運用程式論證特定群試設計之存在性

國 立 交 通 大 學

應用數學系

數學建模與科學計算碩士班

碩

士

論

文

運用程式論證特定群試設計之存在性

The Existence of Certain Pooling Designs by

Programming

研 究 生：劉家安

指導老師：翁志文 教授

運用程式論證特定群試設計之存在性

The Existence of Certain Pooling Designs by

Programming

研 究 生：劉家安 Student：Chia-An Liu

指導教授：翁志文 Advisor：Chih-Wen Weng

國 立 交 通 大 學

應用數學系數學建模與科學計算碩士班

碩 士 論 文

A Thesis

Submitted to Department of Applied Mathematics College of Science, Institute of Mathematical Modeling and Scientific Computing

National Chiao Tung University in Partial Fulfillment of the Requirements

for the Degree of Master

In

Applied Mathematics June 2010

Hsinchu, Taiwan, Republic of China

運用程式論證特定群試設計之存在性

學生：劉家安 指導老師：翁志文 教授

國立交通大學應用數學系數學建模與科學計算碩士班

摘 要

本文先介紹一種特定群試設計方法，討論此設計方法所具有的性

質，並提出三個程式，來論證此種設計方法可應用的情況。程式內容包括

論證此設計方法的存在性、原根(primitive root)的列表、以及找尋此設計方

法存在的最佳情況。

The Existence of Certain Pooling Designs by Programming

Student: Chia-An Liu

Advisor: Chih-Wen Weng

Institute of Mathematical Modeling and Scientific Computing

National Chiao Tung University

Abstract

This thesis introduces a certain pooling design first, including the properties

it has. Then proposes three programs to identify the existence of this pooling

design, list the primitive roots, and optimize the conditions of this pooling

design.

誌 謝

感謝我的指導教授翁志文老師。老師面對問題的嚴謹態度，不論在學術研究，或是 待人處事的方面，都使我獲益良多。也感謝系上給我如此優良的學習環境，交大予我自 由寬廣的學術風氣。他日若有所成，我必定飲水思源。 感謝黃喻培學長和吳欣融學長。若不是你們的承先，也不會有這篇論文的啟後。感 謝應數同學們。(女士優先)玉雯、士軒、葉彬、逸軒、以及育生，不管是學習上的磨練 切磋，或是生活中的守望相助，你們是最好的依靠。也感謝建模所的學弟妹們。(照座 號排)芳竹、啟豪、明虔、淑娟、訓利、玠旼、玉峰、以及芷萱，你們或許不知道幫過 我什麼，但我想說，隨心所欲，行善於舉手投足間，才正是幫助之最高境界。 最後感謝父母親長久以來的細心栽培，哥哥的關懷打氣。日常生活中，給我無微不 至的照料；求學過程中，給我盡其所能的支持，讓我毫無後顧之憂。也感謝我的女朋友 佳玲，是未婚妻了，總能體諒我忙碌的研究生活，同時給我關心與鼓勵。感謝所有在期 間幫助過我的人，僅以此論文的呈獻，表達我最誠摯的感激。 家安 謹誌 2010 年 6 月 於新竹交大

q

Contents

1 Introduction ………..1

2 Our Construction ………..1

3 Testing program of our construction ………6

4 Generators of each prime less than 100 ………...7

5 The minimal elements set _mF based on our construction ………..11

6 Conclusions and future works ………14

1. Introduction

A binary matrix M is called d-disjunct if any column of M is not covered by the boolean sum of d other columns. We construct t n d-disjunct matrices for

, where d is a prime power,

( , ) ((t n  d1) ,m

(d1)m )1 m2d4, m2d 3 , or [1]. The details of this construction are introduced in the chapter 2.

2 1

m d

We proposed an algorithm in each chapter 3 to 5. They have different functions, but the main purpose is the same: to find the existence of the certain pooling designs based on our construction introduced in chapter 2. We also applied some theorems of the Number Theory [2] to certify the correctness of the algorithm. Especially, in chapter 5 we have some new conclusions beyond the thesis [1]. It might be the future work of this research.

2. Our construction

This construction is operated in the sense of finite geometry. Let P be a set of m n

elements. In this chapter we call an element point, and a n-subset of P a line. Our object is to find a class  of lines in P such that | | | | 1  P  , and any two lines in have at most one point in common.



Let q be a prime power and be an integer. Let denote the finite field of q elements. Let

m q : m 0 1 2 : {0, , , , q } q F  a a  a  {0,1, ,m 1}     the elements of

be the addition group of integers modulo m. Our construction starts from _mF_q as points. Then we try to properly pick subsets such that any two lines intersect at at most one point. The followings are

efinition 2.1. (Forward Difference Distinct Property)

erence distinct property if the set

consists of

the foundations of our construction.

D

For T _mF_q, T is said to have the forward diff

: {( , ) ( , ) | ( , ),( , ) with } T FD  j y  i x i x j y T i j | | (| | 1) 2 T T  elements. Lemma 2.2.

Let : {( , ) |i ,0 1}. Then has the forward difference distinct

m T  i a i  i q mFq  . , m q Tm q, property in

(pf)

Given pair ( , )c d _mF_q, solve the equation , for . , then i 0

( , ) ( , ) ( , )_{c d  j a}j _{ i a}i _{0   i j q 1}

If 1c q   and j q  . If c q 11   , then _ai _{d a}/ ( c_{ and j c i1)   . In}

each case the ( , )_{i a and ( , )}i _{j a are uniquely determined. It follows that} j

, m q T consists of , , |T_{m q}| (|T_{m q}| 1) 2  elements. □

We can view T _{m q,} as a line in the plane _mF_q as Figure 1 shows.

Figure 1: T_{m q,} in _mF_q

Definition 2.3. (Difference Distin y)

For , T is said to have the difference distinct property if the set

j

Lemma 2.4.

 . If , then has the difference

istinct p By Lemma 2.2, we have ct Propert m q T  F : {( , ) ( , ) | ( , T D  j y  i x i x),( , )j y T with i } consists of | | (| |T T 1) elements. Let T_{m q,} : {( , ) |_ i _ai _im,0_{ i q} 1} roperty in _mF_q. 2 1 m q T_{m q,} d (pf) , , ( 1) | | | | 2 m q m q T T q q

FD  FD   . The first coordinate of an element in runs from to , and the first coordinate of an element in runs from

, m q T FD 1 q1 , m q T FD 

1

m q to m1. The assumption m2q1 implies that

, ( , ) m q m q T T FD  FD . m Fq □ has the in Lemm The a 2.5.

set T_{m q,} difference distinct property   for m2q and 3

2 4 m q ) . (pf By Lemma 2.2, we have |FD_{T ,} | | FD_{T ,} | q q( 1) 2 m q   m q  . Given ( , )c d FDTm q, :

(i) If m2q3 , then 1   and c q 1 q   2 c 2q4

,

( 1, 0)

m q

T

q F

. The repletion of differences can D

only occur at c q 1 or c q 2. Since  and (q2, 0) F _T

, m q , D  , ( 2, 0) m q T q FD and (q F , m q T D  . 3 1,0)  

(ii) If m2q4, then 1  c q 1 and q   c 2q . The repletion of differences 5 can only occur at c q  1 or c q 2 c q 3

, 0) m q T FD (q1,   or   . Since  and D ,q, (q  (q3, 0) F m T  D , 3,0) _T m q FD and , ( 1, 0) m q T

q  F . Now focus on case c q  . 2 The on ents of

,

m q

T

FD with the fi 2

ly two elem rst coordinate q is _(q2,aq2_{ and 2)} 1

 

(_q 2,_aq _{a , where ) a is a generator for} *

q F . If _aq2 2_aq1_{ , then a a 1} , and ' { | , which  is a cont Lemma 2.6. Supp radi □ os h operty ction.

e that T as the difference dis ct pr_{m q,} tin B  u T _{m q,} u _mF_q}.

Then |L₁L₂| 1 , for  1, 2 ' L L B   ,L₁L₂. 2 L (pf)

Suppose not. Then L L₁, ₂B L', ₁ and _{1 2 1} such that 2 1 2 , , 1 2 |L L |2 . Suppose L₁( ,u₁ v₁)T_{m q,} , 2 ( ,2 L  u v₂)T_{m q,} p p L  L p  p . Let p₁ ( ,u v_{1 1}) ( , ) ( , ) c d_{1 1}  u v_{2 2} ( ,c d_{2 2}), 2 ( 1, 1) ( ,3 3) ( , ) ( ,2 2 p  u v  c d  u v  c₄ d . Then ₄) ( ,u v_{1 1}) ( , u v_{2 2}) ( , ) ( , ) c d_{1 1}  c d_{2 2} ( , )c d_{3 3} 4 (c

 , )d₄ , and it is true only when ( , ) ( ,c d_{1 1}  c₃ d₃) and ( ,c d_{2 2}) ( , ) c d_{4 4} . Hence p₁  p₂, which is a contradiction. □

Note that there are mq lines and mq points in _mF_q

s the fra

, and a line has points with q different first coordinates. This i me of our work. Now, add more points and lines in , | _{m q}| q T ' B . Since , , (0, ) m q m q T T

x  FD FD , L((0, )x L) for any nonzero x F _q and L B '. We add a common point (i q  , ) _m(F_q{}) to each line L u T_{m q,}

to forms a new set B where '' i is the first coord te of _m ina u. Note that the points set of ''

B becomes _m(F_q{ }) . To show that any two lines in B also interse''

the following Lemma 2 first.

Suppose has the difference distinct property in . Let , _, be two distinct lines in

ct at at most one point, we prove .7

Lemma 2.7. that T_{m q,} _mF_q , L2 ( , )c d2  mFq  1 ( , )1 m q L  c d T T _{m q} B . Then ' L₁L₂ . (pf) 2

Suppose ( , )e f   , then L₁ L₂ ( , ) ( , ) ( ,e f  c d₁  x y_{1 1}) ( , ) ( , ) c d₂  x y₂ for some

1 1 2 2

( , ),( , )x y x y T_{m q,} . Thus e c x   . Since each element in T1 x2 m q, has distinct first coordinate, we can conclude that ( , ) ( , )c d₁  c d₂ and hence L₁L₂. It is a contradiction. □

Lemma 2.8.

Any two distinct lines in B intersect at at most one point. '' (pf)

''

B ( , )c 

It is easy to see that contains exactly one point of the form . Let L L₁, ₂ distinct lines be two in B containing '' ( , )c₁  , ( , )c₂  , respectively. If c₁c₂, L₁\ ( ,c₁ ) tinct lin and 2\ ( , )2

L c  are two dis es in B and have at most one point in common by Lemma '' 2.6. If c₁ c , the set of the first coordinates of ₂ L₁\ ( , )c₁  and L c , ) be the same. Thus L₁\ ( , ) ( , )c₁ e f₁ T

2\ ( 2  must ,

m q

   and L₂ \ (c₂, ) ( , )  e f₂ T_{m q,} for some e , _m

1 2 q

f f,  . By Lemma 2.7, F L₁\ ( , )c₁  L₂ \ (c , )₂   , so  L L only intersect at c₁, ₂ ( , )₁  .□ 

Let {( , ) |V  i j j F { for 0  i m 1, and V is called the i-th vertical line. e add thes

}}

i q i

Let {( , ) | 0H  i   i q}, and H is called the infinite line. W e to B and '' complete our construction.

0 1 m1}

Lemma 2.9.

Set :B B'' { , , , , H V V V _ as the set of lines with underground point set _m (F_q{}) . Then any two lines in B intersect at at most one point.

(pf)

t V_iV_j  for i , j and V_iH  ( , )i

It is easily seen tha .It re

1 any

mains to show that and |L H | 1

|L V _i| for L B '', 1  i m 1

ns only one point of the type

. Since each point in L has distinct first coordinate and contai ( , )c  , the result follows. □

Note that |_m(F_q{ }) | m q( 1) and | |B m q(   , which is our final result. 1) 1

Theorem 2.10.

has the difference distinct property. Let M be the incidence matrix of

Suppose that T_{m q,} _mF_q

{ })  

_m (F_q and B . Then M is a nontrivial q-disjunct matrix with m q(  1) rows and constant column weight (q . 1)

(pf)

Applying Lemma 2.4 and Lemma 2.5 to Theorem ry 3.11 also follows.

2.11.

Let M be the incidence matrix of _m (F_q { })

2.10. Corolla □

Corollary

 

  and B where m2q4, 2q , 3 or 2m q1. Then M is a nontrivial q-disjunct matrix with m q 1)(  rows and constant

(q1) column weight . xample 2.12. (A construction of E 36 37 5-disjunct matrix) , and Take 5q , 6 2m  q4 a2 3),(4,1 is a generator of . Then . We write 5 { 6,5 {( , ) |i T  i a 24,33, 41} 6,0 4),(3, )}   4} i i {(0,1),(1, 2),(2, T_6,5 01,12, for simplifying the notation.

et , where is the first coordinate of . Then (1) L L u( ) ( u T _6,5) ( i 5, ) i u

L(00) {01,12, 24,33, 41,5 }  , (01) {02,13, 20,34, 42,5 }L    , … , (54) {50,01,13, 22,30, 4L

, (10) {11, 22,34, 43,51,0 }L   , }

(11) {12, 23,35

L  , 44,52,0 }   . There are 30 lines.

(2) Let V_i {( , ) |i j j F _q{ }} for 0 i 5. V is called_i rtical line.

0 {00,01,02,04 V   the i-th ve , , ,03,0 } V₁{10,11,12,14,13,1 } V₂ {20, 21, 22, 24, 23, 2 } , , V₅ {50,51,52,54,53,5 } 3 {30,31,32,34,33,3 } , V V₄ {40, 41, 42, 44, 43, 4}   . , an is called . . There is 1 line.

d (3) are the 37 lines based on out construction. □

3. Testing program of our construction

An important work after the construction of a type of pooling at roperties it has. Here we provides a way to verify the existence of difference distinct

. The existence of this pr can ma can be applied into the pooling design.

Algorithm 3.1.

Step 1: Input (q,a,m), where q is a prime power, a is a generator of q, and m>q is an integer. Step 2: Construct the matrix of order

There are 6 lines.

(3) Let {( d H the infinite line

The above (1), (2), an , ) | 0 } H  i   i q {0 ,1 , 2 ,3 , 4 ,5 } H        design is to know wh p

property operty ke sure the construction in chapter 2

, m q T q by 2 1 q , ( ( ) ( , )i , 1, 2, , m q i T _{ 1) th row}  i a _mF i_q    .

Step 3: Construct another “checking matrix” of size (q q 1) 4. The 4 components of each row is minuend term, subtrahend term, and the results.

ix.

xample 3.2.

atrix:

Step 4: Check the repetition of each row after the construction of the checking matr

E

Input ( , , ) (7,3,12)q a m  , then construct T m_{m q,}

Tmq =

1 3 - term 2 2 2 - term 3 3 6 - term 4 5 - m 6 6 1 - term 7 hecking m 4 4 - term 5 5 ter

The T matrix is a _{m q,} 7 2 matrix. Now construct the “c atrix” of size 42 4 , in

each row stores the minuend term, subtrahend term, an

whic d the results in

g sults:

1 7 6 0 7 1 6 0

It means the result of term 1 minus als to the result of term 7 minus term 1. Additionally, since th

cannot have the difference distinct p uction. In f proved that will not ha rty

4. Generators of each prime

In th prop prime less

than 100, and then show the results as a table of generator database. Also showing is the emmas proposed to help the program be faster as finding the generators of large prime.

m 4.1. (See if a is a generator of prime p or not.)

put prime p and generator a

=1, count=1; while count≤p-2 temp=temp×a (mod p); if temp=1 a+1 h ₅F . Then, ₇

check the repetition of the checking matrix. In this example, it will run out the followin re

ans =

term 7 is (6,0) , which equ

ere are some results run out, this case ( , , ) (7,3,12)q a m  act, it is easy to based on our construction.

roperty based on our constr

2 2

m q ve difference distinct prope

less than 100

is chapter we ose an algorithm for finding the all generators of each relation between the Euler's phi function and the number of generators. Two l are

Algorith

In Set temp

break the while loop and try next a= ; end

if count=p-2

print a and try next a=a+1; end

count=count+1;

Tabl

) end

e 4.2. (The generators of each prime less than 100.)

Prime p Generators a ( p 1 3 2 1 5 2,3 2 7 3,5 2 11 2,6,7,8 4 13 2,6,7,11 4 17 3,5,6,7,10,11,12,14 8 19 2,3,10,13,14,15 6 23 5,7,10,11,14,15,17,19,20,21 10 29 2,3,8,10,11,14,15,18,19,21,26,27 12 31 3,11,12,13,17,21,22,24 8 37 2,5,13,15,17,18,19,20,22,24,32,35 12 41 6,7,11,12,13,15,17,19,22,24,26,28,29,30,34,35 16 43 3,5,12,18,19,20,26,28,29,30,33,34 12 47 5,10,11,13,15,19,20,22,23,26,29,30,31,33,35,38,39,40, 41,43,44,45 22 53 2,3,5,8,12,14,18,19,20,21,22,26,27,31,32,33,34,35,39, 24 41,45,48,50,51 59 2,6,8,10,11,13,14,18,23,24,30,31,32,33,34,37,38,39,40, 28 42,43,44,47,50,52,54,55,56 61 2,6,7,10,17,18,26,30,31,35,43,44,51,54,55,59 16 67 2,7,11,12,13,18,20,28,31,32,34,41,44,46,48,50,51,57,61,63 20 71 7,11,13,21,22,28,31,33,35,42,44,47, 52,53,55,56,59,61,62,63,65,67,68,69 24 73 5,11,13,14,15,20,26,28,29,31,33,34,39,40,42,44,45,47, 53,58,59,60,62,68 24 79 3,6,7,28,29,30,34,35,37,39,43,47,48, 53,54,59,60,63,66,68,70,74,75,77 24 83 2,5,6,8,13,14,15,18,19,20,22,24,32,34,35,39,42,43,45,46,47,50, ,62,66,67,71,72,73,74,76,79,80 52,53,54,55,56,57,58,60 40 89 3,6,7,13,14,15,19,23,24,26,27,28,29,30,31,33,35,38,41,43,46,48, 40

51,54,56,58,59,60,61,62,63,65,66,70,74,75,76,82,83,86 97 5,7,10,13,14,15,17,21,23,26,29,37,38,39,40,41,

56,57,58,59,60,68,71,74,76,80,82,83,84,87,90,92

32 In fact, the number of generators is equal to (p , where 1)  is the Euler’s phi

function.

De tion

The n and some positive integer m that are relatively prime

to an ame:

fini 4.3. (Euler’s Phi Function)

umber of integers between 0

m is important quantity, so we give this quantity a n

( ) |{ |1m a a m gcd a m, ( , ) 1}|

     .

Theorem

(a) is

4.4. (Euler’s Phi Function Formulas)

If p a prime and k1, then _(pk_{) p}k _pk1_.

(b) If gcd( , ) 1m n  , then ( mn)( ) (m  n). (pf

The verifi a) is easy, so we need to check the formula (b).

He e d : COUNT !

Briefly, w )

cation of the prime power formula (

re, w id this by using one of the most powerful tools in number theory ING e are going to find a set contains ( mn) elements, and find another set contains ( ) ( )m n

  elements. Then, show that the two sets contains the same number of elem . The first s ents et is: A{ |1a  a mn, and gcd( ,a mn) 1} .  Th con  . Clearly th

e se d set is: B{( , ) |1b c  b m, and gcd( , ) 1,b m  and 1 c n , and gcd( ,c n) 1} at A has ( mn) elements and B has ( ) ( ) m n elements. Then, find a function f from A to B in the following way:

( ) ( , ), if (mod ) and (mod )

f a  b c a b m a c n .

ow, check that f is one-to-one and onto:

A, such that N

(i) Take two numbers a and ₁ a from₂ f a( )₁  f a( )₂ . Then a₁  b a₂(mod

m) and a₁ c a₂ (mod )n . Thus, a₁ is divisible by both m and n, in other words, a₂

1 2 (mod )

a a mn , which means a and ₁ a are the same elements in A. ₂

(ii) Clearly that for any given pairs (b,c) from B, we can always find a integer n ,

tisfying □

Lemma 4.5. (Euler’s Phi Function Summation Formula)

e div n. Then n

, 1

a  a m

(mod ) and (mod )

a b m a c n .

sa

Let d d1, , ,2  dr be th isors of ( )d1 (d2)  ( )dr  .

(pf)

Let F n( )( )d1 ( )d2   ( )dr , and from Euler’s Phi Function Multiplication Formula

we can get that F mn( )F m F n( ) ( ) if gcd( , ) 1m n  . Check the value of ( )F p for prime

powers: _{F p(} k_{)(1)( )p ( )p}2 _(pk_{) 1 (  p 1) (p}2_{p)  (p}k_pk1_{) p}k_.

Now, factor n into a product of prime power 1 2

k

s, say ₁k ₂k ks

s

1 2 ks 1 2 ks 1 2 ks

k k k k k k

1 2 1 2 1 2

( ) ( _s ) ( ) ( ) ( _s )

F n F p p p F p F p F p  p p p_s n

Hence we verify that always equals n. □

(1)   

ax ( )

F n Definition 4.6. (Primitive Root)

( ) the smallest exponent 1 so that e 1(mod ), for is prime and 1 1.

e a_p  e a  p p a p

(2) A number g with m imum exponent e g_p( ) p 1 primitive ulo p.

Note tha in Numb ory is so called the generator in this thesis. mitive Root The rem)

(p 1)

is called a root mod

t the primitive root er The

Theorem 4.7. (Pri o

There are exactly   primitive roots modulo p.

t by using one of the most powerfu tools in number theory: COUNTING! Define a

nction: a d

(pf)

We prove i l

( ) (the number of ' with 1d a s a p and ( )e_p )

     . In particular, (p 1) fu

is the number of primitive roots modulo p. be any number that dividing

Let n p , say, 1 p 1 nk. Then, any roots these polynomials have mo

1 _{1 1}

p nk

X    X  _(Xn _1)((Xn k₎ 1_(Xn k₎ 2_{  X}n_1)

and count how m dulo p.

First, _Xp1_{ 1 0 (mod )p has exactly p1 solutions X 1, 2, , p1}

hand, 1

. On the other 0 (mod )

n

X   p has at most n solutions and _(Xn k₎ 1_(Xn k₎ 2_{ 1 0 (mod p})

1 0 (mod )

n

has at most (n k solutions. Hence the o ly way is 1) n X   p has exactly n solutions and _(Xn k₎ 1_(Xn)k2_ 1 0 (mod _{p) solutions.}

1 od

n

 

0 (m

has at exactly (n k1) Now, count the number of solutions to X   p) using another way. Let

1 0 (mod )

1, , ,2 r

d d  d

n

be the divisors of n. Then the number of solutions to X   p is equal to

1 2

( )d ( )d ( )dr

     , and we have the formula: ( )d₁ ( )d₂   ( )dr n.

) As is a prime,

(i n q (1)( )q  q (1)( )q . (1)(1) 1 , so ( )q ( )q . (ii) As _{n q} 2_{, (1)( )q (q}2_)q2 _{(1)( )q ( )q}2 _{. So, ( )q}2 _{( )q}2 _.

By induction method, _( )_qk _( )_qk _{, as n q} k

(iii) is a prime power.

for two different primes , _{1 2} (iv) As n q q _{1 2} q q1, 2 (1)( )q1 ( )q2 (q q1 2)q q

1 2 1 2

(1) ( )q ( )q (q q )

   

    . So, (q q1 2)q q1 2 (q q_{1 2}).

(v) By induction method, assume ( )d ( )d , for all numbers d n . We may also assume

r

1 i, 2,3, ,

n d d i  . From r ( )n ( )d₂   ( )dr  n ( )n ( )d₂   (d ),

we can get the equality ( )n ( )n .

We also noticed the following two lemmas from the table so that the performance of

Lemma 4.8.

program can be enhanced as finding the generators of larger primes.

For prime p1 (mod 4), if g were a generator of p, then -g is also a generator of p. Suppose not, i.e., g is a generator of p, but there exists 1)

(pf)

2 b (p2), | (b p , such that

(_g)b_1 (mod )_{p .}

(i) if b were even, then _gb_{ }( _g)b_1 (mod )_{p , which is clearly a contradiction.}

b were odd: im

(ii) if 4 | (p1) plies that 2 | (b p1) and 2b(p1), and hence

2b _{( )}2b _{1 (mod )}

g  g p □

For prim 3 (mod 4)

 , which is a contradiction.

Lemma 4.9.

e p , if g were a generator of p.

Clearly that

p, then -g is not a generator of

(pf) 1 2 _{( 1) (mod )} p g p    . Hence 1 2 ( ) 1 (mod 1 2 p is odd and ) p g p   . □

the generators of prime 19.



Note that g and (-g) may both not be the generators of p. For example, 7 and 12 are both not

nimal elements set



m



F

q

5. The mi

based on our construction

rime less than 100, we are interested in the minimal that can make have ifference distinct proper based on our construction.

orithm first, and then get the conclus correspond can be less than

After finding the generators of each p

the d ty mFq  size of , m q T ing

In this chapter we introduce an alg ion that the minimal to the

m

 F _q m2q , which is one lower bound 4 that we proposed in our paper.

ecall the lgorithm 3.1 in th e input , here q is a prime power, a is a generator of q, and m is the size of the addition g

The

1. (Find the minimal corresponding to the prime q and generator a)

( , , )q a m

roup . _m R A e previous chapter. In the algorithm w

w

results shows the repetition between the differences of every two terms.

Inpu d generator a t m from q+1 to 2q-5 if there are sults run out

1;

output (

break the for loop;

e 5.2 (The minimal corresponding to every generator of each prime less than 100.) Genera

t prime q an for in

do Alorithm 3.1 with the input ( , , )q a m ;

re

try the next m+

else (there are no results run out) , )m ; , q a end Tabl  _m tors a Prime q

Corresponding minimal _mF_q based on our construction

2 3

5 ( , ) (5,6)m q  is the only case for m q  1 6 6 3 5 7 10 10 2 6 7 8 11 15 15 16 16

Note that the minimal  _m is less that 2q-4.

2 6 7 11 13 20 18 20 18 3 5 6 7 10 11 12 14 17 26 27 26 27 28 27 28 27 2 3 10 13 14 15 19 30 31 30 31 30 30 5 7 10 11 14 15 17 19 20 21 23 36 37 37 38 36 38 37 37 38 38 2 3 8 10 11 14 15 18 19 21 26 27 29 45 44 47 44 47 46 45 48 44 48 44 46 3 11 12 13 17 21 22 24 31 52 52 50 50 52 52 48 48 2 5 13 15 17 18 19 20 22 24 32 35 37 63 59 62 63 62 63 59 63 62 63 62 63 6 7 11 12 13 15 17 19 22 24 26 28 29 30 34 35 41 66 66 68 68 66 68 68 66 74 68 66 74 68 66 73 73 3 5 12 18 19 20 26 28 29 30 33 34 43 72 74 72 72 73 73 74 73 72 74 74 73

5 10 11 13 15 19 20 22 23 26 29 30 31 33 35 38 81 79 74 81 79 81 76 79 80 77 81 74 79 79 80 77 39 40 41 43 44 45 47 6 78 79 80 78 7 80 2 3 5 8 12 14 18 19 20 21 22 26 27 31 32 33 91 88 90 90 92 92 88 92 90 90 87 87 91 92 90 92 34 35 39 41 45 48 50 51 53 92 96 92 87 92 90 96 87 2 6 8 10 11 13 14 18 23 24 30 31 32 33 100 99 104 99 102 101 102 101 101 96 100 97 96 98 34 37 38 39 40 42 43 44 47 50 52 54 55 56 59 98 104 102 103 97 97 102 95 98 101 97 98 95 103 2 6 7 10 17 18 26 30 31 35 43 44 51 54 103 104 105 104 107 107 107 105 103 105 101 101 104 107 55 59 61 104 105 2 7 1 2 131 1 18 20 28 31 32 34 41 44 46 115 118 11 115 7 117 110 118 117 117 114 115 110 114 115 48 50 51 7 615 63 67 118 118 11 115 8 115 118 7 11 13 12 22 28 31 33 35 42 44 47 52 53 125 125 12 11 1 45 8 2 126 122 126 126 124 118 119 121 124 55 56 59 16 26 63 65 67 68 69 71 122 121 12 12 1 64 5 2 126 124 124 119 126 5 11 13 41 51 20 26 28 29 31 33 34 39 40 120 128 12 12 1 63 6 2 128 123 122 124 123 123 125 126 129 42 44 45 74 35 58 59 60 62 68 73 129 120 12 12 1 03 6 3 125 123 122 130 124 3 6 7 82 92 30 34 35 37 39 43 47 48 53 144 141 13 13 1 39 9 3 133 139 143 133 134 141 133 139 144 54 59 60 36 66 68 70 74 75 77 79 133 138 13 13 1 13 8 4 141 143 138 138 134 2 5 16 8 3 14 15 18 19 20 2 2 24 23 34 144 138 14 14 1 80 4 4 140 141 141 144 147 14 1 4 486 4 1 146 35 39 42 34 54 46 47 50 52 53 54 55 65 57 144 142 14 13 1 44 6 4 143 144 138 144 144 147 147 136 140 83 58 60 62 66 76 71 72 73 74 76 7 80 9

4 2 4 139 141 151 143 13 144 147 151 141 14 14 1 0 9 3 6 7 31 41 15 19 23 24 26 27 28 29 30 152 154 154 591 1 55 154 156 158 156 156 15 1 6 576 5 1 152 31 33 35 83 14 43 46 48 51 54 5 58 96 5 60 158 156 15 15 1 76 4 5 157 154 159 154 151 148 155 157 154 61 62 63 56 66 70 74 75 76 82 83 86 89 151 148 154 154 155 155 153 156 157 154 153 157 5 7 10 13 14 15 17 21 23 26 29 37 38 39 168 166 175 172 166 172 169 169 170 166 167 169 170 168 40 41 56 57 58 59 60 68 71 74 76 80 82 83 169 172 166 172 167 170 161 175 172 170 161 172 172 170 84 87 90 92 97 170 167 172 167

We can give the table a brief conclusion that we find the minimal size of can be les an 4 r e p a ion y, dis ce between m

of a s s er You m al ot ha m  i s th 2 n

q fo very rime p . In ddit11 all the tan nimal size d 2q get longer as the prime get larg .

ay so n ice t t (m

m



,q)(5,6) is the only case for m q 1. In fact, it is the Example 2 i t onstruction in chapter 2.

6. Conclusions and future w

s

e ap ed co uc t pl ent a cer po g ign th e e o tried to fin y im p rt f c u . u e r i shows that

.12 which s introduced to fi our c

ork

W pli our nstr tion o im em tain olin des . In is th sis w als d wa s to prove the rope ies o this onstr ction Thro gh th prog amm ng, it

m

 can be less than 2q 4 for every prime p11, and this result is better than the re w o i a 1

ght be lower if we keep running the program in chapter 5 through

every gen s w o e n k f a

not get the lt proving the a rith and ath atical deduction will be the following challenge of this research.

References

[1] Yu-pei Huang, Hsin-jung W , a e d-disjunc tr column ig +1 ar 0

[2] Joseph H. Silverm r ti n e o nd c p 5 Prentic all 04

[3] Ding-Z u P g ig n a iv ting, World ientif 20

sults e pr posed i

on the orig nal p per [ ]. The bound of  m_m

erator . Ho ever, due to the c mpl xity a d lac ing o memories, so far we h ve resu s. Im lgo m m em

u ch, 2

nd Chih-wen W 10.

ng, t Ma ices with constant we ht d , M

an, A friendly int oduc on to umb r the ry, 2 Ed., h20, p12 -134, e-H , 20 .

hu D and Frank K Hwang, oolin des ns a d non dapt e group tes Sc ic, 06.