在m乘n陣列裡的橫截

國 立 交 通 大 學

應用數學系

碩 士 論 文

在 m 乘 n 陣列裡的橫截

Transversals in m × n Arrays

研 究 生：李張圳

指導教授：傅恆霖 教授

中 華 民 國 九 十 六 年 六 月

在 m 乘 n 陣列裡的橫截

Transversals in m × n Arrays

研 究 生：李張圳 Student：Chang-Chun Lee

指導教授：傅恆霖 Advisor：Hung-Lin Fu

國 立 交 通 大 學

應 用 數 學 系

碩 士 論 文

A Thesis

Submitted to Department of Applied Mathematics

College of Science

National Chiao Tung University

in Partial Fulfillment of the Requirements

for the Degree of

Master

in

Applied Mathematics

June 2006

Hsinchu, Taiwan, Republic of China

謝誌

首先誠摯的感謝我的指導教授傅恆霖老師，在傅老師的悉心指導

下，使我對組合數學可以有更深一層的了解，老師也教導我做研究應有

的態度與方向，在寫論文上，老師也給我很多寫作上的意見，使我能完

成我的碩士論文，老師不僅在課業上幫助良多，有時也會教導我們一些

做人處事的道理，這兩年從老師身邊學到很多，真的很感謝老師，讓我

可以開心收穫良多的順利畢業。

在學校的這兩年，不僅傅老師對我幫助甚多，還要感謝黃大原老師、

翁志文老師以及陳秋媛老師，這三位老師在課業上的指導，讓我對組合

數學更加有興趣。

另外我還要感謝我研究所的同學，肌肉澍仁、老闆國安、美女ＲＥ、

嘴砲老吳、帥哥柏澍、歌王宜庭、文強、妙妙、強者皜文、威雄、帥哥

怡中、好友老謝以及所有的學長姐跟學弟妹讓我在交大的這兩年生活可

以更多彩多姿更加的開心，也希望大家畢業後要記得保持聯絡喔！

最後，謹以此文獻給我摯愛的雙親。

在 m 乘 n 陣列裡的橫截

研 究 生：李張圳

指導老師：傅恆霖 教授

國 立 交 通 大 學

應 用 數 學 系

摘

要

當2≤m≤n，一個m乘n的陣列是由m個列和n個行組成的mn個格子。在m乘n的陣列裡 的一個部分橫截是收集m個格子的集合，這些格子是來自不同行不同列。在m乘n的 陣列裡的一個橫截是一個部分橫截，這個部分橫截裡的m個符號都是不一樣的。定 義L(m,n)是一個最大的整數使得如果每一個符號在m乘n的陣列裡出現最多L(m,n) 次，則這個陣列一定會有一個橫截。在本篇論文，我們把找拉丁方陣的橫截的研究 延伸到找m乘n陣列的橫截的研究。大體上，我們對於對某些正整數m和n的L(m,n)值 感到興趣。

中 華 民 國 九 十 六 年 六 月

Transversals in m × n Arrays

Student: Chang-Chun Lee

Advisor: Hung-Lin Fu

Department of Applied Mathematics National Chiao Tung University

Hsinchu, Taiwan 30050

Abstract

An m by n array consists of mn cells in m rows and n columns, where 2 ≤ m ≤ n. A partial transversal in an m by n array is a set of m cells, one from each row and no two from the same column. A transversal in an m by n array is a partial transversal which m symbols are distinct. Define L(m, n) as the largest integer such that if each symbol in an m by n array appears at most L(m, n) times, then the array must have a transversal. In this thesis, we extend the study of finding transversals in a Latin square to find transversals in m × n arrays. Mainly, we are interested in determining the value L(m, n) for certain pairs of positive integers m and n.

Contents

Abstract (in Chinese) i

Abstract (in English) ii

Contents iii

1 Introduction and Preliminaries 1

1.1 Introduction . . . 1

1.2 Preliminaries . . . 4

1.2.1 Probabilistic method: Lov´asz Local Lemma . . . 4

1.2.2 Ideas in direct argument . . . 5

2 Known Results 7

3 Main Result 13

1

Introduction and Preliminaries

1.1

Introduction

A Latin square M of order n based on an n-set S is an n × n array such that each symbol of S occurs in each row and each column exactly once. For convenience, we may use S = {1, 2, 3, ..., n} and the symbol appears in the i-th row and j-th column is called the (i, j)-entry of the Latin square, denoted by M (i, j). Then, the following figures are examples of a Latin square of order 4 and a Latin square of order 5 re-spectively. M1 = 1 2 3 4 4 3 2 1 2 1 4 3 3 4 1 2 M2 = 1 2 3 4 5 5 1 2 3 4 4 5 1 2 3 3 4 5 1 2 2 3 4 5 1

A transversal T of a Latin square is a set of n cells such that no two are in the same row and the same column and the symbols occur in T are distinct. It is not difficult to see that the above squares have transversals respectively. For exam-ples, {(1, 1), (2, 2), (3, 3), (4, 4)} and {(1, 1), (2, 3), (3, 5), (4, 2), (5, 4)}. These two sets are the transversals of M1 and M2 respectively. But, not every Latin square has a transversal. For example,

M3 = 1 2 3 4 5 6 2 3 1 5 6 4 3 1 2 6 4 5 4 5 6 1 2 3 5 6 4 2 3 1 6 4 5 3 1 2

It is easy to check that M3 has no transversal. Therefore, to determine whether a Latin square has a transversal or not is an interesting problem. More than 250 years

ago, Euler conjectured that there do not exist two orthogonal Latin squares of order 4k + 2 for each positive integer k. It is believed that the idea is mainly originated from the fact that there exists a Latin square of order 4k + 2 which does not have a transversal. This is easy to see from M3.

Now, we known that a pair of orthogonal Latin squares of order 4k+2, k ≥ 2 , does not exist [9]. But, for a given Latin square, to determine whether a transversal exists is still an open problem. Toward solving this problem, in 1967, Ryser [7] conjectured that every Latin square of odd order has a transversal, and the number of transversals of a Latin square has the same parity as the order of the square. But, Parker pointed out that many Latin square of order 7 have an even number of transversals in 1989. Balasubramanian [2] proved that a Latin square of even order has an even number of transversals in 1990.

Unfortunately, the above results do not provide any assistance in determining whether there exists a transversal in a given Latin square or not. An intuitive ap-proach is to find as many distinct elements from distinct rows and columns as possi-ble. A partial transversal of a Latin square is a set of n cells from distinct rows and columns. The size of a partial transversal is the number of distinct symbols which appears in the partial transversal. For example, P1 = {(1, 1), (2, 3), (3, 2), (4, 4)} is a partial transversal of M1 of size 2. P2 = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)} is a par-tial transversal of M2 of size 1. It is easy to see that we can always find a partial transversal of size at least n/2 in a Latin square of order n. (Pick any cell in the first row, then a cell in the second row with a different symbol, and so on.) But, for larger size, it takes a while to get to the best known result today. First, in 1969, Koksma [6] showed that the length of a partial transversal in a Latin square is at least n − (1/3)n. Later Drake [3] showed that the lower bound is n − (1/4)n in 1977. Then,

by using the idea of matchings in the bipartite graph Kn,n, Woolbright [11] improved this lower bound to n −√n in 1978. Four years later, 1982, Shor [10] gave a better bound n − (5.53)(ln n)2_{. Finally, by using a careful calculation in Shor’s technique,} Fu et al. [5] improved this the lower bound to n − (5.518)(ln n)2 in 2002.

Recently, the notion ”transversals in Latin square” has been converted to that of arrays where we allow common symbols in both rows and columns. For positive integers m and n, where 2 ≤ m ≤ n, an m by n array contains m rows and n columns. An m by n array A consists of mn cells and each cell contains one symbol and for 1 ≤ i ≤ m and 1 ≤ j ≤ n, we use A(i, j) to denote the symbol which appears in the row i and column j. A partial transversal in an m by n array is a set of m cells such that no two are in the same row and the same column. A partial transversal of size k contains exactly k distinct symbols which appears in the partial transversal. A transversal is a partial transversal of size m. Let L(m, n) be the largest integer such that if each symbol in an m by n array appears at most L(m, n) times, then the array must have a transversal. For example,

A = 1 1 2 3 4 2 4 1 2 5 3 2 B = 1 1 2 2 2 2 3 3 3 3 1 1

Then A and B are 3 by 4 arrays. Each symbol in A appears at most 4 times. Each symbol in B appears at most 4 times. T = {(1, 1), (2, 2), (3, 3)} is a transversal of A. P = {(1, 1), (2, 2), (3, 3)} is a partial transversal of B of size 2. It is easy to check that B has no transversal. By the array B, L(3, 4) < 4. In 1991, P. Erd˝os and J. Spencer [4] showed that an array of order n in which each symbol appears at most (n − 1)/16 times has a transversal. This implies L(n, n) ≥ b(n − 1)/16c. Recently, S. Akbari. et al. [1] proved that L(m, n) = b(mn − 1)/(m − 1)c for m ≥ 2 and

n ≥ 2m3_{− 6m}2_{+ 6m − 1. In this thesis, we study the value L(m, n) for certain pairs} of positive integers m and n.

1.2

Preliminaries

1.2.1 Probabilistic method: Lov´asz Local Lemma

Let A1, A2, ..., An be events in an arbitrary probability space. Let ¯Ai denote the complement of event Ai. Then the probability of A1 given A2 is P r(A1|A2) =

P r(A1∩ A2) P r(A2)

. If P r(A1|A2) = P r(A1), we say that A1 and A2 are mutually inde-pendent. Let S be a set of events. In general, Ai is mutually independent of S if P r(Ai|

T

Aj∈TAj) = P r(Ai) for all T ⊆ {Aj|Aj ∈ S or ¯Aj ∈ S}.

Definition 1.1. Let A1, A2, ..., An be events in an arbitrary probability space. A graph G = (V, E) on the set of vertices V = {1, 2, ..., n} is called a lopsidependency graph for the events A1, A2, ..., An if P r(Ai|

T

j∈SA¯j) ≤ P r(Ai) for each i ∈ V and each S ⊆ V \ NG[i].

Definition 1.2. Let A1, A2, ..., An be events in an arbitrary probability space. A di-rected graph D = (V, E) on the set of vertices V = {1, 2, ..., n} is called a dependency digraph for the events A1, A2, ..., An if for each i, 1 ≤ i ≤ n, the event Ai is mutually independent of all the events {Aj : (i, j) /∈ E}.

Theorem 1.3. [Lopsided Lov´asz Local Lemma] Let A1, A2,...,An be events with lopsidependency graph G and suppose all the events have probability at most p and that each i ∈ G has degree at most d. Assume 4pd ≤ 1. Then P r(Tn

i=1A¯i) > 0. The following lemma, first proved in Erd˝os and Lov´asz in 1975, is an extremely powerful tool.

Theorem 1.4. [Lov´asz Local Lemma; General Case] Let A1, A2, ..., An be events in an arbitrary probability space. Suppose that D = (V, E) is a dependency

digraph for the above events and suppose there are real numbers x1, x2, ..., xn such that 0 ≤ xi < 1 and P r(Ai) ≤ xi

Q

(i,j)∈E(1 − xi) for all 1 ≤ i ≤ n. Then

P r(Tn

i=1A¯i) ≥ Qn

i=1(1 − xi). In particular, with positive probability for no event Ai holds.

Theorem 1.5. [Lov´asz Local Lemma; Symmetric Case] Let A1, A2, ..., An be events in an arbitrary probability space. Suppose that each event Ai is mutually inde-pendent of a set of all the other events Aj but at most d, and that P r(Ai) ≤ p for all 1 ≤ i ≤ n. If ep(d + 1) ≤ 1 then P r(Tn

i=1A¯i) > 0.

In 1985, Shearer proved that the constant ”e” is the best possible constant in the above lemma. In Lov´asz Local Lemma of general case, we can replace the two assumptions that each ”Ai is mutually independent of {Aj : (i, j) /∈ E} ” and that ”P r(Ai) ≤ xiQ_(i,j)∈E(1 − xi) ” by the weaker assumption that ”for each i and each S ⊂ {1, 2, ..., n}\{j : (i, j) ∈ E}, P r(Ai|T_j∈SA¯j) ≤ xiQ_(i,j)∈E(1 − xi) ”.

1.2.2 Ideas in direct argument

Besides probabilistic method, we also use a direct argument to find the lower bound of L(m, n). The idea is based on the following fact which is easy to see.

Proposition 1.6. Let A be an m by n array such that A has a transversal. Then, the new array A0 obtained by the following three operations also has a transversal.

1. a permutation of rows

2. a permutation of columns

So, without loss of generality, we may assume the transversal of an m by n array A lies on the following set of cells: {(1, 1), (2, 2), ..., (m, m)}. For convenience, we also use A(1, 1), A(2, 2), ..., A(m, m) to denote the transversal of A.

2

Known Results

For completeness, we also include their proofs.

Theorem 2.1. [4] Given an n × n array A. Let k ≤ (n − 1)/16 and suppose that no entry of A appears more than k times. Then A has a transversal.

Proof. We use Lopsided Lov´asz Local Lemma. Let Sn be a set of permutations on an n-set. Let V = {(s, t, u, v)| s < u, t 6= v and A(s, t) = A(u, v)}. For each (s, t, u, v) ∈ T , let Astuv = {σ| σ ∈ Sn, σ(s) = t and σ(u) = v}. Then A has a transversal if and only if P r(T

(s,t,u,v)∈V A¯stuv) 6= 0. Hence we will show that P r(T

(s,t,u,v)∈V A¯stuv) 6= 0.

Note that P r(Astuv) = (n − 2)!/n! = 1/n(n − 1).

Define a graph G with vertex set V and (s, t, u, v) adjacent to (x, y, z, w) if and only if {s, u} ∩ {x, z} 6= ∅ or {t, v} ∩ {y, w} 6= ∅. Then we can count the maximal degree of G. Given (s, t, u, v) ∈ V , there are at most 4n choices of (x, y) with either x ∈ {s, u} or y ∈ {t, v} and k choices for (z, w) with A(x, y) = A(z, w). Either (x, y, z, w) adjacent to (s, t, u, v) or (z, w, x, y) adjacent to (s, t, u, v). Thus G has maximal degree at most 4nk. Then 4 · 4nk · (1/n(n − 1)) ≤ 1.

To show G is a lopsidependency graph. By symmetric, it suffices to show P r(A1122|T_{(s,t,u,v)∈S}A¯stuv) ≤ 1/n(n − 1) where s, t, u, v 6= 1, 2.

Let Nij = {σ| σ(1) = i, σ(2) = j and σ ∈T_{(s,t,u,v)∈S}A¯stuv} Claim: |N12| ≤ |Nij| for all i 6= j.

subpf : If i, j > 2. Let σ ∈ N12. There exist a, b with σ(a) = i, σ(b) = j. Define σ∗ by σ∗(1) = i, σ∗(2) = j, σ∗(a) = 1, σ∗(b) = 2, and σ∗(x) = σ(x) for all x 6= 1, 2, a, b. Since (1, i), (2, j), (a, 1), (b, 2) are not part of any element in S, σ∗ is in Nij. Then f : N12→ Nij is injective. Thus |N12| ≤ |Nij|. The case {1, 2} ∩ {i, j} 6= ∅ is similar.

Hence, P r(A1122| \ (s,t,u,v)∈S ¯ Astuv) = |N12|/ X i6=j |Nij| ≤ |N12|/ X i6=j |N12| = 1/n(n − 1).

By Lopsided Lov´asz Local Lemma, P r(T

(s,t,u,v)∈V A¯stuv) 6= 0. So A has a transversal. The followings are direct proofs

Lemma 2.2. [8] (1) L(m + 1, n) ≤ L(m, n) and (2) L(m, n) ≤ L(m, n + 1).

Proof. (1) Suppose that L(m + 1, n) = k. Consider an m by n array A in which each symbols appears at most k times. Without loss of generality, the symbols in A are positive integers. Then we add a row to get an (m + 1) × n array B and the symbols in that row are negative integers and each symbol in that row appears at most k times. Hence B has a transversal. This implies that A must have a transversal.

(2) Suppose that L(m, n) = k. Consider an m by (n + 1) array A in which each symbols appears at most k times. Deleting the first column, then we get an m × n array B. Hence B has a transversal. This implies that A must have a transversal.

Theorem 2.3. [8] If n ≤ 2m − 2, then L(m, n) ≤ n − 1.

Proof. We illustrated for the cases when (m, n) = (3, 3), and (m, n) = (3, 4):

1 1 3 2 2 1 3 3 2 1 1 3 3 2 2 1 1 3 3 2 2

It is easy to check that the above arrays have no tranversals.

Theorem 2.4. [8] L(m, n) < mn/(m − 1).

Proof. If only m − 1 distinct symbols appear in an m × n array, the array has no transversal. Hence, if each of (m − 1) symbols appears at most mn/(m − 1) times, the symbols can fill all the cells.

Theorem 2.5. [8] L(2, n) = 2n − 1 for n ≥ 3.

Proof. Consider a 2 by n array A in which each symbol appears at most 2n − 1 times. Suppose A has no transversal. Then A is equivalent to the following array:

1 b b b b b ...

a 1 a a a a ...

It is easy to check that a, b stand for 1. Then 1 appears 2n times, a contradiction.

Lemma 2.6. [8] Assume that in a 3 by n array, n ≥ 4, some symbol occurs at most three times. Then, if there is no transversal some symbol occurs at least 2n − 2 times, hence at least 3n/2 times.

Proof. There are 10 inequivalent cases when one symbol appears at most three times. We list the 10 cases.

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

We illustrate the case when 1 appears one time. Then we have the following ar-ray:

1

2 b b b b b ...

It is easy to check that a and b stand for 2. Hence the symbol 2 appears at least 2n − 2 times. The other cases are similar.

Theorem 2.7. [8] (a) L(3, 3) = 2 and L(3, 4) = 3. (b) For n ≥ 5, L(3, n) = b(3n − 1)/2c.

Proof. Exhaustive computer calculations shows that L(3, 3) = 2, L(3, 4) = 3, L(3, 5) = 7.

By induction on n. Assume that the induction holds for a particular odd n. i.e. L(3, n) = (3n − 1)/2. We will show that it holds for n + 1, that is, L(3, n + 1) = (3n + 1)/2.

Consider a 3 by n + 1 array A in which each symbol appears at most (3n + 1)/2 times. If each symbol appears at most (3n − 1)/2 times, then deleting one column to obtain a 3 by n array. By induction hypothesis, the 3 by n array has a transversal. Hence A has a transversal.

Suppose there is at least one symbol appears at least (3n + 1)/2 times. If there are two such symbols, they appear at least 3n + 1 times. Hence some symbol appears at most three times. By Lemma 2.6, if there is no transversal, then some symbol occurs at least 3(n + 1)/2 times. So A has a transversal.

Hence there is only one symbol that appears at least (3n + 1)/2 times. There must be a column in which it appears at least twice. Deleting that column, we get a 3 by n array in which each symbol appears at most (3n − 1)/2 times. By induction hypothesis, the 3 by n array has a transversal. Hence A has a transversal. Thus L(3, n + 1) ≥ (3n + 1)/2. By Theorem 2.4, L(3, n + 1) < (3n + 3)/2. So, L(3, n + 1) = (3n + 1)/2. When n is even, the argument is similar.

Proof. We use induction on m to prove the assertion.

The theorem is true for m = 2 or m = 3. Assume that it is true for m − 1. We will show that it holds for m.

Assume that L(m − 1, n) ≥ n − m + 2. Consider an m by n array A in which each symbol appears at most n − m + 1 times. Deleting the last row of A, we get an m − 1 by n array. The m − 1 by n array has a transversal. Suppose that A has no transversal. Then A is equivalent to the following array:

1 a a a a ... 2 ... 3 ... . .. ... m − 1 ... 1 a a a a ...

An a stands for 1, 2, ..., m − 1. Then there are at least 2(n − m) + 2 cells containing a or 1. Since 1 appears at most n − m + 1 times in A and 2(n − m) + 2 > n − m + 1, there must be an element in {2, 3, ..., m−1} occuring in some cells marked a. Without loss of generality, we take the symbol to be 2. Then we have the following array:

1 2 a a a ... a 2 a a a ... 3 ... . .. ... m − 1 ... 1 a a a a ...

Then there are at least 3(n − m) + 3 cells containing a, 1 or 2. Since 1 and 2 appear at most 2n − 2m + 2 times inA and 3(n − m) + 3 > 2n − 2m + 2, there be an element in {3, ..., m − 1} occuring in some cells marked a. Without loss of generality, we take the symbol to be 3.

1 2 3 a a ... a 2 a a a ... a 3 a a a ... . .. m − 1 ... 1 a a a a ...

Continuing the analysis, the symbols 1, 2, ..., m − 1 appear at least (m − 1)(n − m) + m times. But, (m − 1)(n − m) + m > (m − 1)(n − m + 1), a contradiction. This concludes the proof.

3

Main Result

Theorem 2.1 implies L(n, n) ≥ b(n − 1)/16c. We improve this lower bound.

Theorem 3.1. L(n, n) ≥ b(n + 4e)/4ec.

Proof. Let k = b(n + 4e)/4ec.

Consider an n by n array A in which each symbol appears at most k times. We use Lov´asz Local Lemma. Let Sn be a set of permutations on an n-set. Let V = {(s, t, u, v)| s < u, t 6= v and A(s, t) = A(u, v)}. For each (s, t, u, v) ∈ T , let Astuv = {σ| σ ∈ Sn, σ(s) = t and σ(u) = v}. Then A has a transversal if and only if P r(T

(s,t,u,v)∈V A¯stuv) 6= 0. Hence we will show that P r( T

(s,t,u,v)∈V A¯stuv) 6= 0. Note that P r(Astuv) = (n − 2)!/n! = 1/n(n − 1).

Define a graph G with vertex set V and (s, t, u, v) adjacent to (x, y, z, w) if and only if {s, u} ∩ {x, z} 6= ∅ or {t, v} ∩ {y, w} 6= ∅. Then we can count the maximal degree of G. Given (s, t, u, v) ∈ V , there are at most 4n − 4 choices of (x, y) with either x ∈ {s, u} or y ∈ {t, v} and k − 1 choices for (z, w) with A(x, y) = A(z, w). Either (x, y, z, w) adjacent to (s, t, u, v) or (z, w, x, y) adjacent to (s, t, u, v). Thus G has maximal degree at most (4n − 4)(k − 1) − 1. Then e · ((4n − 4)(k − 1) − 1 + 1) · (1/n(n − 1)) ≤ e · 4(n − 1)(n/4e) · (1/n(n − 1)) = 1.

To show G is a lopsidependency graph. By symmetric, it suffices to show P r(A1122|T_{(s,t,u,v)∈S}A¯stuv) ≤ 1/n(n − 1) where s, t, u, v 6= 1, 2.

Let Nij = {σ| σ(1) = i, σ(2) = j and σ ∈ T

(s,t,u,v)∈SA¯stuv} Claim: |N12| ≤ |Nij| for all i 6= j.

subpf : If i, j > 2. Let σ ∈ N12. There exist a, b with σ(a) = i, σ(b) = j. Define σ∗ by σ∗(1) = i, σ∗(2) = j, σ∗(a) = 1, σ∗(b) = 2, and σ∗(x) = σ(x) for all x 6= 1, 2, a, b. Since (1, i), (2, j), (a, 1), (b, 2) are not part of any element in S, σ∗ is in Nij. Then

f : N12→ Nij is injective. Thus |N12| ≤ |Nij|. The case {1, 2} ∩ {i, j} 6= ∅ is similar. Hence, P r(A1122| \ (s,t,u,v)∈S ¯ Astuv) = |N12|/ X i6=j |Nij| ≤ |N12|/ X i6=j |N12| = 1/n(n − 1).

By Lov´asz Local Lemma, P r(T

(s,t,u,v)∈V A¯stuv) 6= 0. So A has a transversal. The following results obtain from direct argument.

Lemma 3.2. L(m, n) ≤ b(mn − 1)/(m − 1)c

Proof. _{Suppose mn = k(m − 1) + r where k, r ∈ Z, 0 ≤ r < m − 1. If r = 0. By} Theorem 2.4, L(m, n) < mn/(m−1) = k. Then L(m, n) ≤ k−1 = b(mn−1)/(m−1)c. If 1 ≤ r < m − 1, L(m, n) < mn/(m − 1) = k + r/(m − 1) < k + 1. Then L(m, n) ≤ k = b(mn − 1)/(m − 1)c.

By above Lemma, if we can show that L(m, n) ≥ b(mn − 1)/(m − 1)c, then L(m, n) = b(mn − 1)/(m − 1)c. The following results use the idea.

Theorem 3.3. For n ≥ 43, L(4, n) = b(4n − 1)/3c.

Proof. Consider a 4 by n array A in which each symbol appears at most b(4n−1)/3c times. Since L(3, n) = b(3n − 1)/2c ≥ b(4n − 1)/3c, the 3 by n array consisting of the first three rows of A has a transversal. Suppose that A has no transversal. Then A is equivalent to the following array:

1 x1 x2 x3 x4 x5 ... xn−4

2 ...

3 ...

1 xn−3 xn−2 xn−1 xn xn+1 ... x2n−8

Then there are at least 2(n − 4) + 2 cells containing xi or 1. Since 1 appears at most b(4n − 1)/3c times in A and 2(n − 4) + 2 > b(4n − 1)/3c, there must be a 2 or 3 in some cells marked xi. Without loss of generality, we take x1 to be 2. Then we have the following array:

1 2 x2 x3 x4 x5 ... xn−4

y1 2 y2 y3 y4 y5 ... yn−4

3 ...

1 xn−3 xn−2 xn−1 xn xn+1 ... x2n−8

where xi, yj ∈ {1,2,3}, for all 2 ≤ i ≤ 2n − 8 and 1 ≤ j ≤ n − 4.

Then there are at least 3(n − 4) + 3 cells containing xi, yj, 1, or 2. Since 1 and 2 appear at most 2b(4n − 1)/3c times in A and 3(n − 4) + 3 > 2b(4n − 1)/3c, there must be a 3 in some cell marked xi or yj. There are 5 inequivalent cases, x2 = 3, xn−3 = 3, xn−2 = 3, y1 = 3 or y2 = 3.

If x2 = 3, then we have the following array:

1 2 3 x3 x4 x5 ... xn−4

y1 2 y2 y3 y4 y5 ... yn−4

z1 3 z2 z3 z4 z5 ... zn−4

1 xn−3 xn−2 xn−1 xn xn+1 ... x2n−8

where xi, yj, zk ∈ {1,2,3}, for all 3 ≤ i ≤ 2n − 8 and 1 ≤ j ≤ n − 4 and 1 ≤ k ≤ n − 4. Deleting the first six columns and deleting the last row we get a 3 by n − 6 array B in which each symbol appears at most b(4n − 1)/3c − 2 times. Since b(3(n − 6) − 1/2c ≥ b(4n − 1)/3c − 2, B has a transversal T . Note that the symbols

occur in T are 1, 2, 3. Hence A(4, 1), A(4, 2), A(4, 3) ∈ {1, 2, 3}. Otherwise, A has a transversal. Similarly, all cells contain 1, 2, 3. Then the symbols 1, 2, 3 appear 4n times. But 4n > 3b(4n − 1)/3c, a contradiction. Then A has a transversal. Since the argument of the other cases are similar, we omit the details. In fact, no matter

which case, we can get an 4 by n − 6 array consisting of the last n − 6 columns of A in which symbols in the array are 1, 2, 3.

Thus, L(4, n) ≥ b(4n − 1)/3c. By Lemma 3.2, L(4, n) ≤ b(4n − 1)/3c. So, L(4, n) = b(4n − 1)/3c.

We can use the same technique for general case.

Theorem 3.4. For m ≥ 2 and n ≥ 2m3_{− 8m}2 _{+ 12m − 5,} L(m, n) = b(mn − 1)/(m − 1)c.

Proof. We use induction on m to prove the assertion.

If m = 2. Then n ≥ 3. By Theorem 2.5, L(2, n) = 2n − 1. Assume that it is true for m − 1. That is L(m − 1, n) = b((m − 1)n − 1)/(m − 2)c for n ≥ 2(m − 1)3− 8(m − 1)2_{+ 12(m − 1) − 5 = 2m}3_{− 14m}2_{+ 34m − 27. To show it holds} that for n ≥ 2m3− 8m2_{+ 12m − 5, L(m, n) = b(mn − 1)/(m − 1)c.}

For n ≥ 2m3_{− 8m}2 _{+ 12m − 5, consider an m by n array A in which each} sym-bol appears at most b(mn − 1)/(m − 1)c times. Since 2m3 _{− 8m}2 _{+ 12m − 5 ≥} 2m3_{− 14m}2_{+ 34m − 27 and b((m − 1)n − 1)/(m − 2)c ≥ b(mn − 1)/(m − 1)c, then the} m − 1 by n array consisting of the first m − 1 rows of A has a transversal. Suppose that A has no transversal. Then A is equivalent to the following array:

1 a a a a ... a

2 . ..

m − 1

1 a a a a ... a

where an a stands for 1, 2, ..., m − 1.

There are at least 2(n − m) + 2 cells containing a or 1. Since 1 appears at most b(mn − 1)/(m − 1)c times and 2(n − m) + 2 > b(mn − 1)/(m − 1)c, there must

be an element in {2, 3, ..., m − 1} occuring in some cells marked a. Without loss of generality, we take the symbol to be 2. Then A is equivalent to the following array:

1 2 a a a ... a

a 2 a a a ... a

. ..

m − 1

1 a a a a ... a

where an a stands for 1, 2, ..., m − 1.

If k(n−m)+k > (k−1)b(mn−1)/(m−1)c for 2 ≤ k ≤ m−1, then we can continue the argument. It is enough to show k(m − 1)(n − m) + k(m − 1) > (k − 1)(mn − 1). k(m − 1)(n − m) + k(m − 1) − (k − 1)(mn − 1) = (m − k)n − km2 + 2km − 1. Since m − k ≥ 1, it is enough to show that n > km2 − 2km + 1 = m(m − 2)k + 1. When k is getting larger, m(m − 2)k + 1 is getting larger. It is enough to show that n > (m−1)m2_{−2(m−1)m+1. Since 2m}3_−8m2_{+12m−5 > (m−1)m}2_{−2(m−1)m+1,} then n > (m − 1)m2_{− 2(m − 1)m + 1. Hence, we can continue the argument. Then} we can get an m by n − (2m − 2) array B consisting of the last n − (2m − 2) columns of A. The symbols in B are 1, 2, ..., m − 1. And each symbol in B appears at most b(mn − 1)/(m − 1)c − 2 times. Since b((m − 1)(n − 2m + 2) − 1)/(m − 2)c ≥ b(mn−1)/(m−1)c−2, the array obtained from deleting any row in B has a transversal T . Note that the symbols occur in T are 1, 2, ..., m − 1. Then all cells contain 1, 2, ..., m − 1. Otherwise, A has a transversal. Therefore, in total 1, 2, ..., m − 1 appear mn times. But, mn > (m − 1)b(mn − 1)/(m − 1)c, a contradiction. Thus, A has a transversal. Hence L(m, n) ≥ b(mn − 1)/(m − 1)c.

4

Conclusion

From the study of the ”transversal problem” of an m by n array, we notice that the most difficult part remains in the situation when m is not that far from n. That is why the transversal problem of a Latin square is still one of the most difficult problem in combinatorial designs. So, for future study, we should focus on determining L(n, n) or L(m, n) where m is a linear function of n instead the bound we obtain in this thesis which is in cubic order.

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