Then we have a contradiction as 7 � r � 11. The lemma is proved. (

Proof of Theorem 2. (1)(2)Wemay assume k � 7 fromLemma 17. Suppose t � 3 toderive a contradiction. Lemma 17 (1) and Lemma 16 (3) imply that r � 3t � 5 and

ðqffiffiffi

6pÞ3t�5 � ðq


k � 1p

Þr < k2

2qðk � 1Þ2t þ k � 1

k � 2

� �

� 49

72qt þ 6

5

� �

;

where q ¼ 0:622. This is a contradiction.

(3) Suppose k � 9 and t � 2 to derive a contradiction. Then t ¼ 2 and r � 8 from(2) and Corollary 10 (4). It follows, by Lemma 16 (3), that

ðqffiffiffi

8pÞ8 � ðq


k � 1p

Þr < k2

2qðk � 1Þ22þ k � 1

k � 2

� �

<81

128q� 22

7;

where q ¼ 0:483. This is a contradiction. The theorem is proved. (


Remark. There are only two known examples of a distance-regular graph ofvalency k � 5 with r :¼ ‘ð1; 0; k � 1Þ � 1 and ðcrþ1; arþ1; brþ1Þ ¼ ð1; k � 2; 1Þ.They are the Wells graph and a subgraph of the Hoffman-Singleton graph whoseintersection arrays are

iðCÞ ¼� 1 1 4 5

0 0 3 0 0

5 4 1 1 �

8

>

<

>

:

9

>

=

>

;

; iðCÞ ¼� 1 1 6

0 0 4 0

6 5 1 �

8

>

<

>

:

9

>

=

>

;

;

respectively. (See [9, Theorem 9.2.9, Theorem 13.1.1 (ii)].)Suppose k ¼ n2 þ 1 for some integer n � 2 with n 6� 0ðmod 4Þ. Then the array

� 1 1 k � 1 k

0 0 k � 2 0 0

k k � 1 1 1 �

8

>

<

>

:

9

>

=

>

;

is feasible. The eigenvalues are fk;ffiffiffi

kp

; n� 1;�ffiffiffi

kp

;�n� 1g with multiplicities

mðkÞ ¼ 1;mð�ffiffiffi

kpÞ ¼ n4 þ 3n2 þ 4

4; mð�1� nÞ ¼ ðn

2 þ 1Þðn2 þ 2� nÞ4

:

When n ¼ 2 we have the Wells graph. No other examples are known.

References

1. Bannai, E., Ito, T.: Algebraic Combinatorics I, California: Benjamin-Cummings 19842. Bannai, E., Ito, T.: On distance-regular graph with fixed valency. Graphs Comb. 3, 95–

109 (1987)3. Bannai, E., Ito, T.: On distance-regular graph with fixed valency, II. Graphs Comb. 4,

219–228 (1988)4. Bannai, E., Ito, T.: On distance-regular graph with fixed valency, III. J. Algebra 107,

43–52 (1987)5. Bannai, E., Ito, T.: On distance-regular graph with fixed valency, IV. Eur. J. Comb. 10,

137–148 (1989)6. Biggs, N.L.: Algebraic Graph Theory, Camb. Tracts Math. 67, Cambridge Univ. Press

19747. Biggs, N.L., Boshier, A.G., Shawe-Taylor, J.: Cubic distance-regular graphs. J. Lond.

Math Soc., II. 33, 385–394 (1986)8. Boshier, A., Nomura, K.: A remark on the intersection arrays of distance-regular

graphs. J. Comb. Theory, Ser. B 44, 147–153 (1988)9. Brouwer, A.E., Cohen, A.M., Neumaier, A.: Distance-Regular Graphs, Berlin, Hei-

delberg: Springer Verlag 198910. Brouwer, A.E., Koolen, J.H.: The distance-regular graphs of valency four. J. Algebr.

Comb. 10, 5–24 (1999)11. Weng, C.-W.: Weak geodetically closed subgraphs in distance-regular graphs. Graphs

Comb. 14, 275–304 (1998)12. Higashitani, N., Suzuki, H.: Bounding the number of columns ð1; k � 2; 1Þ in the

intersection array of a distance-regular graph. Math. Jap. 37, 487–494 (1992)

386 A. Hiraki

13. Hiraki, A.: Distance-regular subgraphs in a distance-regular graph, I. Eur. J. Comb. 16,589–602 (1995)

14. Hiraki, A.: Distance-regular subgraphs in a distance-regular graph, III. Eur. J. Comb.17, 629–636 (1996)

15. Hiraki, A.: A constant bound on the number of columns ð1; k � 2; 1Þ in the intersectionarray of distance-regular graphs. Graphs Comb. 12, 23–37 (1996)

16. Hiraki, A.: Retracing argument for distance-regular graphs. J. Comb. Theory, Ser. B79, 211–220 (2000)

17. Hiraki, A.: A distance-regular graph with strongly closed subgraphs. J. Algebr. Comb.14, 127–131 (2001)

18. Hiraki, A., Koolen, J.H.: An improvement of the Godsil bound. Ann. Comb. 6, 33–44(2002)

19. Nomura, K.: Some inequalities in distance-regular graphs. J. Comb. Theory, Ser. B. 58,243–247 (1993)

20. Terwilliger, P.: Eigenvalue multiplicities of highly symmetric graphs. Discrete Math.41, 295–302 (1982)

Received: December 13, 2001Final version received: November 20, 2002


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The Number of Columns (1,k-2,1) in the Intersection Array of a Distance-Regular Graph