ORTHOGONAL SETS OF LATIN SQUARES AND. CLASS-r HYPERCUBES GENERATED BY FINITE ALGEBRAIC SYSTEMS. A Dissertation in. Mathematics. Daniel R.

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1 The Pennsylvania State University The Graduate School Eberly College of Science ORTHOGONAL SETS OF LATIN SQUARES AND CLASS-r HYPERCUBES GENERATED BY FINITE ALGEBRAIC SYSTEMS A Dissertation in Mathematics by Daniel R. Droz c 2016 Daniel R. Droz Submitted in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy May 2016

2 The dissertation of Daniel R. Droz was reviewed and approved by the following: Gary L. Mullen Professor of Mathematics Dissertation Adviser Chair of Committee W. Dale Brownawell Distinguished Professor Emeritus of Mathematics James L. Rosenberger Professor of Statistics, Director of SCC and Online Programs James A. Sellers Professor of Mathematics, Assoc. Head for Undergraduate Studies Yuxi Zheng Head of the Department of Mathematics Signatures are on file in the Graduate School ii

3 ABSTRACT Latin squares are combinatorial objects which have applications in some various and slightly surprising settings. A latin square of order n is a square array on n symbols such that each symbol occurs once in each row and column. Two latin squares are called orthogonal when superimposing them gives each of the n 2 ordered pairs of symbols exactly once. It is well known that if q is a prime power, the squares formed from the polynomials ax + y, a F q form q 1 latin squares of order q which are mutually orthogonal (each pair of squares is orthogonal). In this dissertation, we explore four problems relating to latin squares and other objects with similar properties, especially focusing on constructing large mutually orthogonal sets. We explore the extent to which sets of mutually orthogonal latin squares, hypercubes, and frequency squares can be obtained by polynomials over finite fields. We are able to rescue two classical conjectures of Euler and MacNeish which are false for general latin squares but which are true when out attention is restricted to polynomial-generated squares only. We also introduce the theory of the finite algebraic structures called uniform cyclic neofields, and explore the construction of sets of latin squares which are nearly orthogonal. Our main result with be to give a simple construction of large sets of such nearly orthogonal squares for all even orders n where n 1 is prime. We then examine a new generalization of latin squares called class-r hypercubes which feature a larger alphabet (n r rather than n symbols). We give solutions to several open problems in this area, most notably the construction of large mutually orthogonal sets for r 3. As our last topic, we give some partial progress toward solutions about a long-standing problem on the computability of partially filled latin cubes. Although the immediate extension of the famous Evans conjecture seems fail for latin cubes of type 1 although it is true for latin squares, we explore what weaker versions of this conjecture can be said to hold. iii

4 Contents List of Figures vi 1 Introductory Material Latin Squares - Motivation Latin Squares - Definitons and Examples Finite Fields Latin Hypercubes Constant Frequency Squares Précis of New Results Generation of Orthogonal Latin Objects By Finite Rings Prior Results Higher Dimensional Objects over Z/ n Frequency Squares over Z/ n Latin Squares over Finite Rings Generation of Latin Squares by Neofields Uniform Cyclic Neofields - Definition and Examples Existence and Construction of N (u) q Latin Squares over Neofields Aggregate Neofields Latin Squares over Aggregate Neofields Future Directions Class-r Hypercubes Definition Basic Orthogonality Results Basics of PDNS Sets PDNS Sets at r = 1, PDNS Sets for General r: First Case iv

5 4.6 Counting the Size of PDNS* Sets for r = 3, PDNS Sets for General r: Second Case (Frobenius) PDNS Sets for General r: Third Case Directions for Further Study Blocking Sets in Partial Latin Cubes Partial Latin Squares Blocking Sets Blocking Sets for d = 3, j = Extending Evans to d = 3, j = Bibliography 71 v

6 List of Figures 3.1 The neofields N (2) 10 and N(5) Suitable characters for small q vi

7 Chapter 1 Introductory Material In this first chapter, we lay out some basic definitions and standard theorems about latin squares and finite fields to which we will refer throughout the paper. After this chapter, the main bulk of the new work, in Chapters 2 through 5, are essentially independent of each other and may be read in any order. Throughout the entire paper, all variables will be assumed to be positive integers unless otherwise specified. The letter p always denotes a prime, and q is typically a prime power (except in chapter 4), unless otherwise specified. In this chapter, we will mostly omit proofs since all these results are considered standard. 1.1 Latin Squares - Motivation Latin squares have been studied since the classical age of mathematics, at first merely as a curiosity. Recently, we have seen that these combinatorial designs have applications in several areas; see a standard reference like [12] or [4] for more examples and in-depth discussion of these. Sets of orthogonal latin squares are used in the designs of large-scale experiments, where several different factors, each of which is allowed to vary over a range of values, are to be compared pairwise; latin-square constructions allow that not every permutation of experimental factors needs to be tested, but by selecting a subset of certain permutations we can still run all possible pairwise comparisons. For example, let us say that we wish to test the growth rates of eleven kinds of plants using eleven different soil compositions, eleven watering schedules, eleven sources of light, and eleven climates. A brute-force setup would require 11 5 (over 123 million) plants, which would 1

8 carry a prohibitive price tag. Doing any pair of factors would require only 11 2 = 121 setups, so we might also run ten separate experiments of 121 setups each. However, using a set of mutually orthogonal latin squares, we can use only 121 setups, if carefully chosen, to test all possible pairs of all five factors. If we wanted to test triple correlations, we can use a similar construction to form 11 3 which would test all possible triples simultaneously. Latin squares are also used in the field of coding theory, where they provide one excellent way to formulate a class of error-correcting codes. This can be applied to signals sent over a noisy channel, where even if some parts of the message are corrupted the original clear message can be reconstructed a large fraction of the time. In this paper we are not really directly concerned with either of these applications; rather we look at the theory lying behind them and extend some of the standard results of the theory to other related contexts. Let us lay the groundwork for these extensions by a review of the basic theory. 1.2 Latin Squares - Definitons and Examples Definition 1.1 A latin square of order n is an n n square array on n symbols such that in each row and each column, each of the symbols occurs exactly once. Typically, we use the integers 0 through n 1 for the n symbols. Here is an example of a latin square of order 6: Definition 1.2 Given two latin squares of order n, form the list of n 2 ordered pairs consisting of the entries in the same location from the first and second square respectively. (We refer to this procedure as superimposing the two squares.) The two squares are called orthogonal if this list consists of all n 2 possible pairs on the n symbols each occurring exactly once. Two squares are called r-orthogonal if the total number of distinct pairs in this list is r. We must have have n r n 2 and the squares are orthogonal if r = n 2. 2

9 This is best illustrated by an example. First, these two squares are not orthogonal: , The list of pairs starts with (working across the top rows): (0, 0), (1, 3), (2, 1), (3, 2),... but in this list, the pair (0, 0) occurs twice and the pair (0, 3) never occurs. In fact these squares are only 8-orthogonal. However, each pair of squares in the following set is an orthogonal pair: , , , A set of squares like this are called mutually orthogonal latin squares, which is usually shortened to the acronym MOLS for convenience. We would say the above is a set of four MOLS of order 5. The great classical problem for latin squares asks: given an integer n, what is the largest possible size of a set of MOLS of order n? We use N(n) to denote the answer to this question. We can prove combinatorially that: Proposition 1.3 If n 2, N(n) n 1. Proof. (Sketch.) Suppose we have a set of MOLS of order n. In each square, send the symbols through a permutation so that the top row consists of the symbols 0, 1, 2,... n 1 in order [In the example above, this was already done.] This permutation will not alter the mutual orthogonality of the squares. Now consider the symbols in the first cell of the second row of all the squares. None of these symbols can be 0, for the first cell in the first row of all squares is 0. Also, no two of these symbols can be the same, for between any two squares the pairs of the form (a, a) occur in the top row, so they cannot occur again elsewhere. Therefore, we cannot have more than n 1 of these symbols in the first cell of the second row, and hence no more than n 1 squares. 3

10 Determining the value of N(n) in general is one of the most important open problems in combinatorics at the moment. Right now, what we observe is that it seems to be possible to achieve n 1 MOLS of order n only when n is a prime power; when n is not a prime power we believe that it is never possible to achieve n 1 MOLS. This conjecture is called the Prime Power Conjecture, and has been compared to a new Fermat problem for modern mathematics (see [16]). Despite the interest in this problem, the value of N(n) is known only for prime powers and n = 6; even n = 10 is unknown although it is conjectured to be 2 or 3. Conjecture 1.4 (Prime Power) If n 2, N(n) = n 1 if and only if n is a prime power. In the next section, we will exhibit the construction that shows N(q) = q 1 for prime powers q. 1.3 Finite Fields Finite fields underlie much of the basic theory of latin squares and their relatives. In this work we will cite most algebraic facts about finite fields without proof; here we give a very brief summary of the terminology and basic propositions which we will need below. Proposition If a field is finite, its order is a prime power. 2. For each prime power q, any two finite fields of order q are isomorphic. 3. If p is prime, the ring of integers modulo p is a field, which we will call F p. 4. If q = p k is a prime power, we may form the finite field of size q as the splitting field of the polynomial x q x over F p. We call this field F q. 5. The multiplicative group F q is cyclic of order q 1. The generators of this group are called primitive elements of the field. 6. If q = p k, the field F q is a vector space of dimension k over F p ; if α is a primitive element of F q, the set {1, α, α 2,..., α k 1 } forms a basis for this vector space. 7. If q r are prime powers, the field F q is a subfield of F r if and only if q = p k and r = p l where k l. 4

11 8. If q = p k, the function F (x) = x p is a linear transformation on the vector space F q over F p ; this is called the Frobenius map. 9. If α F q, we have α q = α and in fact α q 1 = 1 if α If q 1 (mod k), the polynomial x k 1 splits completely over F q ; so that q 1 k nonzero elements have k distinct k-th roots in F q, and the remaining elements have none. In particular, all primitive elements have no k-th roots. 11. If (q 1, k) = 1, the polynomial (x k 1)/(x 1) is irreducible in F q so that each element of F q has exactly one k-th root in F q. 12. Lagrange Interpolation. If f : F q F q is any function, there is a unique polynomial p(x) F q [x] of degree at most q 1 for which p(α) = f(α) for all α F q. For the proofs of these results, we refer to any standard reference; the most complete of which are [14] and [17]. We can use finite fields to construct latin squares. Suppose q is a prime power; label the rows and columns of a q q array with the elements of the field F q so that each cell of the array is associated to a pair of coordinates (x, y) with x, y F q. Then we select some suitable function f : (F q ) 2 F q, which by Lagrange interpolation we may as well assume is a polynomial. Then in cell (x, y) we place the field element f(x, y). This will form a latin square if f(x, y) is a so-called local permutation polynomial; that is, the actions x f(x, α) and y f(α, y) are both bijections for any choice of α F q. It is not necessary to look at very complicated polynomials to form latin squares; in fact, the addition table of a finite field must be a latin square by definition so the polynomial f(x, y) = x + y always forms a latin square. In fact, so does f(x, y) = ax + y for any a F q. Even better, these linear polynomials not only form many examples of latin squares, they also form a set of mutually orthogonal latin squares: 5

12 Theorem 1.6 If q is a prime power, and a b F q, the two polynomials ax + y and bx + y from F q [x, y] generate orthogonal latin squares of order q. Proof. We can accomplish this by noting that for any pair (α, β) (F q ) 2, the simultaneous equations have precisely one solution, namely ax + y = α, bx + y = β, x = α β a b, aβ bα y = a b. These are the coordinates of the cell at which the pair (α, β) occurs when the two squares are superimposed. This means that each such pair occurs at exactly one place in the squares, meaning they fulfill the definiton of orthogonality. Another way [ of saying ] this is to note that the linear mapping given a 1 by the matrix is invertible, meaning there is a bijective relation b 1 between locations (x, y) and pairs (α, β). We will have more to say about this notion in Chapter 4. Corollary 1.7 If q is a prime power, let f a (x, y) = ax + y. Then the set: {f a a F q}, gives a set of q 1 MOLS of order q. Therefore, N(q) = q 1. This method of construction achieves the maximum possible size of a set of MOLS of order q. The ideas of the last two sections foreshadow much of the theory of related objects: we use combinatorics to prove an upper bound on the size of a set of mutually orthogonal objects, then use finite fields and linear polynomials to achieve that bound. 1.4 Latin Hypercubes We will now extend the definition of latin squares to higher-dimensional arrays which are called latin hypercubes. 6

13 Definition 1.8 A latin hypercube of dimension d, type j, and order n is an n n (d times) array on n symbols in which, fixing any j of the coordinates, each symbol occurs exactly n d j 1 times in this subarray. Two such hypercubes are orthogonal when superimposing them has each possible pair occurring exactly n d 2 times; a set of mutually orthogonal hypercubes is usually shortened to MOHC. The most intuitive case of this definition comes when the type j = d 1; then in each row in any direction, each symbol occurs once. Here is a cube (d = 3) of type 2: (By this, we mean that the three squares are to be stacked atop each other in the third dimension.) This cube also has the property that if we exchange the second and third squares given here, the resulting cube is orthogonal to the original (each pair occurs three times when the two are superimposed). When j is smaller, we only need each symbol to occur an equal number of times in each subarray of dimension d j; here is a cube with type j = 1: Notice that a hypercube of type j automatically fulfills the criteria for types less than j also. Latin squares fit this definition with d = 2, j = 1. We will denote the size of the largest possible set of MOHC of dimension d, type j, and order n by N(d, j; n). We can acheive the following upper bound by combinatorial means (see [12, p.45]): Proposition 1.9 If n, d 2 and 0 j d 1, ( N(d, j; n) 1 j ( ) d n d )(n 1) k. n 1 k In particular, if j = d 1, k=0 N(d, d 1; n) (n 1) d 1. Working over finite fields, we construct sets of MOHC that achieve these bound when n = q is a prime power: 7

14 Theorem 1.10 Suppose q is a prime power and d and j with 1 j d 1 are integers. We let a = (a 1, a 2,... a d ) be a d-tuple of elements of F q and define: f a (x 1, x 2,..., x d ) = a 1 x 1 + a 2 x a d x d. Then if S is the set of the polynomials f a where a is restricted to those d-tuples whose last nonzero entry is 1 and which have at most d j 1 zero entries; then S generates a set of MOHC of dimension d and type j which achieve the bound given in Prop Again, see [12] for a detailed proof of this fact; also, a proof rather similar to this occurs below in Chapter 2 (see the proof of Thm. 2.6). Essentially, we must check that each such a yields a hypercube of the correct type, that each pair of hypercubes is orthogonal, and lastly count the number of d-tuples that fulfill the given conditions. As part of a theme we will see continuing especially in Chapter 4, the counting part of the argument may well be the most involved step. As the complexity of our definitions increase, proving the latin properties and the orthogonality properties is usually fairly direct, but accomplishing the counting gets more difficult quickly. 1.5 Constant Frequency Squares Frequency squares are a generalization of latin squares which do not involve an extra dimension; instead, the size of the square array is larger than the alphabet of symbols and each symbol is then required to occur a prescribed number of times in each row and column. In this work, we will focus on a special case of this idea: Definition 1.11 A (constant) frequency square of type F (rn; r) is an rn rn array on n distinct symbols in which each symbol occurs exactly r times in each row and column. Two such squares are orthogonal if, when superimposing them, each pair occurs exactly r 2 times. A set of mutually orthogonal frequency squares is abbreviated MOFS. Here is a frequency square with r = n = 3: 8

15 Finite fields allow us to easily produce sets of MOFS of the type F (n k ; n k 1 ) in those cases where n = q is a prime power. We will denote the largest size of a set of such MOFS by N F (n k, n k 1, n). Now we recall the standard combinatorial result (from [12, p.65]): Proposition 1.12 For any integer n 2, ( n N F (n k, n k 1 k 1 ) 2, n). n 1 If n = q is a prime power, we may achieve this bound using finite fields. Since the frequency square is a q k q k array, we will label the rows and columns with k-tuples of elements of F q, meaning that each location is the frequency square is described by a 2k-tuple, the first k entries determining the row, the last k determining the column. Theorem 1.13 If q is a prime power, and k 1 is an integer, let a = (a 1, a 2,... a 2k ) be a 2k-tuple, and define f a (x 1, x 2,..., x 2k ) = a 1 x 1 + a 2 x a 2k x 2k. Then if S is the set of the polynomials f a where a is restricted to those 2k-tuples whose last nonzero entry is 1 and which have at least one nonzero entry among the first k and among the last k entries; then S generates a set of MOFS of type F (q k ; q k 1 ) which achieve the bound given in Prop We will again omit the proof of this theorem, although the proof of Thm. 2.9 below is quite similar. In this particular case, the counting argument is not quite so delicate; there are q k 1 of each k-tuple which is not all zero; joining two of these we have (q k 1) 2 of the 2k-tuples which fulfill the condition except for the restriction that the last nonzero entry is 1; of these, exactly 1 q 1 of them have each possible last entry, so multiplying we achieve the correct number. 9

16 1.6 Précis of New Results In the rest of this work, we provide new results on four different problems which arise from generalizations of the constructions given above. Each of these four chapters can be read independently of one another. In Chapters 2 and 3, we investigate how algebraic systems other than finite fields can be used to construct latin squares and their relatives, and to what extent they can produce orthogonal sets. In Chapter 4, we introduce a new relative of the latin hypercube and discuss its construction over finite fields. In Chapter 5, we discuss some partial results in the area of completion of partial latin squares. In Chapter 2, we explore the extent to which finite rings which are not also fields can be used to construct MOLS. We will conclude by rescuing some false conjectures of Euler and MacNeish by restricting them to finite rings, where they are true. In Chapter 3, we introduce the concept of a neofield and notice some interesting new orthogonality results. Here, we are able to construct large sets of mutually nearly-orthogonal latin squares (only 2n 2 missing pairs) for n where n 1 is prime. In Chapter 4, we define a new version of the latin hypercube, by extending the alphabet. We explore the properties of these so-called class-r hypercubes, whose theory has a bit of extra subtlety than the fairly clean theory of the latin objects described above. We are able to substantially extend previously published constructions of these objects to work in more generality and in more cases. In Chapter 5, we introduce the theory of completion of partial latin squares, and give several preliminary results on extending Evans conjecture to latin cubes. In general, throughout the rest of the paper, propositions and theorems given without a citation are new results. In those cases where a citation is given but the proof is also present, the proof is by a new method unless explicitly stated that the given argument is a summary of the cited proof method. 10

17 Chapter 2 Generation of Orthogonal Latin Objects By Finite Rings As we have seen, using polynomials over finite fields to generate latin squares is a standard, well-understood method of producing complete sets of MOLS. When we extend to hypercubes and frequency squares, polynomials over finite fields extend in a natural way in these cases also. However, when the desired order of latin objects is not a prime power, finite fields have little to contribute. In this chapter, we explore the exact extent to which polynomials over finite rings can generate sets of MOLS and related objects. We begin by focusing on the most obvious ring of size n, namely the integers modulo n. We start here since this has been an area of past research; indeed, the most recent result is due to Baliff and featured in his doctoral thesis. In the final section, we will extend these results to general finite rings. This material has been published separately, see [6]. 2.1 Prior Results We denote the integers modulo n by Z/ n. We begin by posing the question: what is the maximum number of MOLS of order n that can be generated by polynomials in Z/ n [x, y]? We denote the answer to this question N P (n). Rivest [18] first considered this problem for n = 2 w, and determined that there is not even a pair of orthogonal latin squares that may be formed in 11

18 such a way; in our notation, N P (2 w ) = 1. Ballif [1] extended the result by proving that if p is the smallest prime dividing n, N P (n) = p 1. The method used in [1] does not extend immediately to either higher dimensional objects or to frequency squares. We proceed to give a slightly different proof which will extend to these cases. First a lemma: Lemma 2.1 If m n, N P (n) N(m). Proof. Suppose f(x, y) generates a latin square, L n, modulo n. Consider mapping the entries in L n modulo m. Because f(x + k 1 m, y + k 2 m) f(x, y) (mod m), we will obtain identical m m subsquares repeated n/m times in each direction; call this smaller square L m. Now under this mapping, each element of Z/ m is the image of exactly n/m elements of Z/ n, so considering the rows and columns of L n (mod m) each entry in L m must occur exactly once in each row and column, implying that L m is itself latin. We will call L m the m-square of L n. What is more, if we have two orthogonal latin squares L 1 n, L 2 n of order n generated by polynomials, we claim their m-squares, L 1 m, L 2 m must also be orthogonal. Suppose that some pair of elements (α, β) does not occur in the m-squares. Then L 1 n and L 2 n (mod m) do not have any instance of this pair (since they consist of copies of L 1 m and L 2 m), which means that no pair of the form (α + k 1 m, β + k 2 m) appears across L 1 n and L 2 n, contradicting our assumption that they were orthogonal. Hence any set of MOLS generated by polynomials modulo n creates by its m-squares a set of MOLS of order m, implying that N P (n) N(m) for any m n. Now by taking m = p in Lemma 2.1 we obtain: Theorem 2.2 If p is the smallest prime dividing n, N P (n) = p 1. Proof. First, we claim that the polynomials ax + y for a = 1, 2,..., p 1 form MOLS. To see this, suppose that we take two distinct polynomials from this group, ax + y and bx + y. Suppose that the same pair occurs more than once, at the coordinates (x 1, y 1 ) and (x 2, y 2 ). That is: ax 1 + y 1 ax 2 + y 2 (mod n). bx 1 + y 1 bx 2 + y 2 (mod n). 12

19 However, subtracting these imply (a b)(x 1 x 2 ) 0(mod n). Since x 1 x 2 (mod n), we must have (a b) a nontrivial divisor of n. But a and b are both between 1 and p 1, so a b < p, which is a contradiction since p is the smallest prime dividing n. Therefore no pair occurs more than once between the two squares; since there are only n 2 possible pairs and n 2 pairs are actually involved, each one occurs exactly once, implying that the two squares are orthogonal. Letting m = p in Lemma 2.1 and recalling that N(p) = p 1, we see that this attains an upper bound. We can find these m-squares in the upper-left-hand corner of the order-n squares after reducing mod m, and as we shall see, this idea will carry over into higher dimensions. 2.2 Higher Dimensional Objects over Z/ n We now proceed to extend our result from MOLS to MOHC of dimension d and type j. (Recall Definiton 1.8.) We label all the coordinates of a latin hypercube of dimension d and order n with the integers 0 to n 1. We denote the maximum size of an orthogonal set of any such hypercubes by N(d, j; n). If we restrict our attention to just those hypercubes generated by polynomials in Z/ n [x 1, x 2,..., x d ], we denote the maximum size of such an orthogonal set by N P (d, j; n). We shall proceed as above, first establishing an upper bound and then exhibiting a set of polynomials giving orthogonal hypercubes which attains it. Lemma 2.3 If m n, N P (d, j; n) N(d, j; m). Proof. Now the cells are labeled by d-tuples, (x 1, x 2,..., x d ) where each x i runs from 0 to n 1. With notation as above, H n is a latin hypercube of type j generated by the polynomial f(x 1,... x d ); we reduce H n modulo m and since f(x 1 + k 1 m,..., x d + k d m) f(x 1,..., x d ) (mod m), we again have identical order-m subhypercubes, each repeated n/m times in each direction; we call this smaller hypercube H m. By considering H n (mod m) and fixing j coordinates, we see that again H m is itself a latin hypercube of type j. This we will call the m-hypercube. 13

20 Also, if Hn, 1 Hn 2 are two such hypercubes generated by polynomials which are orthogonal, we claim their m-hypercubes are also orthogonal. Suppose that the pair (α, β) occurs t times across Hm 1 and Hm. 2 Then this pair occurs t (n/m) d times in (Hn, 1 Hn) 2 (mod m), so pairs of the type (α+k 1 m, β+k 2 m) occur this many times across Hn 1 and Hn. 2 Also, there are (n/m) 2 such pairs, each of which occur n d 2 times since the two are assumed orthogonal. We ) d = n d 2 ( n m) 2, so t = m d 2, which is the correct number for have t ( n m the m-hypercubes to be orthogonal. Therefore given a set of MOHC of dimension d and type j and generated by polynomials modulo n, we generate another set of MOHC of the same type of order m, implying N P (d, j; n) N(d, j; m) for any m n. Now we recall the following standard result (also given above as Prop. 1.9): Proposition 2.4 If n, d 2 and 0 j d 1, In particular, if j = d 1, ( N(d, j; n) 1 n d n 1 j k=0 ( ) d )(n 1) k. (2.1) k N(d, d 1; n) (n 1) d 1. (2.2) We have equality at least when n is a prime power. We will begin with the special case j = d 1. Theorem 2.5 If p is the smallest prime dividing n, We consider the (p 1) d 1 hypercubes described by the poly- Proof. nomials N P (d, d 1; n) = (p 1) d 1. a 1 x 1 + a 2 x a d 1 x d 1 + x d, where we allow a i = 1, 2,..., p 1 for each i = 1, 2,..., d 1 and evaluate modulo n. These form latin hypercubes of type d 1 since if we fix any d 1 coordinates, we have essentially a linear polynomial in the remaining one such as a i x i + b (we understand that a d = 1); since by construction a i is relatively prime to n, this will run through a complete sets of residues modulo n as x i does. 14

21 We next claim these are orthogonal to each other. Consider two hypercubes, one formed with the coefficients (a 1,..., a d 1, 1) and the other by the coefficients (b 1,..., b d 1, 1). To be orthogonal, each pair of symbols must occur exactly n d 2 times. Choose any pair (α, β) and consider the number of solutions to the congruences a 1 x 1 + a 2 x a d 1 x d 1 + x d α (mod n), b 1 x 1 + b 2 x b d 1 x d 1 + x d β (mod n). Since we know (a 1,..., a d 1 ) and (b 1,..., b d 1 ) are not identical, choose any index i at which they differ and choose any of the n d 2 possibilities for the values of each of x 1,..., x d 1, with x i excluded. Fix these and subtract to the other side; that is let α = α a 1 x 1 (a i x i ) a d 1 x d 1. and similarly for β. Now we have congruences of the form: a i x i + x d α (mod n), b i x i + x d β (mod n). We see by subtracting the equations and noting that 0 < a i b i < p so that a i b i is invertible that these have exactly one solution in x i and x d. This gives exactly n d 2 solutions to the original congruences. We therefore have exhibited a set of (p 1) d 1 MOHC, so that N P (d, d 1; n) (p 1) d 1. Now, setting m = p in Lemma 2.3 and applying equation (2.2) we have N P (d, d 1; n) N(d, d 1; p) = (p 1) d 1, which together with the above set of polynomials establishes the theorem. 15

22 Now we extend this result to general type j, 0 j d 1. Theorem 2.6 If p is the smallest prime dividing n, we have ( N P (d, j; n) = 1 j ( ) d p d )(p 1) k. p 1 k Proof. First, consider the set of polynomials given by a 1 x 1 + +a d x d, where we have k=0 (a 1,..., a d ) = (,,...,, 1, 0, 0,..., 0), where each represents one of 0, 1,..., p 1, and where the 1 may occur at any place, including the first or the last. This set is designed so that no two polynomials will be constant multiples of each other. By considering the position of the 1, we see that there are a total of 1+p+p 2 + +p d 1 = pd 1 p 1 of these polynomials. For the sake of convenience, we will denote this set by S. Also, let S k be the subset of this set with exactly k + 1 nonzero coefficients. Note that S d 1 is exactly the set of polynomials in the last proof, and that the S k are distinct. Any polynomial in S k will produce a hypercube of type k and no higher. The set S k consists of exactly ( d k+1) (p 1) k polynomials. Given type j, we will consider the set d 1 k=j S k, which will produce hypercubes of type j (or possibly higher), which consists of j 1 S k=0 S k = pd j 1 1 p 1 k=0 ( ) ( d (p 1) k = 1 p d 1 k + 1 p 1 j k=1 ( ) d )(p 1) k k hypercubes, which is exactly the right number as stated in the theorem. We now need only show that these are mutually orthogonal to complete the proof. (Note that these are the same polynomials generating the standard set of MOHC of order p, see Thm. 1.10) Let (a 1,..., a d ) and (b 1,..., b d ) be the coefficients of two distinct polynomials from the above set, and solve the system of congruences d a i x i α (mod n), i=1 d b i x i β (mod n). i=1 16

23 We consider two cases. CASE I: The last nonzero entry occurs at the same place for both (a 1,..., a d ) and (b 1,..., b d ), call this place a j = b j = 1 (it equals 1 by the construction of the set). Then for some j < j, a j b j ; fix the other d 2 of the x i in any of the n d 2 ways and move these constants to the other side so we have a j x j + x j α (mod n), b j x j + x j β (mod n), which we know has a unique solution as in the proof of Theorem 2.5. Thus there are exactly n d 2 instances of each pair and so the hypercubes are orthogonal. CASE II: The last nonzero entries in (a 1,..., a d ) and (b 1,..., b d ) occur at different places; suppose this place is later in the first than in the second and let these places be a j = 1 and b j = 1, j < j. Then b j = 0, so again fixing the other d 2 places, we have a j x j + x j α (mod n), x j β (mod n), which also clearly has a unique solution. Therefore again the two squares are orthogonal. By utilizing the bound in Lemma 2.3 and combining with equation (2.1), we know that this set of MOHC is the largest possible over polynomials modulo n, proving the theorem. 2.3 Frequency Squares over Z/ n We will now turn our attention to those (constant) frequency squares generated over Z/ n. It turns out that the most natural extension of our result to frequency squares is, in analogy to the standard result for prime powers, to consider frequency squares of type F (n k ; n k 1 ), so that there are still exactly n symbols. We may label the rows and columns with elements of (Z/ n ) k, and evaluate with polynomials in 2k variables, mod n. We shall denote the size of the largest set of MOFS of this type by N F (n k, n k 1, n). If we restrict to those sets of MOFS generated by polynomials modulo n, we shall use the notation N P F (n k, n k 1, n). 17

24 Lemma 2.7 If m n, N P F (n k, n k 1, n) N F (m k, m k 1, m). Proof. We proceed exactly as before: label both rows and columns with k-tuples over Z/ n. We note that reducing the entries of a polynomial generated frequency square over the 2k variables modulo m produces an m- F-square in the region where each coordinate is less than m; orthogonality of large squares implies the orthogonality of the m-f-squares (by the same proof as in Lemma 2.3 replacing d by 2k), so we have N P F (n k, n k 1, n) N F (m k, m k 1, m). We restate the result of Chapter 1 in Prop and Thm for convenience: Proposition 2.8 If q is a prime power, N F (q k, q k 1, q) = ( q k 1 ) 2. q 1 Theorem 2.9 If p is the smallest prime dividing n then N P F (n k, n k 1, n) = (pk 1) 2 p 1. Proof. Consider the polynomials where we restrict and a 1 x a k x k + a k+1 x k a 2k x 2k, (a 1,..., a k ) {0, 1,... p 1} k \ {(0, 0,..., 0)}, (a k+1,..., a 2k ) = (,,...,, 1, 0, 0,..., 0), where each represents an element of {0, 1,..., p 1}, and the 1 may occur at any place, including the last or the first, (just as with the set S in the hypercubes proof above). Note that there are therefore (pk 1) 2 p 1 such polynomials overall. These are frequency squares since at least one of the elements in each coordinate k-tuple for both the rows and columns is a unit in Z/ n, so holding the other variables fixed the polynomial will run through all the values modulo n as this variable does, meaning each row and column takes on each possible value an equal number of times. 18

25 Lastly, we claim these squares are orthogonal. The proof goes in the same way as the one for hypercubes: taking cases over the position of the last nonzero entry. Combining Lemma 2.7 and Proposition 1.12 with m = p, we have N P F (n k, n k 1, n) (pk 1) 2 p 1, and this set attains this bound, so the proof is complete. 2.4 Latin Squares over Finite Rings We have established that if we restrict our attention to those latin squares which are constructed by bivariate polynomials over a specific ring: Z/ n, we can form sets of MOLS which are as large as possible over this particular ring. A natural question is then if we widen our attention to other finite rings of cardinality n, perhaps we can do better than p 1 (p being the smallest prime dividing n). Throughout this section, we assume R is a finite commutative ring with identity. We introduce the following notation: label the rows and columns of a latin square with the elements of R, and denote by N P (R) the size of the largest possible set of MOLS constructed by polynomials over R. For R = F q, the finite field of cardinality q, we know that N P (F q ) = q 1. Above we showed N P (Z/ n ) = p 1. The result below includes both of these. First we must establish some basic facts about finite commutative rings: Proposition 2.10 If R is a finite commutative ring with identity, and r R, the following are equivalent: (i) r is invertible; (ii) r is not a zero divisor; (iii) r does not lie in any of the maximal ideals of R. Proof. (i) (ii) is immediate. (ii) (iii): Assume r is in one of the maximal ideals of R; then the action x rx is not onto; we have rx = ry for some x y so r(x y) = 0 and r is a zero divisor. (iii) (i): Suppose r is not invertible. Then the ideal generated by r does not include 1 and so is proper; hence r lies inside some maximal ideal of R. We also introduce what we shall call the minimal index of R; we define this to be the smallest index of any proper ideal of R. Note that the ideal A 19

26 for which [R : A] is as small as possible must be a maximal ideal, implying that the quotient ring R/A is a field, and since finite, must be of prime power cardinality, implying that the minimal index of a ring is always a prime power dividing n = R. Since the decomposition of the additive group of a finite ring into p- groups induces a ring direct sum decomposition (see [15, p.2]), we must have an ideal of index q where q is the smallest prime power exactly dividing n, so that this q is the largest possible value of the minimal index for any finite ring of cardinality n. We now apply this notion of minimal index to the search for MOLS generated over R[x, y]: Theorem 2.11 If m is the minimal index of R, we have N P (R) = m 1. Proof. Let A = A 1, A 2,..., A k be the maximal ideals of R (since R is finite, there are only a finite number of them). Let m i = [R : A i ], and we will assume that m = m 1 m 2 m k. Now order the elements of R in such a way that the first m elements are drawn from distinct cosets modulo A, and each set of m elements after that are also drawn from all the cosets of A in the same order. Then use this order to label the rows and columns of a latin square. Suppose f(x, y) R[x, y] generates a latin square; then for any a 1, a 2 A, f(x + a 1, y + a 2 ) f(x, y) (mod A). This means that when the entries in the latin square are reduced by mapping onto the quotient ring R/A, the same m m array is repeated n/m times in both directions. This means that this m-square is itself latin. If we have two such latin squares which are orthogonal, again by mapping them onto R/A, we can see that their m-squares must also be orthogonal (if no pair (α + A, β + A) appears across the two m-squares, the pair (α, β) could not have occurred across the original squares). This means that sets of MOLS from polynomials over R generate MOLS of order m, of which there cannot be more than m 1. Now we need to build a set of m 1 MOLS over R. To do this, for each A i take m 1 nonzero representatives of distinct cosets b i,1,..., b i,m 1. Now, invoking the Chinese Remainder Theorem, we find c j such that c j b i,j (mod A i ) for each A i. Then we claim that the polynomials c j x + y generate a set of MOLS. Each c j is in a nonzero coset for each maximal ideal, so by Proposition 2.10 it is invertible and c j x + y certainly produces a latin square. Suppose 20

27 the polynomials c j x + y and c j x + y produce the same pair occurring twice, at the coordinates (x 1, y 1 ) and (x 2, y 2 ). Then and subtracting, c j x 1 + y 1 = c j x 2 + y 2, c j x 1 + y 1 = c j x 2 + y 2, (c j c j )(x 1 x 2 ) = 0. Since x 1 x 2, we have that c j c j is a zero divisor; however, we see that since c j and c j are in different cosets for every maximal ideal, their difference cannot lie in any maximal ideal, so by Proposition 2.10 it cannot be a zero divisor, a contradiction. Therefore, these squares are orthogonal. This concludes the proof, since we have shown there cannot be more than m 1 MOLS and have constructed a set of exactly m 1 MOLS. We note that for n not a prime power, we may (possibly) improve the number of MOLS produced by taking a different ring to Z/ n. If n = q 1 q k where each q i is a prime power, and q 1 < < q k, then taking R = F q1 F qk, we have that m = q 1 so that N P (R) = q 1 1 which is (possibly) an improvement over p 1. In fact, this construction is entirely equivalent to the Kronecker product construction of latin squares of order n. Moreover, by the discussion above, the minimal index m of any ring of cardinality n cannot be bigger then q 1 so this ring is optimal. This discussion establishes: Corollary 2.12 Let q be the smallest prime power exactly dividing n. Then for any ring R of cardinality n, N P (R) q 1. This looks quite similar to the MacNeish conjecture: Conjecture 2.13 (MacNeish) Let q be the smallest prime power exactly dividing n. Then N(n) = q 1. This was itself a generalization of a conjecture made by Euler: Conjecture 2.14 (Euler) If n 2 (mod 4), then N(n) = 1. Both are now known to be false (since we can exhibit a pair of orthogonal squares of order 10, implying N(10) 2), and are actually suspected to be false for all n except prime powers and 6. However, Corollary 2.12 implies that these conjectures are weakly true if our attention is restricted to those latin squares generated by polynomials over finite rings. In short, when these conjectures fail, counterexamples cannot be generated by polynomials. The extensions of the results of sections 2 and 3 are: 21

28 Theorem 2.15 If R has cardinality n and minimal index m, (a) the maximum possible number of hypercubes of order n, dimension d, and type j, generated from elements of R[x 1, x 2,..., x d ], is ( 1 m d m 1 j k=0 ( ) d )(m 1) k. k (b) the maximum possible number of frequency squares of type F (n k ; n k 1 ) generated over R[x 1, x 2,..., x 2k ], is (mk 1) 2 m 1. The proofs use the elements c j from the proof of Theorem 2.11 in place of 1, 2,..., p 1 in the proofs of the corresponding results on hypercubes or frequency squares. Part (a) implies that the d-dimensional versions of the Euler and Mac- Neish conjectures, generally known to be false for all n where the ordinary two-dimensional versions fail (see [13]), also have their counterexamples generated in ways other than by polynomials over finite rings. 22

29 Chapter 3 Generation of Latin Squares by Neofields In this chapter, we introduce a finite algebraic object called a neofield which is quite new to the mathematical literature and explore the ways in which we can construct sets of latin squares using an analogue of polynomials. Neofields are much like fields except we do not require that addition be associative or commutative. This allows to to have sizes which are not prime powers. See [9] for an exposition on the construction of finite neofields in more generality than is given here, and see [8] for a lengthy exposition of cyclic neofields generally. The paper [10] explored the extent to which sets of polynomials over a certain class of neofields gave sets of latin squares which, though not orthogonal, had a property of being partially orthogonal. The proof given there was descriptive, but not algebraic; and therefore was not easily generalized to higher-dimensional objects. We will here provide a new algebraic proof (in section 3) which should lay the groundwork for future explorations. Also, using the algebraic language we have built up, we will be able to slightly generalize the construction (section 4) to make a distinct method which gives sets of latin squares with a vastly improved partial orthogonality condition (section 5). 3.1 Uniform Cyclic Neofields - Definition and Examples A neofield is an algebraic structure with many of the same assumptions as a field except that we remove what is often considered to be the most basic 23

30 assumption: associativity of addition. To be precise: Definition 3.1 A set N equipped with two operations (called addition) and (called multiplication) is a neofield if: Addition has a two-sided identity called 0 and each element has a two-sided additive inverse. For any a N, the actions x a x and x x a are bijections. The non-zero elements of N form a group under multiplication; the multiplicative identity is called 1 and 1 0. (In all our examples below, the multiplicative group will be cyclic, but this is not required.) Multiplication distributes over addition (from both sides, if the multiplication is noncommutative). A neofield N (with cyclic multiplication) will be called commutative if its addition is commutative. In this work, however, we confine our attention only to a certain class of finite neofields of even order. Definition 3.2 Let q be even, and let u be chosen so that 2 u q 2 and (u, q 1) = (u 1, q 1) = 1. A uniform cyclic neofield of order q and character u is the neofield with the following properties: There are q elements, incuding additive and multiplicative identities 0 and 1. The multiplicative group of non-zero elements is cyclic of order q 1; we will call the generator θ so that the the elements of the neofield may be listed as {0, 1, θ, θ 2,..., θ q 2 }. The characteristic is 2; that is a a = 0 for any element a. We have, for all 1 k q 2, 1 θ k = θ uk. The fact that multiplication distributes over addition uniquely defines all other additions. This neofield will be denoted N (u) q. 24

31 We now give two examples of neofields when q = 10. In this case, the only possible choices for u are 2, 5, and 8. Given in Figure 3.1 (on page 25) are two addition tables. These tables are given in logarithmic notation, that is, the entry of a number a means the neofield element θ a ; the entry represents the neofield element 0. In general, we define the discrete logarithm of an element α N (u) q : { a if θ log θ (α) = a = α, 0 a q 2, if α = Figure 3.1: The addition tables for the neofields N (2) 10 and N(5) 10 (respectively) in logarithmic form. Notice that this means is the additive identity; while 0 represents the multiplicative identity θ 0 = 1. Notice that the table for N (5) 10 is commutative, while that for N(2) is not.

32 In fact, the addition table for N (8) 10 is the transpose of that of N(2) 10, so we are now fully equipped to do arithmetic in any uniform cyclic neofield of order 10. For example, within N (5) 10, if we were asked to compute (θ θ4 )(θ 5 1), we would look the two additions up in the table; the first comes out θ 7, the second θ 7 also; multiplying, we get θ 14 = θ 5 since θ 9 = 1; see below. Asked to do the same computation within N (2) 10, we get θ7 for the first addition and θ 4 for the second; their product is then θ 2. Also, notice that both addition tables form latin squares. We will return to this idea in section 3 below. 3.2 Existence and Construction of N (u) q. We need to notice some basic properties of N (u) q. First we will be a bit clearer about the last part of the previous definition. Given two distinct nonzero elements of N (u) q, denote them as θ a, θ b. Then: ( θ a θ b = θ a 1 θ b a) = θ a θ u(b a) = θ (1 u)a+ub. (3.1) Next we notice a property identical to that for finite fields: Proposition 3.3 For any α N (u) q, If α 0, α q 1 = 1. α q = α. If α 0 and θ is a generator of the multiplicative group, α is also a generator if and only if α = θ k where (k, q 1) = 1. Proof. Each of these statements relies only on the fact that the multiplicative group of N (u) q is cyclic of order q 1; they are proved in exactly the same fashion as the corresponding finite-fields results. Next, we note that the criteria given for choosing u at the beginning of Definition 3.2 are necessary. In order to streamline our future discussion, we introduce the following terminology: Definition 3.4 If q is even, a u with 2 u q 2 is called a suitable character for q if (u, q 1) = (u 1, q 1) = 1. We will use the notation u q. 26

33 Another way to look at this is to look more carefully at equation (3.1). Consider the action x θ a x; we know it must be transitive. Selecting x = 0 or x = θ a, we produce the two images θ a and 0 respectively. Therefore, for any b a, 0 b q 2, we must have that or in short (1 u)a + ub a (mod q 1), u(b a) 0 (mod q 1). which is definitely true if (u, q 1) = 1 and false otherwise. Similarly, considering the action x x θ b, allowing x = θ a to vary we have that for any a b, 0 a q 2, or: (1 u)a + ub b (mod q 1), (u 1)(b a) 0 (mod q 1). which is definitely true if (u 1, q 1) = 1 and false otherwise. This proves that for any u q, we certainly have that N (u) q satisfies all the criteria to be a neofield. Moreover, there are always neofields for any even q 4: Proposition 3.5 Given any even q 4, we may form the neofields N (u) q for u = 2 and for u = q/2. Proof. We need only prove that 2 q and q/2 q for any even q. It is clear that 2 and 1 are both prime to q 1; as are q/2 and q/2 1, so the definition is satisfied. Another result about existence of various suitable characters u: Proposition 3.6 Given any even q 4 and u = u 0 q, we also have (q u 0 ) q. Moreover, the two neofields N (u 0) q and N (q u 0) q are identical except for the order of addition; to be precise, if α β = γ in N (u 0) q, then β α = γ in N (q u 0) q. Proof. We have, by the linearity of the g.c.d., (q u 0, q 1) = ((q 1) (q u 0 ), q 1) = (u 0 1, q 1) = 1, 27