國 立 交 通 大 學
應用數學系
碩 士 論 文
二階投影平面架構能提供最好的群試設計
- - -
對七物件及最多兩感染物而論
A projective plane of order 2
offers the best group tests
- - - -for 7 items and at most 2 defectives
研 究 生：陳建文
指導教授：翁志文 教授
中 華 民 國 九 十 九 年 六 月
二階投影平面架構能提供最好的群試設計
- - - -
對七物件及最多兩感染物而論
研究生
:
陳建文
指導教授
:
翁志文 教授
國立交通大學
應用數學系
中文摘要 我們證明一個由二階投影平面的關係矩陣,刪去一列所得到大小為6乘以 7的二元矩陣其分離性為二。 我們同時證明不存在分離性為二, 行數小於七, 列數小於六的二元矩陣。A projective plane of order 2 offers
the best group tests
- - - -for 7 items and at most 2 defectives
Student: Chien-Wen Chen Advisor: Chih-Wen Weng
Department of Applied Mathematics National Chiao T ung University
Hsinchu, T aiwan 30050
Abstract
We prove that the matrix obtained by deleting a row of the incidence matrix of the projective plane of order 2 is 2-separable. We also prove that there is no 2-separable s × t matrix with s < t < 7.
謝誌
首先感謝翁志文老師在這兩年間細心的指導, 雖然我程度不太好, 老師還是耐 心的跟我討論問題, 從老師身上學習到如何思考問題的方法, 也學到一些人生觀 念, 除此之外, 真覺得老師的修養很好, 從沒看他發脾氣, 總是不厭其煩的幫助學 生,是我學習的對象, 真的很感激他。 在這兩年也謝謝學長和同學們的陪伴, 包含黃喻培、 黃皜文、 李光祥、 陳巧玲、 葉彬、 林育生、 林志嘉、 黃思綸、 林家銘、 劉侖欣· · ·等人, 不僅在寫論文的事上, 也在修課上, 常常受到你們的幫助和指導, 藉著和你們的討論, 不足我所短缺的點, 實在由衷的感謝你們。 也感謝在交大求學過程中, 有教導過我的老師們, 有翁志文、 陳秋媛、 黃大原、 莊重、 林琦琨, 及在淡大的王筱蘭、 胡守仁、 高金美、 余成義、 李武炎、 吳孟年, 帶 我一步一步的認識數學, 使我慢慢的學得思考數學的技巧和方法。 在我大學及研究所的日子裡, 謝謝淡水及新竹召會的弟兄姊妹一路關心我, 扶 持我, 請我吃飯, 找我交通, 一同經歷高山, 一同經歷低谷, 使我能一直留在美地 裡, 享受生命的交通, 使我們在一裡被建造在一起, 其中包含陳啟文一家、 盧正五 一家、 譚昌琳一家、 吳耀銓一家、 黃世賢一家、 蔡明隆一家、 陳富強一家、 張忠斌 一家、 周浩忠一家、 楊龍杰一家、 賴緯杰一家、 李天政一家、 蕭宏志一家、 林四皓 一家、 馮藝鴻一家、 李林彩一家、 陶逖一家、 冉龍台一家、 劉雲楷、 劉錢亞滿、 王李 秀珊、 尤姊、 淑雲姊、 李苑佳、 陳深泓、 張士毓、 吳緯中、 嚴熯軍、 李佳興、 李禧年、 林韶烈、 王得安、 何家齊、 朱永恩、 李耀主、 鄭子恆、 薛漢偉、 顏以理、 黃湘傑、 陳 俊元、 何振泰、 黃逸齊、 鄧展政、 葉承恩、 李彥德、 李銘倫、 蔡承安、 蔣岳鋒、 王柏 凱、 王詩婷、 葉軍政、 羅博文、 王鍾宣恩、 龔昭維、 鄭家胤、 丁俊宏、 陳達恩、 劉家 豪、 林帝宇、 羅晟、 劉浩皿、 林同心、 潘建綱、 黃振銘、 李文凡、 詹博全、 祝偉倫、陳彥興、 鍾宗穎、 詹哲賢、 陳建町、 李彥廷、 許時挺、 周節、 徐傳源、 林卓誠、 陳育
廷、 林韋任· · · , 還有許多別的人,若是一一都寫出來,我想所寫的書就是世界也容
不下了。
最後還要謝謝我的父母和家人, 謝謝你們的鼓勵和養育之恩, 使我在生活的需
Contents
1 Introduction 1 2 The Matrix Representation 2 3 Basic Definitions of BIBD 3 4 The main result 6 5 References 20
1
Introduction
In combinatorial group testing, a prototype problem called (d, n) problem is to assume that there are up to d defectives among n given items, and the problem is to separate the good items from the defective ones by group tests. A group test is administered on an arbitrary subset S of the items with two possible outcomes; a negative outcome means S contains no defectives and a positive outcome means S contains at least one defective, not knowing exactly how many or which ones. A group testing algorithm is nonadaptive if all tests must be specified at once. A nonadaptive algorithm can be rep-resented by a 0-1 matrix where columns are items, rows are tests, and a 1-entry in cell (i, j) means item j is contained in test i. Note that a column can be viewed as a subset whose elements are indices of the rows incident to the column. Thus we can talk about the union of columns. S.H.Hung and F.K.Hwang [1] prove that what values of n, given d, individual testing is optimal on nonadaptive group testing.
Group testing has applications to biological experiments, DNA sequenc-ing, electrical and chemical testsequenc-ing, codsequenc-ing, etc. The binary matrices have three types: d-separable, d-separable and d-disjunct which have been found to be major tools in understanding and constructing a nonadaptive group testing. Hong-Bin Chen and Frank K. Hwang [3] proved that M is a d-separable matrix and 1 ≤ k ≤ d − 1, then M is k + 1-d-separable, if and only if M is k-disjunct. We use the property to prove that the matrix obtained by deleting a row of the incidence matrix of a projective plane of order n is n-separable. In particular, n = 2, the matrix obtained by deleting a row of the incidence matrix of the projective plane of order 2 is 2-separable. In this
paper, we want to show there is no 2-separable s × t matrix with s < t < 7. For example, there is a 2-separable matrix M5×7. Now, we get a matrix M5×6 by deleting a column from the matrix M5×7. Then, the matrix M5×6 must be not 2-separable.
2
The Matrix Representation
Consider a s × t 0-1 matrix M where Ri and Cj denote row i and column j, respectively. M is called d-separable if the boolean sums of d columns are all distinct. M is called d-separable if the boolean sums of ≤ d columns are all distinct. M is called d-disjunct if the boolean sum of any d columns does not contain any other column. It is clear to know that d-separable implies d-separable and d-separable implies k-separable for every 1 ≤ k ≤ d.
Let B(S) denote the boolean sum of a set S of columns.
Lemma 1. [2] If the matrix M is d-disjunct then M is d-separable. Proof. Suppose that M is not d-separable, i.e., there exist a set K of k columns and another set K′
of k′ columns, 1 ≤ k ≤ k′ ≤ d, such that B(K) = B(K′ ). Let Cj be a column in K ′ \ K. Then Cj ⊆ B(K) and M is not k-disjunct, hence not d-disjunct.
Lemma 2. [2] Deleting any row Ri from a d-disjunct matrix M yields a d-separable matrix Mi.
Proof. Let S be a set of d columns and S′
be an another set of d columns. We claim that B(S) and B(S′
) must differ in at least 2 rows. Suppose not, B(S) and B(S′
) differ in one row. Assume B(S) ⊆ B(S′
cloumn Ci in S\S ′
such that Ci ⊆ B(S) ⊆ B(S ′
). Since, M is d-disjunct. This is a contradiction. Hence, they are different even after the deletion of a row.
Theorem 3. [3] Let M be a d-separable matrix and 1 ≤ k ≤ d − 1. Then M is k + 1-separable, if and only if M is k-disjunct.
Proof. Suf f iciency:
Suppose to the contrary that there exist two distinct sets S and S′ of columns in M, | S |≤ k + 1, | S′
|≤ k + 1, such that B(S) = B(S′
). By the d-separable property of M, we may assume | S |<| S′
|≤ k + 1. Then there exist a column C ∈ S′
\S. Since C ⊆ B(S′
), we obtain C ⊆ B(S), which violates the k-disjunct property of M.
Necessity:
Suppose M is not k-disjunct, i.e., there exist a column C and a set S of k other columns such that C ⊆ B(S). Then B(S) = B(S′
) where S′
= SS{C} and | S |, | S′
|≤ k + 1. Hence M is not k + 1-separable.
3
Basic Definitions of BIBD
Definition 4. A design is a pair (X, B) such that the following properties are satisfied:
1. X is a set of elements called points, and
2. B is a collection of nonempty subsets of X called blocks.
Let v, k, and λ be positive integers such that v > k ≥ 2. A (v, k, λ)-balanced incomplete block design (which we abbreviate to (v, k, λ)-BIBD) is
a design (X, B) such that the following properties are satisfied: 1. | X |= v,
2. each block contains exactly k points, and
3. every pair of distinct points is contained in exactly λ blocks. Example 5. A (7, 3, 1)-BIBD
X = {1, 2, 3, 4, 5, 6, 7},
B = {123, 145, 167, 246, 257, 347, 356}.
We will use the notation that b =| B | and rx is the number of blocks containing x, for all x ∈ X. In a (v, k, λ)-BIBD, every point has the same number of blocks which pass it. So, we called rx = r.
Definition 6. The incidence matrix of (X, B) is the v × b 0-1 matrix M = (mi,j) defined by the rule
mi,j = 1 if xi ∈ Aj, 0 if xi ∈ A/ j,
where A1, · · ·, Av are blocks. The incidence matrix, M of a (v, k, λ)-BIBD satisfies the following properties:
1. every column of M contains exactly k 1’s , 2. every row of M contains exactly r = λ(v−1)k−1 1’s,
3. two distinct rows of M both contain 1 in exactly λ columns.
An (n2+ n + 1, n + 1, 1)-BIBD with n ≥ 2 is called a projective plane of order n and it is a symmetric BIBD (b = v, r = k).
Corollary 7. The incidence matrix of a projective plane of order n is n-disjunct.
Proof. In the incidence matrix of a projective plane of order n, any two columns intersect in exactly 1 point and every column contains exactly n + 1 1’s. Now, we take a set S of n columns. Suppose there exists another cloumn Cj with weight n + 1 such that Cj ⊆ B(S). By P igeonhole P rinciple, the column Cj and one of columns in S have two 1’s in their intersection. This is a contradiction.
Corollary 8. The matrix obtained by deleting a row of the incidence matrix of the projective plane of order n is n-separable.
Proof. Now, we delete a point from a projective plane, i.e.,deleting a row from the incidence matrix M. Let it be M′
. By Lemma 2, M′
is n-separable. Every column in M′
contains n+1 or n 1’s. Now, we claim that M′
is (n−1)-disjunct. We take a set S of n − 1 columns. Suppose there exists another cloumn Cj with weight n such that Cj ⊆ B(S). By P igeonhole P rinciple, the column Cj and one of columns in S have two 1’s in their intersection. This is a contradiction. M′
is (n − 1)-disjunct. By Theorem 3, M′ is n-separable.
In particular, when n = 2, this is a (7,3,1)-BIBD, i.e., this is a projective plane of order 2. The 6×7 matrix obtained by deleting a row of the incidence matrix of the projective plane of order 2 is 2-separable. Now, we prove that there is no 2-separable s × t matrix with s < t < 7.
4
The main result
Theorem 9. There is no 2-separable s × t matrix with s < t < 7.
Proof. If the s×t matrix is not 2-separable, the k×t matrix is not 2-separable for k < s, either. So, we just consider the condition s = t − 1. Suppose to the contrary that there exists a 2-separable matrix Ms×t = [mij]. So, any two columns in Ms×t are different. Let (s, t) be such a pair of Ms×t that t is smallest.
First, we have two claims:
Suppose there is a zero column in Ms×t. Any column union with the zero column is still itself. This is a contradiction.
Suppose there is a column with one 1 in Ms×t. Then the other elements of the row corresponding to this 1 are all 0. Otherwise, Ms×t is not a 2-separable matrix. So, Ms×t has the following form.
0 . .. ... 0 0 . . . 0 1 0 . . . 0 0 .. . . .. 0
But we can get a 2-separable matrix Ms−1×t−1 by deleting the row and the column corresponding to this 1. This is a contradiction to t be the smallest.
2. Each column in Ms×t has at most s-2 1’s
Suppose there is a column with all 1’s in Ms×t. Any column union with the this column is this column. This is a contradiction.
column as follows. 1 1 .. . 1 1 0
Then msj = 1, where 1 ≤ j ≤ t − 1. Otherwise, if a msk = 0 for some 1 ≤ k ≤ t − 1, the union of column k and column t is identical with
column t. So, Ms×t= 1 1 ... 1 1 1 1 · · · 1 1 0
But the union of any column from 1 to t-1 and the column t is a column with all 1’s. This is a contradiction to Ms×t be 2-separable.
Since Ms×t is 2-separable. The columns which we choosed have t 1 + t 2
conditions. Since each column in M5×6 has at least 2 1’s, the boolean sum of the columns which we choosed have
s 2 + s 3 + · · · + s s
results. Since the number of results is more than the number of conditions, the 2-separable matrix Ms×t satisfies
t 1 + t 2 ≤ s 2 + s 3 + · · · + s s . Now we want to discuss the conditions for t < 7.
1. When s=2, t=3; LHS= 6, RHS= 1. This is a contradiction. 2. When s=3, t=4; LHS= 10, RHS= 4. This is a contradiction. 3. When s=4, t=5; LHS= 15, RHS= 11. This is a contradiction. 4. When s=5, t=6; LHS= 21, RHS= 26. This case satisfies the neccessary condition. Hence, we just consider the matrix M5×6 .
By two claims, the weight of a column in M5×6 is 2 or 3, so we have 5 conditions.
1. There are at least four columns with weight 3.
2. There are three columns with weight 3 and three columns with weight 2 in M5×6.
3. There are two columns with weight 3 and four columns with weight 2 in M5×6.
4. There are only one column with weight 3 in M5×6. 5. The weight of every column in M5×6 is 2.
Now we discuss the cases step-by-step.
First, we define N = (n1, n2, n3, n4, n5) where ni is the number of zeros at the ith row in M5×6.
Case 1: There are at least four columns with weight 3.
We just consider the matrix M5×4 which consists of four columns with weight 3. In other word, every column in this M5×4 has 2 0’s. So, there are 8 0’s in this M5×4. Now, we want to discuss the conditions of N. If N = (4, 4, 0, 0, 0), then M5×4 as follows. 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1
But there are two columns identical. This is a contradiction. Thus, we
find N such that any two columns are different. So, when N = (4, 4, 0, 0, 0), (4, 3, 1, 0, 0), (4, 2, 2, 0, 0), (4, 2, 1, 1, 0), (3, 3, 2, 0, 0) and (3, 3, 1, 1, 0), they do not satisfy
Case 1.1: N = (4, 1, 1, 1, 1). W.L.O.G, we take M5×4 as follows. A B C D 0 0 0 0 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0
But the unions of any two columns are identical. Hence, M5×4 is not 2-separable. Thus, M5×6 is not 2-separable in this case.
Case 1.2: N = (3, 2, 2, 1, 0). W.L.O.G, we take M5×4 as follows. A B C D 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 1 1 1 1
But the union of column A and column B is identical with the union of column A and column C. Hence, M5×4is not 2-separable. Thus, M5×6 is not 2-separable in this case.
Case 1.3: N = (3, 2, 1, 1, 1). W.L.O.G, we take M5×4 as follows. A B C D 0 0 0 1 0 1 1 0 1 0 1 1 1 1 0 1 1 1 1 0
But the union of column A and column B is identical with the union of column A and column C. Hence, M5×4is not 2-separable. Thus, M5×6 is not 2-separable in this case.
Case 1.4: N = (2, 2, 2, 1, 1). W.L.O.G, we take M5×4 as follows. A B C D 0 0 1 1 1 0 0 1 1 1 0 0 0 1 1 1 1 1 1 0
But the union of column A and column C is identical with the union of column A and column D. Hence, M5×4 is not 2-separable. Thus, M5×6 is not 2-separable in this case.
Case 1.5: N = (2, 2, 2, 2, 0). W.L.O.G, we take M5×4 as follows. A B C D 0 0 1 1 1 0 0 1 1 1 0 0 0 1 1 0 1 1 1 1
But the union of column A and column C is identical with the union of column B and column D. Hence, M5×4is not 2-separable. Thus, M5×6 is not 2-separable in this case.
Note: A column with weight 2 has ten conditions. A B C D E F G H I J 1 1 1 1 0 0 0 0 0 0 1 0 0 0 1 1 1 0 0 0 0 1 0 0 1 0 0 1 1 0 0 0 1 0 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 In the following cases, we will use it.
Case 2: There are the three columns with weight 3 and three columns with weight 2 in M5×6.
First, we take three columns with weight 3. There are 6 0’s in these cloumns. Now, we find N such that two columns are different. So, when N = (3, 3, 0, 0, 0) and (3, 2, 1, 0, 0), they do not satisfy the condition. And we find four cases for N which satisfy the condition.
Case 2.1: N = (3, 1, 1, 1, 0). W.L.O.G, we take the three columns as follows. 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1
But the union of any two columns are identical. Thus, M5×6 is not 2-separable in this case.
Case 2.2: N = (2, 2, 1, 1, 0). W.L.O.G, we take the three columns as follows. X Y Z 0 0 1 0 1 0 1 0 1 1 1 0 1 1 1
Since M5×6is 2-separable matrix, we can’t take columns B, D, F, G, H, I, J. Hence, we have M5×6 as follows.
X Y Z A C E 0 0 1 1 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 0 0
But the union of column X and column A is identical with the union of column Y and column Z. Thus, M5×6 is not 2-separable in this case.
Case 2.3: N = (2, 2, 2, 0, 0). W.L.O.G, we take the three columns as follows. 0 0 1 0 1 0 1 0 0 1 1 1 1 1 1
Since M5×6is 2-separable, we can’t take columns C, D, F, G, H, I, J. Hence, we have M5×6 as follows. A B E 0 0 1 1 1 0 0 1 0 1 0 1 1 0 0 0 1 1 1 1 1 0 0 0 1 1 1 0 0 0
But the union of column A and column B is identical with the union of column A and column E. Thus, M5×6 is not 2-separable in this case.
follows. X Y Z 0 0 1 1 1 0 1 0 1 0 1 1 1 1 0
But the union of column X and column Z is identical with the union of column Y and column Z. Thus, M5×6 is not 2-separable in this case.
Case 3: There are two columns with weight 3 and four columns with weight 2 in M5×6.
First, we take two columns with weight 3. There are 4 0’s in these cloumns. Now, we find N such that two columns are different. So, when N = (2, 2, 0, 0, 0), it do not satisfy the condition. And we find two cases for N which satisfy the condition.
Case 3.1: N = (1, 1, 1, 1, 0). W.L.O.G, we take the two columns as follows. 0 1 0 1 1 0 1 0 1 1
we have M5×6 as follows. B C E F 0 1 1 1 0 0 0 1 0 0 1 1 1 0 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 0
But the union of column B and column F is identical with the union of column C and column E. Thus, M5×6 is not 2-separable in this case.
Case 3.2: N = (2, 1, 1, 0, 0). W.L.O.G, we take the two columns as follows. 0 0 0 1 1 0 1 1 1 1
Since M5×6 is 2-separable, we can’t take columns F, G, H, I, J and we just can take one of columns B, C, D. But we only have five columns. This is a contradiction. Thus, M5×6 is not 2-separable in this case.
Case 4: There are only one column with weight 3 in M5×6.
as follow. 1 1 1 0 0
Since M5×6 is 2-separable, we can’t take columns A, B, E. And we just can take one of columns C, F, H, we just can take one of columns D, G and we just can take two of columns H, I, J. But we only have five columns. This is a contradiction. Thus, M5×6 is not 2-separable in this case.
Case 5: The weight of every column in M5×6 is 2.
There are 2 × 6 = 12 1’s in M5×6. But there are five rows. By P igeonhole P rinciple, there must be a row that contains 3 1’s in M5×6.
1 1 1
P rinciple, there must be a row that contains 3 0’s. 1 1 1 0 0 0 W.L.O.G, the three columns are
B C D 1 1 1 0 0 0 1 0 0 0 1 0 0 0 1
M5×6 is 2-separable. But the union of column C and column H is identical with the union of column B and column H, the union of column B and column I is identical with the union of column D and column I and the union of column C and column J is identical with the union of column D and column J. So, we can not take cloumns H, I and J. Since the union of column A and column D is identical with the union of column G and column D, we just can take one of columns A, G. Since the union of column C and column E is identical with the union of column B and column F, we just can take one of columns E, F. But we only have five columns. This is a contradiction. Thus, M5×6 is not 2-separable in this case.
Through above discussion, M5×6 is not 2-separable. Thus, there is no 2-separable s × t matrix with s < t < 7.
Conjecture 10. There is no d-separable matrix of size s × t with s < t < d2+ d + 1.
5
References
[1] S. H. Huang, Frank K. Hwang, When is Individual Testing Optimal for Nonadaptive Group Testing?, Department of Applied Mathematics Na-tional Chiao Tung University, Hsinchu 30050, Taiwan, R.O.C.
[2] Du, Ding-Zhu, Frank K. Hwang, Combinatorial group testing and its applications, 2nd ed, pp135.
[3] Hong-Bin Chen, Frank K. Hwang, Exploring the missing link among d-separable, d-separable and d-disjunct matrices, Discrete Applied Mathematics 155(2007) 662-664.
