• 沒有找到結果。

# 一個群可分設計的構造

N/A
N/A
Protected

Share "一個群可分設計的構造"

Copied!
28
0
0

(1)

(2)
(3)

i

(4)

ii

### A construction of group divisible designs

Advisor: Professor Yao-Tsu Chang Department of Applied Mathematics

I-Shou University

Student: Pai-Jen Tsai

Department of Applied Mathematics National University of Kaohsiung

ABSTRACT

This thesis gives a construction of group divisible designs with block sizes 3; 4; 5 and 6 respectively. This work is motivated from the decoding of binary quadratic residue codes. Furthermore, we present these results in a table and propose a conjecture for this construction of group divisible designs with larger block sizes.

(5)

## Acknowledgement

(6)

Abstract (in Chinese) i

Abstract (in English) ii

Acknowledgement iii

1 Introduction 1

2 Preliminary 3

3 Theorem and proof 6

4 Conclusion remark 20

(7)

## Introduction

In 2003, Chang et al. [2] developed three new decoders for the binary quadratic residue codes with irreducible polynomials. For that kind of quadratic residue codes, there is a one-to-one correspondence between the syndromes and the correctable error pat-terns. Hence, if the syndrome is zero the received bit stream is a codeword, that is, either there is no error happened or the error pattern is also a codeword and then not correctable. However, if the generator polynomial is not irreducible, for example, the binary quadratic residue code of length 31 whose generator polynomial has three irreducible factors, the zero value of the first syndrome does not imply the received bit stream is a codeword.

There are a lot of error patterns with zero first syndrome. When investigating these phenomena, the set of error patterns with zero first syndrome suggests a construction of the group divisible design (GDD). The GDD is a topic generalized from the pairwise balanced design (well-known as PBD) [1, Definition 1.4.1]. Many authors proposed dif-ferent versions of the construction of a GDD, for example, see [3, 5], [1, Definition 1.4.2] and [6, Definition 5.5], since GDD is widely applied with graphs [3] and matrices [5]. We proposed a construction of group divisible designs with block sizes 3, 4, 5 and 6 respectively. The correctness and parameters of our construction are obtained by using the inclusion-exclusion principle.

This thesis is organized as follows. In Chapter 2 we introduce the background, definitions and properties concerning the group divisible design, finite field and our construction. In Chapter 3 we present some new group divisible designs with block sizes at most 6 including their parameters. In Chapter 4 we present a table of the

(8)

results given in Chapter 3, and propose a conjecture for the general cases by observing the table.

(9)

## Preliminary

In this section we give the basic knowledge of group divisible design (or GDD, for short,) and finite field. The notations and definitions of a GDD can be referred to [1, Definition 1.4.2].

Definition 2.1. A group divisible design GDD(v, m, k) is a triple (X, G, B), where G is a collection of m-subsets of v-set X and B is a collection of k-subsets of v-set X; we say that G is a group set and each element in G is a group, and B is a block set and each element in B is a block, such that:

(i) G forms a partition of X;

(ii) For B ∈ B and u, v ∈ B, there does not exist G ∈ G such that u, v ∈ G;

(iii) Every pair of distinct elements x and y from different groups occur together in exactly λ blocks.

In particular, the condition (iii) is called the balance condition, and λ is called balance parameter of (X, G, B).

Proposition 2.2. Let (X, G, B) be a GDD(v, m, k) with balance parameter λ. Then

each element in X occurs in λ(v−m)k−1 blocks.

Proof. Consider the element x ∈ X, and suppose that it occurs in rx blocks. In each of

these blocks x makes a pair with k − 1 other elements; so altogether there are rx(k − 1)

(10)

elements exactly λ times, so rx(k − 1) = λ(v − m), which implies rx = λ(v−m)k−1 . In fact,

this number is independent of the choice of x, so every elements in X occurs in λ(v−m)k−1

blocks.

Note that each element x ∈ X occurs in λ(v−m)k−1 blocks, which is independent of the

choice of x. Hence we denote the repetition number by rk. That is, rk = λ(v−m)k−1 .

Let (X, G, B) be a GDD(v, m, k) with balance parameter λ, and each elements in X occurs in r blocks. We call r repetition number. We denote the cardinality of B is b. In the design theorem, we have the identity bk = vr. That is, the cardinality of B is

b = vr

k . (2.1)

In this thesis, we consider X = GF (2m)\{0, 1} which is a subset of the Galois

field of order 2m. Note that the cardinality of X is 2m− 2. We explain the finite field

properties in the following which are referred to [4, Sec 4.2].

(1) Every finite field has a primitive element.

(2) Every finite field has pm elements for some prime p.

(3) If a field of q = pm elements exists, then it is unique up to isomorphism. We

denote this field by GF (pm).

(4) For any prime p and positive integer m, there is a unique field of pm element,

GF (pm).

We know that the multiplicative group GF (2m)\{0} is a cyclic group. Let α be a

generator of GF (2m)\{0}. We call that α is a primitive element.

Finite field is an important topic in Abstract Algebra. One can see [7, Chapter 6] for more details.

Definition 2.3. Let α be a primitive element of the finite field GF (2m), and X =

GF (2m)\{0, 1}. Let W

2, W3, W4, W5 and W6 be subsets of X, which are defined as

follows.

W2 := {{αx1, αx2} | αx1 + αx2 = 1}

(11)

W4 := {{αx1, αx2, αx3, αx4} | αx1 + αx2 + αx3+ αx4 = 1} W5 := {{αx1, αx2, αx3, αx4, αx5} | 5 P i=1 αxi = 1; ∀1 ≤ i < j ≤ 5, αxi+ αxj 6= 1} W6 :=      {αx1, αx2, αx3, αx4, αx5, αx6} 6 P i=1 αxi = 1; ∀1 ≤ i < j ≤ 6, αxi + αxj 6= 1, and ∀1 ≤ i < j < k ≤ 6, αxi+ αxj+ αxk 6= 1     

(12)

## Theorem and proof

Four GDDs shall be constructed in this chapter. We first prove that W2 is a partition

of X.

Lemma 3.1. W2 is a partition of X.

Proof. For each αa ∈ X, there is a unique αb ∈ X\{αa} such that 1 + αa = αb, that

is, αa+ αb = 1. Hence, {αa, αb} ∈ W2.

Next, for each αc ∈ X\{αa, αb}, there is a unique αd ∈ X\{αa, αb, αc} such that

1 + αc = αd. Hence {αc, αd} ∈ W2\{αa, αb}. Continuing this procedure, since the

set X contains even number of elements, a partition of X will be obtained and this

partition is W2.

W2 is used for the construction of GDDs (X, W2, Wk) with parameters (2m− 2, 2, k)

for k = 3, 4, 5, 6.

Theorem 3.2. For m ≥ 3, (X, W2, W3) is a GDD(2m− 2, 2, 3) with balance parameter

λ3 = 1.

Proof. We first show that each group in W2is not a subset of any block in W3. Suppose

to the contrary that there exists {αa, αb} ∈ W

2and {αa, αb, αc} ∈ W3, hence αa+αb = 1

and αa+ αb+ αc= 1, which implies αc= 0, a contradiction.

We next show that for any {αd, αe} /∈ W

2, there exists a unique B ∈ W3 containing

{αd, αe}. Let αf = 1 + αd+ αe, which implies {αd, αe, αf} ∈ W

(13)

If there are two blocks in W3 containing both αd and αe, that is, {αd, αe, αx},

{αd, αe, αy} ∈ W

3, then αd+ αe+ αx = 1 = αd+ αe+ αy, which implies αx = αy.

To sum up, we conclude that (X, W2, W3) is a GDD(2m−2, 2, 3) with balance parameter

λ = 1.

In Theorem 3.2 we know that λ3 = 1. By further computation, one can get the

following corollary for k = 3.

Corollary 3.3. For the GDD (X, W2, W3), the repetition number

r3 =

(2m− 4)

2 ,

and the number of blocks

|W3| = b3 =

(2m− 2)(2m− 4)

3 × 2 .

We next show that the triple (X, W2, W4) is a GDD(2m− 2, 2, 4).

Theorem 3.4. For m ≥ 4, (X, W2, W4) is a GDD(2m− 2, 2, 4) with balance parameter

λ4 =

2m− 8

2 .

Proof. We first show that each group in W2is not a subset of any block in W4. Suppose

to the contrary that there exists {αa, αb} ∈ W

2 and {αa, αb, αc, αd} ∈ W4, where

αa+ αb = 1 and αa+ αb + αc+ αd = 1. However, it implies αc+ αd = 0 and hence

αc= αd, which is a contradiction.

We next show that for any pair of distinct elements {αd, αe} /∈ W

2, there exists

B ∈ W4 containing {αd, αe}. Since {αd, αe} /∈ W2, by Theorem 3.2 there exists a

unique αx ∈ X \ {αd, αe} such that αd + αe + αx = 1. Let {αi, αj} ∈ W

2. Then

αi+ αj = 1 and αxi+ αj) = αx+i+ αx+j = αx, which imply

αd+ αe+ αx+i+ αx+j = 1. (3.1)

Note that {αx+i, αx+j}∩{αd, αe, 1} 6= φ for some {αi, αj} ∈ W

2. If αd∈ {αx+i, αx+j}

(14)

{αx+i, αx+j} = {αe, 1 + αd}, and if 1 ∈ {αx+i, αx+j} then {αx+i, αx+j} = {1, αd+ αe}.

Clearly that {αd, 1 + αe}, {αe, 1 + αd}, and {1, αd+ αe} are distinct. Hence we conclude

that

λ4 =| W2\ {{αd−x, α−x+ αe−x}, {αe−x, α−x+ αd−x}, {αx, αd−x+ αe−x}} | .

That is, λ4 = 2

m−2

2 − 3 =

2m−8

2 and we have the proof.

In Theorem 3.4 we know that λ4 = 2

m−8

2 . By further computation, one can get the

following corollary for k = 4.

Corollary 3.5. For the GDD (X, W2, W4), the repetition number

r4 =

(2m− 4)(2m− 8)

3 × 2 ,

and the number of blocks

b4 = |W4| =

(2m− 2)(2m− 4)(2m− 8)

4 × 3 × 2 .

We next show that the triple (X, W2, W5) is a GDD(2m− 2, 2, 5).

Theorem 3.6. For m ≥ 5, (X, W2, W5) is a GDD(2m− 2, 2, 5) with balance parameter

λ5 =

(2m− 8)(2m− 16)

3 × 2 .

Proof. By Definition 2.3 each group in W2 is not a subset of any block in W5. For any

{αa, αb} /∈ W

2, we count the number of blocks B ∈ W5 that contain {αa, αb}. Since

{αa, αb} /∈ W

2, by Theorem 3.2 there exists a unique αx ∈ X \ {αa, αb} such that

αa+ αb+ αx= 1. Let {αi, αj, αk} ∈ W3. Then αi+ αj+ αk = 1 and αx(αi+ αj+ αk) =

αx+i+ αx+j + αx+k= αx, which implies

αa+ αb+ αx+i+ αx+j + αx+k = 1. (3.2)

Let λ5(αa, αb) denote the number of the blocks in W5 that contain αa and αb. That is,

λ5(αa, αb) := |{B ∈ W5 | {αa, αb} ⊆ B}|.

By Definition 2.3, for any {αi, αj, αk} ∈ W

3, {αa, αb, αx+i, αx+j, αx+k} ∈ W5 unless

(15)

(I) αa∈ {αx+i, αx+j, αx+k}.

(II) αb ∈ {αx+i, αx+j, αx+k}.

(III) 1 ∈ {αx+i, αx+j, αx+k}.

(IV) There exist αy, αz ∈ {αx+i, αx+j, αx+k} such that αy+ αz = 1. It is equivalent to

αa+ αb ∈ {αx+i, αx+j, αx+k}.

(V) 1 + αa ∈ {αx+i, αx+j, αx+k}.

(VI) 1 + αb ∈ {αx+i, αx+j, αx+k}.

The reason for (I) and (II) is that the elements in each block of W5 are distinct.

Since 1 6∈ X we have (III). (IV), (V) and (VI) are directly from Definition 2.3.

The above conditions (I) to (VI) are equivalent to the following conditions (i) to (vi), respectively. (i) αa−x∈ {αi, αj, αk}. (ii) αb−x ∈ {αi, αj, αk}. (iii) α−x ∈ {αi, αj, αk}. (iv) αa−x+ αb−x ∈ {αi, αj, αk}. (v) α−x+ αa−x∈ {αi, αj, αk}. (vi) α−x+ αb−x ∈ {αi, αj, αk}.

Let A1 = {B ∈ W3 | B contains αa−x}, A2 = {B ∈ W3 | B contains αb−x},

A3 = {B ∈ W3 | B contains α−x}, A4 = {B ∈ W3 | B contains αa−x + αb−x},

A5 = {B ∈ W3 | B contains α−x+ αa−x}, and A6 = {B ∈ W3 | B contains α−x+ αb−x}.

Since W3 is a GDD(2m − 2, 2, 3), |A1| = |A2| = |A3| = |A4| = |A5| = |A6| = r3.

Note that αa−x+ αb−x+ α−x = 1. From Definition 2.1,

|Ai∩ Aj| =        0, if {i, j} = {1, 6} or {2, 5} or {3, 4} λ3, otherwise .

(16)

By a routine check, for the cases involving intersections of 3 or more sets among

A1, A2, A3, A4, A5, A6, only |A1∩A2∩A3|, |A1∩A4∩A5|, |A2∩A4∩A6| and |A3∩A5∩A6|

are nonzero and equal to 1.

Suppose there exist distinct αi, αj, αk such that {αi, αj, αk} 6∈ W3 but

{αa, αb, αx+i, αx+j, αx+k} ∈ W

5. From (3.2) we know that αi + αj + αk = 1. However,

{αi, αj, αk} 6∈ W

3 implies {0, 1} ∩ {αi, αj, αk} 6= φ. If 0 ∈ {αi, αj, αk}, then 0 ∈

{αx+i, αx+j, αx+k}, which is a contradiction. If 1 ∈ {αi, αj, αk}, then αi+ αj + αk= 1

implies that two of αi, αj, αk are equal, which is a contradiction. Hence such block

does not exist.

By the inclusion-exclusion principle,

λ5(αa, αb) = |W3| − 6 X i=1 |Ai| + X 1≤i<j≤6 |Ai∩ Aj| − X 1≤i<j<k≤6 |Ai∩ Aj∩ Ak| = |W3| − 6r3+ 6 2  − 3  λ3− 4 = (2 m− 2)(2m− 4) 3 × 2 − 6 × 2m− 4 2 + 12 − 4 = (2 m− 8)(2m− 16) 3 × 2 .

We can see that this number is independent of the choice of αa and αb. That is,

λ5 = (2

m−8)(2m−16)

3×2 .

Theorem 3.6 tells that λ5 =

(2m−8)(2m−16)

3×2 . Then by Proposition 2.2 and (2.1) we

have the following corollary for the case k = 5.

Corollary 3.7. For the GDD (X, W2, W5), the repetition number

r5 =

(2m− 4)(2m− 8)(2m− 16)

4 × 3 × 2 ,

and the number of blocks

|W5| = b5 =

(2m− 2)(2m− 4)(2m− 8)(2m− 16)

5 × 4 × 3 × 2 .

(17)

Theorem 3.8. For m ≥ 6, (X, W2, W6) is a GDD(2m− 2, 2, 6) with balance parameter

λ6 =

(2m− 8)(2m− 16)(2m− 32)

4 × 3 × 2 .

Proof. By Definition 2.3 each group in W2 is not a subset of any block in W6. For

any {αa, αb} /∈ W

2, we count the number of blocks B ∈ W6 that contain {αa, αb}.

Since {αa, αb} /∈ W

2, by Theorem 3.2 there exists a unique αx ∈ X \ {αa, αb} such

that αa+ αb + αx = 1. Let {αi, αj, αk, αh} ∈ W

4. Then αi + αj + αk+ αh = 1 and

αxi+ αj + αk+ αh) = αx+i+ αx+j + αx+k+ αx+h = αx, which implies

αa+ αb+ αx+i+ αx+j + αx+k+ αx+h = 1. (3.3)

Let λ6(αa, αb) denote the number of the blocks in W6 that contain αa and αb. That is,

λ6(αa, αb) := |{B ∈ W6 | {αa, αb} ⊆ B}|.

However, (3.3) does not imply {αa, αb, αx+i, αx+j, αx+k, αx+h} ∈ W

6. That is, there

are some blocks {αi, αj, αk, αh} ∈ W

4 that should not be counted. We considered those

blocks into 10 families as listed in the following. Give a block B ∈ W6. The families

(I) and (II) are from |B| = 6. Since 0, 1 6∈ B, we have (V). From the construction for

a block B ∈ W6, we have the following condition.

(1) For all distinct x, y ∈ B, x + y 6= 1.

(2) For all distinct x, y, z ∈ B, x + y + z 6= 1.

Condition (1) implies (III), (IV) and (IX), and Condition (2) implies (VI), (VII), (VIII) and (X). (I) αa∈ {αx+i, αx+j, αx+k, αx+h}. (II) αb ∈ {αx+i, αx+j, αx+k, αx+h}. (III) 1 + αa ∈ {αx+i, αx+j, αx+k, αx+h}. (IV) 1 + αb ∈ {αx+i, αx+j, αx+k, αx+h}. (V) 1 ∈ {αx+i, αx+j, αx+k, αx+h}.

(18)

(VI) There exists {αp, αq, αr} ⊆ {αx+i, αx+j, αx+k, αx+h} such that αp+ αq+ αr = 1.

It is equivalent to αa+ αb ∈ {αx+i, αx+j, αx+k, αx+h}.

(VII) There exists {αp, αq} ⊆ {αx+i, αx+j, αx+k, αx+h} such that αb+ αp+ αq = 1. It is

equivalent to there exist αy, αz ∈ {αx+i, αx+j, αx+k, αx+h} such that αyz = αa.

(VIII) There exists {αp, αq} ⊆ {αx+i, αx+j, αx+k, αx+h} such that αa+ αp+ αq = 1. It is

equivalent to there exist αy, αz ∈ {αx+i, αx+j, αx+k, αx+h} such that αyz = αb.

(IX) There exist αy, αz ∈ {αx+i, αx+j, αx+k, αx+h} such that αy+ αz = 1.

(X) There exists αy ∈ {αx+i, αx+j, αx+k, αx+h} such that αa+ αb + αy = 1. It is

equivalent to αx ∈ {αx+i, αx+j, αx+k, αx+h}.

The condition (X) is equivalent to 1 ∈ {αi, αj, αk, αh}. However, it does not occur,

since W4 is a GDD(2m− 2, 2, 4). The remained conditions (I) to (IX) are equivalent to

the following conditions (i) to (ix) respectively. Note that the elements a, b, x are fixed as stated in 3.1. (i) αa−x∈ {αi, αj, αk, αh}. (ii) αb−x ∈ {αi, αj, αk, αh}. (iii) α−x+ αa−x∈ {αi, αj, αk, αh}. (iv) α−x+ αb−x ∈ {αi, αj, αk, αh}. (v) α−x ∈ {αi, αj, αk, αh}. (vi) αa−x+ αb−x ∈ {αi, αj, αk, αh}.

(vii) There exist αr, αs ∈ {αi, αj, αk, αh} such that αr+ αs = αa−x.

(viii) There exist αr, αs ∈ {αi, αj, αk, αh} such that αr+ αs = αb−x.

(19)

Let A1 to A9 be the subsets of W4 corresponding to the descriptions (i) to (ix) , respectively. Then A1 = {B ∈ W4 | B contains αa−x}, A2 = {B ∈ W4 | B contains αb−x}, A3 = {B ∈ W4 | B contains α−x+ αa−x}, A4 = {B ∈ W4 | B contains α−x+ αb−x}, A5 = {B ∈ W4 | B contains α−x}, A6 = {B ∈ W4 | B contains αa−x+ αb−x},

A7 = {B ∈ W4 | There exists {αy1, αy2} ⊆ B such that αy1 + αy2 = αa−x},

A8 = {B ∈ W4 | There exists {αy1, αy2} ⊆ B such that αy1 + αy2 = αb−x}, and

A9 = {B ∈ W4 | There exists {αy1, αy2} ⊆ B such that αy1 + αy2 = α−x}.

We use the inclusion-exclusion principle to obtain λ6(αa, αb). The terms involved

in the expression of inclusion-exclusion process are considered in the following.

Type 1. We consider the cardinality of Ai for 1 ≤ i ≤ 9.

• Case 1. |A1| = |A2| = |A3| = |A4| = |A5| = |A6| = r4.

It is directly from Proposition 2.2.

• Case 2. |A7| = |A8| = |A9| = (2

m−4

2 )λ4.

We show that |A7| = (2

m−4

2 )λ4 in the following. Let

A = GF (2m) \ {0, αa−x, 1, 1 + αa−x}

be a subset of X, and

S = {{αy1, αy2} ⊆ A | αy1 + αy2 = αa−x}

forms a partition of A. Then we have |S| = 2m2−4. By Theorem 3.4, {αy1, αy2} is

contained in exactly λ4 blocks in W4 for any {αy1, αy2} ∈ S. Note that any

block in W4 contains at most one pair in S, since |B| = 4 and the sum of

elements in B is 1 (instead of 0). Hence |A7| = |S|λ4 = (2

m−4

2 )λ4. Similarly

|A8| = |A9| = (2

m−4

(20)

Type 2. We consider the cardinality of Ai ∩ Aj for 1 ≤ i < j ≤ 9. There are 9

2 = 36 cases.

• Case 1. |A1∩ A4| = |A2∩ A3| = |A5∩ A6| = 0.

We prove |A1 ∩ A4| = 0 in the following. Let B ∈ A1 ∩ A4. That is, B =

{αa−x, α−xb−x, αd, αe} for some αd, αe ∈ X. Then αa−x−xb−xde=

1, which implies αd = αe since αa−x+ α−x + αb−x = 1. However, the elements

in B must be distinct. Hence A1 ∩ A4 = φ and the result follows. Similarly

|A2∩ A3| = |A5∩ A6| = 0.

• Case 2. |A1∩ A7| = |A2∩ A8| = |A5∩ A9| = 0.

We prove |A1 ∩ A7| = 0 in the following. Let B = {αy1, αy2, αy3, αy4} and B ∈

A1∩ A7. Then there are two possibilities.

(i) If αy1 = αa−x and αy1 + αy2 = αa−x, then αy2 = 0. However, any block

B ∈ W4 has no element 0, which is a contradiction.

(ii) If αy1 = αa−xand αy2y3 = αa−x, then αy1y2y3 = 0 and thus αy4 = 1.

However, any block B ∈ W4 has no element 1, which is a contradiction.

Hence A1∩ A7 = φ and the result follows. Similarly |A2∩ A8| = |A5∩ A9| = 0.

• Case 3. |A3∩ A8| = |A4∩ A7| = |A6∩ A9| = 0.

We prove |A3 ∩ A8| = 0 in the following. Let B = {αy1, αy2, αy3, αy4} and B ∈

A3∩ A8. Then there are two possibilities.

(i) If αy1 = α−x+ αa−x and αy1+ αy2 = αb−x, then αy2 = 1. However, any block

B ∈ W4 has no element 1, which is a contradiction.

(ii) If αy1 = α−x+ αa−x and αy2 + αy3 = αb−x, then αy1 + αy2 + αy3 = 1 and

thus αy4 = 0. However, any block B ∈ W

4 has no element 0, which is a

Hence A3∩ A8 = φ and the result follows. Similarly |A4∩ A7| = |A6∩ A9| = 0.

• Case 4. |A1∩ A8| = |A1∩ A9| = |A2∩ A7| = |A2∩ A9| = |A3∩ A7| = |A3∩ A9| =

(21)

We prove |A1 ∩ A8| = 2λ4 in the following. Let B = {αy1, αy2, αy3, αy4} and

B ∈ A1∩ A8. Then there are two possibilities.

(i) If αy1 = αa−xand αy1y2 = αb−x, then αy2 = αa−xb−x. By Definition 2.1

and Theorem 3.4, the number of blocks B ∈ W4 that contains {αy1, αy2} is

λ4.

(ii) If αy1 = αa−xand αy2+ αy3 = αb−x, then αy1+ αy2+ αy3 = αa−x+ αb−x, and

thus αy4 = α−x. By Definition 2.1 and Theorem 3.4, the number of blocks

B ∈ W4 that contains {αy1, αy4} is λ4.

The above two cases are clearly disjoint. Hence |A1∩A8| = 2λ4. By the similarity,

|A1∩A9| = |A2∩A7| = |A2∩A9| = |A3∩A7| = |A3∩A9| = |A4∩A8| = |A4∩A9| =

|A5∩ A7| = |A5∩ A8| = |A6∩ A7| = |A6∩ A8| = 2λ4.

• Case 5. |A7∩ A8| = |A7∩ A9| = |A8∩ A9| = 2m− 8.

We prove |A7∩ A8| = 2m− 8 in the following. Let B = {αy1, αy2, αy3, αy4} and

B ∈ A7∩ A8. Then there are two possibilities.

(i) If αy1 + αy2 = αa−x and αy1 + αy2 = αb−x, then αa−x = αb−x, which is a

(ii) If αy1 + αy2 = αa−x and αy3 + αy4 = αb−x, then αy1 + αy2 + αy3 + αy4 =

1 + α−x 6= 1. However, the sum of elements of B ∈ W4 must be 1, which is

(iii) If αy1 + αy2 = αa−x and αy1+ αy3 = αb−x, then

αy2 = αy1 + αa−x,

αy3 = αy1 + αb−x, and hence

αy4 = αy1 + αy2 + αy3 + 1 = αy1 + α−x. (3.4)

Note that B ∩ {0, 1} = φ, since B ∈ W4. Therefore

αy1 ∈ GF (2m) \ {0, 1, αa−x, αb−x, α−x, 1 + αa−x, 1 + αb−x, 1 + α−x}.

(22)

From the above argument, |A7∩ A8| = 2m− 8. Similarly |A7∩ A9| = |A8∩ A9| =

2m− 8.

Given B ∈ A7∩ A8 and αy1 ∈ B. Then from (3.4) we have αy1 + α−x∈ B. Thus

there exist two elements in B such that the sum of them equals α−x, and B ∈ A9.

By the similarity we conclude that

B ∈ A7∩ A8 implies B ∈ A9; (3.5)

B ∈ A7∩ A9 implies B ∈ A8; (3.6)

B ∈ A8∩ A9 implies B ∈ A7. (3.7)

• Case 6. For the remaining 12 cases not mentioned in Case 1 to Case 5, we have

|A1∩A2| = |A1∩A3| = |A1∩A5| = |A1∩A6| = |A2∩A4| = |A2∩A5| = |A2∩A6| =

|A3∩ A4| = |A3∩ A5| = |A3∩ A6| = |A4∩ A5| = |A4∩ A6| = λ4.

Note that the cases involved with A7, A8, A9 are already considered. Hence for

each pair (i, j) discussed here, 1 ≤ i < j ≤ 6 and each block B ∈ Ai ∩ Aj

contains two fixed elements that do not form a block in W2. By Definition 2.1

and Theorem 3.4, |Ai∩ Aj| = λ4.

Type 3. We consider the cardinality of Ai∩ Aj∩ Ak for 1 ≤ i < j < k ≤ 9. There

are 93 = 84 cases.

• Case 1. |A1 ∩ A2∩ A9| = |A1 ∩ A5∩ A8| = |A2∩ A5∩ A7| = |A3∩ A4∩ A9| =

|A3∩ A6∩ A7| = |A4∩ A6∩ A8| = λ4.

We prove |A1∩ A2 ∩ A9| = λ4 in the following. Let B = {αy1, αy2, αy3, αy4} and

B ∈ A1∩ A2∩ A9. Then there are three possibilities.

(i) If αy1 = αa−x, αy2 = αb−x and αy1+ αy2 = α−x, then αy1+ αy2 = 1 + αa−x6=

(ii) If αy1 = αa−x, αy2 = αb−x and αy3y4 = α−x, then αy1y2y3y4 = 1.

Hence we just consider the number of blocks B ∈ W4that contain {αy1, αy2}.

(23)

(iii) If αy1 = αa−x, αy2 = αb−x and αy1 + αy3 = α−x, then αy1 + αy2 + αy3 =

α−x+ αb−x, which implies αy4 = αa−x = αy1. However, all elements in B

have to be distinct, which is a contradiction.

(iv) If αy1 = αa−x, αy2 = αb−x and αy2 + αy3 = α−x, then αy1 + αy2 + αy3 =

α−x+ αa−x, which implies αy4 = αb−x = αy2. However, all elements in B

have to be distinct, which is a contradiction.

Hence |A1∩ A2∩ A9| = λ4. Similarly, |A1∩ A5∩ A8| = |A2∩ A5∩ A7| = |A3 ∩

A4∩ A9| = |A3∩ A6∩ A7| = |A4∩ A6∩ A8| = λ4.

• Case 2. |A1 ∩ A3∩ A9| = |A1 ∩ A6∩ A8| = |A2∩ A4∩ A9| = |A2∩ A6∩ A7| =

|A3∩ A5∩ A7| = |A4∩ A5∩ A8| = λ4.

We prove |A1 ∩ A3 ∩ A9| = λ4 in the following. Let B = {αy1, αy2, αy3, αy4}

and B ∈ A1∩ A3 ∩ A9. Since B ∈ A1 ∩ A3, there exist αy1, αy2 ∈ B such that

αy1 = αa−x and αy2 = α−x + αa−x. Observing that αy1 + αy2 = α−x, we get

B ∈ A9. Hence A1∩ A3∩ A9 = A1 ∩ A3, and their cardinalities are both λ4 by

the results in Type 2, Case 6. Similarly |A1 ∩ A6 ∩ A8| = |A2 ∩ A4 ∩ A9| =

|A2∩ A6∩ A7| = |A3∩ A5∩ A7| = |A4∩ A5 ∩ A8| = λ4.

• Case 3. |A1∩ A2∩ A6| = |A1∩ A3∩ A5| = |A2∩ A4∩ A5| = |A3∩ A4∩ A6| = 0.

We prove |A1 ∩ A2 ∩ A6| = 0 in the following. Let B ∈ A1 ∩ A2 ∩ A6. Then

B = {αa−x, αb−x, αa−x + αb−x, 1} since the sum of elements of B is 1, which

is a contradiction. Hence A1 ∩ A2 ∩ A6 = φ and the result follows. Similarly

|A1∩ A3∩ A5| = |A2∩ A4∩ A5| = |A3∩ A4 ∩ A6| = 0.

• Case 4. |A1∩ A2∩ A5| = |A1∩ A3∩ A6| = |A2∩ A4∩ A6| = |A3∩ A4∩ A5| = 0.

We prove |A1∩ A2∩ A5| = 0 in the following. Let B ∈ A1∩ A2∩ A5. Then B =

{αa−x, αb−x, α−x

, 0} since the sum of elements of B is 1, which is a contradiction.

Hence A1 ∩ A2 ∩ A5 = φ and the result follows. Similarly |A1 ∩ A3 ∩ A6| =

|A2∩ A4∩ A6| = |A3∩ A4∩ A5| = 0.

• Case 5. |A7∩ A8∩ A9| = 2m− 8.

(24)

|A7∩ A8∩ A9| = |A7∩ A8| = 2m− 8 by the results in Type 2, Case 5.

• Case 6. |A1 ∩ A8∩ A9| = |A2 ∩ A7∩ A9| = |A3∩ A7∩ A9| = |A4∩ A8∩ A9| =

|A5∩ A7∩ A8| = |A6∩ A7∩ A8| = 0

We prove |A1∩ A8∩ A9| = 0 in the following. From (3.7) we know that A8∩ A9 ⊆

A7. Hence |A1∩ A8∩ A9| ≤ |A1∩ A7| = 0 by the results in Type 2, Case 2, and

the result follows. Similarly |A2∩ A7∩ A9| = |A3∩ A7∩ A9| = |A4∩ A8∩ A9| =

|A5∩ A7∩ A8| = |A6∩ A7∩ A8| = 0.

• Case 7. |Ai∩ Aj∩ Ak| = 0 in the remaining 57 cases.

Note that for each Ai ∩ Aj ∩ Ak there exist h, ` ∈ {i, j, k} such that Ah∩ A` is

mentioned in Type 2, Case 1-3 which is with cardinality 0. The result follows.

(i) In Case 1 of Type 2, |A1∩ A4| = |A2∩ A3| = |A5∩ A6| = 0. This implies

|Ai∩ Aj∩ Ak| = 0 where {i, j, k} is one of the following.

{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {1, 4, 5}, {1, 4, 6}, {1, 4, 7}, {1, 4, 8}, {1, 4, 9}, {1, 5, 6}, {2, 3, 4}, {2, 3, 5}, {2, 3, 6}, {2, 3, 7}, {2, 3, 8}, {2, 3, 9}, {2, 5, 6}, {3, 5, 6}, {4, 5, 6}, {5, 6, 7}, {5, 6, 8}, {5, 6, 9}.

(ii) In Case 2 of Type 2, |A1∩ A7| = |A2∩ A8| = |A5∩ A9| = 0. This implies

|Ai∩ Aj∩ Ak| = 0 where {i, j, k} is one of the following.

{1, 2, 7}, {1, 2, 8}, {1, 3, 7}, {1, 5, 7}, {1, 5, 9}, {1, 6, 7}, {1, 7, 8}, {1, 7, 9}, {2, 4, 8}, {2, 5, 8}, {2, 5, 9}, {2, 6, 8}, {2, 7, 8}, {2, 8, 9}, {3, 5, 9}, {4, 5, 9}, {5, 7, 9}, {5, 8, 9}.

(iii) In Case 3 of Type 2, |A3∩ A8| = |A4∩ A7| = |A6∩ A9| = 0. This implies

|Ai∩ Aj∩ Ak| = 0 where {i, j, k} is one of the following.

{1, 3, 8}, {1, 6, 9}, {2, 4, 7}, {2, 6, 9}, {3, 4, 7}, {3, 4, 8}, {3, 5, 8}, {3, 6, 8}, {3, 6, 9}, {3, 7, 8}, {3, 8, 9}, {4, 6, 7}, {4, 6, 8}, {4, 6, 9}, {4, 7, 8}, {4, 7, 9}, {6, 7, 9}, {6, 8, 9}.

Type 4. We consider the cardinality of the intersection set of 4 or more than 4 sets

(25)

By Type 2, Case 1-3, any two different sets Ai, Aj chosen from each of the following

three collections A1 = {A1, A4, A7}, A2 = {A2, A3, A8} and A3 = {A5, A6, A9} are

disjoint. By Pigeonhole Principle, any four or more than four sets of Ai, i = 1, 2, . . . , 9,

must contain two sets coming from the same Ak for some k = 1, 2, 3, and then have an

empty intersection.

Now we count λ6(αa, αb) by the inclusion-exclusion principle. From the above

argument Type 1-4, λ6(αa, αb) = |W4| − X 1≤i≤9 |Ai| + X 1≤i<j≤9 |Ai∩ Aj| − X 1≤i<j<k≤9 |Ai∩ Aj ∩ Ak| = |W4| − (6r4+ 3 · (2m− 4) 2 λ4) + (12 · 2λ4+ 3(2 m− 8) + 12λ 4) −(6λ4+ 6λ4 + (2m− 8)) = (2 m− 8)(2m− 16)(2m− 32) 4 × 3 × 2

where the values of |W4|, r4 and λ4 are derived from Corollary 3.5 and Theorem 3.4.

Note that this number is independent of the choice of αa and αb. Hence we denote

λ6(αa, αb) by λ6. The proof is complete.

In Theorem 3.8 we know that λ6 = (2

m−8)(2m−16)(2m−32)

4×3×2 for k = 6, and the following

Corollary is immediate from Proposition 2.2 and (2.1).

Corollary 3.9. For the GDD (X, W2, W6), the repetition number

r6 =

(2m− 4)(2m− 8)(2m− 16)(2m− 32)

5 × 34 × 3 × 2 ,

and the number of blocks

|W6| = b6 =

(2m− 2)(2m− 4)(2m− 8)(2m− 16)(2m− 32)

(26)

## Conclusion remark

In this paper, we prove that the triple (X, W2, Wk) is a GDD(2m − 2, 2, k) for k =

3, 4, 5, 6. Related parameters are shown in the following table.

λk rk bk

k = 3 1 2m2−4 (2m−2)(23×2m−4)

k = 4 2m2−8 (2m−4)(23×2m−8) (2m−2)(24×3×2m−4)(2m−8)

k = 5 (2m−8)(23×2m−16) (2m−4)(24×3×2m−8)(2m−16) (2m−2)(2m5×4×3×2−4)(2m−8)(2m−16)

k = 6 (2m−8)(24×3×2m−16)(2m−32) (2m−4)(2m5×4×3×2−8)(2m−16)(2m−32) (2m−2)(2m−4)(26×5×4×3×2m−8)(2m−16)(2m−32)

By the observation of this table, we give a conjecture that the triple (X, W2, Wk)

is also a GDD for 6 < k ≤ m. We also conjecture the exact values for the parameters λk, rk and bk.

Conjecture 4.1. For 3 ≤ k ≤ m, let

Wk =      {αx1, αx2, . . . , αxk} k P i=1 αxi = 1, and P i∈S αxi 6= 1 for each

nonempty proper subset S ⊂ {1, 2, . . . , k}      .

(27)

Then the triple (X, W2, Wk) is a GDD with parameters λk = k−1 Q i=3 (2m− 2i) (k − 2)! , rk = k−1 Q i=2 (2m− 2i) (k − 1)! , and bk = k−1 Q i=1 (2m− 2i) k! .

We believe that the proof of Conjecture 4.1 is also involved with the idea of including-excluding principle. Due to the complication for larger k, the key of proof should contain other counting methods.

(28)

## References

[1] I. Anderon, Combinatorial Designs: Construction Methods, Ellis Horwood, 1990.

[2] Y. Chang, T.K. Truong, I.S. Reed, H.Y. Cheng and C.D. Lee, Algebraic decoding of (71, 36, 11), (79, 40, 15), and (97, 49, 15) quadratic residue codes, IEEE Trans. on Communications, 51 (2003), 1463-1473.

[3] H.L. Fu and C.A. Rodger, Group divisible designs with two associate classes: n = 2 or m = 2, J. Combin. Theory Ser. A 83 (1998) 94-117.

[4] V. Pless, Introduction to the Theory of Error-Correction Codes 3rd ed., John Wiley & Sons Inc., 1998.

[5] D.G. Sarvate and J. Seberry, Group divisible designs, GBRSDS, and generalized weighing matrices, Utilitas Mathematica 54 (1998) 157-174.

[6] Z.X. Wan, Design Theory, Higher Education, 2009.

• By definition, a pseudo-polynomial-time algorithm becomes polynomial-time if each integer parameter is limited to having a value polynomial in the input length.. • Corollary 42

• A language has uniformly polynomial circuits if there is a uniform family of polynomial circuits that decide

The format of the URI in the first line of the header is not specified. For example, it could be empty, a single slash, if the server is only handling XML-RPC calls. However, if the

• As the binary quadratic programming program is NP-hard in general, identifying polynomially solvable subclasses of binary quadratic programming problems not only offers the-

The main tool in our reconstruction method is the complex geometri- cal optics (CGO) solutions with polynomial-type phase functions for the Helmholtz equation.. This type of

Piecewise polynomial interpolation: divide the interval into a collection of subintervals and construct different approximation on each subinterval. The simplest piecewise

Courtesy: Ned Wright’s Cosmology Page Burles, Nolette &amp; Turner, 1999?. Total Mass Density