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Graphs and Combinatorics (1999) 15 : 417±428
Graphs andCombinatorics( Springer-Verlag 1999
Distance-regular Graphs of the Height h
Akira Hiraki*
Division of Mathematical Sciences, Osaka Kyoiku University, Kashiwara, Osaka 582±8582, Japan. e-mail: [email protected]
Abstract. The height of a distance-regular graph of the diameter d is de®ned by h �maxf j j pd
d; j 0 0g. We show that if G is a distance-regular graph of diameter d, height h > 1and every pd
d; h-graph is a clique, then h A fd ÿ 1; dg.
1. Introduction
Distance-regular graphs have a lot of substructures with high regularity. Some-times they have distance-regular subgraphs. For example, there exist a lot ofJohnson subgraphs in a Johnson graph. We are very much interested in thisfact and believe that studying the substructures of distance-regular graphs areimportant.
The simplest distance-regular graph is a clique that is a graph any two of itsvertices are adjacent. It is well known that a local graph G1�x� is a disjoint unionof cliques of size a1 � 1 for all vertices x in a distance-regular graph G without theinduced subgraph K2;1;1. The case does occur if c2 � 1.
On the other hand the following result is known. (See [2, Proposition 5.5.2].)Let G be a distance-regular graph of diameter d and i be an integer with 1 <
i < d. If every ci�1-graph is a clique, then ci�1 � 1.This result might suggest that there is no distance-regular graphs where every
pti; j-graph is a clique of size at least 2 for given i; j; t. However, we have several
interesting examples.For the Johnson graphs J�3d ÿ 1; d� and J�3d � 1; d�, every pt
i; j-graph is aclique of size d for �t; i; j� � �d; d; d ÿ 1� and �t; i; j� � �d; d; d�, respectively.
It is known that the distance 2-graph of a generalized Odd graph, is also dis-tance-regular and its every pt
i; j-graph is a clique for �t; i; j� � �0; d; d�; �d; d; 1�.
Problem 1.1. Find a distance-regular graph such that every pti; j -graph is a clique of
size at least 2 for given i; j; t.
* Supported in part by a grant of Japan Society for Promotion of Science
Let h :� maxf j j pdd; j 0 0g which is called the height of a distance-regular graph.
In the above examples, every pdd;h-graph is a clique. Conversely the author
knows no distance-regular graph such that every pdd;h-graph is a clique, except the
above examples. So we have the following conjecture.
Conjecture 1.2. Let G be a distance-regular graph of diameter d and height h. Ifevery pd
d;h-graph is a clique of size at least 2, then one of the following holds.
(1) h � 1 and G is the distance 2-graph of a generalized Odd graph,(2) h � d ÿ 1 and G is the Johnson graph J�3d ÿ 1; d�,(3) h � d and G is the Johnson graph J�3d � 1; d�.
In [5], Tomiyama showed that the Johnson graph J�8; 3� is the only distance-regular graph such that 2 � h < d and every pd
d;2-graph is a clique. That meansthe conjecture is true for 2 � h < d. In this paper, we consider arbitrary height h.The main results in this paper are Theorem 2.1 and Theorem 2.2. The ®rst oneshows that the �2d � 1�-gon is the only distance-regular graph with 1U h < d andpd
d;h � 1. The second one shows that the height is equal to either d ÿ 1 or d if1 < h and every pd
d;h-graph is a clique. That gives a partial answer to the aboveconjecture.
2. De®nition and Results
A connected undirected ®nite simple graph G with usual distance qG is said to bedistance-regular if the cardinality of the set
Pti; j�x; y� :� fz A G j qG�x; z� � i; qG�y; z� � jg
depends only on i; j and t � qG�x; y�. In this case, we write
pti; j :� jPt
i; j�x; y�j:The diameter and the height of G are de®ned by
d :� maxfqG�x; y� j x; y A Ggand
h :� maxf j j pdd; j 0 0g;
respectively. Let
Ct�x; y� :� Pttÿ1;1�x; y�; At�x; y� :� Pt
t;1�x; y�; Bt�x; y� :� Ptt�1;1�x; y�
and
G t�x� :� P0t; t�x; x� � fz A G j qG�x; z� � tg:
We denote by ct, at, bt and kt the cardinality of them which depend only on t.The numbers ct, at, bt, kt and pt
i; j are called the intersection numbers of G . In
particular, we write k � k1 that is called the valency of G .
418 A. Hiraki
The following are basic properties of the intersection numbers which we useimplicitly in this paper.
(1) ci � ai � bi � k.(2) k � b0 V b1 V � � � V bdÿ2 V bdÿ1 V 1.(3) 1 � c1 U c2 U � � � U cdÿ1 U cd U k.(4) ci U bj if i � j U d:(5) pt
i; j � ptj; i and kt pt
i; j � ki pit; j � kj p
jt; i.
(6) ci�1 pdi�1; j � ai p
di; j � biÿ1 pd
iÿ1; j � cj�1 pdi; j�1 � aj pd
i; j � bjÿ1 pdi; jÿ1.
(7) pi�ji; j �
ci�1 � � � ci�j
c1 � � � cjand pt
i; i�t �bt � � � bt�iÿ1
c1 � � � ci.
The reader is referred to [1], [2] for general theory of distance-regular graphs.A graph is called a clique if any two vertices of it are adjacent.We say every pt
i; j -graph is a clique if the induced subgraph Pti; j�x; y� is a clique
for any pair of vertices x and y at distance t. So do we for ct, at and bt-graphs.In this paper, we prove the following results.
Theorem 2.1. Let G be a distance-regular graph of diameter d and height h. Suppose
pdd;h � 1. Then one of the following holds.
(1) h � d,(2) h � 1 and G is the �2d � 1�-gon,
(3) h � 0 and G is an antipodal 2-cover.
Theorem 2.2. Let G be a distance-regular graph of diameter d and height h > 1.Suppose every pd
d;h-graph is a clique. Then h A fd ÿ 1; dg.
Remarks. Let G be a distance-regular graph of diameter d and height h.(1) h � 0 i¨ G is an antipodal 2-cover. In this case, pd
d;0 � 1.(2) h � 1 i¨ G is the distance 2-graph of a generalized Odd graph. (See
[2, §4.2 D].)In this case, every pd
d;1-graph is a clique of size ad . The Johnson graphJ�2d � 1; d� and the halved �2d � 1�-cube are contained in this class.
(3) k � 2 i¨ G is either 2d-gon or �2d � 1�-gon that has h � 0 or 1, respec-tively.
(4) There exists several distance-regular graphs with pdd;d � 1. For example,
antipodal 3-cover, the Johnson graph J�3d; d� and the Biggs-Smith graph.(5) In [5], Tomiyama showed that J�8; 3� is the only distance-regular graph
such that 2 � h < d and every pdd;2-graph is a clique. This proves the Theorem 2.2
for the case h � 2.
3. Preliminaries
In this section, we collect several results for a distance-regular graph.
Distance-regular Graphs of the Height h 419
Proposition 3.1. (1) If c2 � c3, then c3 � 1.(2) If bi � ci�1 � a1 � 2 for some 2U iU d ÿ 1, then ci�1 � 1.(3) If cs�1 � 1 and b1 > b2, then b1 > b2 > � � � > bs.(4) Suppose �csÿ1; bsÿ1�0 �cs; bs� � �cs�tÿ1; bs�tÿ1�0 �cs�t; bs�t� with 3U s.
Then tU sÿ 1. Moreover the following hold.(4-i) If bsÿ1 � bs, then tU sÿ 2.(4-ii) If bsÿ1 � bs, and cs � cs�t then tU sÿ 3.
Proof. See [2, Theorem 5.4.1], [2, Corollary 5.5.3], [4, Lemma 2.4] and [3, Propo-sition 2.3], respectively. r
Corollary 3.2. If c3 � c4 and b2 � b3, then 1 � c3 � c4 and b1 � b2 � b3.
Proof. If c2 < c3, then we have a contradiction from Proposition 3.1 (4). Hence wehave 1 � c2 � c3 from Proposition 3.1 (1). The assertion follows from Proposition3.1 (3). r
Lemma 3.3. Suppose there exist vertices v; x A G such that qG�v; x� � t and
Ptt�s; s�v; x� is a clique. Then
bt�j U csÿj�1 for all 1U j U sÿ 1:
Proof. Suppose bt�j > csÿj�1 for some 1U j U sÿ 1. Take u A Ptt�s; s�v; x� and w A
Psjÿ1; sÿj�1�x; u�. Then qG�v;w� � t� j ÿ 1. Since bt�jÿ1 V bt�j > csÿj�1; we have y
A Bt�jÿ1�v;w� ÿ Csÿj�1�u;w�: Take any z A Bt�j�v; y� and let z 0 A Pt�j�1t�s; sÿjÿ1�v; z�.
Then z A Pt�st�j�1; sÿjÿ1�v; z 0�; y A Pt�s
t�j; sÿj�v; z 0� and x A Pt�st; s �v; z 0�. Since fu; z 0gJ
Ptt�s; s�v; x�, we have qG�u; z 0�U 1 from our assumption. Thus
�sÿ j � 1� ÿ 1 � qG�u;w� ÿ qG�w; y�U qG�u; y�U qG�u; z 0� � qG�z 0; y�U 1� �sÿ j�:
As y B Csÿj�1�u;w�, we have qG�u; y� � sÿ j � 1 and qG�u; z 0� � 1. Since
�sÿ j � 1� ÿ 1 � qG�u; y� ÿ qG�y; z�U qG�u; z�U qG�u; z 0� � qG�z 0; z�� 1� �sÿ j ÿ 1�;
we have qG�u; z� � sÿ j and hence z A Csÿj�1�u; y�. This implies Bt�j�v; y�JCsÿj�1�u; y� which contradicts initial assumption. r
Lemma 3.4. Suppose bdÿ1 < c2. Then
(1) cdÿ1 < cd .(2) Let u; v; x A G such that v; x A Gd�u�, qG�v; x� � t and Ct�v; x�JGd�u�.
Then Pttÿj; j�v; x�JGd�u� for all 1U j U tÿ 1.
420 A. Hiraki
Proof. (1) Suppose cdÿ1 � cd . Let a; b A G with qG�a; b� � d. Take g A Cd�b; a�and d A Bdÿ1�g; b�. Note that Cdÿ1�g; b� � Cd�a; b� since cdÿ1 � cd . We have d AAd�a; b�JGd�a�. Since b A Cd�g; d� ÿ Cd�a; d�, there exists h A Cd�a; d�ÿCd�g; d�.Note that b B Cdÿ1�a; h� � Cd�g; h�. We have qG�b; h� � 2. Take any x A C2�b; h�.Since qG�g; h� � d and qG�h; x� � 1, we have x B Cdÿ1�g; b� � Cd�a; b�. Thisimplies x A Ad�a; b�JGd�a� and thus C2�b; h�JBdÿ1�a; h� which contradicts ourassumption.
(2) We prove our assertion by induction on j. The case j � 1 follows from ourassumption. Let 2U j U tÿ 1 and y A Pt
tÿj; j�v; x�. Suppose y B Gd�u� and derive acontradiction. For any z A Cj�x; y�, we have z A Pt
tÿj�1; jÿ1�v; x�. From the induc-
tive assumption, we have z A Gd�u� and hence y A Gdÿ1�u�. This implies Cj�x; y�JBdÿ1�u; y� which contradicts bdÿ1 < c2. The lemma is proved. r
4. Distance-regular graphs of height h
In this section, we study basic properties for distance-regular graphs of height h.Let G be a distance-regular graph of diameter d and height h with 2U h < d.
Lemma 4.1. (1) pdi; j � 0 if i � j > d � h,
(2) pdi; j 0 0 if i � j � d � h.
Proof. The case i � d follows from the de®nition of height. Since
ci�1 pdi�1; j � ai p
di; j � biÿ1 pd
iÿ1; j � cj�1 pdi; j�1 � aj pd
i; j � bjÿ1 pdi; jÿ1;
the assertions follow by induction on d ÿ i. r
Lemma 4.2. Let t be an integer with hU tU d. Let a; b A G with qG�a; b� � t. For
any g A Ptd;dÿt�a; b� and any d A Pt
d;d�hÿt�a; b�, the following hold.
(1) qG�g; d� � h,(2) h0Pt
d;d�hÿt�a; b�JPdd;h�a; g� and 0 < pt
d;d�hÿt U pdd;h,
(3) h0Ptd;dÿt�a; b�JPd
d;h�a; d� and 0 < ptd;dÿt U pd
d;h.
Proof. (1) Since d; g A Gd�a�, we have
hV qG�g; d�V qG�b; d� ÿ qG�b; g� � �d � hÿ t� ÿ �d ÿ t� � h:
This is the desired result.(2), (3) These are direct consequences of (1) and Lemma 4.1 (2). r
Lemma 4.3. Let u; v A G with qG�u; v� � d and y A Pddÿi;h�i�u; v� with 1U i U d ÿ h.
(1) Bdÿi�u; y�JCh�i�v; y� and bdÿi U ch�i.(2) jAdÿi�u; y�VAh�i�v; y�j � jCdÿi�u; y�VCh�i�v; y�j � adÿi � bdÿi ÿ ch�i.
Proof. (1) By Lemma 4.1, we have Bdÿi�u; y�VBh�i�v; y�JPddÿi�1;h�i�1�u; v� �h
and Bdÿi�u; y�VAh�i�v; y�JPddÿi�1;h�i�u; v� �h. The assertion follows.
Distance-regular Graphs of the Height h 421
(2) Let A :� Adÿi�u; y� and C :� Ch�i�v; y�. Since Pddÿi;h�i�1�u; v� �h, we have
A � �AVC�U �AVAh�i�v; y��U �AVBh�i�v; y��� �AVC�U �AVAh�i�v; y��Uh
and
C � �Cdÿi�u; y�VC�U �AVC�U �Bdÿi�u; y�VC�� �Cdÿi�u; y�VC�U �AVC�UBdÿi�u; y�
from (1). Hence
adÿi � jAj � jAVAh�i�v; y�j � jAVCj;ch�i � jCj � jCdÿi�u; y�VCj � jAVCj � bdÿi:
We obtain the desired result. r
Lemma 4.4. Suppose there exist vertices u; v A G such that qG�u; v� � d and
Pdd;h�u; v� is a clique. Then for hU tU d, the following hold.
(1) qG�x; y�V d � hÿ tÿ 1 for any x A Pdd;h�u; v� and any y A Pd
t;dÿt�u; v�(2) qG�x;w�U d ÿ t� 1 for any x A Pd
d;h�u; v� and any w A Pdt;d�hÿt�u; v�
(3) cdÿjÿ1 U bh�j for all 0U j U d ÿ hÿ 2.
Proof. (1) Since v A Ptd;dÿt�u; y�, there exists z1 A Pt
d;d�hÿt�u; y�JPdd;h�u; v� from
Lemma 4.2 (2). Thus
d � hÿ t � qG�z1; y�U qG�z1; x� � qG�x; y�U 1� qG�x; y�as z1; x A Pd
d;h�u; v�.(2) Lemma 4.2 (3) implies that there exists z2 A Pt
d;dÿt�u;w�JPdd;h�u; v� since
v A Ptd;d�hÿt�u;w�. The assertion follows from
qG�w; x�U qG�w; z2� � qG�z2; x�U �d ÿ t� � 1:
(3) Suppose cdÿjÿ1 > bh�j for some 0U j U d ÿ hÿ 2. Let d A Pddÿj; j�u; v�.
Since v A Pdÿjd; j �u; d�, there exists x A P
dÿjd;h�j�u; d�JPd
d;h�u; v� from Lemma 4.2 (2).
Then there exists y A Cdÿj�u; d� ÿ Bh�j�x; d� since cdÿj V cdÿjÿ1 > bh�j .Take any z A Cdÿjÿ1�u; y� and let
z 0 A Pdÿjÿ2h;dÿhÿjÿ2�u; z�JPd
h;dÿh�u; v�VGdÿhÿjÿ1�y�:Since x A Pd
d;h�u; v�, we have qG�x; z 0�V d ÿ 1 from (1). Hence
�h� j� � 1V qG�x; d� � qG�d; y�V qG�x; y�V qG�x; z 0� ÿ qG�z 0; y�V �d ÿ 1� ÿ �d ÿ hÿ j ÿ 1�� h� j
422 A. Hiraki
and thus qG�x; y� � h� j as y B Bh�j�x; d�. Since
�h� j� � 1 � qG�x; y� � qG�y; z�V qG�x; z�V qG�x; z 0� ÿ qG�z 0; z�V �d ÿ 1� ÿ �d ÿ hÿ j ÿ 2�� h� j � 1;
we obtain qG�x; z� � h� j � 1 and hence z A Bh�j�x; y�.This implies Cdÿjÿ1�u; y�JBh�j�x; y� which contradicts initial assumption. r
5. Some Inequalities of Intersection Numbers
In this section, we assume G is a distance-regular graph of diameter d, height h
with 1 < hU d ÿ 1 and every pdd;h-graph is a clique.
Lemma 5.1. Let u; v A G with qG�u; v� � d, x A Pdd;h�u; v� and y A Pd
dÿi;h�i�u; v� with
0U iU d ÿ h:
(1) cdÿjÿ1 U bh�j for 0U j U d ÿ hÿ 2.(2) bh�j U cdÿhÿj�1 for 1U j U d ÿ hÿ 1.(3) Cdÿi�u; y�VCh�i�v; y� �h.(4) Ad�u; x�U fxg � Ch�v; x�UPd
d;h�u; v� and ad � 1 � ch � pdd;h.
(5) qG�x; y�U i � 1.(6) qG�x; z�U i for any z A Bdÿi�u; y�.(7) If qG�x; y� � i V 2, then Adÿi�u; y�VAh�i�v; y�JAi�x; y�.Proof. (1) This follows from Lemma 4.4 (3).
(2) As u A Phd;d�v; x�, we have Ph
d;dÿh�v; x�JPdd;h�v; u� from Lemma 4.2 (3).
Thus Phd;dÿh�v; x� is a clique. The assertion is a direct consequence of Lemma 3.3.
(3) By replacing u and v, we may assume i 0 d ÿ h. Since v A Pdÿid;h�i�u; y�,
there exists v 0 A Pdÿid; i �u; y�JPd
d;h�u; v� from Lemma 4.2 (3). Take any d ACdÿi�u; y�. Since v A Pd
d;h�u; v 0� and d A Pddÿiÿ1;i�1�u; v 0�, Lemma 4.4 (1) implies
qG�v; d�V h� i. Thus we have d B Ch�i�v; y�.(4) Since Pd
d;h�u; v� is a clique, we have the assertion from (3).(5) The desired result follows from Lemma 4.4 (2).(6) By Lemma 4.3 (1), z A Pd
dÿi�1;h�iÿ1�u; v�. The assertion follows from (5).(7) Take any w A Adÿi�u; y�VAh�i�v; y�JPd
dÿi;h�i�u; v�. From (5), we have
i � 1V qG�x;w�V qG�u; x� ÿ qG�u;w� � i:
To prove the lemma we assume qG�x;w� � i � 1 and get a contradiction. For anyz A Bdÿi�u;w�, we have
�i � 1� ÿ 1 � qG�x;w� ÿ qG�w; z�U qG�x; z�U i
Distance-regular Graphs of the Height h 423
from (6). So qG�x; z� � i and Bdÿi�u;w�U fygJCi�1�x;w�. This implies bdÿi <ci�1.
On the other hand we obtain ch�iÿ1 U bdÿi from (1). This is a contradiction. r
Lemma 5.2. Suppose 1 < hU d ÿ 2. Then
(1) pdd;h � ph
d;dÿh
(2) bi U ch�i � cdÿi ÿ bdÿi for all 2U i U d ÿ h.
Proof. Let u; v A G with qG�u; v� � d and x A Pdd;h�u; v�.
(1) Since u A Phd;d�v; x�, there exists y A Ph
d;dÿh�v; x�JPdd;h�v; u� from Lemma
4.2 (3). For any z A Pdd;h�v; u� ÿ fyg, we have z A Ah�u; y�VAd�v; y� as Pd
d;h�v; u�is a clique. From Lemma 5.1 (7), z A Adÿh�x; y�. This implies that Pd
d;h�v; u�JPh
d;dÿh�v; x�. Since phd;dÿh U pd
d;h from Lemma 4.2 (3), the assertion follows.
(2) Let 2U i U d ÿ h and take y A Pddÿi;h�i�u; v�. Since v A Pdÿi
d;h�i�u; y�, thereexist x 0 A Pdÿi
d; i �u; y�JPdd;h�u; v� from Lemma 4.2 (3). Since qG�x 0; y� � i V 2, we
have
Bdÿi�u; y�U �Adÿi�u; y�VAh�i�v; y��JAi�x 0; y�UCi�x 0; y�by Lemma 5.1 (6) (7). Therefore we obtain
bdÿi � �adÿi � bdÿi ÿ ch�i�U ai � ci
from lemma 4.3 (2) and Lemma 5.1 (3). The lemma is proved. r
Proposition 5.3. There exist no distance-regular graphs of diameter d, height h with
3U hU d ÿ 3 and every pdd;h-graph is a clique.
Proof. Since 3U hU d ÿ 3, we have cdÿjÿ1 � bh�j � cdÿhÿj�1 for 1U j U d ÿ hÿ2 from Lemma 5.1 (1), (2). In particular,
c3 � c4 � � � � � cdÿ2 � bh�1 � � � � � bdÿ2:
By Lemma 5.2 (2) and Lemma 5.1 (1),
b2 U ch�2 � cdÿ2 ÿ bdÿ2 � ch�2 U cdÿ1 U bh:
Hence ch�2 � cdÿ1 � bh � b2. From Corollary 3.2, we have
b1 � b2 � b3 � � � � � bh � cdÿ1
and1 � c2 � � � � � cdÿ2 � bh�1 � � � � � bdÿ1:
From Lemma 5.1 (4) and Lemma 5.2 (1), we have
ad � pdd;h � ph
d;dÿh �bhbh�1 � � � bdÿ1
c1c2 � � � cdÿh� bh � b1 � cdÿ1:
Hence we obtain cd � k ÿ ad � k ÿ b1 � a1 � 1 and bdÿ1 � cd � a1 � 2. FromLemma 3.1 (2), we have cd � 1 and thus
424 A. Hiraki
k � ad � cd � cdÿ1 � cd U 2cd � 2:
Hence G is an ordinary polygon which contradicts hV 3. The assertion follows. r
6. The Case h � d ÿ 2
In this section, we assume that G denotes a distance-regular graph of diameter d,height h � d ÿ 2V 3 and every pd
d;dÿ2-graph is a clique.From Lemma 5.1 (1) (2), we have bdÿ2 V cdÿ1 and bdÿ1 U c2.Fix u; v A G with qG�u; v� � d and set Pi; j � Pd
i; j�u; v�.
Lemma 6.1. Let x A Pd;dÿ2 and z A Pdÿ2;dÿ1. Suppose bdÿ2 > cdÿ1. Then
(1) G2�z�VPd;dÿ2 0h and qG�x; z�U 3.(2) qG�x;w�U 3 for any w A Bdÿ2�u; z�.Proof. Since bdÿ2 > cdÿ1, there exists z1 A Bdÿ2�u; z� ÿ Cdÿ1�v; z�. Let z2 ABdÿ1�u; z1�. As Pdÿ1;d �h, we have z1 A Pdÿ1;dÿ1 and thus z2 A Cdÿ1�v; z1� fromLemma 4.3 (1). Hence qG�x; z�U 1 as fx; z2gJPd;dÿ2. This implies
qG�x; z�U qG�x; z2� � qG�z2; z�U 1� 2:
The assertion (1) follows. Let w A Bdÿ2�u; z�. Take w2 A Bdÿ1�u;w� and y APdÿ2
d;d �u; z�. Then fz2;w2gJPdÿ2d;2 �u; z�JPd
d;dÿ2�u; y� from Lemma 4.2 (3). Hencez2 and w2 are adjacent and
qG�x;w�U qG�x; z2� � qG�z2;w2� � qG�w2;w� � 3:
This is the desired result. r
Lemma 6.2. bdÿ2 � cdÿ1.
Proof. We assume bdÿ2 > cdÿ1 and derive a contradiction.Note that b2 V b3 V bdÿ2 > cdÿ1 V cdÿ2. Let x A Pd;dÿ2. Then Pdÿ2;d �
Pdd;dÿ2�v; u� � Pdÿ2
d;2 �v; x�JG2�x� from Lemma 4.2 (3) and Lemma 5.2 (1). Takey A Pdÿ2;d . Then there exists z A B2�x; y� ÿ Cdÿ2�u; y�. If z A Bdÿ2�u; y�, thenLemma 5.1 (6) implies qG�x; z�U 2 contradicting z A B2�x; y�. Thus z AAdÿ2�u; y�. As Pdÿ2;d JG2�x�, we have z A Pdÿ2;dÿ1. We can take w A B3�x; z�ÿCdÿ2�u; z� as b3 > cdÿ2. We show qG�u;w� � qG�v;w� � d ÿ 2. From Lemma 6.1(2), B3�x; z�VBdÿ2�u; z� �h and hence w A Adÿ2�u; z�. If qG�v;w�V d ÿ 1, thenw A Pdÿ2;dÿ1 as Pdÿ2;d JG2�x�. This contradict Lemma 6.1 (1). Therefore wehave qG�u;w� � qG�v;w� � d ÿ 2.
Then we have w A Pddÿ2;2�v; y� and u A Pd
d;dÿ2�v; y�. Lemma 4.4 (1) impliesqG�u;w�V d ÿ 1. This is a contradiction. The lemma is proved. r
Lemma 6.3. Let y A Pdÿ1;dÿ1 and z A Pdÿ2;2. If bdÿ1 < c2, then the following hold.
(1) There exists g1 A Gd�z�VPd;dÿ2:(2) There exists g2 A Gd�z�VPdÿ2;d :
Distance-regular Graphs of the Height h 425
(3) qG�z; y�V d ÿ 1.(4) qG�y;w�V d ÿ 1 for any w A Cd�u; v�.
Proof. (1) From Lemma 4.2 (2), there exists g1 A Pdÿ2d;d �u; z�JPd
d;dÿ2�u; v� �Pd;dÿ2.
(2) Let z 0 A Pdÿ22;dÿ4�u; z�JP2;dÿ2. There exists g2 A Gd�z 0�VPd
d;dÿ2�v; u�similar to (1). Note that qG�g2; v� � qG�g2; z
0� � d, qG�v; z 0� � d ÿ 2 and z APdÿ2
2;dÿ4�v; z 0�. From Lemma 5.1 (3), we have Cdÿ2�v; z 0�JGd�g2�. As bdÿ1 < c2
from our assumption, we apply Lemma 3.4 (2) and obtain z A Gd�g2�. Hence g2 AGd�z�VPdÿ2;d .
(3) Let g1; g2 as in (1), (2). Then it is clear that g1 and g2 are not adjacent.Suppose qG�z; y�U d ÿ 2. Then Lemma 5.1 (5) implies
d � qG�z; g1�U qG�z; y� � qG�y; g1�U �d ÿ 2� � 2:
Thus qG�z; y� � d ÿ 2 and qG�y; g1� � 2. Similarly, qG�y; g2� � 2. From Lemma4.2 (3),
fg1; g2gJPdÿ2d;2 �z; y�JPd
d;dÿ2�z; d�where d A Pdÿ2
d;d �z; y�. This contradicts g1 and g2 are not adjacent.(4) Suppose qG�y;w�U d ÿ 2 for some w A Cd�u; v�. Since qG�y; v� � d ÿ 1,
we have qG�y;w� � d ÿ 2. Note that Cdÿ1�u;w�JPdÿ2;2. From (3), we have
fvgUCdÿ1�u;w�JBdÿ2�y;w�:which contradicts Lemma 6.2. The lemma is proved. r
Lemma 6.4. (1) bdÿ1 � c2,(2) adÿ1 � 1U cdÿ1.
Proof. (1) We assume bdÿ1 < c2 and derive a contradiction.Let x A Pd;dÿ2. By Lemma 3.4 (1) and Lemma 6.2, we have cd > cdÿ1 � bdÿ2
and thus there exists w A Cd�u; v� ÿ Bdÿ2�x; v�. Since Cd�u; v�VCdÿ2�x; v� �hfrom Lemma 5.1 (3), we have w A Adÿ2�x; v�. Let z A Cdÿ1�u;w�JPdÿ2;2 andz 0 A Gd�z� VPd;dÿ2 as Lemma 6.3 (1). Since x; z 0 A Pd;dÿ2, we have
d � qG�z 0; z�U qG�z 0; x� � qG�x; z�U qG�z 0; x� � qG�x;w� � qG�w; z�U 1� �d ÿ 2� � 1
� d:
This implies qG�z 0; x� � 1; qG�x; z� � d ÿ 1; qG�x;w� � d ÿ 2 and qG�z 0;w� �d ÿ 1: Since Bdÿ2�v; x�JPdÿ1;dÿ1 from Lemma 4.3 (1), it follows from Lemma6.3 (4) that
fz 0gUBdÿ2�v; x�JBdÿ2�w; x�:This is a contradiction.
426 A. Hiraki
(2) Let y A Pdÿ1;dÿ1 and Y � Adÿ1�u; y�VAdÿ1�v; y�. Take x A Bdÿ1�u; y� andx 0 A Bdÿ1�v; y� Then x A Pd;dÿ2 and x 0 A Pdÿ2;d from Lemma 4.3 (1). We showthat Y U fygJC2�x; x 0�.
Suppose there exists z A Y ÿ G1�x�. As fxgUBdÿ1�u; z�JPd;dÿ2 from Lemma4.3 (1), we have Bdÿ1�u; z�JG1�x� as Pd;dÿ2 is a clique. Thus Bdÿ1�u; z�U fygJC2�z; x� which contradicts (1). Hence Y JG1�x�. By symmetry, we obtain Y JG1�x 0�. Whence this implies Y U fygJC2�x; x 0� and
�adÿ1 � bdÿ1 ÿ cdÿ1� � 1 � jY j � jfygjU c2 � bdÿ1
from (1), Lemma 4.3 (2) and Lemma 5.1 (3). This is the desired result. r
7. Proof of Theorems
In this section, we prove our results.
Proof of Theorem 2.1. We may assume h0 0; d and show that (2) holds.From Lemma 4.2 (3), we have
pdh;dÿh ph
d;d � phd;dÿh pd
d;h U �pdd;h�2 � 1
and hence pdh;dÿh � ph
d;d � 1. Note that
cd Ucdcdÿ1 � � � cdÿh�1
c1c2 � � � ch� pd
h;dÿh � 1:
Since pdd;h � 1, every pd
d;h-graph is a clique. From Lemma 5.1 (4), ad � ch U cd .Hence we have k � ad � cd U 2. Therefore G is the �2d � 1�-gon and (2) holds. r
Proof of Theorem 2.2. The case h � 2 has already proved by Tomiyama in [5]. Toprove the theorem, it is su½cient to show that there exist no distance-regulargraphs of diameter d, height h � d ÿ 2V 3 and every pd
d;h-graph is a clique byProposition 5.3.
Suppose there exists such a distance-regular graph G . Then Lemma 5.2 (1),Lemma 5.1 (4), Lemma 6.2 and Lemma 6.4 imply bdÿ1 � c2; bdÿ2 � cdÿ1,
pdd;dÿ2 � pdÿ2
d;2 �bdÿ2bdÿ1
c2� bdÿ2 and ad � 1 � cdÿ2 � bdÿ2:
We have
bdÿ1 U b2 U cd � cdÿ2 ÿ bdÿ2
� cd � �ad � 1ÿ bdÿ2� ÿ bdÿ2
� k � 1ÿ 2cdÿ1
� �bdÿ1 � adÿ1 � cdÿ1� � 1ÿ 2cdÿ1
� �adÿ1 ÿ cdÿ1 � 1� � bdÿ1
U bdÿ1
Distance-regular Graphs of the Height h 427
from Lemma 5.2 (2) and Lemma 6.4 (2). Hence
bdÿ1 � bdÿ2 � � � � � b2 and adÿ1 � cdÿ1 ÿ 1:
Since bdÿ1 � c2 U cdÿ1 � bdÿ2 � bdÿ1; we have
1 � c2 � c3 � � � � � cdÿ1 � bdÿ1:
from Lemma 3.1 (1). So we have
k � cdÿ1 � adÿ1 � bdÿ1 � cdÿ1 � �cdÿ1 ÿ 1� � bdÿ1 � 2:
This contradicts 3U h. The theorem is proved. r
References
1. Bannai, E., Ito, T.: Algebraic Combinatorics I, Benjamin-Cummings, California (1984)2. Brouwer, A. E., Cohen, A. M., Neumaier, A.: Distance-Regular Graphs, Springer
Verlag, Berlin, Heidelberg (1989)3. Hiraki, A., Suzuki, H., Wajima, M.: On distance-regular graphs with ki � kj , II,
Graphs Comb. 11, 305±317 (1995)4. Hiraki, A.: Distance-regular subgraphs in a distance-regular graph, III, Eur. J. Comb.
17, 629±636 (1996)5. Tomiyama, M.: On distance-regular graphs with height two, J. Algebraic Comb. 5, 57±
76 (1996)
Received: January 8, 1997Revised: November 30, 1998
428 A. Hiraki
