Ghent UniversityFaculty of Sciences

Department of Mathematics

Mutually orthogonal latin squaresand their generalizations

Jordy Vanpoucke

Academic year 2011-2012

Advisor: Prof. Dr. L. Storme

Master thesis submitted to the Faculty of Sciencesto obtain the degree of Master in Sciences, Math-ematics

Contents

1 Preface 4

2 Latin squares 7

2.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Groups and permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2.2 Construction of different reduced latin squares . . . . . . . . . . . . 9

2.3 General theorems and properties . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3.1 On the number of latin squares and reduced latin squares . . . . . . 10

2.3.2 On the number of main classes and isotopy classes . . . . . . . . . . 11

2.3.3 Completion of latin squares and critical sets . . . . . . . . . . . . . 13

3 Sudoku latin squares 16

3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.2 General theorems and properties . . . . . . . . . . . . . . . . . . . . . . . . 17

3.2.1 On the number of sudoku latin squares and inequivalent sudokulatin squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.3 Minimal sudoku latin squares . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.3.1 Unavoidable sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.3.2 First case: a = b = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3.3 Second case: a = 2 and b = 3 . . . . . . . . . . . . . . . . . . . . . 26

3.3.4 Third case: a = 2 and b = 4 . . . . . . . . . . . . . . . . . . . . . . 28

3.3.5 Fourth case: a = 3 and b = 3 . . . . . . . . . . . . . . . . . . . . . 29

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4 Latin squares and projective planes 30

4.1 Projective planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.1.1 Coordinatization of projective planes . . . . . . . . . . . . . . . . . 31

4.1.2 Planar ternary rings . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.2 Orthogonal latin squares and projective planes . . . . . . . . . . . . . . . . 33

5 MOLS and MOSLS 34

5.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5.2 Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.3 Examples of small order . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5.3.1 Latin squares of order n . . . . . . . . . . . . . . . . . . . . . . . . 46

5.3.2 Sudoku latin squares of order n× n . . . . . . . . . . . . . . . . . . 48

5.4 On the number of (n− 1)MOLS(n) . . . . . . . . . . . . . . . . . . . . . . 49

5.4.1 First case: GF (p) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

5.4.2 Second case: GF (q) . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5.4.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.4.4 Further remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

6 Transversals 64

7 Magic squares 69

7.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

7.2 A little history . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

7.3 Construction of magic squares . . . . . . . . . . . . . . . . . . . . . . . . . 72

7.3.1 First case: n = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

7.3.2 Second case: n = 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

7.3.3 General case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

7.3.4 First case (bis): n = 3 . . . . . . . . . . . . . . . . . . . . . . . . . 74

7.3.5 Second case (bis): n = 4 . . . . . . . . . . . . . . . . . . . . . . . . 75

7.3.6 General case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

3

Chapter 1

Preface

Magic squares have long been an interest of mine and in my search to learn more aboutthose magic squares, I stumbled upon latin squares. It fascinated me how it was possiblethat something which seemed so easy, could be connected with so many different anddifficult areas of mathematics. Latin squares are not just some ‘spielerei’ and I got theidea to write a thesis on this subject.On Tuesday, July 12, 2011, I had the opportunity to attend a lecture of Kenneth Hickson the numbers of latin squares of prime power orders with orthogonal mates at the10th International Conference on Finite Fields and their Applications in Ghent [15]. Atthe end of this talk, Kenneth Hicks conjectured that there are (p − 2)! distinct sets of(p − 1)MOLS(p), for p a prime, describing PG(2, p). This conjecture seemed to me aninteresting challenge and my advisor, Prof. Dr. L. Storme, and I decided trying to provethis conjecture. This was not as easy as it seemed, but after a while we managed to provethe conjecture. An interesting idea that followed from this proof was to investigate thecase in which we work with a prime power q = pd instead of a prime number p. Our goalwas to find how many distinct sets of (q − 1)MOLS(q) there are, describing PG(2, q).This was not so easy and it took us some time to find the result. There was a lot ofmathematics involved to achieve our goal and finally we found some new results, whichare all presented in chapter 5. If we want to talk about mutually orthogonal latin squares,we will need some definitions and general theorems. These can be found in chapter 2.Another interesting subject are sudoku latin squares. We will see them as a special caseof a latin square and more specific we will talk about a very interesting recent result ofGary McGuire on minimal sudoku latin squares in chapter 3, but first of all we will startthis chapter with some definitions and some general theorems and properties of sudokulatin squares. Later on, when we will talk about mutually orthogonal latin squares, wewill also discuss mutually orthogonal sudoku latin squares and we will prove the theoremsfound in [14].An important chapter is chapter 4, because the connection between projective planes andlatin squares is of great importance for our results on sets of MOLS. In chapter 6, wewill briefly talk about transversals and finally we will discuss how we can construct magicsquares by using mutually orthogonal latin squares.This thesis has not the intention to discuss everything that is known about the subjects.

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This would be impossible, because there are a lot of theorems and properties known oflatin squares, sudoku latin squares, MOLS, . . . However, these latin squares are veryinteresting and there still are a lot of open problems. The nice part is that new resultson mutually orthogonal latin squares are connected to results on projective planes.Finally, I would like to thank some people for their support. First of all, I would like tothank my advisor, L. Storme, for his support and good assistance. It was not easy, butwe achieved our goals and we found new results. Secondly, I would like to thank KennethHicks and Gary L. Mullen for their ideas on sets of mutually orthogonal latin squares andHans-Dietrich O. F. Gronau for his slides of his presentation in ALCOMA10, Thurnau,Germany, on orthogonal latin squares of sudoku type [14]. A special thank you to TimPenttila for his great help concerning our questions on group theory. I would also liketo thank my in-laws for their support. They always believed in me and they will alwaysbelieve in me. Last but not least I would like to thank Domien Broeckx for his tremendoussupport. I appreciated all of the encouraging words.To end this preface I would like to say this. Whoever you are, reading this, I hope youwill enjoy this thesis and I would like to thank you too for reading this.

Jordy VanpouckeGhent, May 29, 2012

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0

c©by Jordy VanpouckeThe author and the advisor agree this thesis to be available for consultation and forpersonal reference use. Every other use falls within the constraints of the copyright,

particularly concerning the obligation to specially mention the source when citing theresults of this thesis.

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Chapter 2

Latin squares

2.1 Definitions

2.1 Definition. A latin square of order n is an n × n matrix in which n distinctsymbols from a symbol set S are arranged, such that each symbol occurs exactly once ineach row and in each column.

2.2 Definition. We say that a latin square of order n is reduced (or in standardform), if the first row and the first column of this latin square is in the natural order ofthe symbol set we chose.

From now on, we will work mostly with the symbol set S = {0, 1, 2, . . . , n− 1}. An easyexample of a reduced latin square can be obtained as follows.

2.3 Example. A reduced latin square of order n on the symbol set S = {0, 1, 2, . . . , n−1}.

0 1 . . . n− 2 n− 11 2 . . . n− 1 0...

. . ....

n− 1 0 . . . n− 3 n− 2

By giving this example we can see that the following theorem is true.

2.4 Theorem. There is a (reduced) latin square of order n for any positive integer n.

Proof. We can take the integers 0, 1, . . . , n−1 as the first row of an n×n matrix. To buildthe ith row, we simply do a cyclic shift of our first row, where we move the integers i− 1positions to the left. By doing this we obtain for the ith row i−1, i, . . . , n−1, 0, 1, . . . , i−2.By construction, there are n distinct symbols and these symbols occur exactly once ineach row. Also by construction, we can easily see that each symbol occurs exactly once ineach column, so we constructed a latin square. Finally we see that the first row and firstcolumn is in the natural order of the symbol set, so we obtained a reduced latin squareof order n.

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2.5 Example. Another example of a latin square.

2.6 Definition. The transpose of a latin square L of order n on the symbol set S,denoted by LT , is the latin square defined by LT (i, j) = L(j, i) for all 0 ≤ i, j ≤ n− 1.

2.7 Definition. A latin square L of order n is symmetric if L(i, j) = L(j, i) for all0 ≤ i, j ≤ n− 1.

2.8 Definition. A latin square L of order n is idempotent if L(i, i) = i for 0 ≤ i ≤ n−1.

Now that we know from theorem 2.4 that we can find a latin square for any positiveinteger n, it would be nice to know how many distinct latin squares of order n there arefor a positive integer n. By distinct we mean that these latin squares differ in at leastone position. In paragraph 2.3 we will discuss some results but first we have to add asmall paragraph on groups and permutations, because we will use these permutations alot, further in this project. This small paragraph on groups and permutations is onlyintended to be a brief summary of some necessary definitions and theorems. Later on, wewill see some other theorems that build on the content of this paragraph.

2.2 Groups and permutations

2.2.1 Definitions

We use the following definition of a group [19].

2.9 Definition. A group is a nonempty set G with an operation ∗ on G that has thefollowing four properties:

1. (∀a, b ∈ G)(a ∗ b ∈ G).

2. (∀a, b, c ∈ G)(a ∗ (b ∗ c) = (a ∗ b) ∗ c).

3. (∃e ∈ G)(∀a ∈ G)(a ∗ e = e ∗ a = a). This element e is called the identity element.

4. (∀a ∈ G)(∃a−1 ∈ G)(a ∗ a−1 = a−1 ∗ a = e). This element a−1 is called the inverseelement of a.

We will work with the following definition of a permutation.

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2.10 Definition. If S is a set, then we call φ a permutation of S if φ : S → S is abijection.

A first result is the following theorem.

2.11 Theorem. If S is a finite set of n distinct elements, then there are n! permutationson S.

2.2.2 Construction of different reduced latin squares

One of the reasons of this paragraph is that we want to use these permutations to constructdifferent reduced latin squares. Let us take a reduced latin square of order n. We cannow apply three different permutations. First of all we can apply a permutation on thesymbol set. Next we can also apply a permutation on the rows and finally we can apply apermutation on the columns. We will give a small example to show you how it is possibleto construct different reduced latin squares. We start with a reduced latin square L1 oforder 5.

L1 =

0 1 2 3 41 2 3 4 02 3 4 0 13 4 0 1 24 0 1 2 3

.

We can apply the following permutation on the alphabet, σ = (01)(23). This gives us anew latin square L2.

L2 =

1 0 3 2 40 3 2 4 13 2 4 1 02 4 1 0 34 1 0 3 2

.

This is obviously not a reduced latin square, but we can obtain a reduced latin square L3

by applying a permutation ρ = (134) on the rows and a permutation γ = (01)(23) on thecolumns of L2.

L3 =

0 1 2 3 41 4 3 0 22 3 1 4 03 0 4 2 14 2 0 1 3

.

We see that L1 6= L3, so we obtained a different reduced latin square by applying permu-tations. It is very important to mention that this is not a random example. It is perfectlypossible to apply a permutation σ on the alphabet such that after applying a permutationon the rows and columns we would obtain the same latin square as in the beginning. Thismeans that there are permutations stabilizing a given reduced latin square.

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2.3 General theorems and properties

2.3.1 On the number of latin squares and reduced latin squares

2.12 Notation. In what follows, we will write Ln to denote the set of distinct latinsquares of order n and we will write LRn to denote the set of different reduced latin squaresof order n.

First of all we will give an enumeration of numbers of latin squares that we found indifferent sources. In the first table, you can see the number of different reduced latinsquares of order n, n ≤ 11.

n |LRn | references1 12 13 14 45 56 L. Euler [10]6 9408 M. Frolov [13]7 16942080 A. Sade [28]8 535281401856 M. B. Wells [33]9 377597570964258816 S. E. Bammel and J. Rothstein [5]10 7580721483160132811489280 B. D. McKay and E. Rogoyski [22]11 5363937773277371298119673540771840 B. D. McKay and I. M. Wanless [23]

In the following table, you can see the total number of distinct latin squares of order n,n ≤ 11, where we used the same sources as in the previous table.

n |Ln|1 12 23 124 5765 1612806 8128512007 614794199040008 1087760324590829568009 552475149615689284253122560010 998243765821303987172506475692032000011 776966836171770144107444346734230682311065600000

We see that the number of latin squares of order n grows very fast. In fact we did not haveto give both the total number of distinct latin squares and the number of different reducedlatin squares of a given order, because the total number of distinct latin squares of ordern is equal to the number of different reduced latin squares of order n times n!(n− 1)!. Sowe have the following theorem, which can also be found in [19].

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2.13 Theorem. For each n ≥ 2, |Ln| = n!(n− 1)!|LRn |.

Proof. Take the set of all the different reduced latin squares of order n. We can apply apermutation ρ on the rows and a permutation γ on the columns to get a new latin square.There are n! ways to apply a permutation of the columns and clearly we will each timeobtain a new latin square. After applying the permutation on the columns we can stillpermute n− 1 rows and there are (n− 1)! ways to do this. We can’t apply a permutationon all of the rows, because we want to obtain distinct latin squares.For every reduced latin square L, we will now obtain n!(n− 1)! distinct latin squares byapplying these permutations, so we obtain |LRn |n!(n− 1)! distinct latin squares of order nby applying these permutations on every reduced latin square of order n.The question is: will we really obtain all of the latin squares of order n by doing this?The answer to this question is yes.Recall the construction of the set LRn . If we start with a reduced latin square L1, we canobtain a new Latin square L2, by applying a permutation on the alphabet. This newLatin square is normally not reduced. If we want L2 to be a reduced latin square, thenwe will have to apply a permutation on the rows and columns and so we will obtain areduced latin square L3 and it is possible that L3 6= L1. The preceding reasoning learnsus that if we want to obtain all the distinct latin squares of order n, we only have to applya permutation on the rows and/or the columns of the reduced latin squares. If we wouldtake a reduced latin square L and if we would apply a permutation on the alphabet toobtain a latin square L′ and a permutation on the rows and/or the columns to obtain areduced latin square L′′ and if L 6= L′′, then it would be possible to obtain L′ by applyinga permutation on the rows and/or the columns of L′′.

2.3.2 On the number of main classes and isotopy classes

First of all we give some definitions as found in [9].

2.14 Definition. Let us take a latin square L of order n on the symbol set S3, with rowsindexed by the elements of the symbol set S1 = {a0, a1, . . . , an−1} and columns indexed bythe elements of the symbol set S2 = {b0, b1, . . . , bn−1}. Let S = {(x1, x2, x3) : L(x1, x2) =x3}. Let {i, j, k} = {1, 2, 3}. The (i, j, k)−conjugate of L, L(i,j,k), has rows indexed bySi, columns indexed by Sj and symbols indexed by Sk, and is defined by L(i,j,k)(xi, xj) = xkfor each (x1, x2, x3) ∈ S.

There is a different and easier way to find the conjugates of a latin square. We can view alatin square of order n as an (3× n2)−array, called the line array of the latin square,where the entries of the first row in the array are the row indices, the entries of the secondrow in the array are the column entries and finally the entries of the last row in the arrayare the entries of the latin square.

We will now illustrate this with an example.

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2.15 Example. A latin square L of order 4 and its associated line array, where R, Cand E stand for respectively rows, columns and entries.

L =

0 3 1 21 2 0 33 0 2 12 1 3 0

.

The associated line array:

R : 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3C : 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3E : 0 3 1 2 1 2 0 3 3 0 2 1 2 1 3 0

Recall the definition of a latin square. This definition says that each symbol occurs exactlyonce in each row and in each column. This is equivalent to saying that in any two rowsof the associated line array all n2 ordered pairs appear exactly once. We can now applya permutation on the symbol set in each row of the line array and those permutationscorrespond to permutations of the rows, columns or symbols in the latin square. We canalso apply a permutation on the rows of the line array and still preserve the conditionson the line array and hence obtain a latin square. Latin squares that are obtained byapplying a permutation on the rows of the associated line array of a latin square L areconjugates of this latin square L as defined in definition 2.14. We see now that there aremaximum 6 distinct conjugates of a latin square. In fact a latin square has 1, 2, 3 or 6distinct conjugates.

2.16 Example. A latin square of order 4 and its six conjugates.

0 3 1 21 2 0 33 0 2 12 1 3 0

0 1 3 23 2 0 11 0 2 32 3 1 0

0 2 1 31 3 0 23 1 2 02 0 3 1

(1, 2, 3)−conjugate (2, 1, 3)−conjugate (3, 2, 1)−conjugate

0 1 3 22 3 1 01 0 2 33 2 0 1

0 2 3 12 0 1 31 3 2 03 1 0 2

0 2 1 32 0 3 13 1 2 01 3 0 2

(2, 3, 1)−conjugate (1, 3, 2)−conjugate (3, 1, 2)−conjugate

2.17 Example. The following latin square of order 4 has only one distinct conjugate.

0 1 2 31 0 3 22 3 0 13 2 1 0

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2.18 Definition. Two latin squares L and L′ of order n are isotopic or equivalent, ifthere are three bijections from respectively the rows, the columns and the symbols of L, torespectively the rows, the columns and the symbols of L′, that map L to L′.

2.19 Definition. Two latin squares L and L′ of order n are main class isotopic orparatopic if L is isotopic to any conjugate of L′.

2.20 Definition. The set of latin squares paratopic, respectively isotopic to L is the mainclass, respectively the isotopy class of L.

Finally we give some results for the main classes and the isotopy classes of latin squaresof order n. These results can be found in [16] and [24].

n main classes isotopy classes1 1 12 1 13 1 14 2 25 2 26 12 227 147 5648 283657 16762679 19270853541 11561872153310 34817397894749939 20890437135436300611 2036029552582883134196099 12216177315369229261482540

2.3.3 Completion of latin squares and critical sets

Suppose we have an empty matrix of order n. An interesting idea is as follows. Supposewe start to fill in the entries with the numbers from the symbol set S = {0, 1, 2, . . . , n−1}.At some point we stop and we ask ourselves if it is possible to fill in the remaining entriesto obtain a latin square. At this point the first thing we need is the definition of a partiallatin square.

2.21 Definition. A partial latin square of order n is an n×n matrix, where each entryis either empty or contains exactly one of the n distinct symbols from the symbol set S,such that no symbol occurs more than once in any row or column.

2.22 Definition. The size of a partial latin square is its number of filled in entries.

The first thing that came to our mind was the question if it is possible to complete apartial latin square of order n to a latin square of order n. The answer is yes, but thequestion is now, is this always possible and more specific, when exactly is this possible?Suppose we have a partial latin square of order n and size n2 − n, where the first n − 1rows are filled and the last row is empty. It is easy to see that we can complete this partiallatin square of order n in exactly one way to a latin square of order n, because we know

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that every symbol appears exactly n − 1 times in the partial latin square and hence ismissing from exactly one column.Let us now suppose that only the first row is filled. Then we can complete this partiallatin square to a latin square of the same order by applying a cyclic shift on the symbolsof the first row, where we move the symbols one step to the left in each of the subsequentrows. We can now ask ourselves if we can always complete a partial latin square of ordern and size n to a latin square of the same order. The answer is no, because the followingpartial latin square of order n and size n can not be completed to a latin square of thesame order.

0 1 · · · n− 2n− 1

...

Suppose we have a partial latin square of order n and size ≤ n − 1. Is it now alwayspossible to complete this partial latin square? This question was posed by Trevor Evansin 1960 and he thought this was indeed always possible, which became widely known asthe Evans Conjecture. More than 20 years later, in 1981, this conjecture was proved bySmetaniuk. His proof is constructive, thus allowing us to complete a given partial latinsquare by following the construction in the proof.In this proof, we need two results, due to Herbert J. Ryser and to Charles C. Lindner.More information can be found in [3].The first result is a proof on the completion of a latin rectangle.

2.23 Definition. An (r × n) latin rectangle is a partial latin square, where the firstr rows in the latin square of order n are completely filled in and the remaining rows areempty.

2.24 Lemma. Any (r × n) latin rectangle, r < n, can be extended to an ((r + 1) × n)latin rectangle. This means that any (r × n) latin rectangle can be completed to a latinsquare of order n.

To prove this theorem, we need Hall’s theorem, also known as the marriage theorem. Thistheorem is proved in [2].

2.25 Definition. Consider a finite set X and a collection A1, . . . , An of subsets of X,which need not to be distinct. Let us call a sequence x1, · · · , xn a system of distinctrepresentatives of A1, . . . , An if the elements xi are distinct elements of X and if xi ∈ Aifor all i.

2.26 Theorem. (Hall’s theorem) Let A1, . . . , An be a collection of subsets of a finite setX. Then there exists a system of distinct representatives of A1, . . . , An if and only if theunion of any m sets Ai contains at least m elements, for 1 ≤ m ≤ n.

The second result is the following lemma.

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2.27 Lemma. Suppose P is a partial latin square of order n and size ≤ n − 1, with atmost n

2distinct symbols filled in. This partial latin square can be completed to a latin

square of the same order.

The proof of this lemma can be found in [2]. Important to notice is that we may replacethe condition ‘at most n

2distinct symbols’ by the condition that the entries appear in at

most n2

rows. These two results are used to prove Smetaniuk’s theorem.

2.28 Theorem. Any partial latin square of order n and size at most n− 1 can always becompleted to a latin square of the same order.

The proof uses induction.It is very interesting to investigate given a partial latin square that can be completed, ifthis partial latin square can be completed in different ways or if there is only one uniquecompletion of this partial latin square. This brings us to the definition of a critical set.

2.29 Definition. A partial latin square P of order n is a critical set if it is completableto exactly one latin square of the same order, but when we remove one of the filled entriesfrom P , then the completion is no longer unique.

We will investigate these critical sets more in detail in the next chapter on sudoku latinsquares.

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Chapter 3

Sudoku latin squares

Sudoku puzzles are very popular the last decade. We see them in magazines, newspapers,books and on the world wide web. In fact these sudoku puzzles are a special kind of latinsquares.

3.1 Definitions

First of all we will give the definition of a sudoku latin square as found in [9].

3.1 Definition. Let a, b and m be positive integers with a× b = m. Partition an m×marray in the natural way into a × b rectangles. An (a, b)−sudoku latin square on thesymbol set S = {a0, a1, . . . , am−1} is a latin square on the symbol set S such that each(a× b)−rectangle contains all of the symbols of S.A sudoku latin square of order n× n is an (n, n)−sudoku latin square.

3.2 Remark. We will not speak about reduced sudoku latin squares of order n, but lateron we will define a reduced set of mutually orthogonal sudoku latin squares. Important tonotice is that we will always write the first row of our sudoku latin square in the naturalorder of the symbol set we chose.

3.3 Example. A sudoku latin square of order 4.

0 1 2 32 3 0 11 0 3 23 2 1 0

3.4 Example. A (3, 4)−sudoku latin square.

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0 1 2 3 4 5 6 7 8 9 10 114 5 6 7 8 9 10 11 0 1 2 38 9 10 11 0 1 2 3 4 5 6 71 2 3 0 5 6 7 4 9 10 11 85 6 7 4 9 10 11 8 1 2 3 09 10 11 8 1 2 3 0 5 6 7 42 3 0 1 6 7 4 5 10 11 8 96 7 4 5 10 11 8 9 2 3 0 110 11 8 9 2 3 0 1 6 7 4 53 0 1 2 7 4 5 6 11 8 9 107 4 5 6 11 8 9 10 3 0 1 211 8 9 10 3 0 1 2 7 4 5 6

3.2 General theorems and properties

3.2.1 On the number of sudoku latin squares and inequivalentsudoku latin squares

Now that we have defined an (a, b)−sudoku latin square, we will give some results onthe number of (n, n)−sudoku latin squares. First of all we will define some equivalenceoperations on the sudoku latin squares to be able to give the definition of inequivalentsudoku latin squares. The following definitions can also be found in [21].

3.5 Definition. A band of rows is the set of rows ka + 1, ka + 2, . . . , (k + 1)a, wherek ∈ {0, 1, . . . , b−1}. A stack of columns is the set of columns lb+1, lb+2, . . . , (l+1)b,where l ∈ {0, 1, . . . , a− 1}.

3.6 Definition. Let a, b and m be positive integers with a× b = m. Two (a, b)−sudokulatin squares on the symbol set S = {a0, a1, . . . , am−1} are equivalent if one can beobtained from the other by any sequence of the following equivalence operations.

1. A permutation on the symbol set. (m!)

2. A permutation on the rows. There are two different types of permutations on therows permitted:

(a) A permutation of the rows within a given band. (a!b)

(b) A permutation of the bands. (b!)

3. A permutation on the columns. There are two different types of permutations onthe columns permitted:

(a) A permutation of the columns within a given stack. (b!a)

(b) A permutation of the stacks. (a!)

4. A transposition of the sudoku latin square. (2)

17

Rotations and reflections are already included in these equivalence operations. For ex-ample the clockwise rotation of 90 degrees is the same as applying a transposition of thesudoku latin square and a permutation on the columns. The clockwise rotation of 180degrees is the same as applying a permutation on the rows and a permutation on thecolumns. The clockwise rotation of 270 degrees is the same as applying a transposition ofthe sudoku latin square and a permutation on the rows. Reflections can also be obtainedby applying a permutation on the rows and/or columns and/or a transposition of thesudoku latin square.We can now ask ourselves how many (a, b)−sudoku latin squares and how many inequiv-alent (a, b)−sudoku latin squares there are. We see that the set of all the equivalenceoperations forms a group of order m! × (a!b+1) × (b!a+1) × 2. In the next table we givesome results for (a, b)−sudoku latin squares. More information on these numbers can befound in [12], [17] and [26].

(a, b) inequivalent (a, b)−sudoku latin squares all (a, b)−sudoku latin squares(1, 1) 1 1(2, 2) 2 288(2, 3) 49 28200960(2, 4) 1673187 29136487207403520(2, 5) 4743933602050718 1903816047972624930994913280000(3, 3) 5472730538 6670903752021072936960

3.3 Minimal sudoku latin squares

In this section we will investigate the article of Gary McGuire, Bastian Tugemann andGilles Civario, ‘There is no 16-Clue Sudoku: Solving the Sudoku Minimum Number ofClues Problem’ [21].

3.7 Definition. With a, b and n as in definition 3.1, an (a, b)−sudoku critical set isa partial latin square P that is completable in exactly one way to an (a, b)−sudoku latinsquare, but when we remove one of the filled entries from P , then the completion is nolonger unique.The size of an (a, b)− sudoku critical set is the number of elements in this set, wherethe elements are the filled entries of the partial latin square.An (a, b)−sudoku puzzle is an (a, b)−sudoku critical set without the requirement thatwhen we remove one of the filled entries, the completion is no longer unique. When this re-quirement holds we have an (a, b)−sudoku critical set, which is also called an irreducible(a, b)−sudoku puzzle.

It is now very interesting to investigate what the minimum size of a sudoku criticalset is. More specific we will summarize some new results for sudoku latin squares oforder 3 × 3 = 9. A sudoku latin square of order 9 has 81 entries. When we constructa (3, 3)−sudoku puzzle we would like to have a partial latin square which is uniquelycompletable to a (3, 3)−sudoku latin square. In other words, when we use a sudoku puzzle

18

solver, we would like to obtain one unique solution. The big question in the constructionis how many entries, also called clues, do we give to the solver and is there a minimumnumber of clues that can possibly be given such that a (3, 3)−sudoku puzzle still has aunique solution?There are (3, 3)−sudoku puzzles known with only 17 clues. A collection of 49151 distinct(3, 3)−sudoku puzzles with 17 entries can be found in [27].

3.8 Example. A sudoku critical set of size 17.

1 25 4

37 6 4

18

9 2 85 1 7

3

Its unique completion to a sudoku latin square of order 9.

3 6 4 9 7 8 5 1 21 5 2 4 3 6 9 7 88 7 9 1 2 5 6 3 47 3 8 6 5 1 4 2 96 9 1 2 4 7 3 8 52 4 5 3 8 9 1 6 79 2 3 7 6 4 8 5 14 8 6 5 1 2 7 9 35 1 7 8 9 3 2 4 6

However, nobody has ever found a (3, 3)−sudoku critical set of size 16 and that was thereason why they conjectured that the minimum number of clues needed to complete a(3, 3)−sudoku puzzle, with only one unique solution, is 17. We know now that nobodywill ever find such a (3, 3)−sudoku puzzle with only 16 clues, because this conjecture wasproved by Gary McGuire, Bastian Tugemann and Gilles Civario.They performed an exhaustive search for a 16−clue sudoku puzzle to prove this conjecture,where they applied a new hitting set enumeration algorithm.

3.9 Definition. Given a collection of sets, a set which intersects all sets in the collectionin at least one element is called a hitting set.

We will now briefly explain how the method works.There are three big steps in the method. First of all they needed a catalogue of all theinequivalent (3, 3)−sudoku latin squares, where we know that there are 5472730538 dis-tinct inequivalent (3, 3)−sudoku latin squares. Secondly they had to write a program that

19

efficiently searches within a given completed (3, 3)−sudoku latin square for (3, 3)−sudokupuzzles with 16 clues whose solution is the given (3, 3)−sudoku latin square. Finallythey had to run through the catalogue, to apply the program to each (3, 3)−sudoku latinsquare. Their program, named ‘checker’, proved that 16 clues never lead to a unique(3, 3)−sudoku latin square.We need to make a remark on the second step in this method, since there are differentpoints of view on the minimum number of clues problem. First of all one can construct apartial latin square of order m = ab with a certain number of clues and check if this partiallatin square is uniquely completable to an (a, b)−sudoku latin square. One can start withfor example c clues and if it is not possible to complete any partial latin square of orderm with c clues in a unique way to an (a, b)−sudoku latin square, one can start again withc+1 clues until the minimum number of clues necessary for a unique completion is found,but this is very time-consuming.Another method is to consider a completed (a, b)−sudoku latin square S from the setof inequivalent sudoku latin squares and then look at all the partial latin squares withc clues whose solution is S. This means that we think of the partial latin square withc clues as ‘contained in’ that particular (a, b)−sudoku latin square. This method is alsovery time-consuming if we would just try every partial latin square with c clues which iscontained in S. Especially because we will have to run this program 5472730538 times inthe case we are working on, where a = b = 3. Therefore the authors of [21] tried to reducethe number of possibilities to check with some theory. Briefly explained they identifiedcertain regions in the sudoku latin squares, called unavoidable sets, such that any partiallatin square which is completable to a sudoku latin square had to contain at least oneclue from each unavoidable set contained in that sudoku latin square.The strategy presented by the authors in the second step of the method is now as follows.First of all they search a sufficiently powerful collection of unavoidable sets for the given(3, 3)−sudoku latin square. Secondly they enumerate all the hitting sets of size 16 forthis collection. These hitting sets are sets of size 16 that intersect all the unavoidablesets found in the first step. Finally they had to check if any of these hitting sets was avalid (3, 3)−sudoku puzzle with 16 clues, which means that the hitting set is uniquelycompletable to a (3, 3)−sudoku latin square, by running all of the hitting sets through asudoku solver procedure. Let us now take a closer look at these unavoidable sets.

3.3.1 Unavoidable sets

We start this paragraph with an example.

3.10 Example. A sudoku latin square of order 4.

S1 =

0 1 2 32 3 1 01 0 3 23 2 0 1

.

When we interchange 0 and 1 among the four red numbers in this sudoku latin square, weobtain a different sudoku latin square.

20

S ′1 =

0 1 2 32 3 0 11 0 3 23 2 1 0

.

This means that if we would have a (2, 2)−sudoku puzzle with this (2, 2)−sudoku latinsquare as its solution, then one of the four red numbers must be a clue, otherwise wewould not have a unique completion. This explains why we call this set of four numbersan unavoidable set.

3.11 Definition. Consider an (a, b)−sudoku latin square S. A subset of S is a set ofentries contained in S.

3.12 Example. Let us consider the sudoku latin square S1 of example 3.10. The setV = {S1(1, 2), S1(1, 3), S1(3, 2), S1(3, 3)} is the set of red coloured entries. We see thatthis set is contained in S1. In fact we interpret this sudoku latin square as a matrix, thuseach element from a subset is an element from this matrix on a certain row i and a certaincolumn j, S1(i, j).

We can now also define S\V , which is the sudoku latin square without the entries of thesubset V , thus the complement of V .

3.13 Definition. Consider an (a, b)−sudoku latin square S. A subset U of S is called anunavoidable set if S\U has more than one completion to an (a, b)−sudoku latin square.

This means that if a set of clues does not intersect every unavoidable set, then it cannotbe used as an (a, b)−sudoku puzzle, because there is more than one completion. Thisgives us the following lemma.

3.14 Lemma. Suppose that V ⊆ S is a set of clues of an (a, b)−sudoku latin square S,such that V hits (intersects) every unavoidable set of S, then S is the only completion ofV .

Proof. If V had more than one completion, then S\V would be an unavoidable set nothit by V , which is a contradiction.

Finally we say that an unavoidable set U is minimal if there is no proper subset W of Usuch that this subset is itself unavoidable.

3.15 Example. We will now, as an example, look at the unavoidable sets in the following(2, 2)−sudoku latin square.

S1 =

0 1 2 32 3 1 01 0 3 23 2 0 1

.

21

The following 4 sets are all minimal unavoidable sets of this (2, 2)−sudoku latin square.

U1 =

0 1 0 0

1 0, U2 =

0 02 3

3 2

, U3 =

0 0 2 3

3 2, U4 =

0 01 0

0 1

.

The set U1 for example is an unavoidable set, since S1\U1 can be completed to S1, butalso to (S1\U1) ∪ U ′1, where

U ′1 =

1 0 0 0

0 1.

This means that if we want a ‘hitting set’, thus a (2, 2)−sudoku puzzle, this set needs tointersect all of these unavoidable sets, which means there are at least 4 clues necessary.We can find an irreducible (2, 2)−sudoku puzzle of size 4.

0 013

2

This set has a unique completion to a (2, 2)−sudoku latin square.

0 1 013

2

0 11 03

2

0 13 1 0

32

0 12 3 1 0

32

0 12 3 1 0

0 32

0 1 32 3 1 0

0 32

0 1 2 32 3 1 0

0 32

0 1 2 32 3 1 0

0 32 0

0 1 2 32 3 1 0

0 33 2 0

0 1 2 32 3 1 01 0 33 2 0

0 1 2 32 3 1 01 0 3 23 2 0

0 1 2 32 3 1 01 0 3 23 2 0 1

We see that each added element in every step is uniquely determined.

At this point the program to find the unavoidable sets was not fast enough. That is whythe authors of [21] tried to reduce the work with a bit more theory on the unavoidable sets.We will summarize the theorems and lemmas, and finally we will look at some differentexamples and explain how we can find an (a, b)−sudoku critical set for some different aand b.

3.16 Lemma. Consider an (a, b)−sudoku latin square S and suppose that U ⊆ S is aminimal unavoidable set. If S ′ is any other completion of S\U , then S and S ′ differexactly in the entries contained in U . In particular, every element in U occurs at leasttwice.

22

Proof. Suppose that S and S ′ agree in some entries contained in U . This would mean thatU is not minimal, because it would properly contain the unavoidable set S\(S∩S ′). Whenwe move between S and S ′, we see that the contents of the entries in U are permuted.If there was an element contained in only one entry of U , that element could neither moveto a different row nor to a different column, since otherwise this row or column would notcontain this element anymore, which is in conflict with the definition of an (a, b)−sudokulatin square, so this proves that every element in U occurs at least twice.

3.17 Corollary. Consider an (a, b)−sudoku latin square S and suppose that U ⊆ S is aminimal unavoidable set. If S ′ is any other completion of S\U , then S ′ may be obtainedfrom S by a permutation with no fixed points of the cells in each row (column, box) of U .Hence the intersection of U with any row (column, box) is either empty or contains atleast two elements.

The first statement, that S ′ may be obtained from S by a permutation with no fixedpoints of the cells in each row (column, box) of U , is in fact lemma 3.16. Suppose thatthe intersection of U with a certain row (column, box) contains only one element. This isimpossible, because every element of U is permuted with no fixed points, implying thatthis certain row (column, box) would not contain this element anymore.

Until now we only considered unavoidable sets where we needed one clue, but there arealso unavoidable sets that require more than one clue. We will give an example.

3.18 Example. We consider the following (2, 3)−sudoku latin square.

0 1 2 3 4 53 4 5 0 1 21 2 0 4 5 34 5 3 1 2 02 0 1 5 3 45 3 4 2 0 1

The set of red coloured numbers is an unavoidable set, where we need at least two cluesfrom this set to completely determine these nine entries. If we would only have one clueof these nine entries, we would have more than one completion. Suppose we choose theentry 0 in the top left corner as a clue, then we also have the following (2, 3)−sudoku latinsquare as a solution. The reason is in fact that if we only have one clue on a certain rowgiven for this set, we could interchange the other two entries of that row.

0 2 1 3 4 53 4 5 0 1 21 0 2 4 5 34 5 3 1 2 02 1 0 5 3 45 3 4 2 0 1

23

Suppose we choose the entry 0 on the first row and the entry 2 on the third row, then wehave only one unique completion.

0 3 4 53 4 5 0 1 2

2 4 5 34 5 3 1 2 0

5 3 45 3 4 2 0 1

0 3 4 53 4 5 0 1 2

2 0 4 5 34 5 3 1 2 0

5 3 45 3 4 2 0 1

0 3 4 53 4 5 0 1 2

2 0 4 5 34 5 3 1 2 0

0 5 3 45 3 4 2 0 1

0 2 3 4 53 4 5 0 1 2

2 0 4 5 34 5 3 1 2 0

0 5 3 45 3 4 2 0 1

0 1 2 3 4 53 4 5 0 1 2

2 0 4 5 34 5 3 1 2 0

0 5 3 45 3 4 2 0 1

0 1 2 3 4 53 4 5 0 1 21 2 0 4 5 34 5 3 1 2 0

0 5 3 45 3 4 2 0 1

0 1 2 3 4 53 4 5 0 1 21 2 0 4 5 34 5 3 1 2 0

0 1 5 3 45 3 4 2 0 1

0 1 2 3 4 53 4 5 0 1 21 2 0 4 5 34 5 3 1 2 02 0 1 5 3 45 3 4 2 0 1

We will call the unavoidable set from the previous example a degree two unavoidable set,because we need at least two clues from this set to have a unique completion. Let us callthe unavoidable sets that we defined earlier, where we needed one clue, unavoidable setsof degree 1. Now we will recursively define an unavoidable set of degree k for k > 1.

3.19 Definition. A nonempty subset U ⊆ S is called an unavoidable set of degreek > 1, if for all e ∈ U the set U\{e} is an unavoidable set of degree k − 1.

As before, we say that an unavoidable set U of degree k > 1 is minimal if there is noproper subset W of U such that this subset is itself an unavoidable set of the same degree.

3.20 Notation. We will say that U is an (m, k)−unavoidable set if U is an unavoidableset of degree k having m elements.

Finally we have the following theorem.

3.21 Theorem. Consider an (a, b)−sudoku latin square S and suppose that U ⊆ S isan (m, k)−unavoidable set. Then we will need to add at least k elements from U toS\U to obtain an (a, b)−sudoku puzzle with a unique completion. Moreover, if V ⊆ Sis an (m′, k′)−unavoidable set, such that U ∩ V = ∅, then U ∪ V is an (m + m′, k +k′)−unavoidable set.

3.22 Corollary. Suppose that U1, . . . , Ut are unavoidable sets of degree k of an (a, b)−sudoku latin square S that are pairwise disjoint. Then U1 ∪ . . .∪Ut is an unavoidable setof degree tk.

By using all of these theorems the authors of [21] could reduce the number of possibilitiesto check in their algorithm. We will not go into further details, but more information canbe found in the article. The only thing we will do now is finish this paragraph with someexamples, where we will study some unavoidable sets in some (a, b)−sudoku latin squaresand check how many clues we need at least in an (a, b)−sudoku puzzle.

24

3.3.2 First case: a = b = 2

We know there are only two distinct inequivalent (2, 2)−sudoku latin squares.

S1 =

0 1 2 32 3 1 01 0 3 23 2 0 1

and S2 =

0 1 2 32 3 0 11 0 3 23 2 1 0

.

In example 3.15, we found a (2, 2)−sudoku critical set of size 4. We will prove in theorem3.25 that the minimum number of clues needed to complete a (2, 2)−sudoku puzzle, withonly one unique solution, is exactly 4.First of all we need to consider all the distinct inequivalent (2, 2)−sudoku latin squares.We know that there are exactly two distinct inequivalent (2, 2)−sudoku latin squares.Now we will have to look if we can say something about the minimum number of cluesneeded in a (2, 2)−sudoku puzzle to obtain a unique completion which is one of those two(2, 2)−sudoku latin squares.

3.23 Lemma. Any (2, 2)−sudoku puzzle whose solution is the following (2, 2)−sudokulatin square has at least 4 clues.

0 1 2 32 3 1 01 0 3 23 2 0 1

To prove this lemma we will use the theorems on the unavoidable sets.

Proof. We see that this (2, 2)−sudoku latin square is the union of the following fourpairwise disjoint (4, 1) unavoidable sets.

U1 =

0 1 0 0

1 0, U2 =

0 02 3

3 2

, U3 =

0 0 2 3

3 2, U4 =

0 01 0

0 1

.

For instance, U4 occurs in S1, but S1\U4 can also be completed to S2, so U4 is an un-avoidable set.By corollary 3.22, this (2, 2)−sudoku latin square is therefore a (16, 4)−unavoidable set.Using theorem 3.21, we know that we need at least 4 clues for a unique completion.

3.24 Lemma. Any (2, 2)−sudoku puzzle whose solution is the following (2, 2)−sudokulatin square has at least 4 clues.

0 1 2 32 3 0 11 0 3 23 2 1 0

25

Proof. We see that this (2, 2)−sudoku latin square is the union of the following fourpairwise disjoint (4, 1) unavoidable sets.

U ′1 =

0 1 0 0

1 0, U ′2 =

0 02 3

3 2

, U ′3 =

0 0 2 3

3 2, U ′4 =

0 00 1

1 0

.

By corollary 3.22, this (2, 2)−sudoku latin square is therefore a (16, 4)−unavoidable set.Using theorem 3.21, we know that we need at least 4 clues for a unique completion.

3.25 Theorem. The minimum number of clues needed to complete a (2, 2)−sudoku puz-zle, with only one unique solution, is 4.

Proof. We know by lemma 3.23 and lemma 3.24 that we need at least 4 clues to completea (2, 2)−sudoku puzzle. If we can find such a (2, 2)−sudoku critical set, then we haveproved theorem 3.25.In example 3.15, we presented the following (2, 2)−sudoku critical set of size 4.

0 013

2

3.3.3 Second case: a = 2 and b = 3

3.26 Theorem. Any (2, 3)−sudoku puzzle whose solution is the following (2, 3)−sudokulatin square has at least 9 clues.

0 1 2 3 4 53 4 5 0 1 21 2 3 4 5 04 5 0 1 2 32 3 4 5 0 15 0 1 2 3 4

Proof. We see that this (2, 3)−sudoku latin square is the union of the following ninepairwise disjoint (4, 1) unavoidable sets.

U1 =

0 33 0

0 0 0 0 0 0

, U2 =

1 44 1

0 0 0 0 0 0

, U3 =

2 55 2

0 0 0 0 0 0

,

26

U4 =1 44 1

0 0 0 0 0 0

, U5 =2 55 2

0 0 0 0 0 0

, U6 =3 00 3

0 0 0 0 0 0

,

U7 =0 0 0 0 0 02 55 2

, U8 =0 0 0 0 0 0

3 00 3

, U9 =0 0 0 0 0 0

4 11 4

.

By corollary 3.22, this (2, 3)−sudoku latin square is therefore a (36, 9)−unavoidable set.Using theorem 3.21, we know that we need at least 9 clues for a unique completion.

Now we would like to find a (2, 3)−sudoku critical set with only 9 elements. This meansthat this set needs to intersect all of these (4, 1)−unavoidable sets in exactly one element.We found the following (2, 3)−sudoku critical set.

0 1 2 0

1 23

23 4

This set has a unique completion. In the following (2, 3)−sudoku latin square we givethe completion, where the exponent of each entry gives an ordering, how the previous(2, 3)−sudoku critical set can be completed.

0 1 2 31 42 53

319 420 521 018 117 216

1 2 34 45 56 07

422 523 024 115 214 32 38 49 510 011 112

525 026 127 213 3 4

27

3.3.4 Third case: a = 2 and b = 4

3.27 Theorem. Any (2, 4)−sudoku puzzle whose solution is the following (2, 4)−sudokulatin square has at least 16 clues.

0 1 2 3 4 5 6 74 5 6 7 0 1 2 31 2 3 4 5 6 7 05 6 7 0 1 2 3 42 3 4 5 6 7 0 16 7 0 1 2 3 4 53 4 5 6 7 0 1 27 0 1 2 3 4 5 6

Proof. We see that this (2, 4)−sudoku latin square is the union of 16 pairwise disjoint(4, 1) unavoidable sets, just as in theorem 3.26. By corollary 3.22, this (2, 4)−sudokulatin square is therefore a (64, 16)−unavoidable set. Using theorem 3.21, we know thatwe need at least 16 clues for a unique completion.

Now we would like to find a (2, 4)−sudoku critical set with only 16 elements. Similar asin the previous case, we found the following (2, 4)−sudoku critical set.

0 1 2 3 0

1 2 34

2 34 5

34 5 6

This set has a unique completion. In the following (2, 4)−sudoku latin square, we givethe completion, where the exponent of each entry gives an ordering, how the previous(2, 4)−sudoku critical set can be completed.

0 1 2 3 41 52 63 74

433 534 635 736 032 131 230 329

1 2 3 45 56 67 78 09

537 638 739 040 128 227 326 42 3 410 511 612 713 014 115

641 742 043 144 225 324 4 53 416 517 618 719 020 121 222

745 046 147 248 323 4 5 6

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3.3.5 Fourth case: a = 3 and b = 3

We know that the minimum number of clues needed to complete a (3, 3)−sudoku puz-zle, with only one unique solution, is 17. Remark that this does not mean that every(3, 3)−sudoku latin square can be obtained in a unique way from a (3, 3)−sudoku puzzlewith only 17 clues.

3.28 Theorem. Any (3, 3)−sudoku puzzle whose solution is the following (3, 3)−sudokulatin square has at least 18 clues.

0 1 2 3 4 5 6 7 83 4 5 6 7 8 0 1 26 7 8 0 1 2 3 4 51 2 0 4 5 3 7 8 64 5 3 7 8 6 1 2 07 8 6 1 2 0 4 5 32 0 1 5 3 4 8 6 75 3 4 8 6 7 2 0 18 6 7 2 0 1 5 3 4

Proof. We see that this (3, 3)−sudoku latin square is the union of 9 pairwise disjoint(9, 2) unavoidable sets. By corollary 3.22, this (3, 3)−sudoku latin square is therefore an(81, 18)−unavoidable set. Using theorem 3.21, we know that we need at least 18 clues fora unique completion.

29

Chapter 4

Latin squares and projective planes

4.1 Projective planes

We start with some definitions found in [29].

4.1 Definition. A finite incidence structure, or finite geometry, P = (P,B, I) isa finite set of points P , a finite set of lines B, and a relation I between the points and thelines, called the incidence relation.

4.2 Definition. A finite projective plane P is a finite incidence structure such thatthe following properties hold.

1. Any two distinct points are incident with exactly one line.

2. Any two distinct lines are incident with exactly one point.

3. There exists a set of four points such that no three of them are incident with oneline.

4.3 Definition. For a finite projective plane P, there is a positive integer n, such thatany line of P has exactly n+ 1 points. This number n is the order of P.

An important class of finite projective planes can be obtained as follows. Let V be the3−dimensional vector space over the finite field F = GF (q) of order q and let us definea geometry P(V ), where the points are the 1−dimensional subspaces of V and the linesare the 2−dimensional subspaces of V . This geometry P(V ) is a projective plane, whichis denoted by PG(2, F ) or PG(2, q), if F is a finite field of order q.The projective planes PG(2, q) are the classical examples of finite projective planes. Theyare also known as the desarguesian planes since they are the only finite projective planesin which the theorem of Desargues is valid. There are of course many other examples ofprojective planes. Later on we will discuss a very important conjecture on the existenceof non-Desarguesian planes.

30

4.1.1 Coordinatization of projective planes

Consider a projective plane P = (P,B, I) of order n and a set R with cardinality n, where0, 1 ∈ R, 0 6= 1, and where the symbol ∞ /∈ R. Choose a quadrangle (o, x, y, e) in P .Suppose ox = L2, oy = L1, xy = L∞, xe ∩ L1 = e1, ye ∩ L2 = e2 and e1e2 ∩ L∞ = e∞.

We will use R ∪ {∞} to coordinatize P , with respect to the given quadrangle (o, x, y, e).Consider a bijection θ from the set of points on L1, different from y, to the set R, whereoθ = 0 and eθ1 = 1. If pθ = a, pIL1 and p 6= y, then we write the point p as (0, a). Considera point q on L2, q 6= x. Suppose q′ = e∞q ∩ L1. If q′ = (0, b), then we write q as (b, 0).Next, we consider a point r not incident with L∞. If xr∩L1 = (0, c) and yr∩L2 = (d, 0),then we write the point r as (d, c). Every point in the affine plane P\L∞ now has anordered pair of coordinates (d, c), c, d ∈ R. Let us now consider a point l∞IL∞, wherel∞ 6= y. If l∞e2∩L1 = (0,m), then we write l∞ as (m). So we have x = (0) and e∞ = (1).Finally we write y as (∞). In this way the whole set of points P is coordinatized. Remarkthat all the coordinates are only influenced by the choice of the quadrangle (o, x, y, e) andθ. Now we will look at the set of lines B.Suppose that L is a line not passing through y. If L∩L∞ = (m) and L∩L1 = (0, k), thenwe write L as [m, k]. Next, we suppose that the line L passes through y, but L 6= L∞. IfL ∩ L2 = (k, 0), then we write L as [k]. Finally we write L∞ as [∞].

31

The only work left, is to formulate the incidence relation between the points and the lines.

4.1.2 Planar ternary rings

4.4 Definition. A ternary operation T on a set R is a mapping T : R3 → R, wherea triple (a, b, c) ∈ R3 is mapped to T (a, b, c).A ternary ring is an algebraic structure (R, T ), where R is a nonempty set and T is aternary operation on R.

Consider a projective plane P = (P,B, I) that is coordinatized using the set R. Ifa, b, c ∈ R, then we suppose that T (a, b, c) = k ⇔ (b, c)I[a, k]. This means that the point(0, k) is the intersection of L1 with the line through the points (a) and (b, c), thus k isuniquely determined, given (a, b, c).

4.5 Theorem. Consider a projective plane that is coordinatized using the set R. IfT (a, b, c) = k ⇔ (b, c)I[a, k], then we have the following properties:

1. (∀a, b, c ∈ R)(T (a, 0, c) = T (0, b, c) = c)

2. (∀a ∈ R)(T (a, 1, 0) = T (1, a, 0) = a)

3. (∀a, b, c, d ∈ R)(a 6= c)(∃!x ∈ R)(T (x, a, b) = T (x, c, d))

4. (∀a, b, c ∈ R)(∃!x ∈ R)(T (a, b, x) = c)

5. (∀a, b, c, d ∈ R)(a 6= c)(∃!(x, y) ∈ R2)(T (a, x, y) = b ∧ T (c, x, y) = d)

32

4.6 Definition. A planar ternary ring (PTR) is a ternary ring (TR) including theelements 0 and 1, with 0 6= 1, which has the previous 5 properties.

4.7 Theorem. Consider a PTR (R, T ) and the incidence structure P = (P,B, I) whichis defined as follows. The elements of P are the ordered pairs (x, y) ∈ R2, the elements(x), with x ∈ R and an element (∞), where ∞ is a symbol not belonging to R. Theelements of B are the ordered pairs [m, k], m, k ∈ R, the elements [k] with k ∈ R and theelement [∞]. The incidence relation is defined as follows:

• (x, y)I[m, k]⇔ T (m,x, y) = k

• (x, y)I[k]⇔ x = k

• (∀x, y ∈ R)((x, y)6I[∞])

• (x)I[m, k]⇔ x = m

• (∀x, k ∈ R)((x)6I[k])

• (∀x ∈ R)((x)I[∞])

• (∀m, k ∈ R)((∞)6I[m, k])

• (∀k ∈ R)((∞)I[k])

• (∞)I[∞]

This incidence structure P as defined above is a projective plane.

4.2 Orthogonal latin squares and projective planes

4.8 Theorem. If there exists a PTR (R, T ), where R is a set of order n, then there existsa set of n− 1 mutually orthogonal latin squares of order n.

4.9 Theorem. There exists a projective plane of order n if and only if there exists acomplete set of mutually orthogonal latin squares of order n.

Consider a PTR (R, T ), where we suppose that R = {0, 1, . . . , n − 1}, then there existsa set of (n − 1)MOLS(n), {A1, A2, . . . , An−1}, where Ak, k = 1, . . . , n − 1, is defined as(Ak)i,j = T (k, i, j), i, j = 0, 1, . . . , n− 1.

4.10 Remark. It is now possible to prove that there exists a projective plane of order nby proving that there exists a complete set of MOLS(n).

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Chapter 5

MOLS and MOSLS

5.1 Definitions

5.1 Definition. Let us take two latin squares of order n, L and L′, on respectively thesymbol set S = {a0, a1, . . . , an−1} and the symbol set S ′ = {α0, α1, . . . , αn−1}. These twolatin squares are called mutually orthogonal if there exists for each ordered pair (i, j) ∈{a0, a1, . . . , an−1}×{α0, α1, . . . , αn−1}, exactly one ordered pair (k, l) ∈ {0, 1, . . . , n−1}×{0, 1, . . . , n− 1}, such that Lk,l = i and L′k,l = j.

This means that if these latin squares are superimposed, we would find all of the n2

ordered pairs (i, j) ∈ S × S ′.

5.2 Notation. Instead of talking about mutually orthogonal latin squares, we will use theabbreviation MOLS. If the order of the latin squares is equal to n, we will also use thenotation MOLS(n).

We have a similar definition for sudoku latin squares.

5.3 Definition. Let us take two sudoku latin squares of order n×n, L and L′, on respec-tively the symbol set S = {a0, a1, . . . , an2−1} and the symbol set S ′ = {α0, α1, . . . , αn2−1}.These two sudoku latin squares are called mutually orthogonal if there exists for eachordered pair (i, j) ∈ {a0, a1, . . . , an2−1} × {α0, α1, . . . , αn2−1}, exactly one ordered pair(k, l) ∈ {0, 1, . . . , n2 − 1} × {0, 1, . . . , n2 − 1} such that Lk,l = i and L′k,l = j.

5.4 Notation. Instead of talking about mutually orthogonal sudoku latin squares, wewill use the abbreviation MOSLS. If the order of the sudoku latin squares is equal ton× n = n2, we will also use the notation MOSLS(n2).

The previous definitions can also be reformulated.

5.5 Definition. We say that two (sudoku) latin squares L and L′ of order n are orthogonalif Li,j = Lk,l and L′i,j = L′k,l, implies i = k and j = l.

34

5.6 Definition. Finally we say that a (sudoku) latin square L of order n has an orthog-onal mate if there exists a (sudoku) latin square L′ of the same order, such that L andL′ are mutually orthogonal.A bachelor (sudoku) latin square is a (sudoku) latin square which has no orthogonalmate.

5.7 Example. Two mutually orthogonal latin squares of order 5.

0 1 2 3 4 0 1 2 3 41 2 3 4 0 2 3 4 0 12 3 4 0 1 4 0 1 2 33 4 0 1 2 1 2 3 4 04 0 1 2 3 3 4 0 1 2

5.8 Example. Two mutually orthogonal sudoku latin squares of order 2× 2.

0 1 2 3 0 1 2 33 2 1 0 2 3 0 1

2 3 0 1 1 0 3 21 0 3 2 3 2 1 0

The first thing we want to do is to extend our definitions to sets of latin squares.

5.9 Definition. A set of (sudoku) latin squares L1, L2, . . . , Lk is mutually orthogonal ifLi and Lj are orthogonal for all 1 ≤ i < j ≤ k.

5.10 Notation. If we have k different mutually orthogonal latin squares of order n, wewill also use the notation (k)MOLS(n).If we have k different mutually orthogonal sudoku latin squares of order n × n, we willalso use the notation (k)MOSLS(n2).

5.11 Definition. We say that a set of k different MOLS of order n is reduced (or instandard form), if one of the latin squares is reduced and if the first row in every otherlatin square in this set is in the natural order of the symbol set we chose.

5.12 Definition. We say that a set of k different MOSLS of order n is reduced (or instandard form), if the first row in every sudoku latin square in this set is in the naturalorder of the symbol set we chose.

Our goal is now to prove some results on the number of latin squares in a set of MO(S)LS.

5.2 Bounds

We will start this paragraph with a lemma that we will need later in the other theorems.A lot of the theorems that are discussed in this paragraph can also be found without proofin [14].

35

5.13 Lemma. For a set of k MOLS or MOSLS, a random permutation on the alphabetof the (sudoku) latin squares, does not affect the orthogonality of those (sudoku) latinsquares.

Proof. Let us take two random mutually orthogonal latin squares of order n from a set(k)MOLS(n), L is a latin square on the symbol set S = {a0, a1, . . . , an−1} and L′ is alatin square on the symbol set S ′ = {α0, α1, . . . , αn−1}.By definition of orthogonality, we have that each ordered pair (u, v) ∈ S × S ′ appearsexactly once among the n2 pairs (Li,j, L

′i,j), 1 ≤ i, j ≤ n. This means that if Li,j = Lk,l

and L′i,j = L′k,l, then this would imply that i = k and j = l.Next, we define a permutation σ on the symbol set S and a permutation σ′ on thesymbol set S ′. Our goal is to prove that Lσ and L′σ

′are still orthogonal. If they are

not mutually orthogonal this would mean that there exists an ordered pair (i, j) 6= (k, l),such that Lσi,j = Lσk,l and L′σ

′i,j = L′σ

′

k,l . This is impossible, because a permutation is a

bijection and the previous statement means that the ordered pair (Lσi,j, L′σ′i,j ) = (Lσk,l, L

′σ′k,l)

for (i, j) 6= (k, l). But (Li,j, L′i,j) 6= (Lk,l, L

′k,l) if (i, j) 6= (k, l), so this would mean that σ

or σ′ is not a bijection, which is impossible.We can make a similar reasoning for (a, b)−sudoku latin squares. The only questionthat arises here is, if a random permutation on the alphabet leaves the structure ofan (a, b)−sudoku latin square invariant. This is indeed the case, because every (a ×b)−rectangle as defined in definition 3.1 still contains every symbol from our chosen symbolset.

5.14 Definition. A maximal set MOLS(n) is a set (k)MOLS(n) such that it is im-possible to extend this set to a set (k + 1)MOLS(n).

5.15 Definition. • N(n) = max{k : ∃(k)MOLS(n)}.

• NS(n2) = max{k : ∃(k)MOSLS(n2)}.

• NS(a× b) = max{k : ∃(k)MOSLS(ab) in (a× b)−rectangles}.

5.16 Theorem. For each n ≥ 2, N(n) ≤ n− 1.

Proof. Let us take two mutually orthogonal latin squares of order n. We can perform apermutation on the alphabet without affecting the orthogonality of those latin squares,such that the first row in both latin squares is given by (0, 1, . . . , n − 1). Now we canperform this for each latin square in a set of (k)MOLS(n). So we obtain

L1 =

0 1 . . . n− 1a . . ....

, L2 =

0 1 . . . n− 1b . . ....

, . . .

Neither symbol a nor symbol b can be equal to 0 or otherwise this would be in conflictwith the definition of a latin square. It is also impossible that a = b, otherwise if we wouldhave a = b = i, then we would have that ((L1)1,0, (L2)1,0) = (i, i) = ((L1)0,i, (L2)0,i). But

36

(1, 0) 6= (0, i), so we are in conflict with the definition of orthogonal latin squares.This means there are at most n − 1 possible symbols of the alphabet that can appearon the first position of the second row, 1, 2, . . . , n− 1, and it is impossible that two latinsquares have the same symbol in that position, hence N(n) ≤ n− 1,∀n ≥ 2.

5.17 Definition. A set of (n − 1)MOLS(n) is called a complete set of MOLS oforder n.

5.18 Theorem. NS(n2)not= NS(n× n) ≤ n2 − n.

Proof. To prove this theorem, we can follow the same reasoning as before. Let us take twomutually orthogonal sudoku latin squares of order n2. We can perform a permutation onthe alphabet without affecting the orthogonality of those sudoku latin squares, such thatthe first row in both sudoku latin squares is given by (0, 1, . . . , n− 1, . . . , n2− n, n2− n+1, . . . , n2−1). Now we can do this for each sudoku latin square in a set of (k)MOSLS(n2).So we obtain

L1 =

0 1 . . . n− 1 . . . n2 − n n2 − n+ 1 . . . n2 − 1a . . .... . . .

. . .

,

L2 =

0 1 . . . n− 1 . . . n2 − n n2 − n+ 1 . . . n2 − 1b . . .... . . .

. . .

,

. . .

Neither symbol a nor b can be an element of the set of integers modulo n, {0, 1, . . . , n−1},or otherwise this would be in conflict with the definition of a sudoku latin square. It isalso impossible that a = b, otherwise if we would have a = b = i, then we would havethat ((L1)1,0, (L2)1,0) = (i, i) = ((L1)0,i, (L2)0,i). But (1, 0) 6= (0, i), so we are in conflictwith the definition of orthogonal sudoku latin squares.This means there are at most n2 − n possible symbols of the alphabet that can appearon the first position of the second row: n, . . . , n2 − n, n2 − n + 1, . . . , n2 − 1, and it isimpossible that two sudoku latin squares have the same symbol in that position, henceNS(n× n) ≤ n2 − n.

5.19 Definition. A set of (n2 − n)MOSLS(n2) is called a complete set of MOSLSof order n× n.

The previous theorem is in fact a special case of the following theorem.

37

5.20 Theorem. NS(a× b) ≤ min{ab− a, ab− b}.

Proof. First of all we take two mutually orthogonal sudoku latin squares of order ab in(a× b)−rectangles. We can perform a permutation on the alphabet without affecting theorthogonality of those sudoku latin squares, such that the first row in both sudoku latinsquares is given by (0, 1, . . . , b− 1, . . . , ab− b, ab− b+ 1, . . . , ab− 1). Now we can do thisfor each sudoku latin square in a set of (k)MOSLS(ab). So we obtain

L1 =

0 1 . . . b− 1 . . . ab− b ab− b+ 1 . . . ab− 1c . . .... . . .

. . .

,

L2 =

0 1 . . . b− 1 . . . ab− b ab− b+ 1 . . . ab− 1d . . .... . . .

. . .

,

. . .

Neither symbol c nor d can be an element of the set of integers modulo b, {0, 1, . . . , b−1},or otherwise this would be in conflict with the definition of a sudoku latin square of orderab in (a × b)−rectangles. It is also impossible that c = d, otherwise if we would havec = d = i, then we would have that ((L1)1,0, (L2)1,0) = (i, i) = ((L1)0,i, (L2)0,i). But(1, 0) 6= (0, i), so we are in conflict with the definition of orthogonal sudoku latin squaresof order ab in (a× b)−rectangles.This means there are at most ab − b possible symbols of the alphabet that can appearon the first position of the second row and it is impossible that two sudoku latin squareshave the same symbol in that position, hence NS(a× b) ≤ ab− b.Now we can make a similar reasoning to prove that NS(a × b) ≤ ab − a. Let us startagain with two mutually orthogonal sudoku latin squares of order ab in (a×b)−rectangles.We can do a permutation on the alphabet without affecting the orthogonality of thosesudoku latin squares, such that the first column in both sudoku latin squares is given by(0, 1, . . . , a − 1, . . . , ab − a, ab − a + 1, . . . , ab − 1). Now we can do this for each sudokulatin square in a set of (k)MOSLS(ab). So we obtain

L1 =

0 c . . .1 . . .

a− 1 . . .... . . .

ab− a . . .... . . .

ab− 1 . . .

, L2 =

0 d . . .1 . . .

a− 1 . . .... . . .

ab− a . . .... . . .

ab− 1 . . .

, . . .

38

Neither symbol c nor d can be an element of the set of integers modulo a, {0, 1, . . . , a−1},or otherwise this would be in conflict with the definition of a sudoku latin square of orderab in (a × b)−rectangles. It is also impossible that c = d, otherwise if we would havec = d = i, then we would have that ((L1)0,1, (L2)0,1) = (i, i) = ((L1)i,0, (L2)i,0). But(0, 1) 6= (i, 0), so we are in conflict with the definition of orthogonal sudoku latin squaresof order ab in (a× b)−rectangles.This means there are at most ab − a possible symbols of the alphabet that can appearon the second position of the first row and it is impossible that two sudoku latin squareshave the same symbol in that position, hence NS(a× b) ≤ ab− a.

5.21 Definition. A set of (ab − max(a, b))MOSLS(ab) is called a complete set ofMOSLS of order a× b.

5.22 Theorem. (MacNeish) N(n×m) ≥ min{N(n), N(m)}.

Proof. Suppose that N(n) ≤ N(m) without loss of generality. We define k := N(n) andl := N(m). Now we will prove this theorem by constructing a set of (k)MOLS(n×m).First of all we take a set of (k)MOLS(n) {L1, L2, . . . , Lk} on the symbol set S ={a0, a1, . . . , an−1}. We can do a permutation on the symbol set without affecting theorthogonality of those latin squares, such that the first row in each latin square is givenby (a0, a1, . . . , an−1). Secondly we take a set of (l)MOLS(m) {L′1, L′2, . . . , L′l} on thesymbol set S ′ = {α0, α1, . . . , αm−1}. Again, we can do a permutation on the symbol setwithout affecting the orthogonality of those latin squares, such that the first row in eachlatin square is given by (α0, α1, . . . , αm−1).Now we will combine these two sets to obtain a set of (k)MOLS(n×m). To explain howwe do this, we will give an example.Let us take a latin square Lr from the set of (k)MOLS(n) and a latin square L′s from theset of (l)MOLS(m).

Lr =

a0 a1 . . . an−1(Lr)1,0 (Lr)1,1 . . . (Lr)1,n−1

.... . .

(Lr)n−1,0 (Lr)n−1,1 . . . (Lr)n−1,n−1

, L′s =

α0 α1 . . . αm−1(L′s)1,0 (L′s)1,1 . . . (L′s)1,m−1

.... . .

(L′s)m−1,0 (L′s)m−1,1 . . . (L′s)m−1,m−1

.

We can combine these two latin squares to obtain a latin square L of order n×m.

(a0, α0) . . . (a0, αm−1) (a1, α0) . . . (a1, αm−1) . . . (an−1, α0) . . . (an−1, αm−1)(a0, (L

′s)1,0) . . . (a0, (L

′s)1,m−1) (a1, (L

′s)1,0) . . . (a1, (L

′s)1,m−1) . . . (an−1, (L

′s)1,0) . . . (an−1, (L

′s)1,m−1)

.... . .

......

. . ....

. . ....

. . ....

(a0, (L′s)m−1,0) . . . (a0, (L

′s)m−1,m−1) (a1, (L

′s)m−1,0) . . . (a1, (L

′s)m−1,m−1) . . . (an−1, (L

′s)m−1,0) . . . (an−1, (L

′s)m−1,m−1)

((Lr)1,0, α0) . . . ((Lr)1,0, αm−1) ((Lr)1,1, α0) . . . ((Lr)1,1, αm−1) . . . ((Lr)1,n−1, α0) . . . ((Lr)1,n−1, αm−1)((Lr)1,0, (L

′s)1,0) . . . ((Lr)1,0, (L

′s)1,m−1) ((Lr)1,1, (L

′s)1,0) . . . ((Lr)1,1, (L

′s)1,m−1) . . . ((Lr)1,n−1, (L

′s)1,0) . . . ((Lr)1,n−1, (L

′s)1,m−1)

.... . .

......

. . ....

. . ....

. . ....

.... . .

......

. . ....

. . ....

. . ....

((Lr)n−1,0, α0) . . . ((Lr)n−1,0, αm−1) ((Lr)n−1,1, α0) . . . ((Lr)n−1,1, αm−1) . . . ((Lr)n−1,n−1, α0) . . . ((Lr)n−1,n−1, αm−1)...

. . ....

.... . .

.... . .

.... . .

...

This latin square is the direct product of Lr and L′s. We can also say that we superim-posed the second latin square on each entry in the first latin square. Now we will rename

39

each entry in L as follows, im+ j := (ai, αj), where 0 ≤ i ≤ n− 1 and 0 ≤ j ≤ m− 1.Our goal is to construct a set of (k)MOLS(n×m). By following the construction as ex-plained before we can construct k different latin squares Lt = Lt×L′t for t ∈ {1, 2, . . . , k}.The question is, are those latin squares mutually orthogonal?Let us take two random different latin squares of order n ×m obtained by the previousconstruction, Lv = Lv × L′v and Lw = Lw × L′w, where v 6= w.Suppose (Lv)g,h = (Lv)i,j and (Lw)g,h = (Lw)i,j for (g, h) 6= (i, j), where (g, h) and(i, j) are ordered pairs of {0, 1, . . . , nm− 1} × {0, 1, . . . , nm− 1}. This would mean that(Lv × L′v)g,h = (Lv × L′v)i,j and (Lw × L′w)g,h = (Lw × L′w)i,j for (g, h) 6= (i, j). By con-struction this implies that ((Lv)γ,δ, (L

′v)µ,ν) = ((Lv)γ′,δ′ , (L

′v)µ′,ν′) and ((Lw)γ,δ, (L

′w)µ,ν) =

((Lw)γ′,δ′ , (L′w)µ′,ν′) for (g, h) 6= (i, j). We want to translate this last equation ‘(g, h) 6=

(i, j)’ in terms of γ, δ, µ, ν, γ′, δ′, µ′ and ν ′. The ordered pair (g, h) is not equal to theordered pair (i, j), if g 6= i and/or h 6= j. If we look carefully, we see that this is thesame as γ 6= γ′ and/or δ 6= δ′ and/or µ 6= µ′ and/or ν 6= ν ′. This means that the ordered4-tuple (γ, δ, µ, ν) is not equal to the ordered 4-tuple (γ′, δ′, µ′, ν ′).If ((Lv)γ,δ, (L

′v)µ,ν) = ((Lv)γ′,δ′ , (L

′v)µ′,ν′), then we have that (Lv)γ,δ = (Lv)γ′,δ′ and (L′v)µ,ν =

(L′v)µ′,ν′ . If ((Lw)γ,δ, (L′w)µ,ν) = ((Lw)γ′,δ′ , (L

′w)µ′,ν′), then we have that (Lw)γ,δ = (Lw)γ′,δ′

and (L′w)µ,ν = (L′w)µ′,ν′ .Reordering the previous statements, we found that ((Lv)γ,δ, (Lw)γ,δ) = ((Lv)γ′,δ′ , (Lw)γ′,δ′)and ((L′v)µ,ν , (L

′w)µ,ν) = ((L′v)µ′,ν′ , (L

′w)µ′,ν′), for (γ, δ, µ, ν) 6= (γ′, δ′, µ′, ν ′).

Remember that Lv, Lw ∈ (k)MOLS(n) and L′v, L′w ∈ (l)MOLS(m), which means that

Lv and Lw are orthogonal, implying that if ((Lv)γ,δ, (Lw)γ,δ) = ((Lv)γ′,δ′ , (Lw)γ′,δ′), thenwe have that (γ, δ) = (γ′, δ′). We have that L′v and L′w are orthogonal as well, thereforeif ((L′v)µ,ν , (L

′w)µ,ν) = ((L′v)µ′,ν′ , (L

′w)µ′,ν′), then we have that (µ, ν) = (µ′, ν ′).

This is a contradiction, because we supposed that ((Lv)γ,δ, (Lw)γ,δ) = ((Lv)γ′,δ′ , (Lw)γ′,δ′)and ((L′v)µ,ν , (L

′w)µ,ν) = ((L′v)µ′,ν′ , (L

′w)µ′,ν′), for (γ, δ, µ, ν) 6= (γ′, δ′, µ′, ν ′).

So we have that if (Lv)g,h = (Lv)i,j and (Lw)g,h = (Lw)i,j, then (g, h) = (i, j). Conse-quently we found by construction two mutually orthogonal squares of order n×m. BecauseLv and Lw were random latin squares from the set {Lt} = {Lt×L′t} for t ∈ {1, 2, . . . , k},we have found a set of (k)MOLS(n×m), where k = min{N(n), N(m)}. This proves ourtheorem that N(n×m) ≥ min{N(n), N(m)}.

5.23 Definition. In the preceding theorem we defined the direct product of two differentlatin squares. This is also called the Kronecker product. Briefly summarized, the directproduct of two latin squares A and B of respectively order n and order m is the latin squareA⊗B of order n×m given by

A⊗B =

(a00, B) (a01, B) . . . (a0,n−1, B)(a10, B) (a11, B) . . . (a1,n−1, B). . . . . . . . . . . .

(an−1,0, B) (an−1,1, B) . . . (an−1,n−1, B)

.

We would like to formulate now a similar theorem for sudoku latin squares.

40

5.24 Theorem. (∀m,n ≥ 2)(NS((mn)2) ≥ min{NS(m2), NS(n2)}).

Proof. Suppose that NS(n2) ≤ NS(m2) without loss of generality. We define k :=NS(n2) and l := NS(m2). Now we can prove this theorem by constructing a set of(k)MOSLS((mn)2). This construction is the same as before, where we use sudoku latinsquares instead of latin squares.First we define a set of (k)MOSLS(n2) {L1, L2, . . . , Lk} on the symbol set S = {a0, a1, . . . ,an−1, . . . , an2−n, an2−n+1, . . . , an2−1} and a set of (l)MOSLS(m2) {L′1, L′2, . . . , L′l} on thesymbol set S ′ = {α0, α1, . . . , αm−1, . . . , αm2−m, αm2−m+1, . . . , αm2−1}.Let us take a sudoku latin square Lr from the set of (k)MOSLS(n2) and a sudoku latinsquare L′s from the set of (l)MOSLS(m2). We can combine these two sudoku latin squaresto obtain a latin square L = Lr ⊗ L′s of order (mn)2. This latin square is not a sudokulatin square. By following the same reasoning as in the preceding theorem, we obtain aset of (k)MOLS((mn)2).To obtain a set of (k)MOSLS((mn)2), we will have to apply the same permutation onthe rows and the columns of every latin square in our set of (k)MOLS((mn)2). We willillustrate this briefly.The latin square L = Lr ⊗ L′s is given by

((Lr)0,0, L′s) . . . ((Lr)0,n−1, L

′s) . . . ((Lr)0,n2−n, L

′s) . . . ((Lr)0,n2−1, L

′s)

.... . .

.... . .

.... . .

...((Lr)n−1,0, L

′s) . . . ((Lr)n−1,n−1, L

′s) . . . ((Lr)n−1,n2−n, L

′s) . . . ((Lr)n−1,n2−1, L

′s)

.... . .

.... . .

.... . .

...((Lr)n2−n,0, L

′s) . . . ((Lr)n2−n,n−1, L

′s) . . . ((Lr)n2−n,n2−n, L

′s) . . . ((Lr)n2−n,n2−1, L

′s)

.... . .

.... . .

.... . .

...((Lr)n2−1,0, L

′s) . . . ((Lr)n2−1,n−1, L

′s) . . . ((Lr)n2−1,n2−n, L

′s) . . . ((Lr)n2−1,n2−1, L

′s)

where for each entry (Lr)i,j of Lr, ((Lr)i,j, L′s) is given by

((Lr)i,j, (L′s)0,0) . . . ((Lr)i,j, (L

′s)0,m−1) . . . ((Lr)i,j, (L

′s)0,m2−m) . . . ((Lr)i,j, (L

′s)0,m2−1)

.... . .

.... . .

.... . .

...((Lr)i,j, (L

′s)m−1,0) . . . ((Lr)i,j, (L

′s)m−1,m−1) . . . ((Lr)i,j, (L

′s)m−1,m2−m) . . . ((Lr)i,j, (L

′s)m−1,m2−1)

.... . .

.... . .

.... . .

...((Lr)i,j, (L

′s)m2−m,0) . . . ((Lr)i,j, (L

′s)m2−m,m−1) . . . ((Lr)i,j, (L

′s)m2−m,m2−m) . . . ((Lr)i,j, (L

′s)m2−m,m2−1)

.... . .

.... . .

.... . .

...((Lr)i,j, (L

′s)m2−1,0) . . . ((Lr)i,j, (L

′s)m2−1,m−1) . . . ((Lr)i,j, (L

′s)m2−1,m2−m) . . . ((Lr)i,j, (L

′s)m2−1,m2−1)

.

Now we can do a permutation on the rows and the columns, such that the latin squareL has the structure of a sudoku latin square. In fact this permutation gives us a sudokulatin square with the following structure

Lr(1) ⊗ L′s

(1) . . . Lr(1) ⊗ L′s

(m) . . . Lr(n) ⊗ L′s

(1) . . . Lr(n) ⊗ L′s

(m)

.... . .

.... . .

.... . .

...

Lr(1) ⊗ L′s

(m2−m+1) . . . Lr(1) ⊗ L′s

(m2) . . . Lr(n) ⊗ L′s

(m2−m+1) . . . Lr(n) ⊗ L′s

(m2)

.... . .

.... . .

.... . .

...

Lr(n2−n+1) ⊗ L′s

(1) . . . Lr(n2−n+1) ⊗ L′s

(m) . . . Lr(n2) ⊗ L′s

(1) . . . Lr(n2) ⊗ L′s

(m)

.... . .

.... . .

.... . .

...

Lr(n2−n+1) ⊗ L′s

(m2−m+1) . . . Lr(n2−n+1) ⊗ L′s

(m2) . . . Lr(n2) ⊗ L′s

(m2−m+1) . . . Lr(n2) ⊗ L′s

(m2)

41

where we used the structure of our sudoku latin squares

Lr =

Lr(1) Lr

(2) . . . Lr(n)

Lr(n+1) Lr

(n+2) . . . Lr(2n)

......

. . ....

Lr(n2−n+1) Lr

(n2−n+2) . . . Lr(n2)

and

L′s =

L′s(1) L′s

(2) . . . L′s(m)

L′s(m+1) L′s

(m+2) . . . L′s(2m)

......

. . ....

L′s(m2−m+1) L′s

(m2−m+2) . . . L′s(m2)

.

5.25 Theorem. (∀a, b, c, d ≥ 2)(NS(ac× bd) ≥ min{NS(a× b), NS(c× d)}).

Proof. We can prove this theorem by following the same reasoning as before. Insteadof latin squares we work with sudoku latin squares of order ab and cd in respectively(a × b)−rectangles and (c × d)−rectangles. We will have to apply a row and a columnpermutation to our direct product to obtain a sudoku latin square of order abcd in (ac×bd)−rectangles.

5.26 Theorem. (∀m,n ≥ 2)(NS(m× n) ≥ min{N(m), N(n)}).

Proof. To prove this theorem we use the theorem of MacNeish. We know that for allm,n ≥ 2, N(m×n) ≥ min{N(m), N(n)}. In the proof of this theorem we used the directproduct to construct a set of (k)MOLS(m × n), where k = min{N(m), N(n)}. We cannow apply a permutation ρ on the rows, such that every latin square in this set becomesa sudoku latin square of order mn in (m×n)−rectangles. Note the change in the order ofn and m. Our goal is to make a sudoku latin square of order mn in (m× n)−rectangles.Suppose k := N(m), l := N(n) and N(m) ≤ N(n). If a random latin square in ourset of (k)MOLS(m × n) is given by L = Lr ⊗ L′r, where Lr is a latin square of the set(k)MOLS(m) and L′r is a latin square of the set (l)MOLS(n), then this permutation ρis given by:

ρ : in+ j 7→ jm+ i,

where i is an element of {0, 1, . . . ,m−1}, j is an element of {0, 1, . . . , n−1} and the rowsare numbered from 0 to mn− 1.

5.27 Example. Suppose m = 3 and n = 4. Let us firstly take a set of two mutuallyorthogonal latin squares of order 3.

A(1) =0 1 21 2 02 0 1

, A(2) =0 1 22 0 11 2 0

.

42

Now we take a set of two mutually orthogonal latin squares of order 4.

B(1) =

0 1 2 31 0 3 22 3 0 13 2 1 0

, B(2) =

0 1 2 32 3 0 13 2 1 01 0 3 2

.

Using the definition of the direct product of two latin squares, we can construct A(1)⊗B(1)

and A(2) ⊗B(2). We have that A(1) ⊗B(1) =

(0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3) (2, 0) (2, 1) (2, 2) (2, 3)(0, 1) (0, 0) (0, 3) (0, 2) (1, 1) (1, 0) (1, 3) (1, 2) (2, 1) (2, 0) (2, 3) (2, 2)(0, 2) (0, 3) (0, 0) (0, 1) (1, 2) (1, 3) (1, 0) (1, 1) (2, 2) (2, 3) (2, 0) (2, 1)(0, 3) (0, 2) (0, 1) (0, 0) (1, 3) (1, 2) (1, 1) (1, 0) (2, 3) (2, 2) (2, 1) (2, 0)(1, 0) (1, 1) (1, 2) (1, 3) (2, 0) (2, 1) (2, 2) (2, 3) (0, 0) (0, 1) (0, 2) (0, 3)(1, 1) (1, 0) (1, 3) (1, 2) (2, 1) (2, 0) (2, 3) (2, 2) (0, 1) (0, 0) (0, 3) (0, 2)(1, 2) (1, 3) (1, 0) (1, 1) (2, 2) (2, 3) (2, 0) (2, 1) (0, 2) (0, 3) (0, 0) (0, 1)(1, 3) (1, 2) (1, 1) (1, 0) (2, 3) (2, 2) (2, 1) (2, 0) (0, 3) (0, 2) (0, 1) (0, 0)(2, 0) (2, 1) (2, 2) (2, 3) (0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3)(2, 1) (2, 0) (2, 3) (2, 2) (0, 1) (0, 0) (0, 3) (0, 2) (1, 1) (1, 0) (1, 3) (1, 2)(2, 2) (2, 3) (2, 0) (2, 1) (0, 2) (0, 3) (0, 0) (0, 1) (1, 2) (1, 3) (1, 0) (1, 1)(2, 3) (2, 2) (2, 1) (2, 0) (0, 3) (0, 2) (0, 1) (0, 0) (1, 3) (1, 2) (1, 1) (1, 0)

and A(2) ⊗B(2) =

(0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3) (2, 0) (2, 1) (2, 2) (2, 3)(0, 2) (0, 3) (0, 0) (0, 1) (1, 2) (1, 3) (1, 0) (1, 1) (2, 2) (2, 3) (2, 0) (2, 1)(0, 3) (0, 2) (0, 1) (0, 0) (1, 3) (1, 2) (1, 1) (1, 0) (2, 3) (2, 2) (2, 1) (2, 0)(0, 1) (0, 0) (0, 3) (0, 2) (1, 1) (1, 0) (1, 3) (1, 2) (2, 1) (2, 0) (2, 3) (2, 2)(2, 0) (2, 1) (2, 2) (2, 3) (0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3)(2, 2) (2, 3) (2, 0) (2, 1) (0, 2) (0, 3) (0, 0) (0, 1) (1, 2) (1, 3) (1, 0) (1, 1)(2, 3) (2, 2) (2, 1) (2, 0) (0, 3) (0, 2) (0, 1) (0, 0) (1, 3) (1, 2) (1, 1) (1, 0)(2, 1) (2, 0) (2, 3) (2, 2) (0, 1) (0, 0) (0, 3) (0, 2) (1, 1) (1, 0) (1, 3) (1, 2)(1, 0) (1, 1) (1, 2) (1, 3) (2, 0) (2, 1) (2, 2) (2, 3) (0, 0) (0, 1) (0, 2) (0, 3)(1, 2) (1, 3) (1, 0) (1, 1) (2, 2) (2, 3) (2, 0) (2, 1) (0, 2) (0, 3) (0, 0) (0, 1)(1, 3) (1, 2) (1, 1) (1, 0) (2, 3) (2, 2) (2, 1) (2, 0) (0, 3) (0, 2) (0, 1) (0, 0)(1, 1) (1, 0) (1, 3) (1, 2) (2, 1) (2, 0) (2, 3) (2, 2) (0, 1) (0, 0) (0, 3) (0, 2)

.

Presently, A(1) ⊗ B(1) and A(2) ⊗ B(2) are partitioned into n × n = 4 × 4−submatrices.We change the order of the rows to partition A(1) ⊗ B(1) and A(2) ⊗ B(2) into m × n =(3 × 4)−rectangles. We can for example apply the permutation as given in theorem 5.26on the latin square A(1) ⊗B(1). This gives us the following latin square.

(0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3) (2, 0) (2, 1) (2, 2) (2, 3)(1, 0) (1, 1) (1, 2) (1, 3) (2, 0) (2, 1) (2, 2) (2, 3) (0, 0) (0, 1) (0, 2) (0, 3)(2, 0) (2, 1) (2, 2) (2, 3) (0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3)(0, 1) (0, 0) (0, 3) (0, 2) (1, 1) (1, 0) (1, 3) (1, 2) (2, 1) (2, 0) (2, 3) (2, 2)(1, 1) (1, 0) (1, 3) (1, 2) (2, 1) (2, 0) (2, 3) (2, 2) (0, 1) (0, 0) (0, 3) (0, 2)(2, 1) (2, 0) (2, 3) (2, 2) (0, 1) (0, 0) (0, 3) (0, 2) (1, 1) (1, 0) (1, 3) (1, 2)(0, 2) (0, 3) (0, 0) (0, 1) (1, 2) (1, 3) (1, 0) (1, 1) (2, 2) (2, 3) (2, 0) (2, 1)(1, 2) (1, 3) (1, 0) (1, 1) (2, 2) (2, 3) (2, 0) (2, 1) (0, 2) (0, 3) (0, 0) (0, 1)(2, 2) (2, 3) (2, 0) (2, 1) (0, 2) (0, 3) (0, 0) (0, 1) (1, 2) (1, 3) (1, 0) (1, 1)(0, 3) (0, 2) (0, 1) (0, 0) (1, 3) (1, 2) (1, 1) (1, 0) (2, 3) (2, 2) (2, 1) (2, 0)(1, 3) (1, 2) (1, 1) (1, 0) (2, 3) (2, 2) (2, 1) (2, 0) (0, 3) (0, 2) (0, 1) (0, 0)(2, 3) (2, 2) (2, 1) (2, 0) (0, 3) (0, 2) (0, 1) (0, 0) (1, 3) (1, 2) (1, 1) (1, 0)

43

When we replace every (i, j) by 4i+ j, then we obtain the following latin square

S1 =

0 1 2 3 4 5 6 7 8 9 10 114 5 6 7 8 9 10 11 0 1 2 38 9 10 11 0 1 2 3 4 5 6 71 0 3 2 5 4 7 6 9 8 11 105 4 7 6 9 8 11 10 1 0 3 29 8 11 10 1 0 3 2 5 4 7 62 3 0 1 6 7 4 5 10 11 8 96 7 4 5 10 11 8 9 2 3 0 110 11 8 9 2 3 0 1 6 9 4 53 2 1 0 7 6 5 4 11 10 9 87 6 5 4 11 10 9 8 3 2 1 011 10 9 8 3 2 1 0 7 6 5 4

which is a sudoku latin square of order 12 in (3× 4)−rectangles.We can also apply the permutation as given in theorem 5.26 on the latin square A(2)⊗B(2).This gives us the following latin square.

(0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3) (2, 0) (2, 1) (2, 2) (2, 3)(2, 0) (2, 1) (2, 2) (2, 3) (0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3)(1, 0) (1, 1) (1, 2) (1, 3) (2, 0) (2, 1) (2, 2) (2, 3) (0, 0) (0, 1) (0, 2) (0, 3)(0, 2) (0, 3) (0, 0) (0, 1) (1, 2) (1, 3) (1, 0) (1, 1) (2, 2) (2, 3) (2, 0) (2, 1)(2, 2) (2, 3) (2, 0) (2, 1) (0, 2) (0, 3) (0, 0) (0, 1) (1, 2) (1, 3) (1, 0) (1, 1)(1, 2) (1, 3) (1, 0) (1, 1) (2, 2) (2, 3) (2, 0) (2, 1) (0, 2) (0, 3) (0, 0) (0, 1)(0, 3) (0, 2) (0, 1) (0, 0) (1, 3) (1, 2) (1, 1) (1, 0) (2, 3) (2, 2) (2, 1) (2, 0)(2, 3) (2, 2) (2, 1) (2, 0) (0, 3) (0, 2) (0, 1) (0, 0) (1, 3) (1, 2) (1, 1) (1, 0)(1, 3) (1, 2) (1, 1) (1, 0) (2, 3) (2, 2) (2, 1) (2, 0) (0, 3) (0, 2) (0, 1) (0, 0)(0, 1) (0, 0) (0, 3) (0, 2) (1, 1) (1, 0) (1, 3) (1, 2) (2, 1) (2, 0) (2, 3) (2, 2)(2, 1) (2, 0) (2, 3) (2, 2) (0, 1) (0, 0) (0, 3) (0, 2) (1, 1) (1, 0) (1, 3) (1, 2)(1, 1) (1, 0) (1, 3) (1, 2) (2, 1) (2, 0) (2, 3) (2, 2) (0, 1) (0, 0) (0, 3) (0, 2)

.

When we replace every (i, j) by 4i+ j, then we obtain the following latin square

S2 =

0 1 2 3 4 5 6 7 8 9 10 118 9 10 11 0 1 2 3 4 5 6 74 5 6 7 8 9 10 11 0 1 2 32 3 0 1 6 7 4 5 10 11 8 910 11 8 9 2 3 0 1 6 7 4 56 7 4 5 10 11 8 9 2 3 0 13 2 1 0 7 6 5 4 11 10 9 811 10 9 8 3 2 1 0 7 6 5 47 6 5 4 11 10 9 8 3 2 1 01 0 3 2 5 4 7 6 9 8 11 109 8 11 10 1 0 3 2 5 4 7 65 4 7 6 9 8 11 10 1 0 3 2

.

which is a sudoku latin square of order 12 in (3× 4)−rectangles.We see that S1 and S2 are mutually orthogonal (3× 4)−sudoku latin squares.

44

5.28 Theorem. If q = pd is a prime power, then N(q) = q − 1.

Proof. We will prove this theorem by constructing a complete set of MOLS(q).If q = p is a prime number, then we define for each a ∈ GF (p)\{0}, where GF (p) ={0, 1, . . . , p− 1}, a latin square (La)x,y as follows.

(La)x,y = ax+ y mod p,

where x, y ∈ GF (p). This gives us a set of p− 1 different latin squares of order p.

L1 =

0 1 . . . p− 11 . . ....

p− 1

, L2 =

0 1 . . . p− 12 . . ....

p− 2

, . . . , Lp−1 =

0 1 . . . p− 1p− 1 . . .

...1

.

If these latin squares are mutually orthogonal, then we would have that if ((Lb)i,j, (Lb′)i,j) =((Lb)k,l, (Lb′)k,l), for b, b′ ∈ GF (p)\{0} and b 6= b′, then (i, j) = (k, l).Suppose that (Lb)i,j = (Lb)k,l and (Lb′)i,j = (Lb′)k,l. This means that bi+ j = bk + l andb′i+ j = b′k+ l, or bi+ j− (b′i+ j) = bk+ l− (b′k+ l), such that (b− b′)i = (b− b′)k. Weknow that b 6= b′ and b, b′ 6= 0, therefore i = k. Because b and b′ are not equal to 0, wealso have that b′(bi+ j) = b′(bk+ l) and b(b′i+ j) = b(b′k+ l), or b′bi+ b′j − (bb′i+ bj) =b′bk + b′l − (bb′k + bl), such that (b′ − b)j = (b′ − b)l. We know that b 6= b′ and b, b′ 6= 0,therefore j = l. Thus (i, j) = (k, l).This means that we have found a set of (p − 1)MOLS(p) describing PG(2, p), for p aprime number.If q = pd is a prime power, we will define a set of (q−1)MOLS(q) describing PG(2, q), sim-ilar as in the previous case. First we define a primitive polynomial, f(x) = xd+fd−1x

d−1+. . .+f1x+f0 and we call α a root of this polynomial. This means that α is a primitive el-ement of GF (q). Now we can represent GF (q) as follows, GF (q) = {0, 1, α, α2, . . . , αq−2},or GF (q) = {0, 1, 2, . . . , x, . . . , q− 1} where x = xd−1p

d−1 + xd−2pd−2 + . . .+ x1p+ x0 and

xi ∈ GF (p). We also use the notation, X =

x0x1...

xd−1

∈ GF (p)d.

Let us denote the latin squares by La, a ∈ GF (q)\{0}, where

(Aa)x,y = ax+ y mod f(x),

for x, y ∈ GF (q).We can follow a similar reasoning as before to prove that these latin squares are mutuallyorthogonal.

5.29 Corollary. If n = pe11 pe22 . . . pekk , where each number pi is a distinct prime number

and ei ≥ 1, i = 1, . . . , k, then N(n) ≥ min{peii − 1|i = 1, 2, . . . , k}.

45

5.30 Theorem. If q = pd is a prime power, then NS(q2) = q2 − q.

Proof. We know that we can construct a set of (q2 − 1)MOLS(q2) for q = pd a primepower. In fact we only have to take a subset of this set to prove this theorem.Let us consider the finite field GF (q) and the finite field GF (q2). We can write GF (q2) ={k + αl|k, l ∈ GF (q)}. For each a ∈ GF (q2)\GF (q), we define a latin square as follows.

(La)x,y = ax+ y,

where x, y ∈ GF (q2). These latin squares are of sudoku type and are mutually orthogonal.

5.31 Corollary. If n = pe11 pe22 . . . pekk , where each number pi is a distinct prime number

and ei ≥ 1, i = 1, . . . , k, then NS(n2) ≥ min{p2eii − peii |i = 1, 2, . . . , k}.

5.32 Theorem. Suppose that ab = q = pd is a prime power, a = ps, b = pt and t ≥ s.Then NS(a× b) = ab− b.

5.3 Examples of small order

5.3.1 Latin squares of order n

n 2 3 4 5 6 7 8 9 10

N(n) 1 2 3 4 1 6 7 8 ? ≥ 2,≤ 8

5.33 Theorem. N(2) = 1

0 11 0

5.34 Theorem. N(3) = 2

0 1 2 0 1 21 2 0 2 0 12 0 1 1 2 0

5.35 Theorem. N(4) = 3

0 1 2 3 0 1 2 3 0 1 2 31 0 3 2 2 3 0 1 3 2 1 02 3 0 1 3 2 1 0 1 0 3 23 2 1 0 1 0 3 2 2 3 0 1

46

5.36 Theorem. N(5) = 4

0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 41 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 32 3 4 0 1 4 0 1 2 3 1 2 3 4 0 3 4 0 1 23 4 0 1 2 1 2 3 4 0 4 0 1 2 3 2 3 4 0 14 0 1 2 3 3 4 0 1 2 2 3 4 0 1 1 2 3 4 0

5.37 Theorem. N(6) = 1

This was proved by Gaston Tarry in 1900, who was very interested in magic squaresand latin squares. By proving that N(6) = 1, Tarry solved Euler’s 36 Officers Problem.Euler did not only conjecture that there does not exist a pair of mutually orthogonallatin squares of order 6, but he also conjectured that a pair of mutually orthogonal latinsquares does not exist whenever n = 2 mod 4. This conjecture is disproved by R. C.Bose, S. S. Shrikhande and E. T. Parker. We will not go into further details, but moreinformation can be found in [1] and [7].

5.38 Theorem. N(7) = 6

0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 61 2 3 4 5 6 0 2 3 4 5 6 0 1 3 4 5 6 0 1 22 3 4 5 6 0 1 4 5 6 0 1 2 3 6 0 1 2 3 4 53 4 5 6 0 1 2 6 0 1 2 3 4 5 2 3 4 5 6 0 14 5 6 0 1 2 3 1 2 3 4 5 6 0 5 6 0 1 2 3 45 6 0 1 2 3 4 3 4 5 6 0 1 2 1 2 3 4 5 6 06 0 1 2 3 4 5 5 6 0 1 2 3 4 4 5 6 0 1 2 3

0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 64 5 6 0 1 2 3 5 3 0 1 2 3 4 6 0 1 2 3 4 51 2 3 4 5 6 0 3 4 5 6 0 1 2 5 6 0 1 2 3 45 6 0 1 2 3 4 1 2 3 4 5 6 0 4 5 6 0 1 2 32 3 4 5 6 0 1 6 0 1 2 3 4 5 3 4 5 6 0 1 26 0 1 2 3 4 5 4 5 6 0 1 2 3 2 3 4 5 6 0 13 4 5 6 0 1 2 2 3 4 5 6 0 1 1 2 3 4 5 6 0

5.39 Theorem. N(8) = 7

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 71 0 3 2 5 4 7 6 2 3 0 1 6 7 4 5 3 2 1 0 7 6 5 4 4 5 6 7 0 1 2 32 3 0 1 6 7 4 5 4 5 6 7 0 1 2 3 6 7 4 5 2 3 0 1 3 2 1 0 7 6 5 43 2 1 0 7 6 5 4 6 7 4 5 2 3 0 1 5 4 7 6 1 0 3 2 7 6 5 4 3 2 1 04 5 6 7 0 1 2 3 3 2 1 0 7 6 5 4 7 6 5 4 3 2 1 0 6 7 4 5 2 3 0 15 4 7 6 1 0 3 2 1 0 3 2 5 4 7 6 4 5 6 7 0 1 2 3 2 3 0 1 6 7 4 56 7 4 5 2 3 0 1 7 6 5 4 3 2 1 0 1 0 3 2 5 4 7 6 5 4 7 6 1 0 3 27 6 5 4 3 2 1 0 5 4 7 6 1 0 3 2 2 3 0 1 6 7 4 5 1 0 3 2 5 4 7 6

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 75 4 7 6 1 0 3 2 6 7 4 5 2 3 0 1 7 6 5 4 3 2 1 01 0 3 2 5 4 7 6 7 6 5 4 3 2 1 0 5 4 7 6 1 0 3 24 5 6 7 0 1 2 3 1 0 3 2 5 4 7 6 2 3 0 1 6 7 4 52 3 0 1 6 7 4 5 5 4 7 6 1 0 3 2 1 0 3 2 5 4 7 67 6 5 4 3 2 1 0 3 2 1 0 7 6 5 4 6 7 4 5 2 3 0 13 2 1 0 7 6 5 4 2 3 0 1 6 7 4 5 4 5 6 7 0 1 2 36 7 4 5 2 3 0 1 4 5 6 7 0 1 2 3 3 2 1 0 7 6 5 4

47

5.40 Theorem. N(9) = 8

0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 81 2 0 4 5 3 7 8 6 2 0 1 5 3 4 8 6 7 3 4 5 6 7 8 0 1 2 4 5 3 7 8 6 1 2 02 0 1 5 3 4 8 6 7 1 2 0 4 5 3 7 8 6 6 7 8 0 1 2 3 4 5 8 6 7 2 0 1 5 3 43 4 5 6 7 8 0 1 2 6 7 8 0 1 2 3 4 5 7 8 6 1 2 0 4 5 3 1 2 0 4 5 3 7 8 64 5 3 7 8 6 1 2 0 8 6 7 2 0 1 5 3 4 1 2 0 4 5 3 7 8 6 5 3 4 8 6 7 2 0 15 3 4 8 6 7 2 0 1 7 8 6 1 2 0 4 5 3 4 5 3 7 8 6 1 2 0 6 7 8 0 1 2 3 4 56 7 8 0 1 2 3 4 5 3 4 5 6 7 8 0 1 2 5 3 4 8 6 7 2 0 1 2 0 1 5 3 4 8 6 77 8 6 1 2 0 4 5 3 5 3 4 8 6 7 2 0 1 8 6 7 2 0 1 5 3 4 3 4 5 6 7 8 0 1 28 6 7 2 0 1 5 3 4 4 5 3 7 8 6 1 2 0 2 0 1 5 3 4 8 6 7 7 8 6 1 2 0 4 5 3

0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 85 3 4 8 6 7 2 0 1 6 7 8 0 1 2 3 4 5 7 8 6 1 2 0 4 5 3 8 6 7 2 0 1 5 3 47 8 6 1 2 0 4 5 3 3 4 5 6 7 8 0 1 2 5 3 4 8 6 7 2 0 1 4 5 3 7 8 6 1 2 04 5 3 7 8 6 1 2 0 5 3 4 8 6 7 2 0 1 8 6 7 2 0 1 5 3 4 2 0 1 5 3 4 8 6 76 7 8 0 1 2 3 4 5 2 0 1 5 3 4 8 6 7 3 4 5 6 7 8 0 1 2 7 8 6 1 2 0 4 5 32 0 1 5 3 4 8 6 7 8 6 7 2 0 1 5 3 4 1 2 0 4 5 3 7 8 6 3 4 5 6 7 8 0 1 28 6 7 2 0 1 5 3 4 7 8 6 1 2 0 4 5 3 4 5 3 7 8 6 1 2 0 1 2 0 4 5 3 7 8 61 2 0 4 5 3 7 8 6 4 5 3 7 8 6 1 2 0 2 0 1 5 3 4 8 6 7 6 7 8 0 1 2 3 4 53 4 5 6 7 8 0 1 2 1 2 0 4 5 3 7 8 6 6 7 8 0 1 2 3 4 5 5 3 4 8 6 7 2 0 1

5.41 Theorem. N(10) ≥ 2

0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 91 7 4 2 0 6 5 8 9 3 4 6 1 8 5 3 7 9 0 22 3 5 7 8 9 1 0 4 6 7 9 8 2 1 6 3 4 5 03 5 1 4 6 0 8 9 2 7 5 3 6 9 8 7 0 2 1 44 8 9 6 2 3 0 5 7 1 3 5 7 4 6 1 2 0 9 85 6 8 9 7 1 4 2 3 0 2 7 4 5 0 9 8 3 6 16 0 7 1 3 8 9 4 5 2 9 8 3 0 7 2 1 6 4 57 9 6 8 1 4 2 3 0 5 1 4 5 7 2 0 9 8 3 68 2 0 5 9 7 3 1 6 4 6 0 9 1 3 8 4 5 2 79 4 3 0 5 2 7 6 1 8 8 2 0 6 9 4 5 1 7 3

5.3.2 Sudoku latin squares of order n× n

n 2 3

NS(n2) 2 6

5.42 Theorem. NS(4) = 2

0 1 2 3 0 1 2 32 3 0 1 3 2 1 03 2 1 0 1 0 3 21 0 3 2 2 3 0 1

48

5.43 Theorem. NS(9) = 6

0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 83 4 5 6 7 8 0 1 2 4 5 3 7 8 6 1 2 0 5 3 4 8 6 7 2 0 16 7 8 0 1 2 3 4 5 8 6 7 2 0 1 5 3 4 7 8 6 1 2 0 4 5 37 8 6 1 2 0 4 5 3 1 2 0 4 5 3 7 8 6 4 5 3 7 8 6 1 2 01 2 0 4 5 3 7 8 6 5 3 4 8 6 7 2 0 1 6 7 8 0 1 2 3 4 54 5 3 7 8 6 1 2 0 6 7 8 0 1 2 3 4 5 2 0 1 5 3 4 8 6 75 3 4 8 6 7 2 0 1 2 0 1 5 3 4 8 6 7 8 6 7 2 0 1 5 3 48 6 7 2 0 1 5 3 4 3 4 5 6 7 8 0 1 2 1 2 0 4 5 3 7 8 62 0 1 5 3 4 8 6 7 7 8 6 1 2 0 4 5 3 3 4 5 6 7 8 0 1 2

0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 86 7 8 0 1 2 3 4 5 7 8 6 1 2 0 4 5 3 8 6 7 2 0 1 5 3 43 4 5 6 7 8 0 1 2 5 3 4 8 6 7 2 0 1 4 5 3 7 8 6 1 2 05 3 4 8 6 7 2 0 1 8 6 7 2 0 1 5 3 4 2 0 1 5 3 4 8 6 72 0 1 5 3 4 8 6 7 3 4 5 6 7 8 0 1 2 7 8 6 1 2 0 4 5 38 6 7 2 0 1 5 3 4 1 2 0 4 5 3 7 8 6 3 4 5 6 7 8 0 1 27 8 6 1 2 0 4 5 3 4 5 3 7 8 6 1 2 0 1 2 0 4 5 3 7 8 64 5 3 7 8 6 1 2 0 2 0 1 5 3 4 8 6 7 6 7 8 0 1 2 3 4 51 2 0 4 5 3 7 8 6 6 7 8 0 1 2 3 4 5 5 3 4 8 6 7 2 0 1

5.4 On the number of (n− 1)MOLS(n)

5.4.1 First case: GF (p)

Now that we have found some general results on MOLS, we want to know something moreabout sets of MOLS. More specific, we are interested in the number of (n−1)MOLS(n).We will consider two cases, first of all the case n = p, a prime number and then the casen = q = pd, a prime power. Let us start with the case n = p.

5.44 Definition. We consider a set of (p − 1)MOLS(p) describing PG(2, p), for p aprime, where GF (p) = {0, 1, . . . , p − 1} mod p. Let us denote the latin squares by Aa,a ∈ {1, . . . , p− 1} mod p, where

(Aa)x,y = ax+ y mod p,

for x, y ∈ GF (p).This definition gives us

A1 =

0 1 . . . p-11 . . ....

p-1

, A2 =

0 1 . . . p-12 . . ....

p-2

, . . . , Ap−1 =

0 1 . . . p-1p-1 . . .

...1

.

This set of MOLS is reduced. Our goal is to prove that there are (p−2)! different reduced(p − 1)MOLS(p), for p a prime, describing PG(2, p). First of all we want to determinethe group of permutations on the (p− 1)MOLS(p) stabilizing the set {A1, A2, . . . , Ap−1}.To fully understand these permutations on a set of latin squares, we will prove somecharacteristics of the permutations.

49

5.45 Lemma. A permutation on the alphabet commutes with a permutation on the rowsand/or columns of the latin squares.

Proof. We define the following permutations:

1. σ is a permutation on the alphabet: (Aa)x,y 7→ (Aa)σx,y.

2. ρ is a permutation on the rows: x 7→ ρ(x), and we have to use the same permutationon the rows in every latin square in the set of MOLS.

3. γ is a permutation on the columns: y 7→ γ(y), and we have to use the same permu-tation on the columns in every latin square in the set of MOLS.

With these permutations, we have

(σ ◦ ρ ◦ γ)((Aa)x,y) = ((Aa)x,y)σ on row ρ(x) and column γ(y),

(ρ ◦ γ ◦ σ)((Aa)x,y) = ((Aa)x,y)σ on row ρ(x) and column γ(y).

5.46 Lemma. Consider the set of MOLS {A1, A2, . . . , Ap−1} of definition 5.44. Supposewe apply σ1 to A1, σ2 to A2, . . . , and σp−1 to Ap−1, the same column permutations toall of the latin squares in the set and also the same row permutations to all of the latinsquares in the set, whereby row i moves to row 0. Suppose we obtain a new reduced set oflatin squares {C1, C2, . . . , Cp−1}, then

σb = σa ◦ τb, where τb : y 7→ y − bi+ ai mod p.

Proof. After applying σa to Aa and σb to Ab, we find σa(ai + y) and σb(bi + y) on rowi. Now we move row i to row 0. The same permutation of the columns transformseverything on row 0 to 0, 1, . . . , p−1. So necessarily, the same element must be in columny in C1, C2, . . . , Cp−1. So we have

σa(ai+ y) = σb(bi+ y), ∀b and y is the variable.

⇔ σb = σa ◦ τb, where τb : y 7→ y − bi+ ai mod p.

Let us prove this last assumption.Assume σb = σa◦τb, this is if and only if σb(bi+y) = (σa◦τb)(bi+y) = σa(bi+y−bi+ai) =σa(ai+ y).

5.47 Theorem. There are at most (p − 2)! different reduced (p − 1)MOLS(p), for p aprime, describing PG(2, p).

Proof. We work with the alphabet of GF (p) = {0, 1, . . . , p− 1}. We know that there arep! permutations on the alphabet.Define the permutation

ρa′ : GF (p)→ GF (p) : x 7→ a′x, for a′ ∈ GF (p) and a′ 6= 0.

50

We have that ρa′((Aa)x,y) = aa′x+a′y = (Aaa′)x,a′y, but the element aa′x+a′y still standsin row x and column y.If we perform the permutation γa′ : y 7→ a′y on the columns in every latin square of theset of MOLS, we have moved column y to column a′y, so we have transformed Aa inAaa′ . So reordering the latin squares, we get A1, A2, . . . , Ap−1 in the original order.Next, define the permutation

γb : GF (p)→ GF (p) : y 7→ y + b, for b ∈ GF (p).

We have that γb((Aa)x,y) = ax+ y+ b = (Aa)x,y+b, but the element ax+ y+ b still standsin row x and column y.If we perform the permutation γb : y 7→ y + b on the columns in every latin square of theset of MOLS, column y is moved to column y + b and we have transformed Aa in Aa, sowe have the same set of MOLS {A1, A2, . . . , Ap−1}.Now we have found p(p − 1) permutations on the (p − 1)MOLS(p) stabilizing the set{A1, A2, . . . , Ap−1}. Because there are p! permutations on GF (p), we have at most (p−2)!different reduced (p− 1)MOLS(p), for p a prime, describing PG(2, p).

5.48 Theorem. There are (p − 2)! different reduced (p − 1)MOLS(p), for p a prime,describing PG(2, p).

Proof. The p(p − 1) permutations on the alphabet that we defined in the previous the-orem form the group AGL1(p). These permutations on the alphabet that stabilize{A1, A2, . . . , Ap−1}, modulo the same permutations on the rows and columns, form asubgroup of the symmetric group Sp. Now we know from [20] that this group is maximalin Sp. So if there are more permutations stabilizing the set of (p− 1)MOLS(p) describ-ing PG(2, p), we would have the group Sp stabilizing {A1, A2, . . . , Ap−1}. So if we find2 different reduced sets of (p − 1)MOLS(p) describing PG(2, p) we know that we can’textend the group AGL1(p) to Sp, so there are only p(p− 1) permutations stabilizing theset of (p − 1)MOLS(p). We are going to prove this now in the next theorem, whichwill conclude the proof of this theorem that there are exactly (p − 2)! different reduced(p− 1)MOLS(p), for p a prime, describing PG(2, p).

5.49 Theorem. There are at least two different reduced sets of (p − 1)MOLS(p) for pan odd prime > 3.

Proof. In the beginning we defined a standard example of a set of (p − 1)MOLS(p),(Aa)x,y = ax+ y.So, (A1)x,y = x+ y.

A1 =

0 1 . . . p− 3 p− 2 p− 11 2 . . ....

. . .

p− 3 . . . p− 5 p− 4p− 2 . . . p− 5 p− 4 p− 3p− 1 . . . p− 4 p− 3 p− 2

.

51

We apply the permutation (p− 2, p− 1) on the alphabet on the set of (p− 1)MOLS(p),{A1, A2, . . . , Ap−1}. If we do this, we obtain a set of (p−1)MOLS(p) {B′1, B′2, . . . , B′p−1}.

A1 7→ B′1 =

0 1 . . . p− 3 p− 1 p− 21 2 . . ....

. . .

p− 3 . . . p− 5 p− 4p− 1 . . . p− 5 p− 4 p− 3p− 2 . . . p− 4 p− 3 p− 1

.

If we perform the permutation ρ2 = γ2 : (p−2, p−1) on respectively the rows and columnsof every latin square of the set {B′1, B′2, . . . , B′p−1}, we obtain the set {B1, B2, . . . , Bp−1}and this set is reduced. In this set, the first latin square is reduced, so we transformed A1

into B1.

B′1 7→ B1 =

0 1 . . . p− 3 p− 2 p− 11 2 . . ....

. . .

p− 3 . . . p− 4 p− 5p− 2 . . . p− 4 p− 1 p− 3p− 1 . . . p− 5 p− 3 p− 4

.

If we look at the 2 × 2 matrix in the bottom right corner of A1, we see the symbolsp− 4, p− 3 and p− 2. Applying the permutation (p− 2, p− 1) on the alphabet, the rowsand the columns, we obtain the latin square B1. In this latin square, we see the symbolsp− 4, p− 3 and p− 1 in the 2× 2 matrix in the bottom right corner. So we see that thereduced latin square B1 is different from A1, so we have found two different reduced setsof (p− 1)MOLS(p), describing PG(2, p): {A1, A2, . . . , Ap−1} and {B1, B2, . . . , Bp−1}.

5.4.2 Second case: GF (q)

5.50 Definition. Just as in the previous case, we will first define a set of (q−1)MOLS(q)describing PG(2, q), for q = pd a prime power. In this case we define a primitive poly-nomial, f(x) = xd + fd−1x

d−1 + . . . + f1x + f0 and we call α a root of this polynomial.This means that α is a primitive element of GF (q). Now we can represent GF (q) asfollows, GF (q) = {0, 1, α, α2, . . . , αq−2}, or GF (q) = {0, 1, 2, . . . , x, . . . , q − 1} wherex = xd−1p

d−1 + xd−2pd−2 + . . . + x1p + x0 and xi ∈ GF (p). We also use the notation

X =

x0x1...

xd−1

∈ GF (p)d.

Let us denote the latin squares by Aa, a ∈ GF (q)∗ mod f(α), where

(Aa)x,y = ax+ y mod f(α),

52

for x, y ∈ GF (q).

This set of latin squares {A1, A2, . . . , Aq−1} is reduced.First of all, we can identify every element a ∈ GF (q)∗ in our definition with an invertibled× d matrix Ma over GF (p) representing the multiplication by a in GF (q) over GF (p)d.These matrices define a cyclic group of order q − 1, called a Singer cycle. So

ax = MaX, where X =

x0x1...

xd−1

∈ GF (p)d.

5.51 Remark. As a result we have

MαMα = M2α = Mα2 .

The multiplication ax can also be written in terms of permutations as πax, where πa isthe permutation ‘multiplying by a modulo f(α)’. To uniquely determine the permutationπa we have to solve the multiplication ax, ∀x ∈ GF (q).

5.52 Notation. As a result the following notations are equivalent:

• ax mod f(α)

• MaX mod p

• πax

5.53 Remark. The previous equivalence is only the case for ‘multiplying by a modulof(α)’. Not every permutation on GF (q) can be written as an invertible matrix!

5.54 Notation. If we have an invertible matrix M , then we will use the notation (M)πto represent this matrix as a permutation on GF (q).

Our goal is to find how many different reduced (q − 1)MOLS(q) there are, for q = pd

a prime power, describing PG(2, q). We want to follow the same reasoning as for q = pprime, so we are going to determine the group of permutations on the (q − 1)MOLS(q)stabilizing the set {A1, A2, . . . , Aq−1}, describing PG(2, q).

5.55 Lemma. We work in the finite field GF (q), q = pd a prime power. Consider theset of MOLS {A1, A2, . . . , Aq−1} of definition 5.50. Suppose we apply σ1 to A1, σ2 to A2,. . . , and σq−1 to Aq−1, the same column permutations to all of the latin squares in the setand also the same row permutations to all of the latin squares in the set, whereby row imoves to row 0. Suppose we obtain a new reduced set of latin squares {C1, C2, . . . , Cq−1},then

σb = σa ◦ τb, where τb : y 7→ y − bi+ ai mod f(α).

53

Proof. Same reasoning as in the case p prime (Lemma 5.46).

5.56 Lemma. There are at least q permutations on the alphabet leaving the set of MOLS{A1, A2, . . . , Aq−1} invariant and these permutations on the alphabet are the same as apermutation on the columns applied on every latin square in the set.

Proof. We work with the alphabet of GF (q). We know that there are q! permutations onthe alphabet.Define the permutation

γb : GF (q)→ GF (q) : y 7→ y + b, for b ∈ GF (q).

We have that γb((Aa)x,y) = ax+ y+ b = (Aa)x,y+b, but the element ax+ y+ b still standsin row x and column y.If we perform the permutation γb : y 7→ y + b on the columns in every latin square of theset of MOLS, column y is moved to column y + b and we have transformed Aa in Aa, sowe have the same set of MOLS {A1, A2, . . . , Aq−1}.5.57 Theorem. There are at most (q − 2)!/d different reduced (q − 1)MOLS(q), forq = pd a prime power, describing PG(2, q).

Proof. We work with the alphabet of GF (q) and we know that there are q! permutationson the alphabet.Let us define a permutation µM ∈ GLd(p) on the alphabet:

µM : GF (p)d → GF (p)d : X 7→ MX.

We have that µM((Aa)x,y) = MMaX+MY . Now we want to check whether we can obtainthe same result with a permutation on the rows and/or columns, and/or a reordering ofthe latin squares in our set of MOLS. ‘MY ’ is just the same as a column permutation.The question is: what is ‘MMaX’? If this has to be the same as performing the samerow permutation on every latin square of the set of MOLS and/or a reordering of thelatin squares in this set, then we would have that (MMa)πx = (Ma′)ππx, where π is apermutation on the alphabet which will be here the same as a permutation on the rows.We are working in GF (q) so this means we have the following system of equations:

(MMα)π = (Mk1α )ππ = β1

(MMα2)π = (Mk2α )ππ = β2

...

(MMαq−1)π = (Mkq−1α )ππ = βq−1.

Now we have that ∃i : Mkiα = Id. This implies that π = βi. So we obtain the following

system of equations:

MMα = Mk1α MM i

α

MMα2 = Mk2α MM i

α

...

MMαq−1 = Mkq−1α MM i

α.

54

Because M is an invertible matrix, and because the matrices Ma define a cyclic group Cof order q − 1, we have as a result that,

M−1Mkjα M = M j

αM−iα

= M j−iα .

But M was a random matrix of GLd(p), so we have that

MCM−1 ∈ C.This means that M is an element of the normalizer of the Singer cycle of order q − 1defined by the matrices Ma in GLd(p).The order of NGLd(p)(C) is equal to (q − 1)d. This means that we have found q(q − 1)dpermutations on the (q − 1)MOLS(q) stabilizing the set {A1, A2, . . . , Aq−1}. Becausethere are q! permutations on GF (q), we have at most (q − 2)!/d different reduced (q −1)MOLS(q), for q = pd a prime power, describing PG(2, q).

The q(q − 1)d permutations on the alphabet that we obtained in the previous the-orems form the group AΓL1(q). These permutations on the alphabet that stabilize{A1, A2, . . . , Aq−1}, modulo the same permutations on the rows and columns, form asubgroup of AGLd(p). Now we are going to prove that there are exactly (q − 2)!/d dif-ferent reduced (q − 1)MOLS(q), for q = pd a prime power, describing PG(2, q), by usingthe following theorem by Thierry Berger as found in [6].

5.58 Theorem. Let G be a permutation group on the finite field K = GF (pd), d ≥ 2. Ifthe affine group AGL1(p

m) is a subgroup of G, then one of the following assertions holds:

1. G = Sym(pd).

2. p = 2 and G = Alt(2d).

3. There exists a divisor r of d, such that AGLk(pr) ⊂ G ⊂ AΓLk(p

r), where d = rk.

4. p = 3, d = 4, G ⊂ AΓL2(9) and if G0 is the stabilizer of 0 in G, then G0 admits anormal subgroup N isomorphic with SL(2, 5), N = G0 ∩ SL2(9) and [G0 : N ] = 8.

5.59 Lemma. There are at least three different reduced sets of (q − 1)MOLS(q) repre-senting PG(2, q) for q = pd > 4.

Proof. We will prove this lemma in two parts. First of all we suppose that p = 2 and weuse definition 5.50 to construct a set of (q − 1)MOLS(q), describing PG(2, q). We seethat the first element on the second row in each latin square Ai, 1 ≤ i ≤ q − 1, is equalto i and

A1 =

0 1 2 3 . . . q − 2 q − 11 0 3 2 . . . q − 1 q − 22 3 0 1 . . . q − 4 q − 33 2 1 0 . . . q − 3 q − 4...

......

.... . .

......

q − 2 q − 1 q − 4 q − 3 . . . 0 1q − 1 q − 2 q − 3 q − 4 . . . 1 0

.

55

We will now apply the permutation (q−2, q−1) on the alphabet on the set {A1, A2, . . . , Aq−1}.If we do this, we obtain a set of (q − 1)MOLS(q) {B′1, B′2, . . . , B′q−1}.

A1 7→ B′1 =

0 1 2 3 . . . q − 1 q − 21 0 3 2 . . . q − 2 q − 12 3 0 1 . . . q − 4 q − 33 2 1 0 . . . q − 3 q − 4...

......

.... . .

......

q − 1 q − 2 q − 4 q − 3 . . . 0 1q − 2 q − 1 q − 3 q − 4 . . . 1 0

.

If we perform the permutation ρ = γ = (q−2, q−1) on respectively the rows and columnsof every latin square of the set {B′1, B′2, . . . , B′q−1}, we obtain the set {B1, B2, . . . , Bq−1},which is a reduced set of (q − 1)MOLS(q). In this set, the first latin square is reduced,so we transformed A1 into B1.

B′1 7→ B1 =

0 1 2 3 . . . q − 2 q − 11 0 3 2 . . . q − 1 q − 22 3 0 1 . . . q − 3 q − 43 2 1 0 . . . q − 4 q − 3...

......

.... . .

......

q − 2 q − 1 q − 3 q − 4 . . . 0 1q − 1 q − 2 q − 4 q − 3 . . . 1 0

.

We see that the reduced latin squares B1 and A1 are not the same, so we have found twodifferent reduced sets of (q − 1)MOLS(q), describing PG(2, q): {A1, A2, . . . , Aq−1} and{B1, B2, . . . , Bq−1}.We can now apply another permutation on the alphabet, for example (1, 2), on the set{A1, A2, . . . , Aq−1}. If we do this, we obtain a set of (q− 1)MOLS(q) {C ′1, C ′2, . . . , C ′q−1}.

A1 7→ C ′1 =

0 2 1 3 . . . q − 2 q − 12 0 3 1 . . . q − 1 q − 21 3 0 2 . . . q − 4 q − 33 1 2 0 . . . q − 3 q − 4...

......

.... . .

......

q − 2 q − 1 q − 4 q − 3 . . . 0 2q − 1 q − 2 q − 3 q − 4 . . . 2 0

.

If we perform the permutation ρ = γ = (1, 2) on respectively the rows and columnsof every latin square of the set {C ′1, C ′2, . . . , C ′q−1}, we obtain the set {C1, C2, . . . , Cq−1},which is a reduced set of (q − 1)MOLS(q). In this set, the first latin square is reduced,so we transformed A1 into C1.

56

C ′1 7→ C1 =

0 1 2 3 . . . q − 2 q − 11 0 3 2 . . . q − 4 q − 32 3 0 1 . . . q − 1 q − 23 2 1 0 . . . q − 3 q − 4...

......

.... . .

......

q − 2 q − 4 q − 1 q − 3 . . . 0 2q − 1 q − 3 q − 2 q − 4 . . . 2 0

.

We see that A1, B1 and C1 are distinct reduced latin squares. So we have found threedifferent reduced sets of (q − 1)MOLS(q), describing PG(2, q), in the case that p = 2.Now we suppose that p 6= 2 and just as in the case p = 2, we use definition 5.50 toconstruct a set of (q− 1)MOLS(q), describing PG(2, q). We see that the first element onthe second row in each latin square Ai, 1 ≤ i ≤ q − 1, is equal to i and

A1 =

0 1 . . . p− 1 . . . q − p q − p+ 1 . . . q − 11 2 0 . . . q − p+ 1 . . . q − p...

. . ....

. . ....

. . ....

p− 1 0 . . . p− 2 . . . q − 1 q − p . . . q − 2...

.... . .

.... . .

......

. . ....

q − p q − p+ 1 . . . q − 1. . . . . . . . . . . .

q − p+ 1 . . . q − p . . . . . . . . . . . ....

. . ....

......

. . ....

q − 1 q − p . . . q − 2 . . . . . . . . . . . .

.

We will now apply the permutation (q−p, q−1) on the alphabet on the set {A1, A2, . . . , Aq−1}.If we do this, we obtain a set of (q − 1)MOLS(q) {B′1, B′2, . . . , B′q−1}.

A1 7→ B′1 =

0 1 . . . p− 1 . . . q − 1 q − p+ 1 . . . q − p1 2 0 . . . q − p+ 1 . . . q − 1...

. . ....

. . ....

. . ....

p− 1 0 . . . p− 2 . . . q − p q − 1 . . . q − 2...

.... . .

.... . .

......

. . ....

q − 1 q − p+ 1 . . . q − p . . . . . . . . . . . .q − p+ 1 . . . q − 1 . . . . . . . . . . . .

.... . .

......

.... . .

...q − p q − 1 . . . q − 2 . . . . . . . . . . . .

.

If we perform the permutation ρ = γ = (q−p, q−1) on respectively the rows and columnsof every latin square of the set {B′1, B′2, . . . , B′q−1}, we obtain the set {B1, B2, . . . , Bq−1},which is a reduced set of (q − 1)MOLS(q). In this set, the first latin square is reduced,so we transformed A1 into B1.

57

B′1 7→ B1 =

0 1 . . . p− 1 . . . q − p q − p+ 1 . . . q − 11 2 0 . . . q − 1 . . . q − p+ 1...

. . ....

. . ....

. . ....

p− 1 0 . . . p− 2 . . . q − 2 q − 1 . . . q − p...

.... . .

.... . .

......

. . ....

q − p q − 1 . . . q − 2 . . . . . . . . . . . .q − p+ 1 . . . q − 1 . . . . . . . . . . . .

.... . .

......

.... . .

...

q − 1 q − p+ 1 . . . q − p . . . . . . . . . . . .

.

We see that the reduced latin squares B1 and A1 are not the same, so we have found twodifferent reduced sets of (q − 1)MOLS(q), describing PG(2, q): {A1, A2, . . . , Aq−1} and{B1, B2, . . . , Bq−1}.We can now apply another permutation on the alphabet, for example (q−p, p−1), on theset {A1, A2, . . . , Aq−1}. If we do this, we obtain a set of (q−1)MOLS(q) {C ′1, C ′2, . . . , C ′q−1}.

A1 7→ C ′1 =

0 1 . . . q − p . . . p− 1 q − p+ 1 . . . q − 11 2 0 . . . q − p+ 1 . . . p− 1...

. . ....

. . ....

. . ....

q − p 0 . . . p− 2 . . . q − 1 p− 1 . . . q − 2...

.... . .

.... . .

......

. . ....

p− 1 q − p+ 1 . . . q − 1. . . . . . . . . . . .

q − p+ 1 . . . p− 1 . . . . . . . . . . . ....

. . ....

......

. . ....

q − 1 p− 1 . . . q − 2 . . . . . . . . . . . .

.

If we perform the permutation ρ = γ = (q−p, p−1) on respectively the rows and columnsof every latin square of the set {C ′1, C ′2, . . . , C ′q−1}, we obtain the set {C1, C2, . . . , Cq−1},which is a reduced set of (q − 1)MOLS(q). In this set, the first latin square is reduced,so we transformed A1 into C1.

C ′1 7→ C1 =

0 1 . . . p− 1 . . . q − p q − p+ 1 . . . q − 11 2 q − p+ 1 . . . 0 . . . p− 1...

. . ....

. . ....

. . ....

p− 1 q − p+ 1 . . . . . .. . . q − 1 . . . . . .

......

. . ....

. . ....

.... . .

...q − p 0 . . . q − 1 . . . p− 2 p− 1 . . . q − 2

q − p+ 1 . . . . . . . . . p− 1 . . . . . ....

. . ....

......

. . ....

q − 1 p− 1 . . . . . . . . . q − 2 . . . . . .

.

We see that A1, B1 and C1 are distinct reduced latin squares. So we have found threedifferent reduced sets of (q − 1)MOLS(q), describing PG(2, q), in the case that p 6= 2.

58

5.60 Theorem. There are exactly (q−2)!/d different reduced (q−1)MOLS(q), for q = pd

a prime power, describing PG(2, q).

Proof. Suppose there is a larger group G of permutations on the alphabet which stabilizesthe set {A1, A2, . . . , Aq−1}, modulo the same permutations on the rows and the columnsof the latin squares. Then this group G is described as in theorem 5.58. If q = 4, then wehave that |AΓL1(q)| = q(q − 1)d = 24 = |Sq|. So there is only one complete reduced setof MOLS(4), describing PG(2, 4). If q > 4, then the first two cases are not possible sincethere are at least three different complete reduced sets of MOLS(q), describing PG(2, q),as proven in lemma 5.59.For the cases 3 and 4 of theorem 5.58, G always has a normal subgroup of order pd(pr−1),where d = rk, so G is represented by matrices, field automorphisms and translations overthe subfield GF (pk) of GF (pd). But these permutations were all discussed in lemma 5.56and theorem 5.57, so no larger group G of permutations on the alphabet stabilizes the set{A1, A2, . . . , Aq−1}, modulo the same permutations on the rows and the columns of thelatin squares.

5.4.3 Example

To illustrate the previous theorems, we will give an example.Consider the set of MOLS of theorem 5.36.

0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 41 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 32 3 4 0 1 4 0 1 2 3 1 2 3 4 0 3 4 0 1 23 4 0 1 2 1 2 3 4 0 4 0 1 2 3 2 3 4 0 14 0 1 2 3 3 4 0 1 2 2 3 4 0 1 1 2 3 4 0

This is a set of 4 mutually orthogonal latin squares of order 5, describing PG(2, 5). Weknow by theorem 5.47 that there are 20 permutations stabilizing this set. First of all wedefined the following permutation on the alphabet

ρa′ : GF (5)→ GF (5) : x 7→ a′x, for a′ ∈ GF (5) and a′ 6= 0.

Secondly we defined the following permutation on the alphabet

γb : GF (5)→ GF (5) : y 7→ y + b, for b ∈ GF (5).

These two definitions give us the following 9 permutations.

1. ρ1, which is the identity,

2. ρ2 = (1243),

3. ρ3 = (1342),

4. ρ4 = (14)(23),

59

5. γ0, which is the identity,

6. γ1 = (01234),

7. γ2 = (02413),

8. γ3 = (03142),

9. γ4 = (04321).

If we combine these permutations, we will know all of the permutations on the alphabetstabilizing the set of (4)MOLS(5), describing PG(2, p), modulo the same permutationson the rows and columns.

1. ρ1 ◦ γ0 = Id. = γ0 ◦ ρ1,

2. ρ1 ◦ γ1 = γ1 = γ1 ◦ ρ1,

3. ρ1 ◦ γ2 = γ2 = γ2 ◦ ρ1,

4. ρ1 ◦ γ3 = γ3 = γ3 ◦ ρ1,

5. ρ1 ◦ γ4 = γ4 = γ4 ◦ ρ1,

6. ρ2 ◦ γ0 = ρ2 = γ0 ◦ ρ2,

7. ρ2 ◦ γ1 = (0214) = γ2 ◦ ρ2,

8. ρ2 ◦ γ2 = (0423) = γ4 ◦ ρ2,

9. ρ2 ◦ γ3 = (0132) = γ1 ◦ ρ2,

10. ρ2 ◦ γ4 = (0341) = γ3 ◦ ρ2,

11. ρ3 ◦ γ0 = ρ3 = γ0 ◦ ρ3,

12. ρ3 ◦ γ1 = (0324) = γ3 ◦ ρ3,

13. ρ3 ◦ γ2 = (0143) = γ1 ◦ ρ3,

14. ρ3 ◦ γ3 = (0412) = γ4 ◦ ρ3,

15. ρ3 ◦ γ4 = (0231) = γ2 ◦ ρ3,

16. ρ4 ◦ γ0 = ρ4 = γ0 ◦ ρ4,

17. ρ4 ◦ γ1 = (04)(13) = γ4 ◦ ρ4,

18. ρ4 ◦ γ2 = (03)(12) = γ3 ◦ ρ4,

19. ρ4 ◦ γ3 = (02)(34) = γ2 ◦ ρ4,

60

20. ρ4 ◦ γ4 = (01)(24) = γ1 ◦ ρ4.

By theorem 5.48, we know that there are 6 different reduced (4)MOLS(5), describingPG(2, 5). First of all we had the following set of (4)MOLS(5)

0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 41 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 32 3 4 0 1 4 0 1 2 3 1 2 3 4 0 3 4 0 1 23 4 0 1 2 1 2 3 4 0 4 0 1 2 3 2 3 4 0 14 0 1 2 3 3 4 0 1 2 2 3 4 0 1 1 2 3 4 0

.

We will call this set S1. Now we will construct the other 5 different reduced (4)MOLS(5)by applying some permutations. If we apply the permutation (34) on the alphabet onS1, modulo the same permutation on the rows and columns, then we obtain the followingreduced set of (4)MOLS(5)

0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 41 2 4 0 3 2 4 3 1 0 4 3 0 2 1 3 0 1 4 22 4 3 1 0 3 0 1 4 2 1 2 4 0 3 4 3 0 2 13 0 1 4 2 4 3 0 2 1 2 4 3 1 0 1 2 4 0 34 3 0 2 1 1 2 4 0 3 3 0 1 4 2 2 4 3 1 0

.

If we apply the permutation (23) on the alphabet on S1, modulo the same permutationon the rows and columns, then we obtain the following reduced set of (4)MOLS(5)

0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 41 3 4 2 0 3 2 0 4 1 2 4 1 0 3 4 0 3 1 22 4 1 0 3 1 3 4 2 0 4 0 3 1 2 3 2 0 4 13 2 0 4 1 4 0 3 1 2 1 3 4 2 0 2 4 1 0 34 0 3 1 2 2 4 1 0 3 3 2 0 4 1 1 3 4 2 0

.

If we apply the permutation (234) on the alphabet on S1, modulo the same permutationon the rows and columns, then we obtain the following reduced set of (4)MOLS(5)

0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 41 3 0 4 2 3 4 1 2 0 4 2 3 0 1 2 0 4 1 32 0 4 1 3 4 2 3 0 1 3 4 1 2 0 1 3 0 4 23 4 1 2 0 2 0 4 1 3 1 3 0 4 2 4 2 3 0 14 2 3 0 1 1 3 0 4 2 2 0 4 1 3 3 4 1 2 0

.

If we apply the permutation (24) on the alphabet on S1, modulo the same permutationon the rows and columns, then we obtain the following reduced set of (4)MOLS(5)

0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 41 4 0 2 3 4 3 1 0 2 3 2 4 1 0 2 0 3 4 12 0 3 4 1 3 2 4 1 0 4 3 1 0 2 1 4 0 2 33 2 4 1 0 1 4 0 2 3 2 0 3 4 1 4 3 1 0 24 3 1 0 2 2 0 3 4 1 1 4 0 2 3 3 2 4 1 0

.

61

If we apply the permutation (243) on the alphabet on S1, modulo the same permutationon the rows and columns, then we obtain the following reduced set of (4)MOLS(5)

0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 41 4 3 0 2 4 2 0 1 3 2 3 1 4 0 3 0 4 2 12 3 1 4 0 1 4 3 0 2 3 0 4 2 1 4 2 0 1 33 0 4 2 1 2 3 1 4 0 4 2 0 1 3 1 4 3 0 24 2 0 1 3 3 0 4 2 1 1 4 3 0 2 2 3 1 4 0

.

5.4.4 Further remarks

Does there exist a non-Desarguesian projective plane of prime order p? This questionis still unsolved because there is no proof that there does not exist a non-Desarguesianprojective plane of prime order p, but nobody has ever found such a projective plane. Wewould like to sketch a possible way to solve this problem.A first step in the progress is the result of theorem 5.48. This theorems tells us that thereare (p − 2)! different reduced (p − 1)MOLS(p), for p a prime, describing PG(2, p). Bytheorem 4.9 we know that there exists a projective plane of order p if and only if thereexists a complete set of mutually orthogonal latin squares of order p. We also know thatthe projective planes PG(2, p) are Desarguesian projective planes. If we throw all thesetheorems together, we have the following result.

5.61 Corollary. If there are exactly (p− 2)! different reduced (p− 1)MOLS(p), for p aprime, then there does not exist a non-Desarguesian projective plane of prime order p.

To prove that there are exactly (p− 2)! different reduced (p− 1)MOLS(p), for p a prime,we suggest two big steps.First of all we remark that one of the latin squares in a reduced set of (p−1)MOLS(p) isreduced, by definition. In fact, we used reduced sets of MOLS in the previous paragraphs,because we can apply the same permutation on the rows and the columns on each of thelatin squares in the set and a permutation on the symbol set in each latin square to obtaina reduced set of MOLS. We will now take a closer look at the set of reduced latin squaresof order p.

This set can be partitioned in three parts. First of all we have the set B of the bachelor

62

reduced latin squares of order p. These reduced latin squares of order p have no orthogonalmate. Secondly we have the reduced latin squares which have an orthogonal mate. Thisset can be partitioned in the set C of reduced latin squares of order p which are part of aset of (p− 1)MOLS(p) and the set D, which contains the other reduced latin squares oforder p which have an orthogonal mate.If the cardinality of the set C is equal to c, then as a first step we want to prove thatc = (p − 2)!. This would mean that we have already found all the reduced latin squareswhich are part of a set of (p− 1)MOLS(p) in theorem 5.48.To finish the proof we would have to prove that for all L ∈ C there exists a unique reducedset of (p− 1)MOLS(p), because then we would obtain exactly (p− 2)! different reduced(p− 1)MOLS(p), for p a prime.It seems easy to explain what we have to do if we want to prove that there are exactly(p − 2)! different reduced (p − 1)MOLS(p), for p a prime, but we were not able to finda proof at this point. Therefore we did some research on transversals, because this couldbe helpful if we want to prove one of the previous suggested steps.

63

Chapter 6

Transversals

6.1 Definition. A transversal in a latin square of order n on the symbol set S ={a0, a1, . . . , an−1} is a set of n entries, one from each row and column, containing each ofthe n symbols of S exactly once.A partial transversal of length k in a latin square of order n on the symbol set S isa set of k entries, each from a different row and a different column, such that no two ofthem contain the same symbol.

Not every latin square of order n has a transversal. The following latin square is anexample of a latin square without a transversal.

0 1 2 31 2 3 02 3 0 13 0 1 2

Suppose we have a latin square that has a certain number of transversals. We will nowinvestigate if the number of transversals changes if we apply a permutation on the rows,the columns or the alphabet. Firstly we will introduce the following notation.

6.2 Notation. If a certain transversal contains the elements a0, a1, . . . , an−1 in respec-tively column 0, 1, . . . , n− 1 on row i0, i1, . . . , in−1, then we will use the following notationfor the transversal, (ai00 , a

i11 , . . . , a

in−1

n−1 ), where the element in column 0 is placed in the firstposition, the element in column 1 is placed in the second position, . . .

6.3 Example. Consider the following latin square of order 4.

L1 =

0 1 2 31 0 3 22 3 0 13 2 1 0

.

64

This latin square has 8 different transversals: (00, 32, 13, 21), (00, 23, 31, 12), (11, 32, 20, 03),(11, 23, 02, 30), (22, 10, 31, 03), (22, 01, 13, 30), (33, 10, 02, 21), (33, 01, 20, 12).If we apply a permutation on the rows of L1, then we see that this corresponds to a permu-tation of the upper indices in the notation of the transversals. Consider the permutationρ = (0, 1) on the rows of L1, then we obtain the following latin square of order 4.

Lρ1 =

1 0 3 20 1 2 32 3 0 13 2 1 0

.

This latin square has the following 8 different transversals: (01, 32, 13, 20), (01, 23, 30, 12),(10, 32, 21, 03), (10, 23, 02, 31), (22, 11, 30, 03), (22, 00, 13, 31), (33, 11, 02, 20), (33, 00, 21, 12).We see that the number of transversals remains the same.If we apply a permutation on the columns of L1, then we see that this corresponds to apermutation of the elements aikk in the notation of the transversals. Consider the permu-tation γ = (0, 1) on the columns of L1, then we obtain the following latin square of order4.

Lγ1 =

1 0 2 30 1 3 23 2 0 12 3 1 0

.

This latin square has the following 8 different transversals: (32, 00, 13, 21), (23, 00, 31, 12),(32, 11, 20, 03), (23, 11, 02, 30), (10, 22, 31, 03), (01, 22, 13, 30), (10, 33, 02, 21), (01, 33, 20, 12).We see that the number of transversals remains the same.Finally if we apply a permutation on the alphabet, then this corresponds to a permutationof the elements ak in the notation of the transversals. Consider the permutation σ = (0, 1)on the alphabet in L1, then we obtain the following latin square of order 4.

Lσ1 =

1 0 2 30 1 3 22 3 1 03 2 0 1

.

This latin square has 8 different transversals: (10, 32, 03, 21), (10, 23, 31, 02), (01, 32, 20, 13),(01, 23, 12, 30), (22, 00, 31, 13), (22, 11, 03, 30), (33, 00, 12, 21), (33, 11, 20, 02). We see that thenumber of transversals remains the same.

6.4 Theorem. If a latin square L has n transversals, then any latin square isotopic to Lhas the same number of transversals.

Proof. Suppose that L′ is a latin square isotopic to L. This means that there are threebijections from respectively the rows, the columns and the symbols of L, to respectively

65

the rows, the columns and the symbols of L′, that map L to L′.A permutation on the rows corresponds to a permutation of the upper indices in thenotation of the transversals. This means that there is a bijection between the upperindices in the notation of each transversal of L and the upper indices in the notation ofthe corresponding transversal of L′.A permutation on the columns corresponds to a permutation of the elements aikk in thenotation of the transversals. This means that there is a bijection between the elementsaikk in the notation of each transversal of L and the elements aill in the notation of thecorresponding transversal of L′.A permutation on the alphabet corresponds to a permutation of the elements ak in thenotation of the transversals. This means that there is a bijection between the elementsak in the notation of each transversal of L and the elements al in the notation of thecorresponding transversal of L′.The number of transversals remains the same in these three cases, which means that Land L′ have the same number of transversals.

We know by the previous theorem that the number of transversals is isotopy class invari-ant. If we investigate whether the number of transversals is affected if we map a latinsquare L to any conjugate of L, then we will know if the number of transversals is alsomain class invariant.A latin square has at most 6 different conjugates. The (1, 2, 3)−conjugate, the (2, 3, 1)−conjugate, the (3, 1, 2)−conjugate, the (2, 1, 3)−conjugate, the (3, 2, 1)−conjugate andthe (1, 3, 2)−conjugate. In fact, the (2, 1, 3)−conjugate is the transpose of the (1, 2, 3)−conjugate, the (3, 2, 1)−conjugate is the transpose of the (2, 3, 1)−conjugate and the(1, 3, 2)−conjugate is the transpose of the (3, 1, 2)−conjugate.The transpose of a matrix is a conversion of the rows into the columns and the columnsinto the rows, which means that in the notation of the transversals there is a bijectionbetween the position of the elements aikk and the upper indices in the notation of thetransversals. The number of transversals in a latin square and in his transpose is thesame. We can also understand this by looking at the definition of a transversal, which isa set of n entries, one from each row and column, containing each of the n symbols of thesymbol set S exactly once. By transposing the latin square, a transversal is mapped toanother corresponding transversal, because we will still have a set of n entries, one fromeach row and column, containing each of the n symbols of the symbol set S exactly once.We only have to check whether the (1, 2, 3)−conjugate, the (2, 3, 1)−conjugate and the(3, 1, 2)−conjugate have the same number of transversals.

6.5 Example. We will use example 2.16 and we start with the following latin square,which is the (1, 2, 3)−conjugate.

L(1,2,3) =

0 3 1 21 2 0 33 0 2 12 1 3 0

.

66

This latin square has 8 different transversals: (00, 21, 33, 12), (00, 13, 22, 31), (11, 30, 22, 03),(11, 02, 33, 20), (32, 21, 10, 03), (32, 13, 01, 20), (23, 30, 01, 12) and (23, 02, 10, 31).We will now look at the (2, 3, 1)−conjugate.

L(2,3,1) =

0 1 3 22 3 1 01 0 2 33 2 0 1

.

This latin square has 8 different transversals: (00, 31, 22, 13), (00, 23, 11, 32), (21, 10, 03, 32),(21, 02, 30, 13), (12, 31, 03, 20), (12, 23, 11, 20), (33, 10, 22, 01) and (33, 02, 11, 20). We see thatthe number of transversals remains the same.Finally we will now look at the (3, 1, 2)−conjugate.

L(3,1,2) =

0 2 1 32 0 3 13 1 2 01 3 0 2

.

This latin square has 8 different transversals: (00, 12, 31, 23), (00, 33, 22, 11), (21, 12, 03, 30),(21, 33, 10, 02), (32, 20, 03, 11), (32, 01, 10, 23), (13, 20, 31, 02) and (13, 01, 22, 30). We see thatthe number of transversals remains the same.To understand what happens, we will use the line array notation presented in example2.15. This is the line array of L(1,2,3).

R : 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3C : 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3E : 0 3 1 2 1 2 0 3 3 0 2 1 2 1 3 0

What happens to a random transversal? Let us consider the transversal (00, 21, 33, 12) ofL(1,2,3). In this transversal we have on row 0 and column 0 the element 0, on row 1 andcolumn 1 the element 2, on row 3 and column 2 the element 3 and on row 2 and column3 the element 1. A conjugate of L(1,2,3) corresponds with a permutation of the rows in theline array. This means we have a permutation in the notation of the transversals betweenthe upper indices, the position of the elements akii and the elements ai. In fact the roleof rows, columns and elements is permuted. If we consider the (2, 3, 1)−conjugate, ourtransversal (00, 21, 33, 12) of L(1,2,3) is mapped to the transversal (00, 23, 11, 32) of L(2,3,1).

We have now the following theorem.

6.6 Theorem. If a latin square L has n transversals, then any latin square main classisotopic to L has the same number of transversals.

This theorem means that the number of transversals is a main class invariant.It is interesting to know if these permutation have an influence on the number of transver-sals, but it is more useful to know how many transversals a latin square has. We have thefollowing conjecture by Ryser.

67

6.7 Conjecture. Every latin square of odd order has a transversal.

The original conjecture was stronger than that and it said that for every latin square oforder n, the number of transversals is congruent to n mod 2. In 1990, Balasubramanianproved the even case.

6.8 Theorem. Every latin square of even order has an even number of transversals.

There are a lot of counterexamples for the odd case and this is why the original conjectureis weakened to conjecture 6.7.The reason why it is interesting to investigate the number of transversals of a latin squareis the following theorem.

6.9 Theorem. A latin square has an orthogonal mate if and only if it can be decomposedinto pairwise disjoint transversals.

There is a lot more research done on transversals, but we will not go into further detailshere. More information can be found in [4], [9] and [32]. In the final chapter we willdiscuss magic squares and hereby we will use mutually orthogonal latin squares.

68

Chapter 7

Magic squares

7.1 Definitions

In this paragraph we use [18].

7.1 Definition. A magic square of order n > 1 is an n × n matrix A = (ai,j) inwhich n2 distinct numbers from a set S are arranged, such that the sum of the numbers ineach row, each column and both the diagonals is the same. This sum is called the magicsum, which will be denoted by s.

In our definition we have the following properties.

1.∑n−1

i=0 ai,j = s for all j,

2.∑n−1

j=0 ai,j = s for all i,

3.∑n−1

i=0 ai,i = s,

4.∑n−1

i=0 ai,n−1−i = s.

7.2 Definition. A normal magic square of order n is a magic square of order n > 1,which contains the integers from 1 to n2.

We will now calculate the magic sum of a normal magic square of order n. In this casethe magic sum equals the sum of the smallest and the biggest number in the square,multiplied by half of the order. This is also the sum of all the numbers divided by n. Sowe get:

1

n

n2∑i=1

i =n2(n2 + 1)

2n=n(n2 + 1)

2.

7.3 Definition. A semi-magic square of order n > 1 is an n× n matrix A = (ai,j)in which n2 distinct numbers from a set S are arranged, such that the sum of the numbersin each row and each column is the same.

69

7.4 Definition. A magic square of order n > 1 is pandiagonal magic if the sum of allthe numbers on every broken diagonal equals the magic sum. A broken diagonal starts,as we can see in the following figure, in an entry in the first row in a certain column, letus say in column k. From there it goes diagonal to the left or the right side of the square.In the first case the broken diagonal goes further in the entry in the last column on rowk+ 1. In the other case it goes further in the entry in the first column, on row n− k+ 2,where the columns and rows are numbered from 1 to n.

7.2 A little history

The oldest known magic square is called the Lo Shu and was created in China. Thismagic square was the following normal magic square of order 3.

The Lo Shu is mentioned in different old sources, but there is not always explicitly amagic square given. Lo Shu supposedly means ‘river map’ and the legend says that it wasfirst seen by emperor Yu on the back of a turtle. The appearance of the turtle had to dowith sacrifices to the river god. A Shu Ching in 650 BCE1 makes a reference to the ‘rivermap’.This magic square is also mentioned in the I Ching, but the first clear reference is madein the ‘Ta Tai li chi’, which is named after Tai Te, one of the four disciples of Hou Tsang.More information on the history of the Lo Shu, the I Ching or the Ta Tai li chi can be

1Before the Common Era

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found in [18], [25] and [30].The first magic square of order 4 is found in India and is created by the mathematicianNagarajuna in the first century. A well known early magic square of order 4 is also foundin India in Khajuraho in the Parshvanath Jain temple included in an 11th or 12th centuryinscription. This magic square is not only a normal magic square, but also a pandiagonalmagic square. The following picture is taken by Dr. Rainer Typke [31].

We see in this picture the following magic square.

7 12 1 142 13 8 1116 3 10 59 6 15 4

The magic sum of this magic square is equal to 34 and not only the sum of the numbersin each row, each column and both the diagonals is equal to this magic sum, but also thesum of the numbers on each broken diagonal. We notice that the sum of the numbersin all the (2 × 2) subsquares also equals the magic sum. This gives us the possibility tocreate a ‘magic carpet’, by copying the original magic square and put them side to side tocreate a ‘carpet’. Each subsquare of order 4 in this carpet is a normal pandiagonal magicsquare and the sum of the numbers in each subsquare of order 2 is equal to the magicsum 34.

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7 12 1 14 7 12 1 14 7 12 1 142 13 8 11 2 13 8 11 2 13 8 1116 3 10 5 16 3 10 5 16 3 10 59 6 15 4 9 6 15 4 9 6 15 47 12 1 14 7 12 1 14 7 12 1 142 13 8 11 2 13 8 11 2 13 8 1116 3 10 5 16 3 10 5 16 3 10 59 6 15 4 9 6 15 4 9 6 15 47 12 1 14 7 12 1 14 7 12 1 142 13 8 11 2 13 8 11 2 13 8 1116 3 10 5 16 3 10 5 16 3 10 59 6 15 4 9 6 15 4 9 6 15 4

In Europe there are a few mathematicians who investigated magic squares. A few of themare Heinrich Cornelius Agrippa, Albrecht Durer, Leonhard Euler, Pierre de Fermat andBenjamin Franklin. More information on the history of magic squares can be found in[30]. An interesting question in this thesis is whether it is possible to construct certaintypes of magic squares by using mutually orthogonal latin squares.

7.3 Construction of magic squares

It is indeed possible to construct magic squares by using mutually orthogonal latin squares.This method was first presented by Euler on October 17, 1776, and published by Euler in1849 in ‘Commentationes arithmeticae’ [11]. We will first give some examples of differentorders.

7.3.1 First case: n = 3

Let us consider the following two latin squares of order 3,

L1 =a b cb c ac a b

and L2 =γ β αα γ ββ α γ

.

Suppose that a = 0, b = 3, c = 6, α = 3, β = 2 and γ = 1. Now we have that

L1 =0 3 63 6 06 0 3

and L2 =1 2 33 1 22 3 1

.

Let us now consider L1 + L2. This gives us the following result, which is a normal semi-magic square.

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M1 =1 5 96 7 28 3 4

.

If we would have that

N1 =0 1 21 2 02 0 1

and N2 =0 1 22 0 11 2 0

,

then we can also write M = 3N1 +N2 + J3, where

J3 =1 1 11 1 11 1 1

.

If we work with the latin squares N1 and N2, then we can also construct the followingnormal semi-magic square

M2 = N1 + 3N2 + J3 =1 5 98 3 46 7 2

,

which can be obtained by applying a row permutation on M1.

7.3.2 Second case: n = 4

Let us consider the following mutually orthogonal latin squares,

L1 =

0 1 2 31 0 3 22 3 0 13 2 1 0

and L2 =

0 1 2 32 3 0 13 2 1 01 0 3 2

.

It is now possible to construct two different normal semi-magic squares as follows, M1 =4L1 + L2 + J4 and M2 = L1 + 4L2 + J4, where

J4 =

1 1 1 11 1 1 11 1 1 11 1 1 1

.

We obtain the following normal semi-magic squares

M1 =

1 6 11 167 4 13 1012 15 2 514 9 8 3

and M2 =

1 6 11 1610 13 4 715 12 5 28 3 14 9

.

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7.3.3 General case

It is now also possible to construct a normal semi-magic square of order 5, order 7, order8, . . . in the same way as before.

7.5 Theorem. Suppose N(n) ≥ 2, where n is the order of the latin squares, then it ispossible to construct a normal semi-magic square of order n as follows. M = nL1+L2+Jn,where L1 and L2 are different latin squares, on the symbol set S = {0, 1, . . . , n− 1}, fromthe set of MOLS(n) and Jn is the n× n matrix in which each entry is equal to 1.

Proof. Suppose we have two mutually orthogonal latin squares of order n, L1 and L2,on the symbol set S = {0, 1, . . . , n − 1}. By the definition of mutually orthogonal latinsquares, we know that this means that if these two latin squares are superimposed, wewould find all of the n2 ordered pairs (i, j) ∈ S × S.If M is a normal semi-magic square of order n, then it has to contain the integers from 1 ton2. We defined M as M = nL1+L2+Jn, which is a matrix. Let us now look at the elementon row k and column l. We find (M)k,l = (nL1 +L2 +Jn)k,l = ni+ j+1, where i = (L1)k,land j = (L2)k,l. We know that L1 and L2 are mutually orthogonal, which means thatthere does not exist a pair (s, t) ∈ S×S, such that (s, t) 6= ((L1)k,l, (L2)k,l), ∀(k, l) ∈ S×S.We will now look into the solution of ni + j + 1. We have that i = 0, 1, . . . n − 1, thusni = 0, n, 2n, . . . , (n− 1)n and ni+ 1 = 1, n+ 1, 2n+ 1, . . . , (n− 1)n+ 1. Finally, we findthat ni+j+1 = 1, 2, . . . , n, n+1, n+2, . . . , 2n, 2n+1, . . . , (n−1)n+1, (n−1)n+2, . . . , n2.As a result we find that the matrix M contains the integers from 1 to n2.If M is a normal semi-magic square, then it has to contain the integers from 1 to n2, butit also needs to have the property of a semi-magic square that the sum of the numbers ineach row and each column is the same.Let us consider the matrix M . We know that (M)k,l = (nL1 + L2 + Jn)k,l = ni + j + 1,where i = (L1)k,l and j = (L2)k,l. In each row and each column of L1 and L2 we have thenumbers from 0 to n−1 because L1 and L2 are latin squares. This means that the sum ofthe numbers in each row and each column is equal to n(0+1+. . .+n−1)+(0+1+. . .+n−1)+n = n(0+1+. . .+n−1)+(1+2+. . .+n) = n2(n−1)+n(n+1)

2= n(n2−n+n+1)

2= n(n2+1)

2. In

the beginning of this chapter we saw that this is the magic sum of a normal (semi-)magicsquare of order n.We conclude that M = nL1 + L2 + Jn is a normal semi-magic square of order n.

The question is now, is it also possible to construct a normal magic square of order ninstead of a normal semi-magic square of order n? We will first look at some examples.

7.3.4 First case (bis): n = 3

Let us consider the following two mutually orthogonal latin squares of order 3

L1 =0 2 12 1 01 0 2

and L2 =1 0 22 1 00 2 1

.

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We can now construct a normal magic square as follows, M = 3L1 + L2 + J3, where

J3 =1 1 11 1 11 1 1

.

We obtain the following normal magic square

M =2 7 69 5 14 3 8

.

7.3.5 Second case (bis): n = 4

Let us consider the following mutually orthogonal latin squares

L3 =

0 1 2 33 2 1 01 0 3 22 3 0 1

and L4 =

0 1 2 32 3 0 13 2 1 01 0 3 2

.

It is now possible to construct a normal magic square as follows, M = 4L3 + L4 + J4,where

J4 =

1 1 1 11 1 1 11 1 1 11 1 1 1

.

We obtain the following normal magic square

M =

1 6 11 1615 12 5 28 3 14 910 13 4 7

.

7.3.6 General case

7.6 Definition. A latin square of order n on the symbol set S = {0, 1, . . . , n − 1} is adouble-diagonal latin square if both the main diagonal and the antidiagonal containall the elements from the symbol set S.

By theorem 7.5 we know that M = nL1 + L2 + Jn is a normal semi-magic square, whereL1 and L2 are different latin squares from the set of MOLS(n) and Jn is the n×n matrixin which each entry is equal to 1. The following theorem tells us when M is a normalmagic square.

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7.7 Theorem. Suppose N(n) ≥ 2, where n is the order of the latin squares, then it ispossible to construct a normal magic square of order n as follows. Suppose M = nL1+L2+Jn, where L1 and L2 are different latin squares, on the symbol set S = {0, 1, . . . , n − 1},from the set of MOLS(n) and Jn is the n × n matrix in which each entry is equal to 1.Then M is a normal magic square, if either

1. L1 and L2 are both double-diagonal latin squares, or

2. n is odd and two of the main diagonals and antidiagonals of L1 and L2 are constantand equal to n−1

2.

Proof. We know by theorem 7.5 that M is a normal semi-magic square, so the only thingto prove is that the sum of the numbers in both the diagonals of M is equal to the magicsum if L1 and L2 are both double-diagonal latin squares, or if n is odd and two of themain diagonals and antidiagonals of L1 and L2 are constant and equal to n−1

2.

Suppose that L1 and L2 are both double-diagonal latin squares, then we know that boththe main diagonal and the antidiagonal have the numbers from 0 to n−1. This means thatthe sum of the numbers in both the diagonals is equal to n(0+1+. . .+n−1)+(0+1+. . .+

n−1)+n = n(0+1+ . . .+n−1)+(1+2 . . .+n) = n2(n−1)+n(n+1)2

= n(n2−n+n+1)2

= n(n2+1)2

.In the beginning of this chapter we saw that this is the magic sum of a normal magicsquare of order n. As a conclusion we have that if L1 and L2 are both double-diagonallatin squares, then we know that M = nL1 + L2 + Jn will be a normal magic square.Let us now suppose that n is odd and that two of the main diagonals and antidiagonalsof L1 and L2 are constant and equal to n−1

2. Suppose that the main diagonal of L1 is

constant and equal to n−12

, then the main diagonal of L2 can not be constant, becauseL1 and L2 are mutually orthogonal. This means that the antidiagonal of L2 has to beconstant and equal to n−1

2. The antidiagonal of L1 has to contain all the elements of S,

because we know that the antidiagonal of L2 is constant and equal to n−12

and otherwisethere would be a conflict with the definition of mutually orthogonal latin squares. Finallythe main diagonal of L2 also has to contain all the elements of S.We see now that the sum of the numbers in the main diagonal of M is equal to n(nn−1

2)+

(0+1+. . .+n−1)+n = n2(n−1)+n(n+1)2

= n(n2−n+n+1)2

= n(n2+1)2

and the sum of the numbers

in the antidiagonal of M is equal to n(0 + 1 + . . .+ n− 1) + nn−12

+ n = n2(n−1)+n(n+1)2

=n(n2−n+n+1)

2= n(n2+1)

2. This proves our theorem.

This chapter was just a brief chapter on magic squares. There is a lot more theory and itis possible to formulate a lot more theorems on the construction of certain magic squareswith mutually orthogonal latin squares. It is also possible to find magic squares that cannot be constructed with mutually orthogonal latin squares. As a conclusion we can saythat there is still a lot more to discover about magic squares and latin squares, but wewill leave this for the reader.

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