具封閉性質的距離正則圖之研究(III)

(1)

行政院國家科學委員會專題研究計畫成果報告

具封閉性質的距離正則圖之研究(3/3)

研究成果報告(完整版)

計畫類別：個別型

計畫編號： NSC 98-2115-M-009-002-

執行期間： 98 年 08 月 01 日至 99 年 07 月 31 日

執行單位：國立交通大學應用數學系（所）

計畫主持人：翁志文

報告附件：出席國際會議研究心得報告及發表論文

處理方式：本計畫可公開查詢

中華民國 99 年 09 月 03 日

(2)

行政院國家科學委員會補助專題研究計畫

成果報告

期中進度報告

(

具封閉性質的距離正則圖之研究

(3/3))

計畫類別

_:

個別型計畫

整合型計畫

計畫編號

:NSC

98-2115-M-009-002-執行期間

: 98

年

8 月

1 日至

99 年

7 月

31 日

執行機構及系所

:

國立交通大學應用數學系

計畫主持人

:

翁志文

共同主持人

:

計畫參與人員

:

博士生

:

黃喻培、黃皜文。碩士生

:

劉家安、葉彬、林育生、陳建文、林志

嘉、劉侖欣、洪湧昇、施政成

成果報告類型

(

依經費核定清單規定繳交

_):

精簡報告

完整報告

本計畫除繳交成果報告外

,

另須繳交以下出國心得報告

:

赴國外出差或研習心得報告

赴大陸地區出差或研習心得報告

出席國際學術會議心得報告

國際合作研究計畫國外研究報告

處理方式

:

除列管計畫及下列情形者外

,

得立即公開查詢

涉及專利或其他智慧財產權

_,

一年

二年後可公開查詢

中

華

民

國

99 年

9 月

2 日

1

(3)

行政院國家科學委員會補助專題研究計畫結案報告

計畫編號

:NSC 98-2115-M-009-002-

計畫類別

:

個別型計畫

執行期間

: 98

年

8 月

1 日至

99 年

7 月

31 日

計畫名稱

:

具封閉性質的距離正則圖之研究

(3/3)

計畫主持人

:

翁志文

(

國立交通大學應用數學系

)

weng@math.nctu.edu.tw 計畫參與人員: 博士生:黃喻培、黃皜文。碩士生:劉家安、葉彬、林育生、陳建文、林志嘉、劉侖欣、洪湧昇、施政成。 99年9 月3日

一、中文摘要

令Γ為一直徑為D,相交參數a2> a1= 0的距離正則圖且假設Γ無長度為d + 1的平行四邊形,此處1 ≤ d ≤ D − 1。我們證明Γ具d-封閉性。利用此結果,我們證明不存在D ≥ 4, 具古典參數(D, b, α, β) = (D, −2, −2, ((−2)D+1_{− 1)/3)}_且 b < −1的距離正則圖。關鍵詞:距離正則圖、古典參數、平行四邊形、d-封閉性

二、英文摘要

Let Γ denote a distance-regular graph with diameter D and intersection numbers a2 > a1= 0. We show that

for each 1 ≤ d ≤ D − 1, if Γ contains no parallelograms of lengths up to d + 1 then Γ is d-bounded in the sense of the article [D-bounded distance-regular graphs, Eu-ropean Journal of Combinatorics(1997)18, 211-229]. By applying this result we show the nonexistence of distance-regular graphs with classical parameters (D, b, α, β) = (D, −2, −2, ((−2)D+1_{− 1)/3) for any D ≥ 4. In the end,}

we survey the progress on the classification of distance-regular graph with classical parameters (D, b, α, β) and b < −1.

Keywords: Distance-regular graph, classical parame-ters, parallelogram, D-bounded.

三、緣由與目的

The D-bounded distance-regular graphs were intro-duced in 1997[22] by the project investigator. This be-came an important concept in the classification of classi-cal distance-regular graphs of negative type in 1999[23]. There are many interesting geometric structures con-structed from a D-bounded graph that need to be in-vestigated. They also have applications to pooling de-signs [7, 5, 6]. Several authors also devote themselves to the study of -bounded distance-regular graphs as results shown in [25, 26, 27, 28, 17]. This lures the project inves-tigator going back to this line of study.

四、結果與討論

Let Γ = (X, R) be a distance-regular graph with di-ameter D ≥ 3. Recall that a sequence x, z, y of vertices of Γ is geodetic whenever

∂(x, z) + ∂(z, y) = ∂(x, y),

where ∂ is the distance function of Γ. A sequence x, z, y of vertices of Γ is weak-geodetic whenever

(4)

Definition 0.1. A subset ∆ ⊆ X is weak-geodetically closed if for any weak-geodetic sequence x, z, y of Γ,

x, y ∈ ∆ =⇒ z ∈ ∆.

Weak-geodetically closed subgraphs are called strongly closed subgraphs in [15]. If a weak-geodetically closed subgraph ∆ of diameter d is regular then it has valency ad+ cd= b0− bd, where ad, cd, b0, bdare intersection

num-bers of Γ. Furthermore ∆ is distance-regular with inter-section numbers ai(∆) = ai(Γ) and ci(∆) = ci(Γ) for

1 ≤ i ≤ d [21, Theorem 4.5].

Definition 0.2. Γ is said to be i-bounded whenever for all x, y ∈ X with ∂(x, y) ≤ i, there is a regular weak-geodetically closed subgraph of diameter ∂(x, y) which contains x and y.

Note that a (D − 1)-bounded distance-regular graph is clear to be D-bounded. The properties of D-bounded distance-regular graphs were studied in [22], and these properties were used in the classification of classical distance-regular graphs of negative type [23]. Before stat-ing our main result we make one more definition.

By a parallelogram of length i, we mean a 4-tuple xyzw consisting of vertices of Γ such that ∂(x, y) = ∂(z, w) = 1, ∂(x, w) = i, and ∂(x, z) = ∂(y, w) = ∂(y, z) = i − 1. The following theorem is our main result.

Theorem 0.3. Let Γ denote a distance-regular graph with diameter D ≥ 3, and intersection numbers a1 = 0,

a2 6= 0. Fix an integer 1 ≤ d ≤ D − 1 and suppose that

Γ contains no parallelograms of any length up to d + 1. Then Γ is d-bounded.

Theorem 0.3 answers the problem proposed in [21, p. 299]. Many previous results deal with its complement case a16= 0, for examples under an additional assumption

c2 > 1 [21] and under the assumptions a2> a1 > c2 = 1

[16]. More precisely, for the case under the assumptions a2 > a1 and c2 = 1, H. Suzuki proves the case d = 2 in

Theorem 0.3 [16]; in particular Γ contains a regular weak-geodetically closed subgraph Ω of diameter 2. Since the Friendship Theorem [24, Theorem 8.6.39] asserts no such Ω in the case a1= c2= 1, there must be no such

distance-regular graph Γ with a2 > a1 = c2 = 1 and Γ contains

no parallelograms of length 3. Note that the assumption a1 6= 0 implies a2 6= 0 [2, Proposition 5.5.1(i)]. Hence

Theorem 0.3 is also true under the weaker assumptions b1 > b2 and a2 6= 0 (without assuming a1 = 0). Our

method in proving Theorem 0.3 also works for the case b1 > b2 and a2 6= 0 after a slight modification, but we

decide not to duplicate the previous works.

On the other hand we suppose that Γ is d-bounded for d ≥ 2. Let Ω ⊆ ∆ be two regular weak-geodetically closed subgraphs of diameters 1, 2 respectively. Since Ω and ∆ have different valency b0− b1and b0− b2 respectively, we

have b1 > b2. It is also easy to see that Γ contains no

parallelograms of any length up to d + 1 [21, Lemma 6.5]. With these comments, Theorem 0.3 is the final step in the following characterization of d-bounded distance-regular graphs in terms of forbidden parallelograms.

Theorem 0.4. Let Γ denote a distance-regular graph with diameter D ≥ 3. Suppose the intersection number a26= 0. Fix an integer 2 ≤ d ≤ D − 1. Then the following

two conditions (i), (ii) are equivalent: (i) Γ is d-bounded.

(ii) Γ contains no parallelograms of any length up to d + 1 and b1> b2.

Theorem 0.3 is a generalization of [2, Lemma 4.3.13], [13], and is also proved under an additional assumption c2> 1 by A. Hiraki [4]. To prove Theorem 0.3, we need

many previous results of [4]. These will be stated inde-pendently in Section 2. Some applications of Theorem 0.3

(5)

were previously given in [4], [14]. The following is a new application of Theorem 0.3.

Theorem 0.5. There is no distance-regular graph with classical parameters (D, b, α, β) = (D, −2, −2, ((−2)D+1_{− 1)/3), where D ≥ 4.}

A consequence of Theorem 0.5 is the following.

Corollary 0.6. Let Γ denote a distance-regular graph

with classical parameters (D, b, α, β), D ≥ 4 and c2 = 1.

Then a2= a1 and a16= 0.

We prove Theorem 0.3 in Section 3, and prove The-orem 0.5, Corollary 0.6 in Section 4. We survey the progress on the classification of distance-regular graph with classical parameters (D, b, α, β) and b < −1 in Sec-tion 5.

五、附錄

(含出席國際會議報告)

1 Preliminaries

In this section we review some definitions, basic concepts and some previous results concerning distance-regular graphs. See Bannai and Ito [1] or Terwilliger [18] for more background information.

Let Γ=(X, R) denote a finite undirected, connected graph without loops or multiple edges with vertex set X, edge set R, distance function ∂, and diameter D:=max{ ∂(x, y) | x, y ∈ X}. By a pentagon, we mean a 5-tuple u1u2u3u4u5 consisting of distinct vertices in Γ such that ∂(ui, ui+1) = 1 for 1 ≤ i ≤ 4 and ∂(u5, u1) = 1.

For a vertex x ∈ X and an integer 0 ≤ i ≤ D, set Γi(x) := { z ∈ X | ∂(x, z) = i}. The valency k(x) of a vertex

x ∈ X is the cardinality of Γ1(x). The graph Γ is called regular (with valency k) if each vertex in X has valency

k. A graph Γ is said to be distance-regular whenever for all integers 0 ≤ h, i, j ≤ D, and all vertices x, y ∈ X with ∂(x, y) = h, the number

ph_ij = |Γi(x) ∩ Γj(y)|

is independent of x, y. The constants ph

ij are known as the intersection numbers of Γ.

From now on let Γ = (X, R) be a distance-regular graph with diameter D ≥ 3. For two vertices x, y ∈ X, with ∂(x, y) = i, set B(x, y) := Γ1(x) ∩ Γi+1(y), C(x, y) := Γ1(x) ∩ Γi−1(y), A(x, y) := Γ1(x) ∩ Γi(y). Note that |B(x, y)| = pi_{1 i+1}, |C(x, y)| = pi_{1 i−1}, |A(x, y)| = pi1 i

(6)

are independent of x, y. For convenience, set ci := pi1 i−1 for 1 ≤ i ≤ D, ai := pi1 i for 0 ≤ i ≤ D, bi := pi1 i+1 for

0 ≤ i ≤ D − 1 and put bD:= 0, c0:= 0, k := b0. Note that k is the valency of Γ. It is immediate from the definition

of ph

ij that bi6= 0 for 0 ≤ i ≤ D − 1 and ci6= 0 for 1 ≤ i ≤ D. Moreover

k = ai+ bi+ ci for 0 ≤ i ≤ D. (1.1)

A subset Ω of X is weak-geodetically closed with respect to a vertex x ∈ Ω if

C(y, x) ⊆ Ω and A(y, x) ⊆ Ω for all y ∈ Ω. (1.2)

Note that Ω is weak-geodetically closed if and only if for any vertex x ∈ Ω, Ω is weak-geodetically closed with respect to x [21, Lemma 2.3]. We list a few results which will be used later in this paper.

Theorem 1.1. ([21, Theorem 4.6]) Let Γ be a distance-regular graph with diameter D ≥ 3. Let Ω be a regular subgraph of Γ with valency γ and set d := min{i | γ ≤ ci+ ai}. Then the following (i),(ii) are equivalent.

(i) Ω is weak-geodetically closed with respect to at least one vertex x ∈ Ω.

(ii) Ω is weak-geodetically closed with diameter d.

In this case γ = cd+ ad.

Definition 1.2. Fix a vertex x ∈ X. A pentagon u1u2u3u4u5has shape i1, i2, i3, i4, i5with respect to x if ij = ∂(x, uj)

for 1 ≤ j ≤ 5.

Theorem 1.3. ([21, Lemma 6.9],[16, Lemma 4.1]) Let Γ be a distance-regular graph with diameter D ≥ 3. Suppose a1= 0, a2 6= 0 and Γ contains no parallelograms of length up to d + 1 for some integer d ≥ 2. Let x be a vertex of

Γ, and let u1u2u3u4u5 be a pentagon of Γ such that ∂(x, u1) = i − 1 and ∂(x, u3) = i + 1 for 1 ≤ i ≤ d. Then the

pentagon u1u2u3u4u5 has shape i − 1, i, i + 1, i + 1, i with respect to x.

2 A few lemmas

Throughout this section, let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 3, and intersection numbers a1= 0, a26= 0. Such graphs are also studied in [4, 11, 12, 13, 14]. Note that any two vertices at distance 2

are always contained in a pentagon since a26= 0, and two nonconsecutive vertices in a pentagon of Γ have distance

2 since a1 = 0. In this section we give a few lemmas which will be used in the next section. These results were

formulated by A. Hiraki in [4] under an additional assumption c2> 1, but this assumption is essentially not used in

his proofs. For the completeness, we still provide the proofs.

Lemma 2.1. Fix an integer 1 ≤ d ≤ D − 1, and suppose Γ does not contain parallelograms of length up to d + 1. Then for any two vertices z, z0∈ X such that ∂(x, z) ≤ d and z0_{∈ A(z, x), we have B(x, z) = B(x, z}0_).

(7)

Proof. By symmetry, it suffices to show B(x, z) ⊆ B(x, z0_{). Suppose there exists w ∈ B(x, z) \ B(x, z}0_{). Then}

∂(w, z0) 6= ∂(x, z) + 1. Note that ∂(w, z0) ≤ ∂(w, x) + ∂(x, z0) = 1 + ∂(x, z) and ∂(w, z0) ≥ ∂(w, z) − ∂(z, z0) = ∂(x, z). This implies ∂(w, z0) = ∂(x, z) and wxz0z forms a parallelogram of length ∂(x, z) + 1, a contradiction.

Lemma 2.2. Fix integers 1 ≤ i ≤ d ≤ D − 1, and suppose Γ does not contain parallelograms of any length up to d + 1. Let x be a vertex of Γ. Then there is no pentagon of shape i, i, i, i, i + 1 with respect to x for 1 ≤ i ≤ d.

Proof. Let u1u2u3u4u5be a pentagon of shape i, i, i, i, i + 1 with respect to x. We derive a contradiction by induction

on i. The case i = 1 is impossible otherwise u1xu2u3 is a parallelogram of length 2. Suppose i ≥ 2. Note that

B(x, u1) = B(x, u2) = B(x, u3) = B(x, u4) by Lemma 2.1. We shall prove C(x, u1) = C(x, u2) = C(x, u3) = C(x, u4).

First we prove C(x, u1) = C(x, u2). It suffices to show C(x, u2) ⊆ C(x, u1) since both sets have the same size ci. To the

contrary suppose there exists v ∈ C(x, u2)−C(x, u1). Note that v ∈ A(x, u1) as B(x, u1) = B(x, u2). Then B(u1, x) =

B(u1, v) by Lemma 2.1 and hence ∂(v, u5) = i + 1 since u5∈ B(u1, x). Now u2u1u5u4u3has shape i − 1, i, i + 1, i + 1, i

with respect to v by Theorem 1.3, a contradiction since v 6∈ B(x, u4). This proves C(x, u2) ⊆ C(x, u1) as desired.

By symmetry, C(x, u3) = C(x, u4). It remains to show C(x, u2) ⊆ C(x, u4). To the contrary suppose there exists

u ∈ C(x, u2) − C(x, u4). Note that u ∈ A(x, u4) as B(x, u2) = B(x, u4). Then B(u4, x) = B(u4, u) by Lemma 2.1

and hence ∂(u, u5) = i + 1 since u5∈ B(u4, x). Hence u2u1u5u4u3 has shape i − 1, i, i + 1, i + 1, i with respect to u

by Theorem 1.3, a contradiction since u 6∈ B(x, u4). Pick a vertex v ∈ C(x, u1) = C(x, u2) = C(x, u3) = C(x, u4).

Then u1u2u3u4u5 is a pentagon of shape i − 1, i − 1, i − 1, i − 1, i with respect to v, a contradiction to the inductive

hypothesis.

Proposition 2.3. Fix integers 1 ≤ i ≤ d ≤ D − 1, and suppose Γ does not contain parallelograms of any length up to d + 1. Let x be a vertex and u1u2u3u4u5 be a pentagon of shape i, i − 1, i, i − 1, i or of shape i, i − 1, i, i − 1, i − 1

with respect to x for 1 ≤ i ≤ d. Then B(x, u1) = B(x, u3).

Proof. It suffices to show B(x, u3) ⊆ B(x, u1) since both sets have the same size bi. Pick u ∈ B(x, u3). Then

∂(u, u3) = i + 1, ∂(u, u4) = i and ∂(u, u2) = i. Note that ∂(u, u1) 6= i − 1, otherwise by Theorem 1.3, the pentagon

u1u2u3u4u5 has shape i − 1, i, i + 1, i + 1, i with respect to u, a contradiction. Suppose ∂(u, u1) = i for this moment.

Then to avoid obtaining a pentagon u1u2u3u4u5of type i, i, i, i, i + 1 with respect to u we must have ∂(u, u5) = i + 1

by Lemma 2.2. Then ∂(x, u5) = i by construction. Now u5u1xu is a parallelogram of length i + 1, a contradiction.

Hence ∂(u, u1) = i + 1 or equivalently u ∈ B(x, u1). This proves B(x, u3) ⊆ B(x, u1) as desired.

Lemma 2.4. Fix integers 1 ≤ i ≤ d ≤ D − 1, and suppose Γ does not contain parallelograms of any length up to d + 1. Let x be a vertex. Then there is no pentagon of shape i, i, i, i + 1, i + 1 with respect to x for 1 ≤ i ≤ d.

Proof. Suppose that u2u3u4u5u1is a pentagon of shape i, i, i, i+1, i+1 with respect to x. We derive a contradiction by

induction on i. The case i = 1 is impossible otherwise u2xu3u4is a parallelogram of length 2. Suppose i ≥ 2. Pick v ∈

C(x, u2) and note that ∂(v, u1) = i by construction. In particular v 6∈ B(x, u2) and B(x, u2) = B(x, u3) = B(x, u4)

by Lemma 2.1, so v ∈ C(x, u4) ∪ A(x, u4). In fact v ∈ C(x, u4); otherwise ∂(v, u4) = i, ∂(v, u5) = i by Theorem 1.3,

(8)

that ∂(v, u3) = i; otherwise ∂(v, u3) = i − 1 and u2u3u4u5u1is a pentagon of shape i − 1, i − 1, i − 1, i, i with respect

to v, a contradiction to inductive hypothesis. Now as x = v in Proposition 2.3, we have B(v, u1) = B(v, u3), a

contradiction since x ∈ B(v, u1) − B(v, u3).

3 Proof of Theorem 0.3

Let Γ = (X, R) denote a distance-regular graph with intersection numbers a1= 0, a26= 0 and diameter D ≥ 3. Fix

an integer 1 ≤ d ≤ D − 1. Suppose Γ contains no parallelograms of length up to d + 1. We shall prove Γ is d-bounded in this section. We first give a definition.

Definition 3.1. For any vertex x ∈ X and any subset Π ⊆ X, define [x, Π] to be the set

{v ∈ X | there exists y0 ∈ Π, such that the sequence x, v, y0 is geodetic }.

For any x, y ∈ X with ∂(x, y) = d, set

Πxy:= {y0∈ Γd(x) | B(x, y) = B(x, y0)} (3.1)

and

∆(x, y) = [x, Πxy]. (3.2)

We shall prove that for any vertices x, y ∈ X with ∂(x, y) = d the following statement Bd holds.

(Bd) ∆(x, y) is regular weak-geodetically closed with valency ad+ cd.

By referring to Theorem 1.1, (Bd) is equivalent to the following statements (Wd) and (Rd).

(Wd) ∆(x, y) is weak-geodetically closed with respect to x, and

(Rd) the subgraph induced on ∆(x, y) is regular with valency ad+ cd

for any vertices x, y ∈ X with ∂(x, y) = d.

We prove (Wd) and (Rd) by induction on d. Since a1 = 0, there is no edges in Γ1(x) for any vertex x ∈ X.

If d = 1 in Definition 3.1, then Πxy = {y}, and consequently ∆(x, y) = {x, y} is an edge; in particular ∆(x, y) is

regular with valency 1 = a1+ c1 and is weak-geodetically closed with respect to x since a1 = 0. This proves (R1)

and (W1). We now assume d ≥ 2. By inductive hypothesis (Wi), (Ri) and (Bi) are assumed throughout this section

for 1 ≤ i ≤ d − 1. The following proposition proves the statement (Wd).

Proposition 3.2. For any vertices x, y ∈ X with ∂(x, y) = d and for any vertex z ∈ ∆(x, y)∩Γi(x), where 1 ≤ i ≤ d,

we have the following (i), (ii).

(i) A(z, x) ⊆ ∆(x, y).

(ii) For any vertex w ∈ Γi(x) ∩ Γ2(z) with B(x, w) = B(x, z), we have w ∈ ∆(x, y).

(9)

In particular (Wd) holds.

Proof. We prove (i), (ii) by induction on d−i. The case i = d follows from the construction of ∆(x, y) in Definition 3.1 and by Lemma 2.1. Suppose i < d.

To prove (i) we note that if i = 1 then A(z, x) is an empty set, clearly contained in ∆(x, y). Hence we suppose 2 ≤ i < d in this case. We pick a vertex v ∈ A(z, x) and show v ∈ ∆(x, y). Pick u ∈ ∆(x, y) ∩ Γi+1(x) ∩ Γ1(z).

Note that (i), (ii) hold if we use u to replace z by inductive hypothesis. Let uu2u3vz be a pentagon of Γ for some

u2, u3∈ X. Note that uu2u3vz can not have shape i + 1, i, i − 1, i, i, shape i + 1, i + 2, i + 1, i, i by Theorem 1.3, can

not have shape i + 1, i, i, i, i by Lemma 2.2, and can not have shape i + 1, i + 1, i, i, i by Lemma 2.4 with respect to x. Hence uu2u3vz has shape i + 1, i + 1, i + 1, i, i or i + 1, i, i + 1, i, i with respect to x. In the first case we have

u2 ∈ A(u, x), u3 ∈ A(u2, x), and this implies u2, u3 ∈ ∆(x, y) by the inductive hypothesis of (i), and v ∈ ∆(x, y)

by construction. In the latter case we have B(x, u) = B(x, u3) by Lemma 2.3, and consequently u3 ∈ ∆(x, y) by

inductive hypothesis of (ii), v ∈ ∆(x, y) by construction.

To prove (ii) let zv2wv4v5be a pentagon for some v2, v4, v5∈ X. Note that ∆(x, z) is a regular weak-geodetically

closed subgraph of diameter i by (Bi), and ∆(x, z) = ∆(x, w) by construction in Definition 3.1 and since B(x, w) =

B(x, z); in particular v2, v4, v5 ∈ ∆(x, z) and v2, v4, v5 6∈ Γi+1(x). If v2 ∈ A(z, x) then v2, w ∈ ∆(x, y) by (i) that

we just proved. Hence we assume zv2wv4v5 has shape i, i − 1, i, a, b with respect to x for integers a, b ∈ {i − 1, i}.

Pick u ∈ ∆(x, y) ∩ Γi+1(x) ∩ Γ1(z). Let v2zuy3y4 be a pentagon for some y3, y4 ∈ X. Then v2zuy3y4 has shape

i − 1, i, i + 1, i + 1, i with respect to x by Theorem 1.3. Let v2y4w3w4w be a pentagon for some w3, w4 ∈ X. If

w4 ∈ A(w, x) ∪ C(w, x) then w4 ∈ ∆(x, w) = ∆(x, z) and this forces y4 ∈ ∆(x, z) as v2, w4∈ ∆(x, z) and by (Bi).

By the same reason we have y3 ∈ ∆(x, z) as z, y4 ∈ ∆(x, z). We have a contradiction since ∆(x, z) has diameter i

and ∂(x, y3) = i + 1 > i = diam ∆(x, z). Hence ∂(x, w4) = i + 1 and v2y4w3w4w has shape i − 1, i, i + 1, i + 1, i

with respect to x by Theorem 1.3 as shown in Figure 1. Note that B(x, u) = B(x, y3) and B(x, w3) = B(x, w4)

by Lemma 2.1. If B(x, y3) = B(x, w3) then by (i) and the inductive hypothesis of (ii) we have y3, w3, w4∈ ∆(x, y)

in the order, and w ∈ ∆(x, y) by the construction in (3.2) to complete the proof. Suppose B(x, y3) 6= B(x, w3) in

the remaining. Let y4y3p3p4w3 be a pentagon for some p3, p4 ∈ X. By Lemma 2.1, Lemma 2.3 and Theorem 1.3,

the pentagon y4y3p3p4w3 has shape i, i + 1, i + 2, i + 2, i + 1 with respect to x. Now we have three pentagons and

their shapes with respect to x as shown in Figure 1. Note that B(x, y4) 6= B(x, z), otherwise ∆(x, y4) = ∆(x, z)

and y3∈ ∆(x, z), a contradiction as before. Pick p ∈ B(x, y4) − B(x, z). Then ∂(p, y4) = i + 1 and ∂(p, z) = i − 1

or i. Suppose for this moment ∂(p, z) = i − 1. Then zuy3y4v2 is a pentagon of shape i − 1, i, i + 1, i + 1, i with

respect to p by Theorem 1.3. Note that ∂(p, p3) = i + 2, otherwise ∂(p, p3) = i + 1 and xpy3p3 is a parallelogram

of length i + 2 ≤ d + 1, a contradiction. Now by applying Lemma 2.2, Lemma 2.4, we have ∂(p, w3) = i + 2 and

consequently v2y4w3w4w is a pentagon of shape i, i + 1, i + 2, i + 2, i + 1 with respect to p by Theorem 1.3. That is

p ∈ B(x, w), a contradiction to B(x, z) = B(x, w). By symmetry, we also have ∂(p, w) 6= i − 1. We suppose in the last case ∂(p, z) = ∂(p, w) = i. As p ∈ A(x, z), we have B(z, x) = B(z, p) by Lemma 2.1, in particular ∂(p, u) = i + 1. By symmetry, ∂(p, w4) = i + 1. As p 6∈ B(x, u) = B(x, y3), we have ∂(p, y3) = i or i + 1. We shall prove ∂(p, y3) = i,

(10)

∂(p, p3) = i + 2. Applying Lemma 2.2, Lemma 2.4 to the pentagon w3y4y3p3p4and considering its shape with respect

to p, we find ∂(p, w3) 6= i + 1, and applying Theorem 1.3 to find ∂(p, w3) 6= i. Now ∂(p, w3) = i + 2 and pxw4w3

is a parallelogram of length i + 2 ≤ d + 1, a contradiction. We conclude that y4y3p3p4w3 is a pentagon of shape

i + 1, i, a, i + 1, i or of shape i + 1, i, i + 1, b, i with respect to p for a, b ∈ {i, i + 1} by Lemma 1.3, Lemma 2.2, and this implies x ∈ B(p, p4) = B(p, y4) in the first case or x ∈ B(p, p3) = B(p, y4) in the latter case by Proposition 2.3,

a contradiction since x ∈ C(p, y4).

∆(x, y) is clear to be weak-geodetically closed with respect to x by (1.2) and (i).

r r r r r r r r r r r r p p p p p p p p p p p p x p v2 w y4 z w4 w3 y3 u p4 p3 distance to x 0 i − 1 i i + 1 i + 2

Figure 1. Three pentagons in the proof of Proposition 3.2(ii).

The following proposition proves (Rd) and hence completes the proof of Theorem 0.3.

Proposition 3.3. For any vertices x, y ∈ X with ∂(x, y) = d, ∆(x, y) is regular with valency ad+ cd.

Proof. Set ∆ = ∆(x, y). Clearly from the construction and Proposition 3.2, |Γ1(y0) ∩ ∆| = ad+ cd for any y0∈ Πxy.

First we show |Γ1(x) ∩ ∆| = ad+ cd. Note that y ∈ ∆ ∩ Γd(x) by construction of ∆. For any z ∈ C(x, y) ∪ A(x, y),

∂(x, z) + ∂(z, y) ≤ ∂(x, y) + 1.

This implies z ∈ ∆ since ∆ is weak-geodetically closed with respect to x by Proposition 3.2. Hence C(x, y)∪A(x, y) ⊆ ∆. Suppose B(x, y)∩∆ 6= ∅. Choose t ∈ B(x, y)∩∆. Then there exists y0∈ Πxysuch that t ∈ C(x, y0), a contradiction

to B(x, y) = B(x, y0). Hence B(x, y) ∩ ∆ = ∅ and Γ1(x) ∩ ∆ = C(x, y) ∪ A(x, y). This proves |Γ1(x) ∩ ∆| = ad+ cd.

(11)

Since each vertex in ∆ appears in a sequence of vertices x = x0, x1, . . . , xdin ∆, where ∂(x, xj) = j, ∂(xj−1, xj) = 1

for 1 ≤ j ≤ d, and xd∈ Πxy, it suffices to show

|Γ1(xi) ∩ ∆| = ad+ cd (3.3)

for 1 ≤ i ≤ d − 1. For each integer 1 ≤ i ≤ d, we show

|Γ1(xi−1) \ ∆| ≤ |Γ1(xi) \ ∆| (3.4)

by the 2-way counting of the number of the pairs (z, s) for z ∈ Γ1(xi−1) \ ∆, s ∈ Γ1(xi) \ ∆ and ∂(z, s) = 2. For a

fixed s ∈ Γ1(xi) \ ∆, we have ∂(x, s) = i + 1 and ∂(xi−1, s) = 2 since ∆ is weak-geodetically closed with respect to x

by Proposition 3.2. Hence z ∈ A(xi−1, s). The number of such pairs (z, s) is at most |Γ1(xi) \ ∆|a2.

On the other hand, we show this number is |Γ1(xi−1)\∆|a2exactly. Fix an z ∈ Γ1(xi−1)\∆. Note that ∂(x, z) = i

by Proposition 3.2, and ∂(xi, z) = 2 since a1 = 0. Pick any s ∈ A(xi, z). We shall prove s 6∈ ∆. Suppose to the

contrary s ∈ ∆ in the below arguments and choose any w ∈ C(s, z). Note that ∂(x, s) ≤ i, otherwise ∂(x, s) = i + 1 and the pentagon xi−1xiswz has shape i − 1, i, i + 1, i + 1, i with respect to x by Theorem 1.3 to force z ∈ ∆ by

Proposition 3.2(i) and construction of ∆, a contradiction. Similarly ∂(x, w) ≤ i. If s ∈ A(xi, x), w ∈ A(s, x) and

z ∈ A(w, x), then z ∈ ∆ by Proposition 3.2(i), a contradiction. Applying Proposition 2.3 in the remaining cases we have B(x, z) = B(x, xi) and then z ∈ ∆ by Proposition 3.2(ii), a contradiction.

From the above counting, we have

|Γ1(xi−1) \ ∆|a2≤ |Γ1(xi) \ ∆|a2 (3.5)

for 1 ≤ i ≤ d. Eliminating a2 from (3.5), we find (3.4) or equivalently

|Γ1(xi−1) ∩ ∆| ≥ |Γ1(xi) ∩ ∆| (3.6)

for 1 ≤ i ≤ d. We have known previously |Γ1(x0) ∩ ∆| = |Γ1(xd) ∩ ∆| = ad+ cd. Hence (3.3) follows from (3.6).

4 Classical parameters

Let Γ = (X, R) denote a distance-regular graph with diameter D ≥ 3. Γ is said to have classical parameters (D, b, α, β) whenever the intersection numbers of Γ satisfy

ci = i 1 1 + αi − 1 1 for 0 ≤ i ≤ D, (4.1) bi = D 1 − i 1 β − α i 1 for 0 ≤ i ≤ D, (4.2) where i 1 := 1 + b + b2+ · · · + bi−1. (4.3)

(12)

Applying (1.1) with (4.1), (4.2), we have ai = i 1 β − 1 + α(D 1 − i 1 −i − 1 1 ) (4.4) = i 1 a1− α( i 1 +i − 1 1 − 1) (4.5) for 1 ≤ i ≤ D.

Suppose Γ has classical parameters (D, b, α, β) and D ≥ 3. Then b is an integer, b 6= 0 and b 6= −1 [2, p. 195]. To apply Theorem 0.4 we need the following lemma.

Lemma 4.1. ([19, Theorem 2.12], [21, Lemma 7.3(ii)]) Let Γ denote a distance-regular graph with classical param-eters (D, b, α, β), b < −1 and D ≥ 3. Then Γ contains no parallelograms of any length.

More general version of Lemma 4.1 can be found in [20, 11, 12].

Theorem 4.2. ([22, Theorem 4.2]) Let Γ denote a distance-regular graph with classical parameters (D, b, α, β) and b < −1. Suppose that Γ is D-bounded with D ≥ 4. Then

β = α1 + b

D

1 − b . (4.6)

Proof of Theorem 0.5. Let Γ denote a distance-regular graph with classical parameters (D, b, α, β) = (D, −2, −2, ((−2)D+1− 1)/3), where D ≥ 4. Then Γ contains no parallelograms of any length by Lemma 4.1. By (4.1), (4.4), we have c2= 1

and a2= 2 > 0 = a1. Hence Γ is D-bounded by Theorem 0.4 and since b1> b2. By (4.6), β = ((−2)D+1− 2)/3), a

contradiction.

We quote a few previous results in the study of distance-regular graphs with classical parameters and c2= 1 for

later use.

Lemma 4.3. ([22, Corollary 6.3]) There is no distance-regular graph Γ with classical parameters (D, b, α, β), D ≥ 4, c2= 1 and a2> a1> 1.

Lemma 4.4. ([14, Theorem 2.2]) Let Γ denote a distance-regular graph with classical parameters (D, b, α, β) and D ≥ 3. Assume the intersection numbers a1= 0, a26= 0, and c2= 1. Then (b, α, β) = (−2, −2, ((−2)D+1− 1)/3).

Lemma 4.5. ([19, Theorem 2.11], [21, Lemma 7.3(ii)]) Let Γ denote a distance-regular graph with classical param-eters (D, b, α, β) and D ≥ 3. Suppose Γ contains no parallelograms of length 2. Then Γ contains no parallelograms of

any length.

Proof of Corollary 0.6. Since c2= 1, Γ contains no parallelograms of length 2 and then contains no parallelogram

of any length by Lemma 4.5. By Lemma 4.3, Lemma 4.4, Theorem 0.5, only the case a2 > a1 = 1 and the case

a2 = a1 remain. The first case is impossible by Friendship Theorem as mentioned in the introduction. For the

(13)

latter case, we have α = −b/(1 + b) since c2 = 1 and by (4.1). Applying this to (4.5) we find the impossibility of

a2= a1= 0.

We close this section by proposing the following conjecture.

Conjecture 4.6. There is no distance-regular graph Γ with classical parameters (D, b, α, β), D ≥ 4, and c2= 1.

There is a mistake in [2, Proposition 6.1.2] which proves the above conjecture. This mistake is corrected in [3].

Remark 4.7. (See [2, p. 194]) The Triality graph 3_D

4,2(q) is a distance-regular graph with classical parameters

(3, −q, q/(1 − q), q2+ q), c2 = 1 and a1 = a2 = q − 1. Hence the assumption D ≥ 4 in Conjecture 4.6 is necessary.

Note that the Triality graph3D4,2(q) is not 3-bounded by Theorem 0.4 since b1= b2.

5 Classical parameters with b < −1

Let Γ = (X, R) denote a distance-regular graph with classical parameters (D, b, α, β), b < −1 and D ≥ 3. We survey the progress on the classification of such Γ in this section. Two main classes of such examples are the dual polar graphs 2A2D−1(−b) and the Hermitian forms graphs Her−b(D) as listed in [2, Tabel 6.1]. A.A. Ivanov and S.V.

Shpectorov show that if Γ has the same intersection numbers as the dual polar graph 2A2D−1(−b) then Γ is the

dual polar graph2_A

2D−1(−b) [8]. They also show that if Γ does not contain parallelograms of length 2 and has the

same intersection numbers as the Hermitian forms graph Her−b(D) then Γ is the Hermitian forms graph Her−b(D)

[9, 10]. P. Terwilliger shows that in fact Γ does not contains parallelograms of any length [19] as also stated in Lemma 4.1. According to different assumptions on the intersection numbers of Γ, the D-bounded property of Γ are proved by different authors as stated in the introduction. Putting all these results together, if Γ has intersection numbers b1> b2and a26= 0 then Γ is D-bounded as also stated in Theorem 0.4.

We assume b1> b2 and a26= 0 in Γ thereinafter. The third author shows that if D ≥ 4 then

β = α1 + b

D

1 − b (5.1)

in [22] as also stated in (4.6), and use this to conclude in [23] that if Γ is not the dual polar graph2_A

2D−1(−b) and

not the Hermitian forms graph Her−b(D) then

α = (b − 1)/2, β = −(1 + bD)/2, (5.2)

where −b is a power of an odd prime.

There are some results of Γ in the assumption D ≥ 3, a1= 0 and a26= 0. For example in [14], the second author

and the third author show that c2≤ 2, and in the case c2= 1, it must be

(14)

Note that if D ≥ 4, (5.3) does not hold by (5.2). This is essentially the proof of Theorem 0.5. Hence we have the following conjecture about the case D = 3.

Conjecture 5.1. There is no distance-regular graphs with classical parameters (D, b, α, β) = (3, −2, −2, 5).

Also in [4] A. Hiraki assume that D ≥ 3, a1 = 0, a26= 0, c2 > 1 and show that Γ is either the Hermitian forms

graph Her2(D) or α, β satisfy (5.2) with b = −3. Hence the following conjecture is the first step to study the

unknown case of (5.2).

Conjecture 5.2. There is no distance-regular graph with classical parameters (D, b, α, β) = (3, −3, −2, 13).

References

[1] E. Bannai and T. Ito, Algebraic Combinatorics I: Association Schemes, Benjamin/Cummings, Menlo Park, 1984.

[2] A.E. Brouwer, A.M. Cohen, and A. Neumaier, Distance-Regular Graphs, Springer-Verlag, Berlin, 1989.

[3] A.E. Brouwer, A.M. Cohen, and A. Neumaier, Additions and Corrections to the book of BCN, http://www.win.tue.nl/ aeb/drg/index.html

[4] A. Hiraki, Distance-regular graphs with c2> 1 and a1= 0 < a2, Graphs Combin. 25(1)(2009), 65-79.

[5] Tayuan Huang, Kaishun Wang and Chih-wen Weng, Pooling spaces associated with finite geometry, European Journal of Combinatorics, 29(2008) 1483-1491

[6] Tayuan Huang, Kaishun Wang and Chih-wen Weng, A Class of Error-Correcting Pooling Designs over Com-plexes, J Comb Optim, (2010) 19: 486-491 (doi 10.10071 s10878-008-9179-4)

[7] Tayuan Huang and Chih-wen Weng, Pooling Spaces and Non-Adaptive Pooling Designs, Discrete Mathematics, 282(1-3), 163-169, 2004

[8] A.A. Ivanov and S.V. Shpectorov, The association schemes of dual polar spaces of type2_A

2d−1(pf) are

charac-terized by their parameters if d ≥ 3, Linear Algebra Appl., 144/155(1989), 133–139.

[9] A.A. Ivanov and S.V. Shpectorov, Characterization of the association schemes of Hermitian forms over GF (22), Geometriae Dedicata, 30(1989), 23–33.

[10] A.A. Ivanov and S.V. Shpectorov, A characterization of the association schemes of Hermitian forms, J. Math. Soc. Japan, 43(1)(1991), 25-48.

[11] Y. Liang, and C. Weng, Parallelogram-free distance-regular graphs, J. Combin. Theory Ser. B, 71(2)(1997), 231–243.

(15)

[12] Y. Pan, M. Lu, and C. Weng, Triangle-free distance-regular graphs, J. Algebraic Combin., 27(2008), 23–34.

[13] Y. Pan and C. Weng, Three bounded properties in triangle-free distance-regular graphs, European J. Combin., 29(2008), 1634–1642.

[14] Y. Pan and C. Weng, A note on triangle-free distance-regular graphs with a2 6= 0, J. Combin. Theory Ser. B,

99(2009), 266–270.

[15] H. Suzuki, On strongly closed subgraphs of highly regular graphs, European J. Combin., 16(1995), 197–220.

[16] H. Suzuki, Strongly closed subgraphs of a distance-regular graph with geometric girth five, Kyushu Journal of Mathematics, 50(2)(1996), 371–384.

[17] Hiroshi Suzuki, On Strongly Closed Subgraphs with Diameter Two and Q-Polynomial Property, preprint, 2004

[18] P. Terwilliger, The subconstituent algebra of an association scheme (Part I), Journal of Algebraic Combinatorics, 1(1992), 363–388.

[19] P. Terwilliger, Kite-free distance-regular graphs, Europ. J. of Combin., 16(4)(1995), 405–414.

[20] C. Weng, Kite-free P - and Q-polynomial schemes, Graphs and Combinatorics, 11(1995), 201-207.

[21] C. Weng, Weak-geodetically closed subgraphs in distance-regular graphs, Graphs and Combinatorics, 14(1998), 275–304.

[22] C. Weng, D-bounded distance-regular graphs, European Journal of Combinatorics, 18(1997), 211–229.

[23] C. Weng, Classical distance-regular graphs of negative type, J. Combin. Theory Ser. B, 76(1999), 93–116.

[24] D. B. West, Introduction to Graph Theory, Prentice Hall, Upper Saddle River, 1996.

[25] Suogang Gao and Jun Guo, A construction of a Distance-regular Graph with Diameter 2, submitted to Graph and Combinatorics

[26] Jun Guo and Suogagng Gao, Lattices Generated by Join of Strongly Closed Subgraphs in d-bounded Distance-regular Graphs, submitted to Discrete Mathematics

[27] Jun Guo, Wen Liu, Suogang Gao, Some counting formulae of strongly closed subgraphs in distance-regular graphs, preprint

[28] Suogang Gao, Jun Guo, Wen Lin, Lattices Generated by Strongly Closed Subgraphs in d-bounded Distance-regular Graphs, preprint

(16)

參加 2009 年釜山國際大學所舉辦的韓日代數與組合國際學術會議結及順

道訪問浦項大學結案報告

時間: 98 年 2 月 9 日- 97 年 2 月 13 日

地點: 韓國釜山、浦項

報告人:翁志文

計劃編號: NSC 97-2115-M-009-002

承蒙國科會的支助，於上述時間、地點參加韓國釜山國際大學所舉辦的韓日代數

與組合國際學術會議。韓國與日本代數組合學者每年兩次輪流舉辦會議，這是第六

屆。此會議報告主題相當廣泛，有組合矩陣理論、結合方案、代數編碼、距離正則

圖、球面設計、代數表示理論及筆者的翻轉迷宮新主題。

筆者於 2/8 由中正機場搭機前釜山，及至 2/13 晚上返回中正機場，此趟旅程前

後達 5 天之久。筆者主要在 2/10 當天早上以 Flipping Puzzle on a simple Graph

為題演講，這個題目是筆者的博士生黃皜文君的研究第二個成果，前一成果去年已於

荷蘭報告。這一方向的演講每次都獲得相當多回應的，上次在荷蘭學到此研究與

orthogonal groups 有關，而此次聽到此研究的前身 sigma games 與 orthogonal

polynomials—Chebyshev polynomials 有關。

2/11 全天是年輕學者的 Kumjung Seminar，當天結束筆者搭 Jack Koolen 教授

的便車到POSTECH 訪問。 Jack Koolen 教授是荷蘭人，比我小四歲，卻在韓國領導一

代數組合團隊，成果豐碩，相當令人佩服。他介紹 Delsarte clique graphs 給我，

我研究的封閉距離這則圖似乎可推廣到此領域。

此次訪問由於機票直接由網路向航空公司購買，報帳麻煩，所以機票部份就自費

處理，剩餘經費暑假計畫參加中國天津所舉辦的兩岸圖論與組合會議之用。

(17)

參加 2010 年浦項科技大學及浦項數學機構聯合舉辦的代數組合與幾何組合國際學術會議結案報告

時間: 99 年 7 月 11 日- 99 年 7 月 15 日

地點: 韓國慶州

報告人:翁志文

計劃編號: NSC 98-2115-M-009-002

承蒙國科會的支助，於上述時間、地點參加韓國浦項科技大學及浦項數學機構聯合舉辦的代數組合與幾何組合

國際學術會議。此次會議除了邀請許多韓日學者外，也有多位遠從荷蘭、比利時及美國的學者。會議報告主題繞著

結合方案為中心，有距離正則圖、球面設計、區族設計、量子數學、值譜理論、代數編碼等。筆者演講在 7/12 下

午，題目為「Pooling designs with d-disjunct property and block weight d+1」，是此計劃所延生應用方面

的主題。參考文末資料。

此次會議認識一位比利時Ghent 大學的博士生，他在有限幾何 hemisystem 上的研究，也出現一類筆者 10 多年

前提出但懸而未決的距離正則圖，這激發筆者想去學 hemisystem。

(18)

ALGEBRAIC AND GEOMETRIC COMBINATORICS CONFERENCE 2010

Pooling designs with d -disjunct property

and block weight d + 1

Chih-wen Weng (

翁志文

)

(with Yu-pei Huang and Wu, Hsin-Jung)

Department of Applied Mathematics, National Chiao Tung University, Taiwan

July 12, 2010

(19)

Definition

An incidence structure (P, B) is called

d -disjunct

if any block in B is not

covered by the union of d other blocks.

Assume P = {1, 2, . . . , v }, B = {B

1

, B

2

, . . . , B

b

} and M is be the

incidence matrix of (P, B), i.e.

M

ij

=

1,

i ∈ B

j

;

0,

i 6∈ B

j

for 1 ≤ i ≤ v and 0 ≤ j ≤ b.

The incidence matrix M of a d -disjunct incidence structure can be used in

non-adaptive group testing programming, in which v << b is preferred.

(20)

1

_{Let M be a v × b incidence matrix of an incidence structure and set}

F

2

= {0, 1}. Define the

output function

o

M

: F

2b

→ F

2v

by

o

M

(P) := M ? P =

[

Pi=1

M

i

,

where ? is the matrix product by using Boolean sum to replace

addition.

2

If the incidence structure is d -disjunct, then o

_M

F

b

2

(≤ d ) is known

to be injective, where F

₂b

(≤ d ) is the set of binary vectors of length b

and Hamming weight at most d .

3

_{This means that for each element u in the image of o}

_M

_{on F}

b

2

(≤ d ),

we know which P ∈ F

₂b

to have o

M

(P) = u.

4

_{In application, P is interpreted as the unknown infected subset}

{j | P

_j

= 1} of a given set of b items, and u is interpreted as the

sequence of test results. Then the injective property of o

M

implies

that the infected subset can be determined from the sequence of test

results if the number of infected items is known in advance to be at

most d .

(21)

Example

The following 4 × 6 binary matrix is used to detect the infected item in

{1, 2, 3, 4, 5, 6}, if the infected item is known to be at most one in advance

(but do not know which one):







Tests/Items

| 1 2

3

4

5

6 o

M

((0, 0, 1, 0, 0, 0)

T

)

one

| 1 1 1 0 0 0 →

1 Two

| 1 0 0 1 1 0 →

0 Three

| 0 1 0 1 0 1 →

0 Four

| 0 0 1 0 1 1 →

1 





If there are two infected items, the above 4 × 6 matrix does not work for

detecting them. For example, both the infected sets {

3, 4

} and {

1, 6

} have

the same output (1, 1, 1, 1)

T

. So it is impossible to recover the infected set

from the output (1, 1, 1, 1)

T

.

(22)

Relation to t-design

Definition

An incidence structure (P, B) is called a

t-(v , k, λ) design

if

1

|P| = v ,

2

|B| = k for and B ∈ B, and

3

any t-subset of P is contained in exactly λ blocks in B.

Remark

1

_{A 2-(v , k, 1) design is (k − 1)-disjunct since a block has k points and}

it intersects another block in at most one point, so k − 1 other blocks

can cover at most k − 1 points of a block, leaving at least one point

uncovered.

2

_{If any point is incidence in at least two blocks, then any block in a}

d -disjunct matrix has size at least d + 1.

3

_{A d -disjunct incidence structure is called a}

_{pooling design.}

(23)

First result

Theorem

Let (P, B) be a d -disjunct pooling design with constant block size d + 1,

and define v = |P| and b = |B|. Then b ≤ max{v (v − 1)/d (d + 1), v − d }.

Moreover if v − d ≤ v (v − 1)/d (d + 1), then the above upper bound of b

is reached if and only if (P, B) is a 2-(v , d + 1, 1) design.

The v × b incidence matrix

M =

I

b

J

d

satisfies the equality b = v − d , where I

b

is the b × b identity matrix and

(24)

The following example gives the equality in previous theorem for d = q − 1.

Example

(2 − (q

2

, q, 1) design)

Let q be a prime power. The affine plane F

_q2

over

2

+ q and w = q = d + 1 satisfy

b = q

2

+ q = v (v − 1)/d (d + 1).

The first q which is not a prime power is when q = 6 = d + 1. In this case

the equality does not hold by the Bruck-Ryser-Chowla Theorem. Then

there is no 5-disjunct pooling design with 36 points, 42 blocks and

constant bock size 6. We will construct a 5-disjunct pooling design with

36 points, 37 blocks and constant block size 6.

(25)

Forward difference property

1

_{Let q be a prime power and}

_{m ≥ q}

_{be an integer.}

2

_{Let F}

_q

_{:= {0, a}

0

_{, a}

1

_{, . . . , a}

q−2

} denote the finite field of q elements,

where a is a generator of the cyclic multiplication group

F

_q∗

:= F

q

− {0}.

3

Let m ≥ q be an integer. Let Z

_m

:= {0, 1, . . . , m − 1} be the addition

group of integers modulo m. We use the order of integers to order the

elements in Z

m

, e.g. 0 < 1.

4

_{A subset T ⊆ Z}

_m

× F

_q

_{is said to have the}

_{forward difference distinct}

property

m − 1

· · ·

..

(3, 2), (3, 4)

(4, 0)

}.

Since |FD

5T5

| = 10, the set

5

T

5

has the forward difference distinct

(28)

m

T

q

has the forward difference distinct property

Lemma

The set

m

T

q

has the forward difference distinct property in Z

m

× T

q

.

Proof.

), (j , a

j

) is unique determined by the element (c, d ) ∈ Z

m

q

if the

difference set

= {

(1, 1), (1, 2), (1, 4), (1, 3)

(2, 3), (2, 1), (2, 2)

(3, 2), (3, 4)

(4, 0)

}.

Hence

−FD

_5T5

= {

(4, 4), (4, 3), (4, 1), (4, 2)

(3, 2), (3, 4)

, (3, 3)

(2, 3), (2, 1)

(1, 0)

}.

Since |D

5T5

| = 16 6= 20, the set

(2, 3), (2, 1), (2, 2)

(3, 2), (3, 4)

(4, 0)

}.

Hence considering as the negative in Z

6

× F

5

, we have

−FD

_6T5

= {

(5, 4), (5, 3), (5, 1), (5, 2)

(4, 2), (4, 4), (4, 3)

(3, 3), (3, 1)

(2, 0)

}.

Since |D

6T5

| = 20 now, the set

6

T

5

has the difference distinct property in

(32)

2q−1

T

q

has the difference distinct property

Lemma

For m ≥ 2q − 1, the set

m

T

q

has the difference distinct property in

Z

m

× T

q

.

Proof.

We have |FD

mTq

| = | − FD

mTq

| = q(q − 1)/2. The first coordinate of an

element in FD

2q−1Tq

runs from 1 to q − 1, and the first coordinate of an

0

) = (c, d ). Then we have c

0

= q − 2

and d

0

= 0. Hence d = 0, a contradiction. Similarly for c = q − 1, we

have d = 0 but (q − 1, 0) /

∈ −FD

_Tm,q

.

(34)

2q−4

T

q

has the difference distinct property

Lemma

The set

m

T

q

has the difference distinct property for m = 2q − 4.

Proof.

Let (c, d ) ∈ FD

Tm,q

. Since m = 2q − 4, we have 1 ≤ c ≤ q − 1 and

q − 3 ≤ −c ≤ 2q − 5. Thus the repetition of differences occurs only when

c = q − 3, q − 2 or q − 1. Note that d = 0 iff c = q − 1, and −d = 0 iff

−c = q − 3. For c = q − 1 or c = q − 3, similar process as the above

m = 2q − 3 case can be applied to get contradictions. For c = q − 2,

−c = q − 2. Thus a repetition implies that there are

(q − 2, d

1

), (q − 2, d

2

) ∈ FD

Tm,q

such that d

1

= −d

2

. Note that the only

two elements of FD

Tm,q

with the first coordinate q − 2 are (q − 2, a

q−2

− 1)

and (q − 2, a

q−1

− a), where a is the generator chosen for F

_q∗

. So we have

a

q−2

− 1 = −(a

q−1

_{− a) and this implies a = −1, also a contradiction.}

(35)

Lines with any two intersecting in at most a point

Proposition

Suppose that

m

T

q

⊆ Z

m

× F

q

has the difference distinct property in

Z

m

× F

q

. Set B = {u +

m

T

q

| u ∈ Z

m

× F

q

}. Then |L ∩ L

0

| ≤ 1 for any

distinct L, L

0

∈ B.

Proof.

Routine.

1

_{Note that there are mq lines and mq points in Z}

_m

× F

_q

_{, and a line}

has q = |T | points with q different first coordinates.

2

_{Apparently more lines can be added to B still having the conclusion of}

m − 1

· · ·

..

m

× (F

q

∪ {∞})

(37)

Second and final result

Theorem

There exists a q-disjunct pooling design (P, B) with |P| = m(q + 1),

|B| = m(q + 1) + 1 and constant block weight q + 1, where q is a prime

power, and m is an integer at least three satisfying m = 2q − 4,

m = 2q − 3 or m ≥ 2q − 1.

By choosing q = 5 and m = 2q − 4 = 6, there exists a 5-disjunct pooling

design with 36 points, 37 blocks and constant block size 6.

(38)

The end

Thank you for your attention.

(39)

(40)

98 年度專題研究計畫研究成果彙整表

計畫主持人：翁志文

計畫編號：98-2115-M-009-002-

計畫名稱：具封閉性質的距離正則圖之研究(3/3)

量化

成果項目

實際已達成

數（被接受

或已發表）

預期總達成

數(含實際已

達成數)

本計畫實

際貢獻百

分比

單位

備註

（

質化說

明：如數個計畫

共同成果、成果

列為該期刊之

封面故事 ...

等

）

期刊論文

0

0 100%

研究報告/技術報告

0

0 100%

研討會論文

0

0 100%

篇

論文著作

專書

0

0 100%

申請中件數

0

0 100%

專利

已獲得件數

0

0 100%

件

件數

0

0 100%

件

技術移轉

權利金

0

0 100%

千元

碩士生

0

0 100%

博士生

0

0 100%

博士後研究員

0

0 100%

國內

參與計畫人力

（本國籍）

專任助理

0

0 100%

人次

期刊論文

0

2 100%

研究報告/技術報告

0

0 100%

研討會論文

0

0 100%

篇

論文著作

專書

0

0 100%

章/本

申請中件數

0

0 100%

專利

已獲得件數

0

0 100%

件

件數

0

0 100%

件

技術移轉

權利金

0

0 100%

千元

碩士生

0

0 100%

博士生

0

2 100%

博士後研究員

0

0 100%

國外

參與計畫人力

（外國籍）

專任助理

0

0 100%

人次

(41)

其他成果

(

無法以量化表達之成

果如辦理學術活動、獲

得獎項、重要國際合

作、研究成果國際影響

力及其他協助產業技

術發展之具體效益事

項等，請以文字敘述填

列。)

無

成果項目

量化

名稱或內容性質簡述

測驗工具(含質性與量性)

0 課程/模組

0 電腦及網路系統或工具

0 教材

0 舉辦之活動/競賽

0 研討會/工作坊

0 電子報、網站

0 科

教

處

計

畫

加

填

項

目計畫成果推廣之參與（閱聽）人數

0

(42)

(43)

國科會補助專題研究計畫成果報告自評表

請就研究內容與原計畫相符程度、達成預期目標情況、研究成果之學術或應用價

值（簡要敘述成果所代表之意義、價值、影響或進一步發展之可能性）

、是否適

合在學術期刊發表或申請專利、主要發現或其他有關價值等，作一綜合評估。

1. 請就研究內容與原計畫相符程度、達成預期目標情況作一綜合評估

■達成目標

□未達成目標（請說明，以 100 字為限）

□實驗失敗

□因故實驗中斷

□其他原因

說明：

2. 研究成果在學術期刊發表或申請專利等情形：

論文：□已發表 ■未發表之文稿 □撰寫中 □無

專利：□已獲得 □申請中 ■無

技轉：□已技轉 □洽談中 ■無

其他：（以 100 字為限）

3. 請依學術成就、技術創新、社會影響等方面，評估研究成果之學術或應用價

值（簡要敘述成果所代表之意義、價值、影響或進一步發展之可能性）（以

500 字為限）

完成某類距離正則圖的刻劃並進而證明其中一類的不存在性,

能讓後人免於懷疑

(44)

具封閉性質的距離正則圖之研究(III)

行政院國家科學委員會專題研究計畫 成果報告

具封閉性質的距離正則圖之研究(3/3)

研究成果報告(完整版)

計 畫 類 別 ： 個別型

計 畫 編 號 ： NSC 98-2115-M-009-002-

執 行 期 間 ： 98 年 08 月 01 日至 99 年 07 月 31 日

執 行 單 位 ： 國立交通大學應用數學系（所）

計 畫 主 持 人 ： 翁志文

報 告 附 件 ： 出席國際會議研究心得報告及發表論文

處 理 方 式 ： 本計畫可公開查詢

中 華 民 國 99 年 09 月 03 日

行政院國家科學委員會補助專題研究計畫

成果報告

期中進度報告

(

具封閉性質的距離正則圖之研究

(3/3))

計畫類別

:

個別型計畫

整合型計畫

計畫編號

:NSC

98-2115-M-009-002-執行期間

: 98

年

8

月

1

日至

99

年

7

月

31

日

執行機構及系所

:

國立交通大學應用數學系

計畫主持人

:

翁志文

共同主持人

:

計畫參與人員

:

博士生

:

黃喻培、 黃皜文。 碩士生

:

劉家安、 葉彬、 林育生、 陳建文、 林志

嘉、 劉侖欣、 洪湧昇、 施政成

成果報告類型

(

依經費核定清單規定繳交

):

精簡報告

完整報告

本計畫除繳交成果報告外

,

另須繳交以下出國心得報告

:

赴國外出差或研習心得報告

赴大陸地區出差或研習心得報告

出席國際學術會議心得報告

國際合作研究計畫國外研究報告

處理方式

:

除列管計畫及下列情形者外

,

得立即公開查詢

涉及專利或其他智慧財產權

,

一年

二年後可公開查詢

中

華

民

國

行政院國家科學委員會專題研究計畫成果報告

計畫類別：個別型

計畫編號： NSC 98-2115-M-009-002-

執行期間： 98 年 08 月 01 日至 99 年 07 月 31 日

執行單位：國立交通大學應用數學系（所）

計畫主持人：翁志文

報告附件：出席國際會議研究心得報告及發表論文

處理方式：本計畫可公開查詢

中華民國 99 年 09 月 03 日

_:

黃喻培、黃皜文。碩士生

劉家安、葉彬、林育生、陳建文、林志

嘉、劉侖欣、洪湧昇、施政成

_):

_,

一、中文摘要

二、英文摘要

三、緣由與目的

四、結果與討論

五、附錄