2.3.3 Completion of latin squares and critical sets . . . . . . . . . . . . . .... was to find
how many distinct sets of (q − 1)MOLS(q) there are, describing PG(2,q). This was
...
Ghent University Faculty of Sciences Department of Mathematics
Mutually orthogonal latin squares and their generalizations Jordy Vanpoucke
Academic year 2011-2012
Advisor: Prof. Dr. L. Storme Master thesis submitted to the Faculty of Sciences to obtain the degree of Master in Sciences, Mathematics
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
3.3
On the number of sudoku latin squares and inequivalent sudoku latin squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
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
2
. . . . . . . . . . . . . . . . . . . . . 29
4 Latin squares and projective planes 4.1
4.2
30
Projective planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 4.1.1
Coordinatization of projective planes . . . . . . . . . . . . . . . . . 31
4.1.2
Planar ternary rings . . . . . . . . . . . . . . . . . . . . . . . . . . 32
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
5.4
5.3.1
Latin squares of order n . . . . . . . . . . . . . . . . . . . . . . . . 46
5.3.2
Sudoku latin squares of order n × n . . . . . . . . . . . . . . . . . . 48
On the number of (n − 1)M OLS(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 about those magic squares, I stumbled upon latin squares. It fascinated me how it was possible that something which seemed so easy, could be connected with so many different and difficult areas of mathematics. Latin squares are not just some ‘spielerei’ and I got the idea to write a thesis on this subject. On Tuesday, July 12, 2011, I had the opportunity to attend a lecture of Kenneth Hicks on the numbers of latin squares of prime power orders with orthogonal mates at the 10th International Conference on Finite Fields and their Applications in Ghent [15]. At the end of this talk, Kenneth Hicks conjectured that there are (p − 2)! distinct sets of (p − 1)M OLS(p), for p a prime, describing P G(2, p). This conjecture seemed to me an interesting challenge and my advisor, Prof. Dr. L. Storme, and I decided trying to prove this conjecture. This was not as easy as it seemed, but after a while we managed to prove the conjecture. An interesting idea that followed from this proof was to investigate the case in which we work with a prime power q = pd instead of a prime number p. Our goal was to find how many distinct sets of (q − 1)M OLS(q) there are, describing P G(2, q). This was not so easy and it took us some time to find the result. There was a lot of mathematics involved to achieve our goal and finally we found some new results, which are 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 case of a latin square and more specific we will talk about a very interesting recent result of Gary McGuire on minimal sudoku latin squares in chapter 3, but first of all we will start this chapter with some definitions and some general theorems and properties of sudoku latin squares. Later on, when we will talk about mutually orthogonal latin squares, we will also discuss mutually orthogonal sudoku latin squares and we will prove the theorems found in [14]. An important chapter is chapter 4, because the connection between projective planes and latin squares is of great importance for our results on sets of M OLS. In chapter 6, we will briefly talk about transversals and finally we will discuss how we can construct magic squares by using mutually orthogonal latin squares. This thesis has not the intention to discuss everything that is known about the subjects. 4
This would be impossible, because there are a lot of theorems and properties known of latin squares, sudoku latin squares, M OLS, . . . However, these latin squares are very interesting and there still are a lot of open problems. The nice part is that new results on 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 to thank my advisor, L. Storme, for his support and good assistance. It was not easy, but we achieved our goals and we found new results. Secondly, I would like to thank Kenneth Hicks and Gary L. Mullen for their ideas on sets of mutually orthogonal latin squares and Hans-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 Tim Penttila for his great help concerning our questions on group theory. I would also like to thank my in-laws for their support. They always believed in me and they will always believe in me. Last but not least I would like to thank Domien Broeckx for his tremendous support. I appreciated all of the encouraging words. To end this preface I would like to say this. Whoever you are, reading this, I hope you will enjoy this thesis and I would like to thank you too for reading this. Jordy Vanpoucke Ghent, May 29, 2012
5
0
c
by Jordy Vanpoucke The author and the advisor agree this thesis to be available for consultation and for personal reference use. Every other use falls within the constraints of the copyright, particularly concerning the obligation to specially mention the source when citing the results of this thesis.
6
Chapter 2 Latin squares 2.1
Definitions
2.1 Definition. A latin square of order n is an n × n matrix in which n distinct symbols from a symbol set S are arranged, such that each symbol occurs exactly once in each row and in each column. 2.2 Definition. We say that a latin square of order n is reduced (or in standard form), if the first row and the first column of this latin square is in the natural order of the symbol set we chose. From now on, we will work mostly with the symbol set S = {0, 1, 2, . . . , n − 1}. An easy example 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}. 1 2
... ... .. .
n−2 n−1 n−1 0 .. .
n−1 0
...
n−3 n−2
0 1 .. .
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 build the ith row, we simply do a cyclic shift of our first row, where we move the integers i − 1 positions 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 in each row. Also by construction, we can easily see that each symbol occurs exactly once in each column, so we constructed a latin square. Finally we see that the first row and first column is in the natural order of the symbol set, so we obtained a reduced latin square of order n. 7
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 all 0 ≤ 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 positive integer n, it would be nice to know how many distinct latin squares of order n there are for a positive integer n. By distinct we mean that these latin squares differ in at least one position. In paragraph 2.3 we will discuss some results but first we have to add a small paragraph on groups and permutations, because we will use these permutations a lot, further in this project. This small paragraph on groups and permutations is only intended to be a brief summary of some necessary definitions and theorems. Later on, we will see some other theorems that build on the content of this paragraph.
2.2 2.2.1
Groups and permutations 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 the following 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 inverse element of a. We will work with the following definition of a permutation. 8
2.10 Definition. If S is a set, then we call φ a permutation of S if φ : S → S is a bijection. A first result is the following theorem. 2.11 Theorem. If S is a finite set of n distinct elements, then there are n! permutations on 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 construct different reduced latin squares. Let us take a reduced latin square of order n. We can now apply three different permutations. First of all we can apply a permutation on the symbol set. Next we can also apply a permutation on the rows and finally we can apply a permutation on the columns. We will give a small example to show you how it is possible to construct different reduced latin squares. We start with a reduced latin square L1 of order 5. 0 1 2 3 4 1 2 3 4 0 L1 = 2 3 4 0 1 . 3 4 0 1 2 4 0 1 2 3 We can apply the following permutation new latin square L2 . 1 0 L2 = 3 2 4
on the alphabet, σ = (01)(23). This gives us a 0 3 2 4 1
3 2 4 1 0
2 4 1 0 3
4 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 the columns of L2 . 0 1 2 3 4 1 4 3 0 2 L3 = 2 3 1 4 0 . 3 0 4 2 1 4 2 0 1 3 We see that L1 6= L3 , so we obtained a different reduced latin square by applying permutations. It is very important to mention that this is not a random example. It is perfectly possible to apply a permutation σ on the alphabet such that after applying a permutation on the rows and columns we would obtain the same latin square as in the beginning. This means that there are permutations stabilizing a given reduced latin square.
9
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 latin squares of order n and we will write LR n to denote the set of different reduced latin squares of order n. First of all we will give an enumeration of numbers of latin squares that we found in different sources. In the first table, you can see the number of different reduced latin squares of order n, n ≤ 11. n |LR n| 1 1 2 1 3 1 4 4 5 56 6 9408 7 16942080 8 535281401856 9 377597570964258816 10 7580721483160132811489280 11 5363937773277371298119673540771840
references
L. Euler [10] M. Frolov [13] A. Sade [28] M. B. Wells [33] S. E. Bammel and J. Rothstein [5] B. D. McKay and E. Rogoyski [22] 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 1 2 2 3 12 4 576 5 161280 6 812851200 7 61479419904000 8 108776032459082956800 9 5524751496156892842531225600 10 9982437658213039871725064756920320000 11 776966836171770144107444346734230682311065600000
We see that the number of latin squares of order n grows very fast. In fact we did not have to give both the total number of distinct latin squares and the number of different reduced latin squares of a given order, because the total number of distinct latin squares of order n is equal to the number of different reduced latin squares of order n times n!(n − 1)!. So we have the following theorem, which can also be found in [19]. 10
2.13 Theorem. For each n ≥ 2, |Ln | = n!(n − 1)!|LR n |. Proof. Take the set of all the different reduced latin squares of order n. We can apply a permutation 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 time obtain a new latin square. After applying the permutation on the columns we can still permute n − 1 rows and there are (n − 1)! ways to do this. We can’t apply a permutation on 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 by applying these permutations, so we obtain |LR n |n!(n − 1)! distinct latin squares of order n by 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 LR n . If we start with a reduced latin square L1 , we can obtain a new Latin square L2 , by applying a permutation on the alphabet. This new Latin square is normally not reduced. If we want L2 to be a reduced latin square, then we will have to apply a permutation on the rows and columns and so we will obtain a reduced latin square L3 and it is possible that L3 6= L1 . The preceding reasoning learns us that if we want to obtain all the distinct latin squares of order n, we only have to apply a permutation on the rows and/or the columns of the reduced latin squares. If we would take a reduced latin square L and if we would apply a permutation on the alphabet to obtain a latin square L0 and a permutation on the rows and/or the columns to obtain a reduced latin square L00 and if L 6= L00 , then it would be possible to obtain L0 by applying a permutation on the rows and/or the columns of L00 .
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 rows indexed by the elements of the symbol set S1 = {a0 , a1 , . . . , an−1 } and columns indexed by the 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 by Si , columns indexed by Sj and symbols indexed by Sk , and is defined by L(i,j,k) (xi , xj ) = xk for each (x1 , x2 , x3 ) ∈ S. There is a different and easier way to find the conjugates of a latin square. We can view a latin 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 second row in the array are the column entries and finally the entries of the last row in the array are the entries of the latin square. We will now illustrate this with an example.
11
2.15 Example. A latin square L of order 4 and its associated line array, where R, C and E stand for respectively rows, columns and entries. 0 1 L= 3 2
3 2 0 1
1 0 2 3
2 3 . 1 0
The associated line array: R: 0 C: 0 E: 0
0 1 3
0 2 1
0 1 3 0 2 1
1 1 2
1 2 0
1 2 3 0 3 3
2 1 0
2 2 2
2 3 3 0 1 2
3 1 1
3 2 3
3 3 0
Recall the definition of a latin square. This definition says that each symbol occurs exactly once in each row and in each column. This is equivalent to saying that in any two rows of the associated line array all n2 ordered pairs appear exactly once. We can now apply a permutation on the symbol set in each row of the line array and those permutations correspond to permutations of the rows, columns or symbols in the latin square. We can also apply a permutation on the rows of the line array and still preserve the conditions on the line array and hence obtain a latin square. Latin squares that are obtained by applying a permutation on the rows of the associated line array of a latin square L are conjugates of this latin square L as defined in definition 2.14. We see now that there are maximum 6 distinct conjugates of a latin square. In fact a latin square has 1, 2, 3 or 6 distinct conjugates. 2.16 Example. A latin square of order 4 and its six conjugates. 0 3 1 2 1 2 0 3 3 0 2 1 2 1 3 0 (1, 2, 3)−conjugate
0 1 3 2 3 2 0 1 1 0 2 3 2 3 1 0 (2, 1, 3)−conjugate
0 2 1 3 1 3 0 2 3 1 2 0 2 0 3 1 (3, 2, 1)−conjugate
0 1 3 2 2 3 1 0 1 0 2 3 3 2 0 1 (2, 3, 1)−conjugate
0 2 3 1 2 0 1 3 1 3 2 0 3 1 0 2 (1, 3, 2)−conjugate
0 2 1 3 2 0 3 1 3 1 2 0 1 3 0 2 (3, 1, 2)−conjugate
2.17 Example. The following latin square of order 4 has only one distinct conjugate. 0 1 2 3
1 0 3 2
2 3 0 1 12
3 2 1 0
2.18 Definition. Two latin squares L and L0 of order n are isotopic or equivalent, if there are three bijections from respectively the rows, the columns and the symbols of L, to respectively the rows, the columns and the symbols of L0 , that map L to L0 . 2.19 Definition. Two latin squares L and L0 of order n are main class isotopic or paratopic if L is isotopic to any conjugate of L0 . 2.20 Definition. The set of latin squares paratopic, respectively isotopic to L is the main class, respectively the isotopy class of L. Finally we give some results for the main classes and the isotopy classes of latin squares of order n. These results can be found in [16] and [24]. n main classes 1 1 2 1 3 1 4 2 5 2 6 12 7 147 8 283657 9 19270853541 10 34817397894749939 11 2036029552582883134196099
2.3.3
isotopy classes 1 1 1 2 2 22 564 1676267 115618721533 208904371354363006 12216177315369229261482540
Completion of latin squares and critical sets
Suppose we have an empty matrix of order n. An interesting idea is as follows. Suppose we 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 entries to obtain a latin square. At this point the first thing we need is the definition of a partial latin square. 2.21 Definition. A partial latin square of order n is an n × n matrix, where each entry is 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 a partial latin square of order n to a latin square of order n. The answer is yes, but the question 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 − 1 rows are filled and the last row is empty. It is easy to see that we can complete this partial latin square of order n in exactly one way to a latin square of order n, because we know 13
that every symbol appears exactly n − 1 times in the partial latin square and hence is missing from exactly one column. Let us now suppose that only the first row is filled. Then we can complete this partial latin square to a latin square of the same order by applying a cyclic shift on the symbols of the first row, where we move the symbols one step to the left in each of the subsequent rows. We can now ask ourselves if we can always complete a partial latin square of order n and size n to a latin square of the same order. The answer is no, because the following partial latin square of order n and size n can not be completed to a latin square of the same order. 0 1 ···
n−2 n−1
.. .
Suppose we have a partial latin square of order n and size ≤ n − 1. Is it now always possible to complete this partial latin square? This question was posed by Trevor Evans in 1960 and he thought this was indeed always possible, which became widely known as the Evans Conjecture. More than 20 years later, in 1981, this conjecture was proved by Smetaniuk. His proof is constructive, thus allowing us to complete a given partial latin square 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 first r rows in the latin square of order n are completely filled in and the remaining rows are empty. 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 latin square of order n. To prove this theorem, we need Hall’s theorem, also known as the marriage theorem. This theorem 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 distinct representatives of A1 , . . . , An if the elements xi are distinct elements of X and if xi ∈ Ai for all i. 2.26 Theorem. (Hall’s theorem) Let A1 , . . . , An be a collection of subsets of a finite set X. Then there exists a system of distinct representatives of A1 , . . . , An if and only if the union of any m sets Ai contains at least m elements, for 1 ≤ m ≤ n. The second result is the following lemma. 14
2.27 Lemma. Suppose P is a partial latin square of order n and size ≤ n − 1, with at most n2 distinct 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 replace the condition ‘at most n2 distinct 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 be completed 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, if this partial latin square can be completed in different ways or if there is only one unique completion 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 completable to exactly one latin square of the same order, but when we remove one of the filled entries from P , then the completion is no longer unique. We will investigate these critical sets more in detail in the next chapter on sudoku latin squares.
15
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 latin squares.
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 × m array in the natural way into a × b rectangles. An (a, b)−sudoku latin square on the symbol 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 later on we will define a reduced set of mutually orthogonal sudoku latin squares. Important to notice is that we will always write the first row of our sudoku latin square in the natural order of the symbol set we chose. 3.3 Example. A sudoku latin square of order 4. 0 2 1 3
1 3 0 2
2 0 3 1
3.4 Example. A (3, 4)−sudoku latin square.
16
3 1 2 0
0 4 8 1 5 9 2 6 10 3 7 11
3.2 3.2.1
1 2 3 4 5 6 7 8 9 10 11 0 2 3 0 5 6 7 4 9 10 11 8 1 3 0 1 6 7 4 5 10 11 8 9 2 0 1 2 7 4 5 6 11 8 9 10 3
5 6 7 9 10 11 1 2 3 6 7 4 10 11 8 2 3 0 7 4 5 11 8 9 3 0 1 4 5 6 8 9 10 0 1 2
8 0 4 9 1 5 10 2 6 11 3 7
9 10 11 1 2 3 5 6 7 10 11 8 2 3 0 6 7 4 11 8 9 3 0 1 7 4 5 8 9 10 0 1 2 4 5 6
General theorems and properties On the number of sudoku latin squares and inequivalent sudoku latin squares
Now that we have defined an (a, b)−sudoku latin square, we will give some results on the number of (n, n)−sudoku latin squares. First of all we will define some equivalence operations on the sudoku latin squares to be able to give the definition of inequivalent sudoku 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, where k ∈ {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)−sudoku latin squares on the symbol set S = {a0 , a1 , . . . , am−1 } are equivalent if one can be obtained 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 the rows 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 on the 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 example the clockwise rotation of 90 degrees is the same as applying a transposition of the sudoku latin square and a permutation on the columns. The clockwise rotation of 180 degrees is the same as applying a permutation on the rows and a permutation on the columns. The clockwise rotation of 270 degrees is the same as applying a transposition of the sudoku latin square and a permutation on the rows. Reflections can also be obtained by applying a permutation on the rows and/or columns and/or a transposition of the sudoku latin square. We can now ask ourselves how many (a, b)−sudoku latin squares and how many inequivalent (a, b)−sudoku latin squares there are. We see that the set of all the equivalence operations forms a group of order m! × (a!b+1 ) × (b!a+1 ) × 2. In the next table we give some results for (a, b)−sudoku latin squares. More information on these numbers can be found 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 and Gilles Civario, ‘There is no 16-Clue Sudoku: Solving the Sudoku Minimum Number of Clues Problem’ [21]. 3.7 Definition. With a, b and n as in definition 3.1, an (a, b)−sudoku critical set is a partial latin square P that is completable in exactly one way to an (a, b)−sudoku latin square, but when we remove one of the filled entries from P , then the completion is no longer unique. The size of an (a, b)− sudoku critical set is the number of elements in this set, where the 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 that when we remove one of the filled entries, the completion is no longer unique. When this requirement 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 critical set is. More specific we will summarize some new results for sudoku latin squares of order 3 × 3 = 9. A sudoku latin square of order 9 has 81 entries. When we construct a (3, 3)−sudoku puzzle we would like to have a partial latin square which is uniquely completable 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 construction is how many entries, also called clues, do we give to the solver and is there a minimum number of clues that can possibly be given such that a (3, 3)−sudoku puzzle still has a unique 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 5
2
4 3
7
6
4
1 8 9
2 5
8 7
1 3
Its unique completion to a sudoku latin square of order 9. 3 1 8 7 6 2 9 4 5
6 5 7 3 9 4 2 8 1
4 2 9 8 1 5 3 6 7
9 4 1 6 2 3 7 5 8
7 3 2 5 4 8 6 1 9
8 6 5 1 7 9 4 2 3
5 9 6 4 3 1 8 7 2
1 7 3 2 8 6 5 9 4
2 8 4 9 5 7 1 3 6
However, nobody has ever found a (3, 3)−sudoku critical set of size 16 and that was the reason 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 nobody will ever find such a (3, 3)−sudoku puzzle with only 16 clues, because this conjecture was proved 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 collection in 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 the inequivalent (3, 3)−sudoku latin squares, where we know that there are 5472730538 distinct 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)−sudoku puzzles with 16 clues whose solution is the given (3, 3)−sudoku latin square. Finally they had to run through the catalogue, to apply the program to each (3, 3)−sudoku latin square. 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 different points of view on the minimum number of clues problem. First of all one can construct a partial latin square of order m = ab with a certain number of clues and check if this partial latin square is uniquely completable to an (a, b)−sudoku latin square. One can start with for example c clues and if it is not possible to complete any partial latin square of order m with c clues in a unique way to an (a, b)−sudoku latin square, one can start again with c + 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 set of inequivalent sudoku latin squares and then look at all the partial latin squares with c clues whose solution is S. This means that we think of the partial latin square with c clues as ‘contained in’ that particular (a, b)−sudoku latin square. This method is also very time-consuming if we would just try every partial latin square with c clues which is contained in S. Especially because we will have to run this program 5472730538 times in the case we are working on, where a = b = 3. Therefore the authors of [21] tried to reduce the number of possibilities to check with some theory. Briefly explained they identified certain regions in the sudoku latin squares, called unavoidable sets, such that any partial latin square which is completable to a sudoku latin square had to contain at least one clue 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 for this collection. These hitting sets are sets of size 16 that intersect all the unavoidable sets found in the first step. Finally they had to check if any of these hitting sets was a valid (3, 3)−sudoku puzzle with 16 clues, which means that the hitting set is uniquely completable to a (3, 3)−sudoku latin square, by running all of the hitting sets through a sudoku 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. 0 2 S1 = 1 3
1 3 0 2
2 1 3 0
3 0 . 2 1
When we interchange 0 and 1 among the four red numbers in this sudoku latin square, we obtain a different sudoku latin square. 20
0 2 S10 = 1 3
1 3 0 2
2 0 3 1
3 1 . 2 0
This means that if we would have a (2, 2)−sudoku puzzle with this (2, 2)−sudoku latin square as its solution, then one of the four red numbers must be a clue, otherwise we would not have a unique completion. This explains why we call this set of four numbers an unavoidable set. 3.11 Definition. Consider an (a, b)−sudoku latin square S. A subset of S is a set of entries contained in S. 3.12 Example. Let us consider the sudoku latin square S1 of example 3.10. The set V = {S1 (1, 2), S1 (1, 3), S1 (3, 2), S1 (3, 3)} is the set of red coloured entries. We see that this set is contained in S1 . In fact we interpret this sudoku latin square as a matrix, thus each element from a subset is an element from this matrix on a certain row i and a certain column j, S1 (i, j). We can now also define S\V , which is the sudoku latin square without the entries of the subset V , thus the complement of V . 3.13 Definition. Consider an (a, b)−sudoku latin square S. A subset U of S is called an unavoidable 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 cannot be used as an (a, b)−sudoku puzzle, because there is more than one completion. This gives 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 of V. Proof. If V had more than one completion, then S\V would be an unavoidable set not hit by V , which is a contradiction. Finally we say that an unavoidable set U is minimal if there is no proper subset W of U such 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. 0 2 S1 = 1 3
1 3 0 2 21
2 1 3 0
3 0 . 2 1
The following 4 sets are all minimal unavoidable sets of this (2, 2)−sudoku latin square.
U1 =
0
1 0
1
0
0
0 , U2 =
2 3
0
3
0
0 2
3
3
2
, U3 =
0
0
, U4 =
2
1
0
0
1
.
The set U1 for example is an unavoidable set, since S1 \U1 can be completed to S1 , but also to (S1 \U1 ) ∪ U10 , where
U10 =
1
0
0
1
0
0 .
This means that if we want a ‘hitting set’, thus a (2, 2)−sudoku puzzle, this set needs to intersect 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
0 1 3 2
This set has a unique completion to a (2, 2)−sudoku latin square. 0
1
0
0
1
1 3 0 2
2 1 2 3 1 0 3 2
0 1 3
3 0
0 2
2 1 3 0 2
2 1 3 0
0
3 0
0 2 3
1 3 1 3 2 1 2 3 1 0 3 2 0
0
3 0
0 2
1 3
0 2 1 3
2 1 3 0 2
1 3
0
2 1 3 0
3 0
0 2
0 2 1 3
1 3 0 2 1 3 0 2
1 3
0
2 1 3 0
3 0 2
0 2
0 2 1 3
1 3 0 2 1 3 0 2
1 3 2 1 3 0
3 0
3 0 2 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 why the 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 different examples and explain how we can find an (a, b)−sudoku critical set for some different a and b. 3.16 Lemma. Consider an (a, b)−sudoku latin square S and suppose that U ⊆ S is a minimal unavoidable set. If S 0 is any other completion of S\U , then S and S 0 differ exactly in the entries contained in U . In particular, every element in U occurs at least twice. 22
Proof. Suppose that S and S 0 agree in some entries contained in U . This would mean that U is not minimal, because it would properly contain the unavoidable set S\(S ∩S 0 ). When we move between S and S 0 , 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 move to a different row nor to a different column, since otherwise this row or column would not contain this element anymore, which is in conflict with the definition of an (a, b)−sudoku latin 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 a minimal unavoidable set. If S 0 is any other completion of S\U , then S 0 may be obtained from 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 at least two elements. The first statement, that S 0 may be obtained from S by a permutation with no fixed points of the cells in each row (column, box) of U , is in fact lemma 3.16. Suppose that the intersection of U with a certain row (column, box) contains only one element. This is impossible, because every element of U is permuted with no fixed points, implying that this certain row (column, box) would not contain this element anymore. Until now we only considered unavoidable sets where we needed one clue, but there are also 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 3 1 4 2 5
1 4 2 5 0 3
2 5 0 3 1 4
3 0 4 1 5 2
4 1 5 2 3 0
5 2 3 0 4 1
The set of red coloured numbers is an unavoidable set, where we need at least two clues from this set to completely determine these nine entries. If we would only have one clue of these nine entries, we would have more than one completion. Suppose we choose the entry 0 in the top left corner as a clue, then we also have the following (2, 3)−sudoku latin square as a solution. The reason is in fact that if we only have one clue on a certain row given for this set, we could interchange the other two entries of that row. 0 3 1 4 2 5
2 4 0 5 1 3
1 5 2 3 0 4
3 0 4 1 5 2 23
4 1 5 2 3 0
5 2 3 0 4 1
Suppose we choose the entry 0 on the first row and the entry 2 on the third row, then we have only one unique completion. 0 3
4 2 4 5 5 0 3
5 3
3 4 1 2 4 5 2 0 4 5 3 0 5 3 4
3 0 4 1 5 2 3 0 4 1 5 2
4 1 5 2 3 0 4 1 5 2 3 0
5 2 3 0 4 1 5 2 3 0 4 1
0 3
4 5 2 0 4 5 3 5 0 3 1 4
3 1 4 2 5 0 5 3
4 2 5 0 3 4
3 0 4 1 5 2 3 0 4 1 5 2
4 1 5 2 3 0 4 1 5 2 3 0
5 2 3 0 4 1 5 2 3 0 4 1
0 3 4 5 0 3 1 4 5
4 2 5 0 3 1 4 2 5 0 3
5 0 3 4 2 5 0 3 1 4
3 0 4 1 5 2 3 0 4 1 5 2
4 1 5 2 3 0 4 1 5 2 3 0
5 2 3 0 4 1 5 2 3 0 4 1
0 3 4 5 0 3 1 4 2 5
2 4 5 2 0 5 3 0 3 4 1 2 4 5 2 0 5 3 0 1 3 4
3 0 4 1 5 2 3 0 4 1 5 2
4 1 5 2 3 0 4 1 5 2 3 0
5 2 3 0 4 1 5 2 3 0 4 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 call the unavoidable sets that we defined earlier, where we needed one clue, unavoidable sets of 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 degree k > 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 no proper 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 unavoidable set 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 is an (m, k)−unavoidable set. Then we will need to add at least k elements from U to S\U to obtain an (a, b)−sudoku puzzle with a unique completion. Moreover, if V ⊆ S is an (m0 , k 0 )−unavoidable set, such that U ∩ V = ∅, then U ∪ V is an (m + m0 , k + k 0 )−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 set of degree tk. By using all of these theorems the authors of [21] could reduce the number of possibilities to check in their algorithm. We will not go into further details, but more information can be found in the article. The only thing we will do now is finish this paragraph with some examples, where we will study some unavoidable sets in some (a, b)−sudoku latin squares and 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. 0 2 S1 = 1 3
1 3 0 2
2 1 3 0
3 0 0 2 and S2 = 2 1 1 3
1 3 0 2
2 0 3 1
3 1 . 2 0
In example 3.15, we found a (2, 2)−sudoku critical set of size 4. We will prove in theorem 3.25 that the minimum number of clues needed to complete a (2, 2)−sudoku puzzle, with only 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 clues needed 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)−sudoku latin square has at least 4 clues. 0 2 1 3
1 3 0 2
2 1 3 0
3 0 2 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 four pairwise disjoint (4, 1) unavoidable sets. 0 U1 =
1
1 0
0
0
0 , U2 =
2 3
0
3
0 , U3 =
2
0
2 3
3 2
0 , U4 =
0 1
0
0
1
.
For instance, U4 occurs in S1 , but S1 \U4 can also be completed to S2 , so U4 is an unavoidable 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)−sudoku latin square has at least 4 clues. 0 2 1 3
1 3 0 2
2 0 3 1 25
3 1 2 0
Proof. We see that this (2, 2)−sudoku latin square is the union of the following four pairwise disjoint (4, 1) unavoidable sets. 0 U10 =
1
1
0
0
0 , U20 =
0
2 3
0
3
0
0
, U30 =
2 3
3 2
0
0
, U40 =
2
0
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 puzzle, 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 complete a (2, 2)−sudoku puzzle. If we can find such a (2, 2)−sudoku critical set, then we have proved theorem 3.25. In example 3.15, we presented the following (2, 2)−sudoku critical set of size 4. 0
0 1 3 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)−sudoku latin square has at least 9 clues. 0 3 1 4 2 5
1 4 2 5 3 0
2 5 3 0 4 1
3 0 4 1 5 2
4 1 5 2 0 3
5 2 0 3 1 4
Proof. We see that this (2, 3)−sudoku latin square is the union of the following nine pairwise disjoint (4, 1) unavoidable sets. 0 3
3 0
1 4
U1 =
4 1
2 5
, U2 = 0
0
0 0
0
0
5 2
, U3 = 0
0
0 0 26
0
0
, 0
0
0 0
0
0
U4 =
U7 =
1 4
4 1
2 5
, U5 =
0
0
0 0
0
0
0 2 5
0
0 0 5 2
0
0
, U8 =
5 2
3 0
, U6 =
0
0
0 0
0
0
0
0 3 0
0 0
0 0 3
0
, U9 =
0 , 3
0
0
0 0
0
0
0
0
0 0 4 1
0
0 1 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 means that 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
1
2
2 0
3 2 3
4
This set has a unique completion. In the following (2, 3)−sudoku latin square we give the completion, where the exponent of each entry gives an ordering, how the previous (2, 3)−sudoku critical set can be completed. 0 319 1 422 2 525
1 420 2 523 38 026
31 018 45 115 510 213
2 521 34 024 49 127
27
42 117 56 214 011 3
53 216 07 3 112 4
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)−sudoku latin square has at least 16 clues. 0 4 1 5 2 6 3 7
1 5 2 6 3 7 4 0
2 6 3 7 4 0 5 1
3 7 4 0 5 1 6 2
4 0 5 1 6 2 7 3
5 1 6 2 7 3 0 4
6 2 7 3 0 4 1 5
7 3 0 4 1 5 2 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)−sudoku latin square is therefore a (64, 16)−unavoidable set. Using theorem 3.21, we know that we 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 as in the previous case, we found the following (2, 4)−sudoku critical set. 0
1
2
1
2
3
3 0
4 2
3 4
5
5
6
3 4
This set has a unique completion. In the following (2, 4)−sudoku latin square, we give the completion, where the exponent of each entry gives an ordering, how the previous (2, 4)−sudoku critical set can be completed. 0 433 1 537 2 641 3 745
1 534 2 638 3 742 416 046
2 635 3 739 410 043 517 147
41 032 56 128 612 225 719 323
3 736 45 040 511 144 618 248
28
52 131 67 227 713 324 020 4
63 230 78 326 014 4 121 5
74 329 09 4 115 5 222 6
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 puzzle, 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 puzzle with only 17 clues. 3.28 Theorem. Any (3, 3)−sudoku puzzle whose solution is the following (3, 3)−sudoku latin square has at least 18 clues. 0 3 6 1 4 7 2 5 8
1 4 7 2 5 8 0 3 6
2 5 8 0 3 6 1 4 7
3 6 0 4 7 1 5 8 2
4 7 1 5 8 2 3 6 0
5 8 2 3 6 0 4 7 1
6 0 3 7 1 4 8 2 5
7 1 4 8 2 5 6 0 3
8 2 5 6 0 3 7 1 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 for a 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) is a finite set of points P , a finite set of lines B, and a relation I between the points and the lines, called the incidence relation. 4.2 Definition. A finite projective plane P is a finite incidence structure such that the 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 one line. 4.3 Definition. For a finite projective plane P, there is a positive integer n, such that any 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 the 3−dimensional vector space over the finite field F = GF (q) of order q and let us define a geometry P(V ), where the points are the 1−dimensional subspaces of V and the lines are the 2−dimensional subspaces of V . This geometry P(V ) is a projective plane, which is denoted by P G(2, F ) or P G(2, q), if F is a finite field of order q. The projective planes P G(2, q) are the classical examples of finite projective planes. They are also known as the desarguesian planes since they are the only finite projective planes in which the theorem of Desargues is valid. There are of course many other examples of projective planes. Later on we will discuss a very important conjecture on the existence of 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, where 0, 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 e1 e2 ∩ 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, where oθ = 0 and eθ1 = 1. If pθ = a, pIL1 and p 6= y, then we write the point p as (0, a). Consider a point q on L2 , q 6= x. Suppose q 0 = e∞ q ∩ L1 . If q 0 = (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 an ordered pair of coordinates (d, c), c, d ∈ R. Let us now consider a point l∞ IL∞ , where l∞ 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. Remark that all the coordinates are only influenced by the choice of the quadrangle (o, x, y, e) and θ. 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), then we write L as [m, k]. Next, we suppose that the line L passes through y, but L 6= L∞ . If L ∩ 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, where a 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 a ternary operation on R. Consider a projective plane P = (P, B, I) that is coordinatized using the set R. If a, 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 is uniquely determined, given (a, b, c). 4.5 Theorem. Consider a projective plane that is coordinatized using the set R. If T (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 (P T R) is a ternary ring (T R) including the elements 0 and 1, with 0 6= 1, which has the previous 5 properties. 4.7 Theorem. Consider a P T R (R, T ) and the incidence structure P = (P, B, I) which is 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. The elements of B are the ordered pairs [m, k], m, k ∈ R, the elements [k] with k ∈ R and the element [∞]. 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)6 I[∞]) • (x)I[m, k] ⇔ x = m • (∀x, k ∈ R)((x)6 I[k]) • (∀x ∈ R)((x)I[∞]) • (∀m, k ∈ R)((∞)6 I[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 P T R (R, T ), where R is a set of order n, then there exists a 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 a complete set of mutually orthogonal latin squares of order n. Consider a P T R (R, T ), where we suppose that R = {0, 1, . . . , n − 1}, then there exists a set of (n − 1)M OLS(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 n by proving that there exists a complete set of M OLS(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 L0 , on respectively the symbol set S = {a0 , a1 , . . . , an−1 } and the symbol set S 0 = {α0 , α1 , . . . , αn−1 }. These two latin 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 L0k,l = j. This means that if these latin squares are superimposed, we would find all of the n2 ordered pairs (i, j) ∈ S × S 0 . 5.2 Notation. Instead of talking about mutually orthogonal latin squares, we will use the abbreviation M OLS. If the order of the latin squares is equal to n, we will also use the notation M OLS(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 L0 , on respectively the symbol set S = {a0 , a1 , . . . , an2 −1 } and the symbol set S 0 = {α0 , α1 , . . . , αn2 −1 }. These two sudoku latin squares are called mutually orthogonal if there exists for each ordered 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 L0k,l = j. 5.4 Notation. Instead of talking about mutually orthogonal sudoku latin squares, we will use the abbreviation M OSLS. If the order of the sudoku latin squares is equal to n × n = n2 , we will also use the notation M OSLS(n2 ). The previous definitions can also be reformulated. 5.5 Definition. We say that two (sudoku) latin squares L and L0 of order n are orthogonal if Li,j = Lk,l and L0i,j = L0k,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 orthogonal mate if there exists a (sudoku) latin square L0 of the same order, such that L and L0 are mutually orthogonal. A bachelor (sudoku) latin square is a (sudoku) latin square which has no orthogonal mate. 5.7 Example. Two mutually orthogonal latin squares of order 5. 0 1 2 3 4
1 2 3 4 0
2 3 4 0 1
3 4 0 1 2
4 0 1 2 3
0 2 4 1 3
1 3 0 2 4
2 4 1 3 0
3 0 2 4 1
4 1 3 0 2
5.8 Example. Two mutually orthogonal sudoku latin squares of order 2 × 2. 0 3 2 1
1 2 3 0
2 1 0 3
3 0 1 2
0 2 1 3
1 3 0 2
2 0 3 1
3 1 2 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 if Li 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, we will also use the notation (k)M OLS(n). If we have k different mutually orthogonal sudoku latin squares of order n × n, we will also use the notation (k)M OSLS(n2 ). 5.11 Definition. We say that a set of k different M OLS of order n is reduced (or in standard form), if one of the latin squares is reduced and if the first row in every other latin 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 M OSLS of order n is reduced (or in standard form), if the first row in every sudoku latin square in this set is in the natural order of the symbol set we chose. Our goal is now to prove some results on the number of latin squares in a set of M O(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 proof in [14]. 35
5.13 Lemma. For a set of k M OLS or M OSLS, a random permutation on the alphabet of the (sudoku) latin squares, does not affect the orthogonality of those (sudoku) latin squares. Proof. Let us take two random mutually orthogonal latin squares of order n from a set (k)M OLS(n), L is a latin square on the symbol set S = {a0 , a1 , . . . , an−1 } and L0 is a latin square on the symbol set S 0 = {α0 , α1 , . . . , αn−1 }. By definition of orthogonality, we have that each ordered pair (u, v) ∈ S × S 0 appears exactly once among the n2 pairs (Li,j , L0i,j ), 1 ≤ i, j ≤ n. This means that if Li,j = Lk,l and L0i,j = L0k,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 0 on the 0 symbol set S 0 . Our goal is to prove that Lσ and L0σ are still orthogonal. If they are not mutually orthogonal this would mean that there exists an ordered pair (i, j) 6= (k, l), 0 0σ 0 such that Lσi,j = Lσk,l and L0σ i,j = Lk,l . This is impossible, because a permutation is a 0 0σ 0 σ bijection and the previous statement means that the ordered pair (Lσi,j , L0σ i,j ) = (Lk,l , Lk,l ) for (i, j) 6= (k, l). But (Li,j , L0i,j ) 6= (Lk,l , L0k,l ) if (i, j) 6= (k, l), so this would mean that σ or σ 0 is not a bijection, which is impossible. We can make a similar reasoning for (a, b)−sudoku latin squares. The only question that arises here is, if a random permutation on the alphabet leaves the structure of an (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 symbol set. 5.14 Definition. A maximal set M OLS(n) is a set (k)M OLS(n) such that it is impossible to extend this set to a set (k + 1)M OLS(n). 5.15 Definition.
• N (n) = max{k : ∃(k)M OLS(n)}.
• NS (n2 ) = max{k : ∃(k)M OSLS(n2 )}. • NS (a × b) = max{k : ∃(k)M OSLS(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 a permutation 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 can perform this for each latin square in a set of (k)M OLS(n). So we obtain 0 1 ... a ... L1 = .. .
n−1
0 1 ... b ... , L2 = .. .
n−1 ,...
Neither symbol a nor symbol b can be equal to 0 or otherwise this would be in conflict with the definition of a latin square. It is also impossible that a = b, otherwise if we would have 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 appear on the first position of the second row, 1, 2, . . . , n − 1, and it is impossible that two latin squares have the same symbol in that position, hence N (n) ≤ n − 1, ∀n ≥ 2. 5.17 Definition. A set of (n − 1)M OLS(n) is called a complete set of M OLS of order n. not
5.18 Theorem. NS (n2 ) = NS (n × n) ≤ n2 − n. Proof. To prove this theorem, we can follow the same reasoning as before. Let us take two mutually orthogonal sudoku latin squares of order n2 . We can perform a permutation on the alphabet without affecting the orthogonality of those sudoku latin squares, such that the 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)M OSLS(n2 ). So we obtain 0 1 ... a L1 = .. .
n − 1 ... ...
0 1 ... b L2 = .. .
n − 1 ... ...
n2 − n n2 − n + 1
...
n2 − 1 ,
... ... n2 − n n2 − n + 1
...
n2 − 1 ,
... ... ...
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 is also impossible that a = b, otherwise if we would have a = b = 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 squares. This means there are at most n2 − n possible symbols of the alphabet that can appear on the first position of the second row: n, . . . , n2 − n, n2 − n + 1, . . . , n2 − 1, and it is impossible that two sudoku latin squares have the same symbol in that position, hence NS (n × n) ≤ n2 − n. 5.19 Definition. A set of (n2 − n)M OSLS(n2 ) is called a complete set of M OSLS of 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 the orthogonality of those sudoku latin squares, such that the first row in both sudoku latin squares is given by (0, 1, . . . , b − 1, . . . , ab − b, ab − b + 1, . . . , ab − 1). Now we can do this for each sudoku latin square in a set of (k)M OSLS(ab). So we obtain 0 1 ... c L1 = .. .
b − 1 ... ...
0 1 ... d L2 = .. .
b − 1 ... ...
ab − b ab − b + 1
...
ab − 1 ,
... ... ab − b ab − b + 1
...
ab − 1 ,
... ... ...
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 order ab in (a × b)−rectangles. It is also impossible that c = d, otherwise if we would have c = 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 squares of order ab in (a × b)−rectangles. This means there are at most ab − b possible symbols of the alphabet that can appear on the first position of the second row and it is impossible that two sudoku latin squares have 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 start again 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 those sudoku 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 sudoku latin square in a set of (k)M OSLS(ab). So we obtain
L1 =
0 1 a−1 .. . ab − a .. . ab − 1
c
... ... ... ... ...
, L2 =
0 1 a−1 .. . ab − a .. .
... ...
ab − 1
38
d
... ... ... ... ... ... ...
, ...
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 order ab in (a × b)−rectangles. It is also impossible that c = d, otherwise if we would have c = 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 squares of order ab in (a × b)−rectangles. This means there are at most ab − a possible symbols of the alphabet that can appear on the second position of the first row and it is impossible that two sudoku latin squares have the same symbol in that position, hence NS (a × b) ≤ ab − a. 5.21 Definition. A set of (ab − max(a, b))M OSLS(ab) is called a complete set of M OSLS 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) and l := N (m). Now we will prove this theorem by constructing a set of (k)M OLS(n × m). First of all we take a set of (k)M OLS(n) {L1 , L2 , . . . , Lk } on the symbol set S = {a0 , a1 , . . . , an−1 }. We can do a permutation on the symbol set without affecting the orthogonality of those latin squares, such that the first row in each latin square is given by (a0 , a1 , . . . , an−1 ). Secondly we take a set of (l)M OLS(m) {L01 , L02 , . . . , L0l } on the symbol set S 0 = {α0 , α1 , . . . , αm−1 }. Again, we can do a permutation on the symbol set without affecting the orthogonality of those latin squares, such that the first row in each latin square is given by (α0 , α1 , . . . , αm−1 ). Now we will combine these two sets to obtain a set of (k)M OLS(n × m). To explain how we do this, we will give an example. Let us take a latin square Lr from the set of (k)M OLS(n) and a latin square L0s from the set of (l)M OLS(m). a0 (Lr )1,0 .. .
a1 (Lr )1,1
... ... ...
an−1 (Lr )1,n−1
(Lr )n−1,0
(Lr )n−1,1
...
(Lr )n−1,n−1
Lr =
,
α0 (L0s )1,0 .. .
α1 (L0s )1,1
... ... ...
αm−1 (L0s )1,m−1
(L0s )m−1,0
(L0s )m−1,1
...
(L0s )m−1,m−1
L0s =
.
ˆ of order n × m. We can combine these two latin squares to obtain a latin square L (a0 , α0 ) (a0 , (L0s )1,0 ) .. .
... ... ...
(a0 , αm−1 ) (a0 , (L0s )1,m−1 ) .. .
(a1 , α0 ) (a1 , (L0s )1,0 ) .. .
... ... ...
(a1 , αm−1 ) (a1 , (L0s )1,m−1 ) .. .
... ... ...
(an−1 , α0 ) (an−1 , (L0s )1,0 ) .. .
... ... ...
(an−1 , αm−1 ) (an−1 , (L0s )1,m−1 ) .. .
(a0 , (L0s )m−1,0 ) ((Lr )1,0 , α0 ) ((Lr )1,0 , (L0s )1,0 ) .. . .. . ((Lr )n−1,0 , α0 ) .. .
... ... ... ...
(a0 , (L0s )m−1,m−1 ) ((Lr )1,0 , αm−1 ) ((Lr )1,0 , (L0s )1,m−1 ) .. . .. . ((Lr )n−1,0 , αm−1 ) .. .
(a1 , (L0s )m−1,0 ) ((Lr )1,1 , α0 ) ((Lr )1,1 , (L0s )1,0 ) .. . .. . ((Lr )n−1,1 , α0 ) .. .
... ... ... ...
(a1 , (L0s )m−1,m−1 ) ((Lr )1,1 , αm−1 ) ((Lr )1,1 , (L0s )1,m−1 ) .. . .. . ((Lr )n−1,1 , αm−1 ) .. .
... ... ... ...
(an−1 , (L0s )m−1,0 ) ((Lr )1,n−1 , α0 ) ((Lr )1,n−1 , (L0s )1,0 ) .. . .. . ((Lr )n−1,n−1 , α0 ) .. .
... ... ... ...
(an−1 , (L0s )m−1,m−1 ) ((Lr )1,n−1 , αm−1 ) ((Lr )1,n−1 , (L0s )1,m−1 ) .. . .. . ((Lr )n−1,n−1 , αm−1 ) .. .
..
. ... ...
..
. ... ...
..
. ... ...
..
. ... ...
This latin square is the direct product of Lr and L0s . We can also say that we superimposed the second latin square on each entry in the first latin square. Now we will rename 39
ˆ as follows, im + j := (ai , αj ), where 0 ≤ i ≤ n − 1 and 0 ≤ j ≤ m − 1. each entry in L Our goal is to construct a set of (k)M OLS(n × m). By following the construction as exˆ t = Lt × L0t for t ∈ {1, 2, . . . , k}. plained before we can construct k different latin squares L The question is, are those latin squares mutually orthogonal? Let us take two random different latin squares of order n × m obtained by the previous ˆ v = Lv × L0 and L ˆ w = Lw × L0 , where v 6= w. construction, L v w ˆ ˆ ˆ w )g,h = (L ˆ w )i,j for (g, h) 6= (i, j), where (g, h) and Suppose (Lv )g,h = (Lv )i,j and (L (i, j) are ordered pairs of {0, 1, . . . , nm − 1} × {0, 1, . . . , nm − 1}. This would mean that (Lv × L0v )g,h = (Lv × L0v )i,j and (Lw × L0w )g,h = (Lw × L0w )i,j for (g, h) 6= (i, j). By construction this implies that ((Lv )γ,δ , (L0v )µ,ν ) = ((Lv )γ 0 ,δ0 , (L0v )µ0 ,ν 0 ) and ((Lw )γ,δ , (L0w )µ,ν ) = ((Lw )γ 0 ,δ0 , (L0w )µ0 ,ν 0 ) 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 0 , δ 0 , µ0 and ν 0 . The ordered pair (g, h) is not equal to the ordered pair (i, j), if g 6= i and/or h 6= j. If we look carefully, we see that this is the same as γ 6= γ 0 and/or δ 6= δ 0 and/or µ 6= µ0 and/or ν 6= ν 0 . This means that the ordered 4-tuple (γ, δ, µ, ν) is not equal to the ordered 4-tuple (γ 0 , δ 0 , µ0 , ν 0 ). If ((Lv )γ,δ , (L0v )µ,ν ) = ((Lv )γ 0 ,δ0 , (L0v )µ0 ,ν 0 ), then we have that (Lv )γ,δ = (Lv )γ 0 ,δ0 and (L0v )µ,ν = (L0v )µ0 ,ν 0 . If ((Lw )γ,δ , (L0w )µ,ν ) = ((Lw )γ 0 ,δ0 , (L0w )µ0 ,ν 0 ), then we have that (Lw )γ,δ = (Lw )γ 0 ,δ0 and (L0w )µ,ν = (L0w )µ0 ,ν 0 . Reordering the previous statements, we found that ((Lv )γ,δ , (Lw )γ,δ ) = ((Lv )γ 0 ,δ0 , (Lw )γ 0 ,δ0 ) and ((L0v )µ,ν , (L0w )µ,ν ) = ((L0v )µ0 ,ν 0 , (L0w )µ0 ,ν 0 ), for (γ, δ, µ, ν) 6= (γ 0 , δ 0 , µ0 , ν 0 ). Remember that Lv , Lw ∈ (k)M OLS(n) and L0v , L0w ∈ (l)M OLS(m), which means that Lv and Lw are orthogonal, implying that if ((Lv )γ,δ , (Lw )γ,δ ) = ((Lv )γ 0 ,δ0 , (Lw )γ 0 ,δ0 ), then we have that (γ, δ) = (γ 0 , δ 0 ). We have that L0v and L0w are orthogonal as well, therefore if ((L0v )µ,ν , (L0w )µ,ν ) = ((L0v )µ0 ,ν 0 , (L0w )µ0 ,ν 0 ), then we have that (µ, ν) = (µ0 , ν 0 ). This is a contradiction, because we supposed that ((Lv )γ,δ , (Lw )γ,δ ) = ((Lv )γ 0 ,δ0 , (Lw )γ 0 ,δ0 ) and ((L0v )µ,ν , (L0w )µ,ν ) = ((L0v )µ0 ,ν 0 , (L0w )µ0 ,ν 0 ), for (γ, δ, µ, ν) 6= (γ 0 , δ 0 , µ0 , ν 0 ). ˆ v )g,h = (L ˆ v )i,j and (L ˆ w )g,h = (L ˆ w )i,j , then (g, h) = (i, j). ConseSo we have that if (L quently we found by construction two mutually orthogonal squares of order n×m. Because ˆ v and L ˆ w were random latin squares from the set {L ˆ t } = {Lt × L0t } for t ∈ {1, 2, . . . , k}, L we have found a set of (k)M OLS(n × m), where k = min{N (n), N (m)}. This proves our theorem that N (n × m) ≥ min{N (n), N (m)}. 5.23 Definition. In the preceding theorem we defined the direct product of two different latin squares. This is also called the Kronecker product. Briefly summarized, the direct product of two latin squares A and B of respectively order n and order m is the latin square A ⊗ B of order n × m given by (a00 , B) (a10 , B) A⊗B = ... (an−1,0 , B)
(a01 , B) (a11 , B) ... (an−1,1 , B)
... ... ... ...
(a0,n−1 , B) (a1,n−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)M OSLS((mn)2 ). This construction is the same as before, where we use sudoku latin squares instead of latin squares. First we define a set of (k)M OSLS(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)M OSLS(m2 ) {L01 , L02 , . . . , L0l } on the symbol set S 0 = {α0 , α1 , . . . , αm−1 , . . . , αm2 −m , αm2 −m+1 , . . . , αm2 −1 }. Let us take a sudoku latin square Lr from the set of (k)M OSLS(n2 ) and a sudoku latin square L0s from the set of (l)M OSLS(m2 ). We can combine these two sudoku latin squares ˆ = Lr ⊗ L0 of order (mn)2 . This latin square is not a sudoku to obtain a latin square L s latin square. By following the same reasoning as in the preceding theorem, we obtain a set of (k)M OLS((mn)2 ). To obtain a set of (k)M OSLS((mn)2 ), we will have to apply the same permutation on the rows and the columns of every latin square in our set of (k)M OLS((mn)2 ). We will illustrate this briefly. ˆ = Lr ⊗ L0 is given by The latin square L s ((Lr )0,0 , L0s ) .. . ((Lr )n−1,0 , L0s ) .. .
... .. . ... .. .
((Lr )0,n−1 , L0s ) .. .
((Lr )0,n2 −n , L0s ) .. .
... .. . ... .. .
((Lr )n−1,n−1 , L0s ) .. .
((Lr )n−1,n2 −n , L0s ) .. .
((Lr )0,n2 −1 , L0s ) .. .
... .. . ... .. .
((Lr )n−1,n2 −1 , L0s ) .. .
((Lr )n2 −n,0 , L0s ) . . . .. .. . .
((Lr )n2 −n,n−1 , L0s ) . . . .. .. . .
((Lr )n2 −n,n2 −n , L0s ) . . . .. .. . .
((Lr )n2 −n,n2 −1 , L0s ) .. .
((Lr )n2 −1,0 , L0s )
((Lr )n2 −1,n−1 , L0s )
((Lr )n2 −1,n2 −n , L0s )
((Lr )n2 −1,n2 −1 , L0s )
...
...
...
where for each entry (Lr )i,j of Lr , ((Lr )i,j , L0s ) is given by ((Lr )i,j , (L0s )0,0 ) .. .
... .. . ... .. .
((Lr )i,j , (L0s )m−1,0 ) .. . ((Lr )i,j , (L0s )m2 −m,0 ) .. . ((Lr )i,j , (L0s )m2 −1,0 )
... .. . ...
((Lr )i,j , (L0s )0,m−1 ) .. . ((Lr )i,j , (L0s )m−1,m−1 ) .. . ((Lr )i,j , (L0s )m2 −m,m−1 ) .. . ((Lr )i,j , (L0s )m2 −1,m−1 )
((Lr )i,j , (L0s )0,m2 −m ) .. .
... .. . ... .. .
((Lr )i,j , (L0s )m−1,m2 −m ) .. .
... .. . ... .. .
((Lr )i,j , (L0s )m2 −m,m2 −m ) . . . .. .. . . 0 ((Lr )i,j , (Ls )m2 −1,m2 −m ) . . .
... .. . ...
((Lr )i,j , (L0s )0,m2 −1 ) .. . ((Lr )i,j , (L0s )m−1,m2 −1 ) .. .
.
((Lr )i,j , (L0s )m2 −m,m2 −1 ) .. . ((Lr )i,j , (L0s )m2 −1,m2 −1 )
Now we can do a permutation on the rows and the columns, such that the latin square ˆ has the structure of a sudoku latin square. In fact this permutation gives us a sudoku L latin square with the following structure Lr (1) ⊗ L0s (1) .. . Lr (1) ⊗ L0s (m .. . 2 −n+1)
Lr (n
2 −m+1)
⊗ L0s (1)
.. . 2 2 Lr (n −n+1) ⊗ L0s (m −m+1)
... ...
Lr (1) ⊗ L0s (m) .. .
... ...
Lr (1) ⊗ L0s (m .. .
... ...
Lr (n
...
Lr (n
2 −n+1)
.. . 2 −n+1)
... ...
2)
... ...
⊗ L0s (m) ⊗ L0s (m
2)
41
... ... ...
Lr (n) ⊗ L0s (1) .. . Lr (n) ⊗ L0s (m .. .
2 −m+1)
2
Lr (n ) ⊗ L0s (1) .. . 2
Lr (n ) ⊗ L0s (m
2 −m+1)
... ...
Lr (n) ⊗ L0s (m) .. .
... ...
Lr (n) ⊗ L0s (m .. .
... ...
Lr (n ) ⊗ L0s (m) .. .
...
Lr (n ) ⊗ L0s (m
2)
2
2
2)
where we used the structure of our sudoku latin squares Lr (1) Lr (n+1) .. .
Lr =
2 −n+1)
Lr (n
Lr (2) Lr (n+2) .. . Lr (n
2 −n+2)
... ... ...
Lr (n) Lr (2n) .. .
...
Lr (n
... ... ...
L0s (m) L0s (2m) . .. .
...
L0s (m
2)
and L0s (1) L0s (m+1) .. .
L0s =
L0s (m
2 −m+1)
L0s (2) L0s (m+2) .. . 2 −m+2)
L0s (m
2)
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. Instead of 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 column permutation 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 all m, n ≥ 2, N (m × n) ≥ min{N (m), N (n)}. In the proof of this theorem we used the direct product to construct a set of (k)M OLS(m × n), where k = min{N (m), N (n)}. We can now apply a permutation ρ on the rows, such that every latin square in this set becomes a sudoku latin square of order mn in (m × n)−rectangles. Note the change in the order of n 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 our ˆ = Lr ⊗ L0r , where Lr is a latin square of the set set of (k)M OLS(m × n) is given by L 0 (k)M OLS(m) and Lr is a latin square of the set (l)M OLS(n), then this permutation ρ is 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 rows are numbered from 0 to mn − 1. 5.27 Example. Suppose m = 3 and n = 4. Let us firstly take a set of two mutually orthogonal latin squares of order 3.
(1)
A
0 1 2 0 1 2 (2) = 1 2 0 ,A = 2 0 1 . 2 0 1 1 2 0 42
Now we take a set of two mutually orthogonal latin squares of order 4.
B (1)
0 1 = 2 3
1 0 3 2
2 3 0 1
3 0 1 2 2 3 , B (2) = 1 3 2 0 1 0
2 0 1 3
3 1 . 0 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)
(0, 2) (0, 0) (0, 1) (0, 3) (2, 2) (2, 0) (2, 1) (2, 3) (1, 2) (1, 0) (1, 1) (1, 3)
(0, 3) (0, 1) (0, 0) (0, 2) (2, 3) (2, 1) (2, 0) (2, 2) (1, 3) (1, 1) (1, 0) (1, 2)
(1, 0) (1, 2) (1, 3) (1, 1) (0, 0) (0, 2) (0, 3) (0, 1) (2, 0) (2, 2) (2, 3) (2, 1)
(1, 1) (1, 3) (1, 2) (1, 0) (0, 1) (0, 3) (0, 2) (0, 0) (2, 1) (2, 3) (2, 2) (2, 0)
(1, 2) (1, 0) (1, 1) (1, 3) (0, 2) (0, 0) (0, 1) (0, 3) (2, 2) (2, 0) (2, 1) (2, 3)
(1, 3) (1, 1) (1, 0) (1, 2) (0, 3) (0, 1) (0, 0) (0, 2) (2, 3) (2, 1) (2, 0) (2, 2)
(2, 0) (2, 2) (2, 3) (2, 1) (1, 0) (1, 2) (1, 3) (1, 1) (0, 0) (0, 2) (0, 3) (0, 1)
(2, 1) (2, 3) (2, 2) (2, 0) (1, 1) (1, 3) (1, 2) (1, 0) (0, 1) (0, 3) (0, 2) (0, 0)
(2, 2) (2, 0) (2, 1) (2, 3) (1, 2) (1, 0) (1, 1) (1, 3) (0, 2) (0, 0) (0, 1) (0, 3)
(2, 3) (2, 1) (2, 0) (2, 2) (1, 3) (1, 1) (1, 0) (1, 2) (0, 3) (0, 1) (0, 0) (0, 2)
and A(2) ⊗ B (2) = (0, 0) (0, 2) (0, 3) (0, 1) (2, 0) (2, 2) (2, 3) (2, 1) (1, 0) (1, 2) (1, 3) (1, 1)
(0, 1) (0, 3) (0, 2) (0, 0) (2, 1) (2, 3) (2, 2) (2, 0) (1, 1) (1, 3) (1, 2) (1, 0)
.
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 5.26 on the latin square A(1) ⊗ B (1) . This gives us the following latin square. (0, 0) (1, 0) (2, 0) (0, 1) (1, 1) (2, 1) (0, 2) (1, 2) (2, 2) (0, 3) (1, 3) (2, 3)
(0, 1) (1, 1) (2, 1) (0, 0) (1, 0) (2, 0) (0, 3) (1, 3) (2, 3) (0, 2) (1, 2) (2, 2)
(0, 2) (1, 2) (2, 2) (0, 3) (1, 3) (2, 3) (0, 0) (1, 0) (2, 0) (0, 1) (1, 1) (2, 1)
(0, 3) (1, 3) (2, 3) (0, 2) (1, 2) (2, 2) (0, 1) (1, 1) (2, 1) (0, 0) (1, 0) (2, 0)
(1, 0) (2, 0) (0, 0) (1, 1) (2, 1) (0, 1) (1, 2) (2, 2) (0, 2) (1, 3) (2, 3) (0, 3)
(1, 1) (2, 1) (0, 1) (1, 0) (2, 0) (0, 0) (1, 3) (2, 3) (0, 3) (1, 2) (2, 2) (0, 2)
43
(1, 2) (2, 2) (0, 2) (1, 3) (2, 3) (0, 3) (1, 0) (2, 0) (0, 0) (1, 1) (2, 1) (0, 1)
(1, 3) (2, 3) (0, 3) (1, 2) (2, 2) (0, 2) (1, 1) (2, 1) (0, 1) (1, 0) (2, 0) (0, 0)
(2, 0) (0, 0) (1, 0) (2, 1) (0, 1) (1, 1) (2, 2) (0, 2) (1, 2) (2, 3) (0, 3) (1, 3)
(2, 1) (0, 1) (1, 1) (2, 0) (0, 0) (1, 0) (2, 3) (0, 3) (1, 3) (2, 2) (0, 2) (1, 2)
(2, 2) (0, 2) (1, 2) (2, 3) (0, 3) (1, 3) (2, 0) (0, 0) (1, 0) (2, 1) (0, 1) (1, 1)
(2, 3) (0, 3) (1, 3) (2, 2) (0, 2) (1, 2) (2, 1) (0, 1) (1, 1) (2, 0) (0, 0) (1, 0)
When we replace every (i, j) by 4i + j, then we obtain the following latin square 0 1 2 3 4 5 6 7 8 9 10 11 1 0 3 2 5 4 7 6 9 8 11 10 S1 = 2 3 0 1 6 7 4 5 10 11 8 9 3 2 1 0 7 6 5 4 11 10 9 8
4 8 0 5 9 1 6 10 2 7 11 3
5 6 9 10 1 2 4 7 8 11 0 3 7 4 11 8 3 0 6 5 10 9 2 1
7 11 3 6 10 2 5 9 1 4 8 0
8 0 4 9 1 5 10 2 6 11 3 7
9 10 11 1 2 3 5 6 7 8 11 10 0 3 2 4 7 6 11 8 9 3 0 1 9 4 5 10 9 8 2 1 0 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) (2, 0) (1, 0) (0, 2) (2, 2) (1, 2) (0, 3) (2, 3) (1, 3) (0, 1) (2, 1) (1, 1)
(0, 1) (2, 1) (1, 1) (0, 3) (2, 3) (1, 3) (0, 2) (2, 2) (1, 2) (0, 0) (2, 0) (1, 0)
(0, 2) (2, 2) (1, 2) (0, 0) (2, 0) (1, 0) (0, 1) (2, 1) (1, 1) (0, 3) (2, 3) (1, 3)
(0, 3) (2, 3) (1, 3) (0, 1) (2, 1) (1, 1) (0, 0) (2, 0) (1, 0) (0, 2) (2, 2) (1, 2)
(1, 0) (0, 0) (2, 0) (1, 2) (0, 2) (2, 2) (1, 3) (0, 3) (2, 3) (1, 1) (0, 1) (2, 1)
(1, 1) (0, 1) (2, 1) (1, 3) (0, 3) (2, 3) (1, 2) (0, 2) (2, 2) (1, 0) (0, 0) (2, 0)
(1, 2) (0, 2) (2, 2) (1, 0) (0, 0) (2, 0) (1, 1) (0, 1) (2, 1) (1, 3) (0, 3) (2, 3)
(1, 3) (0, 3) (2, 3) (1, 1) (0, 1) (2, 1) (1, 0) (0, 0) (2, 0) (1, 2) (0, 2) (2, 2)
(2, 0) (1, 0) (0, 0) (2, 2) (1, 2) (0, 2) (2, 3) (1, 3) (0, 3) (2, 1) (1, 1) (0, 1)
(2, 1) (1, 1) (0, 1) (2, 3) (1, 3) (0, 3) (2, 2) (1, 2) (0, 2) (2, 0) (1, 0) (0, 0)
(2, 2) (1, 2) (0, 2) (2, 0) (1, 0) (0, 0) (2, 1) (1, 1) (0, 1) (2, 3) (1, 3) (0, 3)
(2, 3) (1, 3) (0, 3) (2, 1) (1, 1) (0, 1) (2, 0) (1, 0) (0, 0) (2, 2) (1, 2) (0, 2)
.
When we replace every (i, j) by 4i + j, then we obtain the following latin square
S2 =
0 8 4 2 10 6 3 11 7 1 9 5
1 2 9 10 5 6 3 0 11 8 7 4 2 1 10 9 6 5 0 3 8 11 4 7
3 11 7 1 9 5 0 8 4 2 10 6
4 5 6 0 1 2 8 9 10 6 7 4 2 3 0 10 11 8 7 6 5 3 2 1 11 10 9 5 4 7 1 0 3 9 8 11
7 3 11 5 1 9 4 0 8 6 2 10
8 4 0 10 6 2 11 7 3 9 5 1
9 10 5 6 1 2 11 8 7 4 3 0 10 9 6 5 2 1 8 11 4 7 0 3
11 7 3 9 5 1 . 8 4 0 10 6 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 M OLS(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−1
1 ... ...
p−1
, L2 =
0 2 .. .
1 ... ...
p−1
, . . . , Lp−1 =
p−2
0 1 ... p−1 ... .. .
p−1
.
1
If these latin squares are mutually orthogonal, then we would have that if ((Lb )i,j , (Lb0 )i,j ) = ((Lb )k,l , (Lb0 )k,l ), for b, b0 ∈ GF (p)\{0} and b 6= b0 , then (i, j) = (k, l). Suppose that (Lb )i,j = (Lb )k,l and (Lb0 )i,j = (Lb0 )k,l . This means that bi + j = bk + l and b0 i + j = b0 k + l, or bi + j − (b0 i + j) = bk + l − (b0 k + l), such that (b − b0 )i = (b − b0 )k. We know that b 6= b0 and b, b0 6= 0, therefore i = k. Because b and b0 are not equal to 0, we also have that b0 (bi + j) = b0 (bk + l) and b(b0 i + j) = b(b0 k + l), or b0 bi + b0 j − (bb0 i + bj) = b0 bk + b0 l − (bb0 k + bl), such that (b0 − b)j = (b0 − b)l. We know that b 6= b0 and b, b0 6= 0, therefore j = l. Thus (i, j) = (k, l). This means that we have found a set of (p − 1)M OLS(p) describing P G(2, p), for p a prime number. If q = pd is a prime power, we will define a set of (q−1)M OLS(q) describing P G(2, q), similar as in the previous case. First we define a primitive polynomial, f (x) = xd +fd−1 xd−1 + . . . + f1 x + 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) as follows, GF (q) = {0, 1, α, α2 , . . . , αq−2 }, d−1 or GF (q) = {0, 1, 2, . . . , x, . . . , q − 1} where x= xd−1 p + xd−2 pd−2 + . . . + x1 p + x0 and x0 x1 xi ∈ GF (p). We also use the notation, X = .. ∈ GF (p)d . . xd−1 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 mutually orthogonal. 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{pei i − 1|i = 1, 2, . . . , k}.
45
5.30 Theorem. If q = pd is a prime power, then NS (q 2 ) = q 2 − q. Proof. We know that we can construct a set of (q 2 − 1)M OLS(q 2 ) for q = pd a prime power. 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 (q 2 ). We can write GF (q 2 ) = {k + αl|k, l ∈ GF (q)}. For each a ∈ GF (q 2 )\GF (q), we define a latin square as follows. (La )x,y = ax + y, where x, y ∈ GF (q 2 ). 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 ei i and ei ≥ 1, i = 1, . . . , k, then NS (n2 ) ≥ min{p2e i − pi |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 5.3.1
Examples of small order Latin squares of order n n N (n)
2 3 4 5 6 7 1 2 3 4 1 6
8 9 10 7 8 ? ≥ 2, ≤ 8
5.33 Theorem. N (2) = 1 0 1 1 0 5.34 Theorem. N (3) = 2 0 1 2 1 2 0 2 0 1
0 1 2 2 0 1 1 2 0
5.35 Theorem. N (4) = 3 0 1 2 3
1 0 3 2
2 3 0 1
3 2 1 0
0 2 3 1
1 3 2 0
2 0 1 3
46
3 1 0 2
0 3 1 2
1 2 0 3
2 1 3 0
3 0 2 1
5.36 Theorem. N (5) = 4 0 1 2 3 4
1 2 3 4 0
2 3 4 0 1
3 4 0 1 2
4 0 1 2 3
0 2 4 1 3
1 3 0 2 4
2 4 1 3 0
3 0 2 4 1
4 1 3 0 2
0 3 1 4 2
1 4 2 0 3
2 0 3 1 4
3 1 4 2 0
4 2 0 3 1
0 4 3 2 1
1 0 4 3 2
2 1 0 4 3
3 2 1 0 4
4 3 2 1 0
5.37 Theorem. N (6) = 1 This was proved by Gaston Tarry in 1900, who was very interested in magic squares and 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 orthogonal latin squares of order 6, but he also conjectured that a pair of mutually orthogonal latin squares 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 more information can be found in [1] and [7]. 5.38 Theorem. N (7) = 6 0 1 2 3 4 5 6
1 2 3 4 5 6 0
2 3 4 5 6 0 1
3 4 5 6 0 1 2
4 5 6 0 1 2 3
5 6 0 1 2 3 4
6 0 1 2 3 4 5
0 2 4 6 1 3 5
1 3 5 0 2 4 6
2 4 6 1 3 5 0
3 5 0 2 4 6 1
4 6 1 3 5 0 2
5 0 2 4 6 1 3
6 1 3 5 0 2 4
0 3 6 2 5 1 4
1 4 0 3 6 2 5
2 5 1 4 0 3 6
3 6 2 5 1 4 0
4 0 3 6 2 5 1
5 1 4 0 3 6 2
6 2 5 1 4 0 3
0 4 1 5 2 6 3
1 5 2 6 3 0 4
2 6 3 0 4 1 5
3 0 4 1 5 2 6
4 1 5 2 6 3 0
5 2 6 3 0 4 1
6 3 0 4 1 5 2
0 5 3 1 6 4 2
1 3 4 2 0 5 3
2 0 5 3 1 6 4
3 1 6 4 2 0 5
4 2 0 5 3 1 6
5 3 1 6 4 2 0
6 4 2 0 5 3 1
0 6 5 4 3 2 1
1 0 6 5 4 3 2
2 1 0 6 5 4 3
3 2 1 0 6 5 4
4 3 2 1 0 6 5
5 4 3 2 1 0 6
6 5 4 3 2 1 0
0 4 3 7 6 2 5 1
1 5 2 6 7 3 4 0
4 0 7 3 2 6 1 5
5 1 6 2 3 7 0 4
5.39 Theorem. N (8) = 7 0 1 2 3 4 5 6 7
1 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 3
5 4 7 6 1 0 3 2 0 5 1 4 2 7 3 6
6 7 4 5 2 3 0 1 1 4 0 5 3 6 2 7
7 6 5 4 3 2 1 0 2 7 3 6 0 5 1 4
0 2 4 6 3 1 7 5 3 6 2 7 1 4 0 5
4 1 5 0 6 3 7 2
1 3 5 7 2 0 6 4 5 0 4 1 7 2 6 3
2 0 6 4 1 3 5 7 6 3 7 2 4 1 5 0
3 1 7 5 0 2 4 6 7 2 6 3 5 0 4 1
4 6 0 2 7 5 3 1
5 7 1 3 6 4 2 0 0 6 7 1 5 3 2 4
6 4 2 0 5 7 1 3 1 7 6 0 4 2 3 5
7 5 3 1 4 6 0 2 2 4 5 3 7 1 0 6
0 3 6 5 7 4 1 2 3 5 4 2 6 0 1 7
4 2 3 5 1 7 6 0
47
1 2 7 4 6 5 0 3 5 3 2 4 0 6 7 1
2 1 4 7 5 6 3 0 6 0 1 7 3 5 4 2
3 0 5 6 4 7 2 1 7 1 0 6 2 4 5 3
4 7 2 1 3 0 5 6
5 6 3 0 2 1 4 7 0 7 5 2 1 6 4 3
6 5 0 3 1 2 7 4 1 6 4 3 0 7 5 2
7 4 1 2 0 3 6 5 2 5 7 0 3 4 6 1
3 4 6 1 2 5 7 0
4 3 1 6 5 2 0 7
5 2 0 7 4 3 1 6
2 6 1 5 4 0 7 3 6 1 3 4 7 0 2 5
3 7 0 4 5 1 6 2 7 0 2 5 6 1 3 4
6 2 5 1 0 4 3 7
7 3 4 0 1 5 2 6
5.40 Theorem. N (9) = 8
0 1 2 3 4 5 6 7 8
1 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 0
5 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 4
0 2 1 6 8 7 3 5 4
1 0 2 7 6 8 4 3 5
2 1 0 8 7 6 5 4 3
3 5 4 0 2 1 6 8 7
4 3 5 1 0 2 7 6 8
5 4 3 2 1 0 8 7 6
6 8 7 3 5 4 0 2 1
7 6 8 4 3 5 1 0 2
8 7 6 5 4 3 2 1 0
0 3 6 7 1 4 5 8 2
1 4 7 8 2 5 3 6 0
2 5 8 6 0 3 4 7 1
3 6 0 1 4 7 8 2 5
4 7 1 2 5 8 6 0 3
5 8 2 0 3 6 7 1 4
6 0 3 4 7 1 2 5 8
7 1 4 5 8 2 0 3 6
8 2 5 3 6 0 1 4 7
0 4 8 1 5 6 2 3 7
1 5 6 2 3 7 0 4 8
2 3 7 0 4 8 1 5 6
3 7 2 4 8 0 5 6 1
4 8 0 5 6 1 3 7 2
5 6 1 3 7 2 4 8 0
6 1 5 7 2 3 8 0 4
7 2 3 8 0 4 6 1 5
8 0 4 6 1 5 7 2 3
0 5 7 4 6 2 8 1 3
1 3 8 5 7 0 6 2 4
2 4 6 3 8 1 7 0 5
3 8 1 7 0 5 2 4 6
4 6 2 8 1 3 0 5 7
5 7 0 6 2 4 1 3 8
6 2 4 1 3 8 5 7 0
7 0 5 2 4 6 3 8 1
8 1 3 0 5 7 4 6 2
0 6 3 5 2 8 7 4 1
1 7 4 3 0 6 8 5 2
2 8 5 4 1 7 6 3 0
3 0 6 8 5 2 1 7 4
4 1 7 6 3 0 2 8 5
5 2 8 7 4 1 0 6 3
6 3 0 2 8 5 4 1 7
7 4 1 0 6 3 5 2 8
8 5 2 1 7 4 3 0 6
0 7 5 8 3 1 4 2 6
1 8 3 6 4 2 5 0 7
2 6 4 7 5 0 3 1 8
3 1 8 2 6 4 7 5 0
4 2 6 0 7 5 8 3 1
5 0 7 1 8 3 6 4 2
6 4 2 5 0 7 1 8 3
7 5 0 3 1 8 2 6 4
8 3 1 4 2 6 0 7 5
0 8 4 2 7 3 1 6 5
1 6 5 0 8 4 2 7 3
2 7 3 1 6 5 0 8 4
3 2 7 5 1 6 4 0 8
4 0 8 3 2 7 5 1 6
5 1 6 4 0 8 3 2 7
6 5 1 8 4 0 7 3 2
7 3 2 6 5 1 8 4 0
8 4 0 7 3 2 6 5 1
5.41 Theorem. N (10) ≥ 2 0 1 2 3 4 5 6 7 8 9
5.3.2
1 7 3 5 8 6 0 9 2 4
2 4 5 1 9 8 7 6 0 3
3 2 7 4 6 9 1 8 5 0
4 0 8 6 2 7 3 1 9 5
5 6 9 0 3 1 8 4 7 2
6 5 1 8 0 4 9 2 3 7
7 8 0 9 5 2 4 3 1 6
8 9 4 2 7 3 5 0 6 1
9 3 6 7 1 0 2 5 4 8
0 4 7 5 3 2 9 1 6 8
1 6 9 3 5 7 8 4 0 2
2 1 8 6 7 4 3 5 9 0
3 8 2 9 4 5 0 7 1 6
4 5 1 8 6 0 7 2 3 9
Sudoku latin squares of order n × n n NS (n2 )
2 3 2 6
5.42 Theorem. NS (4) = 2 0 2 3 1
1 3 2 0
2 0 1 3
3 1 0 2
0 3 1 2
48
1 2 0 3
2 1 3 0
3 0 2 1
5 3 6 7 1 9 2 0 8 4
6 7 3 0 2 8 1 9 4 5
7 9 4 2 0 3 6 8 5 1
8 0 5 1 9 6 4 3 2 7
9 2 0 4 8 1 5 6 7 3
5.43 Theorem. NS (9) = 6
5.4 5.4.1
0 3 6 7 1 4 5 8 2
1 4 7 8 2 5 3 6 0
2 5 8 6 0 3 4 7 1
3 6 0 1 4 7 8 2 5
4 7 1 2 5 8 6 0 3
5 8 2 0 3 6 7 1 4
6 0 3 4 7 1 2 5 8
7 1 4 5 8 2 0 3 6
8 2 5 3 6 0 1 4 7
0 4 8 1 5 6 2 3 7
1 5 6 2 3 7 0 4 8
2 3 7 0 4 8 1 5 6
3 7 2 4 8 0 5 6 1
4 8 0 5 6 1 3 7 2
5 6 1 3 7 2 4 8 0
6 1 5 7 2 3 8 0 4
7 2 3 8 0 4 6 1 5
8 0 4 6 1 5 7 2 3
0 5 7 4 6 2 8 1 3
1 3 8 5 7 0 6 2 4
2 4 6 3 8 1 7 0 5
3 8 1 7 0 5 2 4 6
4 6 2 8 1 3 0 5 7
5 7 0 6 2 4 1 3 8
6 2 4 1 3 8 5 7 0
7 0 5 2 4 6 3 8 1
8 1 3 0 5 7 4 6 2
0 6 3 5 2 8 7 4 1
1 7 4 3 0 6 8 5 2
2 8 5 4 1 7 6 3 0
3 0 6 8 5 2 1 7 4
4 1 7 6 3 0 2 8 5
5 2 8 7 4 1 0 6 3
6 3 0 2 8 5 4 1 7
7 4 1 0 6 3 5 2 8
8 5 2 1 7 4 3 0 6
0 7 5 8 3 1 4 2 6
1 8 3 6 4 2 5 0 7
2 6 4 7 5 0 3 1 8
3 1 8 2 6 4 7 5 0
4 2 6 0 7 5 8 3 1
5 0 7 1 8 3 6 4 2
6 4 2 5 0 7 1 8 3
7 5 0 3 1 8 2 6 4
8 3 1 4 2 6 0 7 5
0 8 4 2 7 3 1 6 5
1 6 5 0 8 4 2 7 3
2 7 3 1 6 5 0 8 4
3 2 7 5 1 6 4 0 8
4 0 8 3 2 7 5 1 6
5 1 6 4 0 8 3 2 7
6 5 1 8 4 0 7 3 2
7 3 2 6 5 1 8 4 0
8 4 0 7 3 2 6 5 1
On the number of (n − 1)M OLS(n) First case: GF (p)
Now that we have found some general results on M OLS, we want to know something more about sets of M OLS. More specific, we are interested in the number of (n − 1)M OLS(n). We will consider two cases, first of all the case n = p, a prime number and then the case n = q = pd , a prime power. Let us start with the case n = p. 5.44 Definition. We consider a set of (p − 1)M OLS(p) describing P G(2, p), for p a prime, 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-1
1
... ...
p-1 , A2 =
0 2 .. .
1
... ...
p-2
p-1 , . . . , Ap−1
0 p-1 = .. .
1
... ...
p-1 .
1
This set of M OLS is reduced. Our goal is to prove that there are (p−2)! different reduced (p − 1)M OLS(p), for p a prime, describing P G(2, p). First of all we want to determine the group of permutations on the (p − 1)M OLS(p) stabilizing the set {A1 , A2 , . . . , Ap−1 }. To fully understand these permutations on a set of latin squares, we will prove some characteristics of the permutations. 49
5.45 Lemma. A permutation on the alphabet commutes with a permutation on the rows and/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 permutation on the rows in every latin square in the set of M OLS. 3. γ is a permutation on the columns: y 7→ γ(y), and we have to use the same permutation on the columns in every latin square in the set of M OLS. 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 M OLS {A1 , A2 , . . . , Ap−1 } of definition 5.44. Suppose we apply σ1 to A1 , σ2 to A2 , . . . , and σp−1 to Ap−1 , the same column permutations to all of the latin squares in the set and also the same row permutations to all of the latin squares in the set, whereby row i moves to row 0. Suppose we obtain a new reduced set of latin 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 row i. Now we move row i to row 0. The same permutation of the columns transforms everything on row 0 to 0, 1, . . . , p − 1. So necessarily, the same element must be in column y 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)M OLS(p), for p a prime, describing P G(2, p). Proof. We work with the alphabet of GF (p) = {0, 1, . . . , p − 1}. We know that there are p! permutations on the alphabet. Define the permutation ρa0 : GF (p) → GF (p) : x 7→ a0 x, for a0 ∈ GF (p) and a0 6= 0. 50
We have that ρa0 ((Aa )x,y ) = aa0 x + a0 y = (Aaa0 )x,a0 y , but the element aa0 x + a0 y still stands in row x and column y. If we perform the permutation γa0 : y 7→ a0 y on the columns in every latin square of the set of M OLS, we have moved column y to column a0 y, so we have transformed Aa in Aaa0 . 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 stands in row x and column y. If we perform the permutation γb : y 7→ y + b on the columns in every latin square of the set of M OLS, column y is moved to column y + b and we have transformed Aa in Aa , so we have the same set of M OLS {A1 , A2 , . . . , Ap−1 }. Now we have found p(p − 1) permutations on the (p − 1)M OLS(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)M OLS(p), for p a prime, describing P G(2, p). 5.48 Theorem. There are (p − 2)! different reduced (p − 1)M OLS(p), for p a prime, describing P G(2, p). Proof. The p(p − 1) permutations on the alphabet that we defined in the previous theorem 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 a subgroup of the symmetric group Sp . Now we know from [20] that this group is maximal in Sp . So if there are more permutations stabilizing the set of (p − 1)M OLS(p) describing P G(2, p), we would have the group Sp stabilizing {A1 , A2 , . . . , Ap−1 }. So if we find 2 different reduced sets of (p − 1)M OLS(p) describing P G(2, p) we know that we can’t extend the group AGL1 (p) to Sp , so there are only p(p − 1) permutations stabilizing the set of (p − 1)M OLS(p). We are going to prove this now in the next theorem, which will conclude the proof of this theorem that there are exactly (p − 2)! different reduced (p − 1)M OLS(p), for p a prime, describing P G(2, p). 5.49 Theorem. There are at least two different reduced sets of (p − 1)M OLS(p) for p an odd prime > 3. Proof. In the beginning we defined a standard example of a set of (p − 1)M OLS(p), (Aa )x,y = ax + y. So, (A1 )x,y = x + y.
A1 =
0 1 .. .
1 2
... ... ...
p − 3 ... p − 2 ... p − 1 ...
p−3 p−2 p−1
p−5 p−4 p−5 p−4 p−3 p−4 p−3 p−2 51
.
We apply the permutation (p − 2, p − 1) on the alphabet on the set of (p − 1)M OLS(p), 0 }. {A1 , A2 , . . . , Ap−1 }. If we do this, we obtain a set of (p − 1)M OLS(p) {B10 , B20 , . . . , Bp−1
A1 7→ B10 =
0 1 .. .
1 2
... ... ...
p − 3 ... p − 1 ... p − 2 ...
p−3 p−1 p−2
p−5 p−4 p−5 p−4 p−3 p−4 p−3 p−1
.
If we perform the permutation ρ2 = γ2 : (p−2, p−1) on respectively the rows and columns 0 }, we obtain the set {B1 , B2 , . . . , Bp−1 } of every latin square of the set {B10 , B20 , . . . , Bp−1 and this set is reduced. In this set, the first latin square is reduced, so we transformed A1 into B1 .
B10 7→ B1 =
0 1 .. .
1 2
... ... .. .
p − 3 ... p − 2 ... p − 1 ...
p−3
p−2 p−1
p−4 p−5
p−4 p−5 p−1 p−3 p−3 p−4
.
If we look at the 2 × 2 matrix in the bottom right corner of A1 , we see the symbols p − 4, p − 3 and p − 2. Applying the permutation (p − 2, p − 1) on the alphabet, the rows and the columns, we obtain the latin square B1 . In this latin square, we see the symbols p − 4, p − 3 and p − 1 in the 2 × 2 matrix in the bottom right corner. So we see that the reduced latin square B1 is different from A1 , so we have found two different reduced sets of (p − 1)M OLS(p), describing P G(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)M OLS(q) describing P G(2, q), for q = pd a prime power. In this case we define a primitive polynomial, f (x) = xd + fd−1 xd−1 + . . . + f1 x + 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) as follows, GF (q) = {0, 1, α, α2 , . . . , αq−2 }, or GF (q) = {0, 1, 2, . . . , x, . . . , q − 1} where d−1 x = x + xd−2 pd−2 + . . . + x1 p + x0 and xi ∈ GF (p). We also use the notation d−1 p x0 x1 X = .. ∈ GF (p)d . . xd−1 Let us denote the latin squares by Aa , a ∈ GF (q)∗ mod f (α), where (Aa )x,y = ax + y 52
mod f (α),
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 invertible d × 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 x0 x1 ax = Ma X, where X = .. ∈ GF (p)d . . xd−1 5.51 Remark. As a result we have Mα Mα = Mα2 = Mα2 . The multiplication ax can also be written in terms of permutations as πa x, where πa is the permutation ‘multiplying by a modulo f (α)’. To uniquely determine the permutation πa 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) • Ma X mod p • πa x 5.53 Remark. The previous equivalence is only the case for ‘multiplying by a modulo f (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 )π to represent this matrix as a permutation on GF (q). Our goal is to find how many different reduced (q − 1)M OLS(q) there are, for q = pd a prime power, describing P G(2, q). We want to follow the same reasoning as for q = p prime, so we are going to determine the group of permutations on the (q − 1)M OLS(q) stabilizing the set {A1 , A2 , . . . , Aq−1 }, describing P G(2, q). 5.55 Lemma. We work in the finite field GF (q), q = pd a prime power. Consider the set of M OLS {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 set and also the same row permutations to all of the latin squares in the set, whereby row i moves 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 M OLS {A1 , A2 , . . . , Aq−1 } invariant and these permutations on the alphabet are the same as a permutation 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 on the 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 stands in row x and column y. If we perform the permutation γb : y 7→ y + b on the columns in every latin square of the set of M OLS, column y is moved to column y + b and we have transformed Aa in Aa , so we have the same set of M OLS {A1 , A2 , . . . , Aq−1 }. 5.57 Theorem. There are at most (q − 2)!/d different reduced (q − 1)M OLS(q), for q = pd a prime power, describing P G(2, q). Proof. We work with the alphabet of GF (q) and we know that there are q! permutations on the alphabet. Let us define a permutation µM˜ ∈ GLd (p) on the alphabet: ˜ X. µ ˜ : GF (p)d → GF (p)d : X 7→ M M
˜ Ma X + M ˜ Y . Now we want to check whether we can obtain We have that µM˜ ((Aa )x,y ) = M the same result with a permutation on the rows and/or columns, and/or a reordering of ˜ Y ’ is just the same as a column permutation. the latin squares in our set of M OLS. ‘M ˜ Ma X’ ? If this has to be the same as performing the same The question is: what is ‘M row permutation on every latin square of the set of M OLS and/or a reordering of the ˜ Ma )π x = (Ma0 )π π latin squares in this set, then we would have that (M ˜ x, where π ˜ is a permutation 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: ˜ Mα )π = (M k1 )π π (M ˜ = β1 α (Mαk2 )π π ˜
˜ Mα2 )π = (M = β2 .. . ˜ Mαq−1 )π = (Mαkq−1 )π π ˜ = βq−1 . (M Now we have that ∃i : Mαki = Id . This implies that π ˜ = βi . So we obtain the following system of equations: ˜ Mα = Mαk1 M ˜ Mαi M ˜ Mα2 = M k2 M ˜ Mi M α α .. . ˜ Mαq−1 = M kq−1 M ˜ Mi . M α
54
α
˜ is an invertible matrix, and because the matrices Ma define a cyclic group C Because M of order q − 1, we have as a result that, ˜ −1 M kj M ˜ = M j M −i M α
=
α α Mαj−i .
˜ was a random matrix of GLd (p), so we have that But M ˜ CM ˜ −1 ∈ C. M ˜ is an element of the normalizer of the Singer cycle of order q − 1 This means that M defined 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)d permutations on the (q − 1)M OLS(q) stabilizing the set {A1 , A2 , . . . , Aq−1 }. Because there are q! permutations on GF (q), we have at most (q − 2)!/d different reduced (q − 1)M OLS(q), for q = pd a prime power, describing P G(2, q). The q(q − 1)d permutations on the alphabet that we obtained in the previous theorems 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 a subgroup of AGLd (p). Now we are going to prove that there are exactly (q − 2)!/d different reduced (q − 1)M OLS(q), for q = pd a prime power, describing P G(2, q), by using the 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. If the affine group AGL1 (pm ) 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 (pr ), where d = rk. 4. p = 3, d = 4, G ⊂ AΓL2 (9) and if G0 is the stabilizer of 0 in G, then G0 admits a normal 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)M OLS(q) representing P G(2, q) for q = pd > 4. Proof. We will prove this lemma in two parts. First of all we suppose that p = 2 and we use definition 5.50 to construct a set of (q − 1)M OLS(q), describing P G(2, q). We see that the first element on the second row in each latin square Ai , 1 ≤ i ≤ q − 1, is equal to i and
A1 =
0 1 2 3 .. .
1 0 3 2 .. .
2 3 0 1 .. .
3 2 1 0 .. .
... ... ... ... ...
q − 2 q − 1 q − 4 q − 3 ... q − 1 q − 2 q − 3 q − 4 ... 55
q−2 q−1 q−4 q−3 .. . 0 1
q−1 q−2 q−3 q−4 . .. . 1 0
We will now apply the permutation (q−2, q−1) on the alphabet on the set {A1 , A2 , . . . , Aq−1 }. 0 }. If we do this, we obtain a set of (q − 1)M OLS(q) {B10 , B20 , . . . , Bq−1
A1 7→ B10 =
0 1 2 3 .. .
1 0 3 2 .. .
2 3 0 1 .. .
3 2 1 0 .. .
... ... ... ... .. .
q − 1 q − 2 q − 4 q − 3 ... q − 2 q − 1 q − 3 q − 4 ...
q−1 q−2 q−4 q−3 .. . 0 1
q−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 columns 0 of every latin square of the set {B10 , B20 , . . . , Bq−1 }, we obtain the set {B1 , B2 , . . . , Bq−1 }, which is a reduced set of (q − 1)M OLS(q). In this set, the first latin square is reduced, so we transformed A1 into B1 .
B10 7→ B1 =
0 1 2 3 .. . q−2 q−1
1 0 3 2 .. .
2 3 0 1 .. .
3 2 1 0 .. .
... ... ... ... ...
q − 1 q − 3 q − 4 ... q − 2 q − 4 q − 3 ...
q−2 q−1 q−3 q−4 .. . 0 1
q−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 two different reduced sets of (q − 1)M OLS(q), describing P G(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 0 {A1 , A2 , . . . , Aq−1 }. If we do this, we obtain a set of (q − 1)M OLS(q) {C10 , C20 , . . . , Cq−1 }.
A1 7→ C10 =
0 2 1 3 .. .
2 0 3 1 .. .
1 3 0 2 .. .
3 1 2 0 .. .
... ... ... ... .. .
q − 2 q − 1 q − 4 q − 3 ... q − 1 q − 2 q − 3 q − 4 ...
q−2 q−1 q−4 q−3 .. . 0 2
q−1 q−2 q−3 q−4 . .. . 2 0
If we perform the permutation ρ = γ = (1, 2) on respectively the rows and columns 0 of every latin square of the set {C10 , C20 , . . . , Cq−1 }, we obtain the set {C1 , C2 , . . . , Cq−1 }, which is a reduced set of (q − 1)M OLS(q). In this set, the first latin square is reduced, so we transformed A1 into C1 .
56
C10 7→ C1 =
0 1 2 3 .. .
1 0 3 2 .. .
q−2 q−1
2 3 0 1 .. .
3 2 1 0 .. .
... ... ... ... ...
q−2 q−4 q−1 q−3 .. .
q − 4 q − 1 q − 3 ... q − 3 q − 2 q − 4 ...
q−1 q−3 q−2 q−4 . .. .
0 2
2 0
We see that A1 , B1 and C1 are distinct reduced latin squares. So we have found three different reduced sets of (q − 1)M OLS(q), describing P G(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 to construct a set of (q − 1)M OLS(q), describing P G(2, q). We see that the first element on the second row in each latin square Ai , 1 ≤ i ≤ q − 1, is equal to i and 0 1 .. .
1 2
p−1 .. .
0 .. .
q−p q−p+1 .. .
q−p+1 ...
q−1
q−p
A1 =
... ..
. ... .. . ... ..
. ...
p−1 0 .. .
... ... .. .
q−p q−p+1 .. .
q−p+1 ...
p−2 .. .
... .. . .. .
q−1 .. .
q−p .. .
... .. .
... ... .. .
... ...
...
...
...
q−1 q−p .. . q−2
... ..
. ... .. .
..
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 }. 0 }. If we do this, we obtain a set of (q − 1)M OLS(q) {B10 , B20 , . . . , Bq−1
A1 7→
B10
=
0 1 .. .
1 2
p−1 .. .
0 .. .
q−1 q−p+1 .. .
q−p+1 ...
q−p
q−1
p − 1 ... 0 ... .. ... ... . ... p − 2 ... .. ... ... . . . . . q − p .. q − 1 ... .. .. ... . . ... q − 2 ...
...
q−1 q−p+1 .. .
q−p+1 ...
q−p .. .
q−1 .. .
... ... .. .
... ...
...
...
... q − p q−1 .. ... . ... ...
...
q−2 .. . . ... ... .. . ...
If we perform the permutation ρ = γ = (q − p, q − 1) on respectively the rows and columns 0 of every latin square of the set {B10 , B20 , . . . , Bq−1 }, we obtain the set {B1 , B2 , . . . , Bq−1 }, which is a reduced set of (q − 1)M OLS(q). In this set, the first latin square is reduced, so we transformed A1 into B1 .
57
B10
7→ B1 =
0 1 .. .
1 2
p−1 .. .
0 .. .
q−p q−p+1 .. .
q−1 ...
q−1
q−p+1
... p − 1 ... 0 ... .. ... ... . ... p − 2 ... .. .. .. . . .
...
q−2 q−1 .. .
...
q−p
...
... ... .. . ...
q−p q−p+1 q−1 ... .. .
... ...
q−1 q−p+1 .. .
q−2 .. .
q−1 .. .
... .. .
q−p .. .
... ... .. .
... ... ...
... ... .. .
...
...
.
...
We see that the reduced latin squares B1 and A1 are not the same, so we have found two different reduced sets of (q − 1)M OLS(q), describing P G(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 the 0 set {A1 , A2 , . . . , Aq−1 }. If we do this, we obtain a set of (q−1)M OLS(q) {C10 , C20 , . . . , Cq−1 }.
A1 7→
C10
=
0 1 .. .
1 2
q−p .. .
0 .. .
p−1 q−p+1 .. .
q−p+1 ...
q−1
p−1
... ..
. ... .. .
q−p 0 .. .
... ... ...
...
p − 2 ... .. .. . . .. . q−1 p − 1 ... .. .. . .
...
q−2
...
p−1 q−p+1 .. .
q−p+1 ...
q−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 columns 0 }, we obtain the set {C1 , C2 , . . . , Cq−1 }, of every latin square of the set {C10 , C20 , . . . , Cq−1 which is a reduced set of (q − 1)M OLS(q). In this set, the first latin square is reduced, so we transformed A1 into C1 .
C10
7→ C1 =
0 1 .. .
1 2
p−1 .. .
q−p+1 .. .
q−p q−p+1 .. .
0 ...
q−1
p−1
... ..
.
... .. . ... ..
. ...
p−1 q−p+1 .. .
q−p 0 .. .
q−p+1 ...
... .. .
... ... .. . .. . .. .
q−1 .. .
... .. .
q−1 ... .. .
... ... .. .
p−2 p−1 .. .
p−1 ...
...
...
q−2
...
... ..
.
q−1 p−1 .. .
..
... .. .
..
q−2 ... .. .
. ... .
.
...
We see that A1 , B1 and C1 are distinct reduced latin squares. So we have found three different reduced sets of (q − 1)M OLS(q), describing P G(2, q), in the case that p 6= 2.
58
5.60 Theorem. There are exactly (q−2)!/d different reduced (q−1)M OLS(q), for q = pd a prime power, describing P G(2, q). Proof. Suppose there is a larger group G of permutations on the alphabet which stabilizes the set {A1 , A2 , . . . , Aq−1 }, modulo the same permutations on the rows and the columns of the latin squares. Then this group G is described as in theorem 5.58. If q = 4, then we have that |AΓL1 (q)| = q(q − 1)d = 24 = |Sq |. So there is only one complete reduced set of M OLS(4), describing P G(2, 4). If q > 4, then the first two cases are not possible since there are at least three different complete reduced sets of M OLS(q), describing P G(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 over the subfield GF (pk ) of GF (pd ). But these permutations were all discussed in lemma 5.56 and 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 the latin squares.
5.4.3
Example
To illustrate the previous theorems, we will give an example. Consider the set of M OLS of theorem 5.36. 0 1 2 3 4
1 2 3 4 0
2 3 4 0 1
3 4 0 1 2
4 0 1 2 3
0 2 4 1 3
1 3 0 2 4
2 4 1 3 0
3 0 2 4 1
4 1 3 0 2
0 3 1 4 2
1 4 2 0 3
2 0 3 1 4
3 1 4 2 0
4 2 0 3 1
0 4 3 2 1
1 0 4 3 2
2 1 0 4 3
3 2 1 0 4
4 3 2 1 0
This is a set of 4 mutually orthogonal latin squares of order 5, describing P G(2, 5). We know by theorem 5.47 that there are 20 permutations stabilizing this set. First of all we defined the following permutation on the alphabet ρa0 : GF (5) → GF (5) : x 7→ a0 x, for a0 ∈ GF (5) and a0 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 alphabet stabilizing the set of (4)M OLS(5), describing P G(2, p), modulo the same permutations on 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)M OLS(5), describing P G(2, 5). First of all we had the following set of (4)M OLS(5) 0 1 2 3 4
1 2 3 4 0
2 3 4 0 1
3 4 0 1 2
4 0 1 2 3
0 2 4 1 3
1 3 0 2 4
2 4 1 3 0
3 0 2 4 1
4 1 3 0 2
0 3 1 4 2
1 4 2 0 3
2 0 3 1 4
3 1 4 2 0
4 2 0 3 1
0 4 3 2 1
1 0 4 3 2
2 1 0 4 3
3 2 1 0 4
4 3 2 . 1 0
We will call this set S1 . Now we will construct the other 5 different reduced (4)M OLS(5) by applying some permutations. If we apply the permutation (34) on the alphabet on S1 , modulo the same permutation on the rows and columns, then we obtain the following reduced set of (4)M OLS(5) 0 1 2 3 4
1 2 4 0 3
2 4 3 1 0
3 0 1 4 2
4 3 0 2 1
0 2 3 4 1
1 4 0 3 2
2 3 1 0 4
3 1 4 2 0
4 0 2 1 3
0 4 1 2 3
1 3 2 4 0
2 0 4 3 1
3 2 0 1 4
4 1 3 0 2
0 3 4 1 2
1 0 3 2 4
2 1 0 4 3
3 4 2 0 1
4 2 1 . 3 0
If we apply the permutation (23) on the alphabet on S1 , modulo the same permutation on the rows and columns, then we obtain the following reduced set of (4)M OLS(5) 0 1 2 3 4
1 3 4 2 0
2 4 1 0 3
3 2 0 4 1
4 0 3 1 2
0 3 1 4 2
1 2 3 0 4
2 0 4 3 1
3 4 2 1 0
4 1 0 2 3
0 2 4 1 3
1 4 0 3 2
2 1 3 4 0
3 0 1 2 4
4 3 2 0 1
0 4 3 2 1
1 0 2 4 3
2 3 0 1 4
3 1 4 0 2
4 2 1 . 3 0
If we apply the permutation (234) on the alphabet on S1 , modulo the same permutation on the rows and columns, then we obtain the following reduced set of (4)M OLS(5) 0 1 2 3 4
1 3 0 4 2
2 0 4 1 3
3 4 1 2 0
4 2 3 0 1
0 3 4 2 1
1 4 2 0 3
2 1 3 4 0
3 2 0 1 4
4 0 1 3 2
0 4 3 1 2
1 2 4 3 0
2 3 1 0 4
3 0 2 4 1
4 1 0 2 3
0 2 1 4 3
1 0 3 2 4
2 4 0 3 1
3 1 4 0 2
4 3 2 . 1 0
If we apply the permutation (24) on the alphabet on S1 , modulo the same permutation on the rows and columns, then we obtain the following reduced set of (4)M OLS(5) 0 1 2 3 4
1 4 0 2 3
2 0 3 4 1
3 2 4 1 0
4 3 1 0 2
0 4 3 1 2
1 3 2 4 0
2 1 4 0 3
3 0 1 2 4
4 2 0 3 1
0 3 4 2 1 61
1 2 3 0 4
2 4 1 3 0
3 1 0 4 2
4 0 2 1 3
0 2 1 4 3
1 0 4 3 2
2 3 0 1 4
3 4 2 0 1
4 1 3 . 2 0
If we apply the permutation (243) on the alphabet on S1 , modulo the same permutation on the rows and columns, then we obtain the following reduced set of (4)M OLS(5) 0 1 2 3 4
5.4.4
1 4 3 0 2
2 3 1 4 0
3 0 4 2 1
4 2 0 1 3
0 4 1 2 3
1 2 4 3 0
2 0 3 1 4
3 1 0 4 2
4 3 2 0 1
0 2 3 4 1
1 3 0 2 4
2 1 4 0 3
3 4 2 1 0
4 0 1 3 2
0 3 4 1 2
1 0 2 4 3
2 4 0 3 1
3 2 1 0 4
4 1 3 . 2 0
Further remarks
Does there exist a non-Desarguesian projective plane of prime order p? This question is still unsolved because there is no proof that there does not exist a non-Desarguesian projective plane of prime order p, but nobody has ever found such a projective plane. We would 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 there are (p − 2)! different reduced (p − 1)M OLS(p), for p a prime, describing P G(2, p). By theorem 4.9 we know that there exists a projective plane of order p if and only if there exists a complete set of mutually orthogonal latin squares of order p. We also know that the projective planes P G(2, p) are Desarguesian projective planes. If we throw all these theorems together, we have the following result. 5.61 Corollary. If there are exactly (p − 2)! different reduced (p − 1)M OLS(p), for p a prime, 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)M OLS(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)M OLS(p) is reduced, by definition. In fact, we used reduced sets of M OLS in the previous paragraphs, because we can apply the same permutation on the rows and the columns on each of the latin squares in the set and a permutation on the symbol set in each latin square to obtain a reduced set of M OLS. We will now take a closer look at the set of reduced latin squares of 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 orthogonal mate. Secondly we have the reduced latin squares which have an orthogonal mate. This set can be partitioned in the set C of reduced latin squares of order p which are part of a set of (p − 1)M OLS(p) and the set D, which contains the other reduced latin squares of order 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 that c = (p − 2)!. This would mean that we have already found all the reduced latin squares which are part of a set of (p − 1)M OLS(p) in theorem 5.48. To finish the proof we would have to prove that for all L ∈ C there exists a unique reduced set of (p − 1)M OLS(p), because then we would obtain exactly (p − 2)! different reduced (p − 1)M OLS(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)M OLS(p), for p a prime, but we were not able to find a proof at this point. Therefore we did some research on transversals, because this could be 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 the n symbols of S exactly once. A partial transversal of length k in a latin square of order n on the symbol set S a set of k entries, each from a different row and a different column, such that no two them contain the same symbol.
= of is of
Not every latin square of order n has a transversal. The following latin square is an example of a latin square without a transversal. 0 1 2 3
1 2 3 0
2 3 0 1
3 0 1 2
Suppose we have a latin square that has a certain number of transversals. We will now investigate 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 respectively column 0, 1, . . . , n − 1 on row i0 , i1 , . . . , in−1 , then we will use the following notation in−1 for the transversal, (ai00 , ai11 , . . . , an−1 ), where the element in column 0 is placed in the first position, the element in column 1 is placed in the second position, . . . 6.3 Example. Consider the following latin square of order 4. 0 1 L1 = 2 3
1 0 3 2
64
2 3 0 1
3 2 . 1 0
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 permutation of the upper indices in the notation of the transversals. Consider the permutation ρ = (0, 1) on the rows of L1 , then we obtain the following latin square of order 4. 1 0 Lρ1 = 2 3
0 1 3 2
3 2 0 1
2 3 . 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 a permutation of the elements aikk in the notation of the transversals. Consider the permutation γ = (0, 1) on the columns of L1 , then we obtain the following latin square of order 4. 1 0 Lγ1 = 3 2
0 1 2 3
2 3 0 1
3 2 . 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 permutation of 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. 1 0 Lσ1 = 2 3
0 1 3 2
2 3 1 0
3 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 the number of transversals remains the same. 6.4 Theorem. If a latin square L has n transversals, then any latin square isotopic to L has the same number of transversals. Proof. Suppose that L0 is a latin square isotopic to L. This means that there are three bijections from respectively the rows, the columns and the symbols of L, to respectively 65
the rows, the columns and the symbols of L0 , that map L to L0 . A permutation on the rows corresponds to a permutation of the upper indices in the notation of the transversals. This means that there is a bijection between the upper indices in the notation of each transversal of L and the upper indices in the notation of the corresponding transversal of L0 . A permutation on the columns corresponds to a permutation of the elements aikk in the notation of the transversals. This means that there is a bijection between the elements aikk in the notation of each transversal of L and the elements aill in the notation of the corresponding transversal of L0 . A permutation on the alphabet corresponds to a permutation of the elements ak in the notation of the transversals. This means that there is a bijection between the elements ak in the notation of each transversal of L and the elements al in the notation of the corresponding transversal of L0 . The number of transversals remains the same in these three cases, which means that L and L0 have the same number of transversals. We know by the previous theorem that the number of transversals is isotopy class invariant. If we investigate whether the number of transversals is affected if we map a latin square L to any conjugate of L, then we will know if the number of transversals is also main 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 and the (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 columns into the rows, which means that in the notation of the transversals there is a bijection between the position of the elements aikk and the upper indices in the notation of the transversals. The number of transversals in a latin square and in his transpose is the same. We can also understand this by looking at the definition of a transversal, which is a set of n entries, one from each row and column, containing each of the n symbols of the symbol set S exactly once. By transposing the latin square, a transversal is mapped to another corresponding transversal, because we will still have a set of n entries, one from each 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 1 = 3 2
66
3 2 0 1
1 0 2 3
2 3 . 1 0
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 2 = 1 3
1 3 0 2
3 1 2 0
2 0 . 3 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 that the number of transversals remains the same. Finally we will now look at the (3, 1, 2)−conjugate.
L(3,1,2)
0 2 = 3 1
2 0 1 3
1 3 2 0
3 1 . 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 that the number of transversals remains the same. To understand what happens, we will use the line array notation presented in example 2.15. This is the line array of L(1,2,3) . R: 0 C: 0 E: 0
0 1 3
0 2 1
0 1 3 0 2 1
1 1 2
1 2 0
1 2 3 0 3 3
2 1 0
2 2 2
2 3 3 0 1 2
3 1 1
3 2 3
3 3 0
What happens to a random transversal? Let us consider the transversal (00 , 21 , 33 , 12 ) of L(1,2,3) . In this transversal we have on row 0 and column 0 the element 0, on row 1 and column 1 the element 2, on row 3 and column 2 the element 3 and on row 2 and column 3 the element 1. A conjugate of L(1,2,3) corresponds with a permutation of the rows in the line array. This means we have a permutation in the notation of the transversals between the upper indices, the position of the elements aki i and the elements ai . In fact the role of rows, columns and elements is permuted. If we consider the (2, 3, 1)−conjugate, our transversal (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 class isotopic 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 transversals, but it is more useful to know how many transversals a latin square has. We have the following 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 of order n, the number of transversals is congruent to n mod 2. In 1990, Balasubramanian proved 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 conjecture is weakened to conjecture 6.7. The reason why it is interesting to investigate the number of transversals of a latin square is the following theorem. 6.9 Theorem. A latin square has an orthogonal mate if and only if it can be decomposed into pairwise disjoint transversals. There is a lot more research done on transversals, but we will not go into further details here. More information can be found in [4], [9] and [32]. In the final chapter we will discuss 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 ) in which n2 distinct numbers from a set S are arranged, such that the sum of the numbers in each row, each column and both the diagonals is the same. This sum is called the magic sum, which will be denoted by s. In our definition we have the following properties. 1.
Pn−1
ai,j = s for all j,
2.
Pn−1
ai,j = s for all i,
3.
Pn−1
ai,i = s,
4.
Pn−1
ai,n−1−i = s.
i=0
j=0
i=0 i=0
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 case the 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. So we get: n2 1X n2 (n2 + 1) n(n2 + 1) i= = . n i=1 2n 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 numbers in 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 all the 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, let us 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 row k + 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. This magic 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 a magic square given. Lo Shu supposedly means ‘river map’ and the legend says that it was first seen by emperor Yu on the back of a turtle. The appearance of the turtle had to do with sacrifices to the river god. A Shu Ching in 650 BCE1 makes a reference to the ‘river map’. This magic square is also mentioned in the I Ching, but the first clear reference is made in 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 1
Before the Common Era
70
found in [18], [25] and [30]. The first magic square of order 4 is found in India and is created by the mathematician Nagarajuna in the first century. A well known early magic square of order 4 is also found in India in Khajuraho in the Parshvanath Jain temple included in an 11th or 12th century inscription. This magic square is not only a normal magic square, but also a pandiagonal magic square. The following picture is taken by Dr. Rainer Typke [31].
We see in this picture the following magic square. 7 12 1 14 2 13 8 11 16 3 10 5 9 6 15 4 The magic sum of this magic square is equal to 34 and not only the sum of the numbers in each row, each column and both the diagonals is equal to this magic sum, but also the sum of the numbers on each broken diagonal. We notice that the sum of the numbers in all the (2 × 2) subsquares also equals the magic sum. This gives us the possibility to create a ‘magic carpet’, by copying the original magic square and put them side to side to create a ‘carpet’. Each subsquare of order 4 in this carpet is a normal pandiagonal magic square and the sum of the numbers in each subsquare of order 2 is equal to the magic sum 34.
71
7 2 16 9 7 2 16 9 7 2 16 9
12 13 3 6 12 13 3 6 12 13 3 6
1 8 10 15 1 8 10 15 1 8 10 15
14 11 5 4 14 11 5 4 14 11 5 4
7 2 16 9 7 2 16 9 7 2 16 9
12 13 3 6 12 13 3 6 12 13 3 6
1 8 10 15 1 8 10 15 1 8 10 15
14 11 5 4 14 11 5 4 14 11 5 4
7 2 16 9 7 2 16 9 7 2 16 9
12 13 3 6 12 13 3 6 12 13 3 6
1 8 10 15 1 8 10 15 1 8 10 15
14 11 5 4 14 11 5 4 14 11 5 4
In Europe there are a few mathematicians who investigated magic squares. A few of them are Heinrich Cornelius Agrippa, Albrecht D¨ urer, Leonhard Euler, Pierre de Fermat and Benjamin 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 certain types 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 in 1849 in ‘Commentationes arithmeticae’ [11]. We will first give some examples of different orders.
7.3.1
First case: n = 3
Let us consider the following two latin squares of order 3, a b c γ L1 = b c a and L2 = α c a b β
β γ α
α β . γ
Suppose that a = 0, b = 3, c = 6, α = 3, β = 2 and γ = 1. Now we have that 0 3 6 1 2 3 L1 = 3 6 0 and L2 = 3 1 2 . 6 0 3 2 3 1 Let us now consider L1 + L2 . This gives us the following result, which is a normal semimagic square. 72
1 5 9 M1 = 6 7 2 . 8 3 4 If we would have that 0 1 2 0 1 2 N1 = 1 2 0 and N2 = 2 0 1 , 2 0 1 1 2 0 then we can also write M = 3N1 + N2 + J3 , where 1 1 1 J3 = 1 1 1 . 1 1 1 If we work with the latin squares N1 and N2 , then we can also construct the following normal semi-magic square 1 5 9 M2 = N1 + 3N2 + J3 = 8 3 4 , 6 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, 0 1 L1 = 2 3
1 0 3 2
2 3 0 1
3 0 1 2 2 3 and L2 = 1 3 2 0 1 0
2 0 1 3
3 1 . 0 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 1 1 J4 = 1 1
1 1 1 1
1 1 1 1
1 1 . 1 1
We obtain the following normal semi-magic squares 1 6 11 16 1 6 11 16 7 4 13 10 10 13 4 7 M1 = and M2 = . 12 15 2 5 15 12 5 2 14 9 8 3 8 3 14 9 73
7.3.3
General case
It is now also possible to construct a normal semi-magic square of order 5, order 7, order 8, . . . in the same way as before. 7.5 Theorem. Suppose N (n) ≥ 2, where n is the order of the latin squares, then it is possible 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}, from the set of M OLS(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 latin squares, we know that this means that if these two latin squares are superimposed, we would 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 to n2 . We defined M as M = nL1 +L2 +Jn , which is a matrix. Let us now look at the element on row k and column l. We find (M )k,l = (nL1 + L2 + Jn )k,l = ni + j + 1, where i = (L1 )k,l and j = (L2 )k,l . We know that L1 and L2 are mutually orthogonal, which means that there 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, thus ni = 0, n, 2n, . . . , (n − 1)n and ni + 1 = 1, n + 1, 2n + 1, . . . , (n − 1)n + 1. Finally, we find that 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 , but it also needs to have the property of a semi-magic square that the sum of the numbers in each 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 the numbers from 0 to n − 1 because L1 and L2 are latin squares. This means that the sum of the numbers in each row and each column is equal to n(0+1+. . .+n−1)+(0+1+. . .+n− 2 2 2 = n(n −n+n+1) = n(n2+1) . In 1)+n = n(0+1+. . .+n−1)+(1+2+. . .+n) = n (n−1)+n(n+1) 2 2 the beginning of this chapter we saw that this is the magic sum of a normal (semi-)magic square 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 n instead 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 0 2 1 1 0 2 L1 = 2 1 0 and L2 = 2 1 0 . 1 0 2 0 2 1 74
We can now construct a normal magic square as follows, M = 3L1 + L2 + J3 , where 1 1 1 J3 = 1 1 1 . 1 1 1 We obtain the following normal magic square 2 7 M = 9 5 4 3
7.3.5
6 1 . 8
Second case (bis): n = 4
Let us consider the following mutually orthogonal latin squares 0 3 L3 = 1 2
1 2 0 3
2 1 3 0
3 0 1 0 2 3 and L4 = 2 3 2 1 1 0
2 0 1 3
3 1 . 0 2
It is now possible to construct a normal magic square as follows, M = 4L3 + L4 + J4 , where 1 1 J4 = 1 1
1 1 1 1
1 1 1 1
1 1 . 1 1
We obtain the following normal magic square 1 6 15 12 M = 8 3 10 13
7.3.6
11 16 5 2 . 14 9 4 7
General case
7.6 Definition. A latin square of order n on the symbol set S = {0, 1, . . . , n − 1} is a double-diagonal latin square if both the main diagonal and the antidiagonal contain all the elements from the symbol set S. By theorem 7.5 we know that M = nL1 + L2 + Jn is a normal semi-magic square, where L1 and L2 are different latin squares from the set of M OLS(n) and Jn is the n × n matrix in which each entry is equal to 1. The following theorem tells us when M is a normal magic square. 75
7.7 Theorem. Suppose N (n) ≥ 2, where n is the order of the latin squares, then it is possible 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 M OLS(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 constant . and equal to n−1 2 Proof. We know by theorem 7.5 that M is a normal semi-magic square, so the only thing to prove is that the sum of the numbers in both the diagonals of M is equal to the magic sum if L1 and L2 are both double-diagonal latin squares, or if n is odd and two of the . main 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 both the main diagonal and the antidiagonal have the numbers from 0 to n−1. This means that the sum of the numbers in both the diagonals is equal to n(0+1+. . .+n−1)+(0+1+. . .+ 2 2 2 n − 1) + n = n(0 + 1 + . . . + n − 1) + (1 + 2 . . . + n) = n (n−1)+n(n+1) = n(n −n+n+1) = n(n2+1) . 2 2 In the beginning of this chapter we saw that this is the magic sum of a normal magic square of order n. As a conclusion we have that if L1 and L2 are both double-diagonal latin 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 antidiagonals . Suppose that the main diagonal of L1 is of L1 and L2 are constant and equal to n−1 2 n−1 constant and equal to 2 , then the main diagonal of L2 can not be constant, because L1 and L2 are mutually orthogonal. This means that the antidiagonal of L2 has to be . The antidiagonal of L1 has to contain all the elements of S, constant and equal to n−1 2 and otherwise because we know that the antidiagonal of L2 is constant and equal to n−1 2 there would be a conflict with the definition of mutually orthogonal latin squares. Finally the 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(n n−1 )+ 2 n(n2 −n+n+1) n(n2 +1) n2 (n−1)+n(n+1) (0+1+. . .+n−1)+n = = = and the sum of the numbers 2 2 2 2 in the antidiagonal of M is equal to n(0 + 1 + . . . + n − 1) + n n−1 + n = n (n−1)+n(n+1) = 2 2 n(n2 −n+n+1) n(n2 +1) = . This proves our theorem. 2 2 This chapter was just a brief chapter on magic squares. There is a lot more theory and it is possible to formulate a lot more theorems on the construction of certain magic squares with mutually orthogonal latin squares. It is also possible to find magic squares that can not be constructed with mutually orthogonal latin squares. As a conclusion we can say that there is still a lot more to discover about magic squares and latin squares, but we will leave this for the reader.
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