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
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
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 n2 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 ∈ Fq 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 (nr 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
Contents
List of Figures
vi
1 Introductory Material
1.1 Latin Squares - Motivation . . . . . . . .
1.2 Latin Squares - Definitons and Examples
1.3 Finite Fields . . . . . . . . . . . . . . . .
1.4 Latin Hypercubes . . . . . . . . . . . . . .
1.5 Constant Frequency Squares . . . . . . . .
1.6 Précis of New Results . . . . . . . . . . .
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1
1
2
4
6
8
10
2 Generation of Orthogonal Latin Objects By Finite
2.1 Prior Results . . . . . . . . . . . . . . . . . . . . . .
2.2 Higher Dimensional Objects over Z/hni . . . . . . .
2.3 Frequency Squares over Z/hni . . . . . . . . . . . . .
2.4 Latin Squares over Finite Rings . . . . . . . . . . . .
Rings
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11
11
13
17
19
3 Generation of Latin Squares by Neofields
3.1 Uniform Cyclic Neofields - Definition and Examples
(u)
3.2 Existence and Construction of Nq . . . . . . . . . .
3.3 Latin Squares over Neofields . . . . . . . . . . . . . .
3.4 Aggregate Neofields . . . . . . . . . . . . . . . . . .
3.5 Latin Squares over Aggregate Neofields . . . . . . .
3.6 Future Directions . . . . . . . . . . . . . . . . . . . .
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23
23
26
30
33
34
37
4 Class-r Hypercubes.
4.1 Definition . . . . . . . . . . . . . . .
4.2 Basic Orthogonality Results . . . . .
4.3 Basics of PDNS Sets . . . . . . . . .
4.4 PDNS Sets at r = 1, 2 . . . . . . . .
4.5 PDNS Sets for General r: First Case
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39
39
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43
44
46
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4.6
4.7
4.8
4.9
Counting the Size of PDNS* Sets for r = 3, 4 . . .
PDNS Sets for General r: Second Case (Frobenius)
PDNS Sets for General r: Third Case. . . . . . . .
Directions for Further Study . . . . . . . . . . . . .
5 Blocking Sets in Partial Latin Cubes
5.1 Partial Latin Squares . . . . . . . . .
5.2 Blocking Sets . . . . . . . . . . . . .
5.3 Blocking Sets for d = 3, j = 1. . . .
5.4 Extending Evans to d = 3, j = 2. . .
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56
56
59
60
63
71
v
List of Figures
3.1
3.2
(2)
(5)
The neofields N10 and N10 . . . . . . . . . . . . . . . . . . .
Suitable characters for small q . . . . . . . . . . . . . . . . . .
vi
25
29
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 115 (over 123 million) plants, which would
1
carry a prohibitive price tag. Doing any pair of factors would require only
112 = 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 113 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:
0
3
2
1
4
5
1
4
5
2
0
3
3
1
0
5
2
4
2
0
4
3
5
1
4
5
1
0
3
2
5
2
3
4
1
0
Definition 1.2 Given two latin squares of order n, form the list of n2 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 n2 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 ≤ n2 and the squares are
orthogonal if r = n2 .
2
This is best illustrated by an example. First, these two squares are not
orthogonal:
0
1
2
3
1
2
3
0
2
3
0
1
0
3
2
1
3
0
,
1
2
3
0
1
2
1
2
0
3
2
1
3
0
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:
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
0
3
1
4
2
4
1
3 ,
0
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
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
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 1.5
1. 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 Fp .
4. If q = pk is a prime power, we may form the finite field of size q as
the splitting field of the polynomial xq − x over Fp . We call this field
Fq .
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 = pk , the field Fq is a vector space of dimension k over Fp ; if α
is a primitive element of Fq , the set {1, α, α2 , . . . , αk−1 } forms a basis
for this vector space.
7. If q ≤ r are prime powers, the field Fq is a subfield of Fr if and only
if q = pk and r = pl where k|l.
4
8. If q = pk , the function F (x) = xp is a linear transformation on the
vector space Fq over Fp ; this is called the Frobenius map.
9. If α ∈ Fq , we have αq = α and in fact αq−1 = 1 if α 6= 0.
10. If q ≡ 1 (mod k), the polynomial xk − 1 splits completely over Fq ; so
that q−1
k nonzero elements have k distinct k-th roots in Fq , and the
remaining elements have none. In particular, all primitive elements
have no k-th roots.
11. If (q − 1, k) = 1, the polynomial (xk − 1)/(x − 1) is irreducible in Fq
so that each element of Fq has exactly one k-th root in Fq .
12. Lagrange Interpolation. If f : Fq → Fq is any function, there is
a unique polynomial p(x) ∈ Fq [x] of degree at most q − 1 for which
p(α) = f (α) for all α ∈ Fq .
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 Fq so that each cell of the array is associated to a pair of coordinates
(x, y) with x, y ∈ Fq . Then we select some suitable function f : (Fq )2 → Fq ,
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 7→ f (x, α) and y 7→ f (α, y) are both bijections for any choice of
α ∈ Fq .
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
Theorem 1.6 If q is a prime power, and a 6= b ∈ Fq , the two polynomials
ax + y and bx + y from Fq [x, y] generate orthogonal latin squares of order q.
Proof.
We can accomplish this by noting that for any pair (α, β) ∈
2
(Fq ) , the simultaneous equations
ax + y = α,
bx + y = β,
have precisely one solution, namely
x=
α−β
,
a−b
y=
aβ − bα
.
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
this is to note that the linear mapping given
of saying
a 1
is invertible, meaning there is a bijective relation
by the matrix
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 fa (x, y) = ax + y. Then the set:
{fa |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
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 nd−j−1 times in this subarray. Two
such hypercubes are orthogonal when superimposing them has each possible
pair occurring exactly nd−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:
0 1 2 2 0 1 1 2 0
1 2 0 0 1 2 2 0 1
2 0 1 1 2 0 0 1 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:
0 0 1 2 1 0 2 2 1
1 2 0 1 2 0 0 1 2
2 1 2 1 0 2 0 0 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,
1
N (d, j; n) ≤
n−1
j X
d
d
n −
(n − 1)k
k
!
.
k=0
In particular, if j = d − 1,
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
Theorem 1.10 Suppose q is a prime power and d and j with 1 ≤ j ≤ d − 1
are integers. We let a = (a1 , a2 , . . . ad ) be a d-tuple of elements of Fq and
define:
fa (x1 , x2 , . . . , xd ) = a1 x1 + a2 x2 + · · · + ad xd .
Then if S is the set of the polynomials fa 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. 1.9.
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 r2 times. A set of mutually orthogonal
frequency squares is abbreviated MOFS.
Here is a frequency square with r = n = 3:
8
0
0
1
2
0
1
2
1
2
0
1
2
0
1
0
2
2
1
1
1
0
1
0
2
2
0
2
1
1
2
2
0
2
0
1
0
2
2
0
0
1
2
1
0
1
0
0
1
2
1
0
1
2
2
2
2
1
0
2
1
0
1
0
2
2
0
1
2
1
1
0
0
1
0
2
1
2
0
0
2
1
Finite fields allow us to easily produce sets of MOFS of the type F (nk ; nk−1 )
in those cases where n = q is a prime power. We will denote the largest size
of a set of such MOFS by NF (nk , nk−1 , n).
Now we recall the standard combinatorial result (from [12, p.65]):
Proposition 1.12 For any integer n ≥ 2,
k
k−1
NF (n , n
2
nk − 1
, 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 Fq , 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 =
(a1 , a2 , . . . a2k ) be a 2k-tuple, and define
fa (x1 , x2 , . . . , x2k ) = a1 x1 + a2 x2 + · · · + a2k x2k .
Then if S is the set of the polynomials fa 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. 1.12.
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,
1
exactly q−1
of them have each possible last entry, so multiplying we achieve
the correct number.
9
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
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/hni.
We begin by posing the question: what is the maximum number of
MOLS of order n that can be generated by polynomials in Z/hni[x, y]? We
denote the answer to this question NP (n).
Rivest [18] first considered this problem for n = 2w , and determined that
there is not even a pair of orthogonal latin squares that may be formed in
11
such a way; in our notation, NP (2w ) = 1. Ballif [1] extended the result by
proving that if p is the smallest prime dividing n, NP (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, NP (n) ≤ N (m).
Proof. Suppose f (x, y) generates a latin square, Ln , modulo n. Consider mapping the entries in Ln modulo m. Because
f (x + k1 m, y + k2 m) ≡ f (x, y) (mod m) ,
we will obtain identical m × m subsquares repeated n/m times in each direction; call this smaller square Lm . Now under this mapping, each element
of Z/hmi is the image of exactly n/m elements of Z/hni, so considering the
rows and columns of Ln (mod m) each entry in Lm must occur exactly once
in each row and column, implying that Lm is itself latin. We will call Lm
the m-square of Ln .
What is more, if we have two orthogonal latin squares L1n , L2n of order
n generated by polynomials, we claim their m-squares, L1m , L2m must also
be orthogonal. Suppose that some pair of elements (α, β) does not occur in
the m-squares. Then L1n and L2n (mod m) do not have any instance of this
pair (since they consist of copies of L1m and L2m ), which means that no pair
of the form (α + k1 m, β + k2 m) appears across L1n and L2n , 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 NP (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, NP (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 (x1 , y1 ) and (x2 , y2 ). That is:
ax1 + y1 ≡ ax2 + y2 (mod n).
bx1 + y1 ≡ bx2 + y2 (mod n).
12
However, subtracting these imply
(a − b)(x1 − x2 ) ≡ 0(mod n).
Since x1 6≡ x2 (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 n2 possible pairs and n2
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/hni
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/hni[x1 , x2 , . . . , xd ], we denote the maximum size of such an
orthogonal set by NP (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, NP (d, j; n) ≤ N (d, j; m).
Proof. Now the cells are labeled by d-tuples, (x1 , x2 , . . . , xd ) where each
xi runs from 0 to n − 1. With notation as above, Hn is a latin hypercube of
type j generated by the polynomial f (x1 , . . . xd ); we reduce Hn modulo m
and since
f (x1 + k1 m, . . . , xd + kd m) ≡ f (x1 , . . . , xd ) (mod m) ,
we again have identical order-m subhypercubes, each repeated n/m times in
each direction; we call this smaller hypercube Hm . By considering Hn (mod m)
and fixing j coordinates, we see that again Hm is itself a latin hypercube of
type j. This we will call the m-hypercube.
13
Also, if Hn1 , Hn2 are two such hypercubes generated by polynomials which
are orthogonal, we claim their m-hypercubes are also orthogonal. Suppose
1 and H 2 . Then this pair occurs
that the pair (α, β) occurs t times across Hm
m
t·(n/m)d times in (Hn1 , Hn2 ) (mod m), so pairs of the type (α+k1 m, β +k2 m)
occur this many times across Hn1 and Hn2 . Also, there are (n/m)2 such pairs,
each of which occur nd−2 times since the two are assumed orthogonal. We
n d
n 2
have t · m
= nd−2 · m
, so t = md−2 , which is the correct number for
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 NP (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,
1
N (d, j; n) ≤
n−1
j X
d
d
n −
(n − 1)k
k
!
.
(2.1)
k=0
In particular, if j = d − 1,
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,
NP (d, d − 1; n) = (p − 1)d−1 .
Proof.
nomials
We consider the (p − 1)d−1 hypercubes described by the polya1 x1 + a2 x2 + · · · + ad−1 xd−1 + xd ,
where we allow ai = 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 ai xi + b (we understand that ad = 1); since by construction ai
is relatively prime to n, this will run through a complete sets of residues
modulo n as xi does.
14
We next claim these are orthogonal to each other. Consider two hypercubes, one formed with the coefficients (a1 , . . . , ad−1 , 1) and the other by the
coefficients (b1 , . . . , bd−1 , 1). To be orthogonal, each pair of symbols must
occur exactly nd−2 times. Choose any pair (α, β) and consider the number
of solutions to the congruences
a1 x1 + a2 x2 + · · · + ad−1 xd−1 + xd ≡ α (mod n) ,
b1 x1 + b2 x2 + · · · + bd−1 xd−1 + xd ≡ β (mod n) .
Since we know (a1 , . . . , ad−1 ) and (b1 , . . . , bd−1 ) are not identical, choose any
index i at which they differ and choose any of the nd−2 possibilities for the
values of each of x1 , . . . , xd−1 , with xi excluded. Fix these and subtract to
the other side; that is let
α0 = α − a1 x1 − · · · − (ai xi )∗ − · · · − ad−1 xd−1 .
and similarly for β 0 . Now we have congruences of the form:
ai xi + xd ≡ α0 (mod n) ,
bi xi + xd ≡ β 0 (mod n) .
We see by subtracting the equations and noting that 0 < |ai −bi | < p so that
ai − bi is invertible that these have exactly one solution in xi and xd . This
gives exactly nd−2 solutions to the original congruences. We therefore have
exhibited a set of (p − 1)d−1 MOHC, so that NP (d, d − 1; n) ≥ (p − 1)d−1 .
Now, setting m = p in Lemma 2.3 and applying equation (2.2) we have
NP (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
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
!
j X
1
d
k
d
NP (d, j; n) =
(p − 1) .
p −
k
p−1
k=0
Proof. First, consider the set of polynomials given by a1 x1 + · · · + ad xd ,
where we have
(a1 , . . . , ad ) = (∗, ∗, . . . , ∗, 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
d −1
position of the 1, we see that there are a total of 1+p+p2 +· · ·+pd−1 = pp−1
of these polynomials. For the sake of convenience, we will denote this set
by S. Also, let Sk be the subset of this set with exactly k + 1 nonzero
coefficients. Note that Sd−1 is exactly the set of polynomials in the last
proof, and that the Sk are distinct.
Any polynomial in Sk will produce
of type k and no higher.
a hypercube
d
(p − 1)k polynomials. Given type j, we
The set Sk consists of exactly k+1
S
will consider the set d−1
k=j Sk , which will produce hypercubes of type j (or
possibly higher), which consists of
j−1 !
j−1 j [ pd − 1 X
X
d
1
d
|S|−
Sk =
−
(p−1)k =
pd − 1 −
(p − 1)k
p−1
k+1
p−1
k
k=0
k=0
k=1
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 (a1 , . . . , ad ) and (b1 , . . . , bd ) be the coefficients of two distinct polynomials from the above set, and solve the system of congruences
d
X
ai xi ≡ α (mod n) ,
i=1
d
X
bi xi ≡ β (mod n) .
i=1
16
We consider two cases.
CASE I: The last nonzero entry occurs at the same place for both
(a1 , . . . , ad ) and (b1 , . . . , bd ), call this place aj = bj = 1 (it equals 1 by
the construction of the set). Then for some j 0 < j, aj 0 6= bj 0 ; fix the other
d − 2 of the xi in any of the nd−2 ways and move these constants to the
other side so we have
aj 0 xj 0 + xj ≡ α0 (mod n) ,
bj 0 xj 0 + xj ≡ β 0 (mod n) ,
which we know has a unique solution as in the proof of Theorem 2.5. Thus
there are exactly nd−2 instances of each pair and so the hypercubes are
orthogonal.
CASE II: The last nonzero entries in (a1 , . . . , ad ) and (b1 , . . . , bd ) occur
at different places; suppose this place is later in the first than in the second
and let these places be aj = 1 and bj 0 = 1, j 0 < j. Then bj = 0, so again
fixing the other d − 2 places, we have
aj 0 xj 0 + xj ≡ α0 (mod n) ,
xj 0 ≡ β 0 (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/hni
We will now turn our attention to those (constant) frequency squares generated over Z/hni.
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 (nk ; nk−1 ), so that there are still exactly n symbols. We may label the rows and columns with elements of (Z/hni)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 NF (nk , nk−1 , n). If we restrict to
those sets of MOFS generated by polynomials modulo n, we shall use the
notation NP F (nk , nk−1 , n).
17
Lemma 2.7 If m|n, NP F (nk , nk−1 , n) ≤ NF (mk , mk−1 , m).
Proof.
We proceed exactly as before: label both rows and columns
with k-tuples over Z/hni. We note that reducing the entries of a polynomial
generated frequency square over the 2k variables modulo m produces an mF-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 NP F (nk , nk−1 , n) ≤
NF (mk , mk−1 , m). We restate the result of Chapter 1 in Prop. 1.12 and Thm. 1.13 for
convenience:
Proposition 2.8 If q is a prime power,
k
NF (q , q
k−1
qk − 1
, q) =
q−1
2
.
Theorem 2.9 If p is the smallest prime dividing n then
NP F (nk , nk−1 , n) =
Proof.
(pk − 1)2
.
p−1
Consider the polynomials
a1 x1 + · · · + ak xk + ak+1 xk+1 + · · · + a2k x2k ,
where we restrict
(a1 , . . . , ak ) ∈ {0, 1, . . . p − 1}k \ {(0, 0, . . . , 0)},
and
(ak+1 , . . . , a2k ) = (∗, ∗, . . . , ∗, 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
k −1)2
the hypercubes proof above). Note that there are therefore (pp−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/hni, 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
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
k −1)2
NP F (nk , nk−1 , n) ≤ (pp−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/hni,
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 NP (R) the size of the largest possible set of MOLS constructed
by polynomials over R. For R = Fq , the finite field of cardinality q, we know
that NP (Fq ) = q − 1. Above we showed NP (Z/hni) = 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 7→ rx is not onto; we have rx = ry for some x 6= 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
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 pgroups 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 NP (R) = m − 1.
Proof. Let A = A1 , A2 , . . . , Ak be the maximal ideals of R (since R is
finite, there are only a finite number of them). Let mi = [R : Ai ], and we
will assume that m = m1 ≤ m2 ≤ · · · ≤ mk .
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 a1 , a2 ∈ A,
f (x + a1 , y + a2 ) ≡ 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 Ai take m − 1 nonzero representatives of distinct cosets bi,1 , . . . , bi,m−1 .
Now, invoking the Chinese Remainder Theorem, we find cj such that cj ≡
bi,j (mod Ai ) for each Ai . Then we claim that the polynomials cj x + y
generate a set of MOLS.
Each cj is in a nonzero coset for each maximal ideal, so by Proposition
2.10 it is invertible and cj x + y certainly produces a latin square. Suppose
20
the polynomials cj x + y and cj 0 x + y produce the same pair occurring twice,
at the coordinates (x1 , y1 ) and (x2 , y2 ). Then
cj x1 + y1 = cj x2 + y2 ,
cj 0 x1 + y1 = cj 0 x2 + y2 ,
and subtracting,
(cj − cj 0 )(x1 − x2 ) = 0.
Since x1 6= x2 , we have that cj − cj 0 is a zero divisor; however, we see
that since cj and cj 0 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/hni. If n =
q1 · · · qk where each qi is a prime power, and q1 < · · · < qk , then taking
R = Fq1 ⊕ · · · ⊕ Fqk , we have that m = q1 so that NP (R) = q1 − 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 q1 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, NP (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
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[x1 , x2 , . . . , xd ], is
1
m−1
j X
d
(m − 1)k
m −
k
d
!
.
k=0
(b) the maximum possible number of frequency squares of type F (nk ; nk−1 )
k −1)2
generated over R[x1 , x2 , . . . , x2k ], is (mm−1
.
The proofs use the elements cj 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 MacNeish 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
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
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 7→ a ⊕ x and x 7→ x ⊕ a are bijections.
• The non-zero elements of N form a group under multiplication; the
multiplicative identity is called 1 and 1 6= 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, θ2 , . . . , θq−2 }.
• The characteristic is 2; that is a ⊕ a = 0 for any element a.
• We have, for all 1 ≤ k ≤ q − 2, 1 ⊕ θk = θuk .
• The fact that multiplication distributes over addition uniquely defines
all other additions.
(u)
This neofield will be denoted Nq .
24
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 θa ; the entry ∞ represents the neofield
element 0. In general, we define the discrete logarithm of an element α ∈
(u)
Nq :
a if θa = α, 0 ≤ a ≤ q − 2,
logθ (α) =
∞ if α = 0.
⊕2 ∞ 0 1 2 3 4 5 6 7 8
∞ ∞ 0 1 2 3 4 5 6 7 8
0
0 ∞ 2 4 6 8 1 3 5 7
1
1 8 ∞ 3 5 7 0 2 4 6
2
2 7 0 ∞ 4 6 8 1 3 5
3
3 6 8 1 ∞ 5 7 0 2 4
4
4 5 7 0 2 ∞ 6 8 1 3
5
5 4 6 8 1 3 ∞ 7 0 2
6 3 5 7 0 2 4 ∞ 8 1
6
7
7 2 4 6 8 1 3 5 ∞ 0
8
8 1 3 5 7 0 2 4 6 ∞
⊕5 ∞ 0 1 2 3 4 5 6 7 8
∞ ∞ 0 1 2 3 4 5 6 7 8
0
0 ∞ 5 1 6 2 7 3 8 4
1
1 5 ∞ 6 2 7 3 8 4 0
2
2 1 6 ∞ 7 3 8 4 0 5
3
3 6 2 7 ∞ 8 4 0 5 1
4
4 2 7 3 8 ∞ 0 5 1 6
5
5 7 3 8 4 0 ∞ 1 6 2
6
6 3 8 4 0 5 1 ∞ 2 7
7
7 8 4 0 5 1 6 2 ∞ 3
8
8 4 0 5 1 6 2 7 3 ∞
(2)
(5)
Figure 3.1: The addition tables for the neofields N10 and N10 (respectively)
in logarithmic form. Notice that this means ∞ is the additive identity; while
0 represents the multiplicative identity θ0 = 1.
(5)
(2)
Notice that the table for N10 is commutative, while that for N10 is not.
25
(8)
(2)
In fact, the addition table for N10 is the transpose of that of N10 , so we are
now fully equipped to do arithmetic in any uniform cyclic neofield of order
10.
(5)
For example, within N10 , if we were asked to compute (θ ⊕ θ4 )(θ5 ⊕ 1),
we would look the two additions up in the table; the first comes out θ7 , the
second θ7 also; multiplying, we get θ14 = θ5 since θ9 = 1; see below.
(2)
Asked to do the same computation within N10 , we get θ7 for the first
addition and θ4 for the second; their product is then θ2 .
Also, notice that both addition tables form latin squares. We will return
to this idea in section 3 below.
3.2
(u)
Existence and Construction of Nq .
(u)
We need to notice some basic properties of Nq . First we will be a bit clearer
about the last part of the previous definition.
(u)
Given two distinct nonzero elements of Nq , denote them as θa , θb .
Then:
(3.1)
θa ⊕ θb = θa 1 ⊕ θb−a = θa · θu(b−a) = θ(1−u)a+ub .
Next we notice a property identical to that for finite fields:
(u)
Proposition 3.3 For any α ∈ Nq ,
• If α 6= 0, αq−1 = 1.
• αq = α.
• If α 6= 0 and θ is a generator of the multiplicative group, α is also a
generator if and only if α = θk where (k, q − 1) = 1.
Proof. Each of these statements relies only on the fact that the mul(u)
tiplicative group of Nq 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
Another way to look at this is to look more carefully at equation (3.1).
Consider the action x 7→ θa ⊕ x; we know it must be transitive. Selecting
x = 0 or x = θa , we produce the two images θa and 0 respectively. Therefore,
for any b 6= a, 0 ≤ b ≤ q − 2, we must have that
(1 − u)a + ub 6≡ a (mod q − 1) ,
or in short
u(b − a) 6≡ 0 (mod q − 1) .
which is definitely true if (u, q − 1) = 1 and false otherwise.
Similarly, considering the action x 7→ x ⊕ θb , allowing x = θa to vary we
have that for any a 6= b, 0 ≤ a ≤ q − 2,
(1 − u)a + ub 6≡ b (mod q − 1) ,
or:
(u − 1)(b − a) 6≡ 0 (mod q − 1) .
which is definitely true if (u − 1, q − 1) = 1 and false otherwise.
(u)
This proves that for any u / q, we certainly have that Nq satisfies all
the criteria to be a neofield.
Moreover, there are always neofields for any even q ≥ 4:
(u)
Proposition 3.5 Given any even q ≥ 4, we may form the neofields Nq
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 = u0 / q, we also have
(u )
(q−u )
(q − u0 ) / q. Moreover, the two neofields Nq 0 and Nq 0 are identical
(u )
except for the order of addition; to be precise, if α ⊕ β = γ in Nq 0 , then
(q−u )
β ⊕ α = γ in Nq 0 .
Proof.
We have, by the linearity of the g.c.d.,
(q − u0 , q − 1) = ((q − 1) − (q − u0 ), q − 1) = (u0 − 1, q − 1) = 1,
27
and also:
(q − u0 − 1, q − 1) = ((q − 1) − (q − u0 − 1), q − 1) = (u0 , q − 1) = 1,
so that both criteria for (q − u0 ) / q are satisfied.
For the second part of the proposition, if α and β are equal or if either
are equal to 0, the given statement is certainly true.
(u )
Otherwise, let α = θa and β = θb . Then within Nq 0 :
α ⊕ β = θ(1−u0 )a+u0 b ,
(q−u0 )
while within Nq
:
β ⊕ α = θ(1−(q−u0 ))b+(q−u0 )a = θ(1−u0 )a+u0 b+(q−1)(a−b) ,
which are equal by Prop. 3.3.
(u)
Corollary 3.7 If q is even and u / q, the neofield Nq
and only if u = q/2.
is commutative if
Lastly we observe:
Proposition 3.8 If q is even and q − 1 is prime, we have u / q for any
2 ≤ u ≤ q − 2.
We will define a counting function similar to Euler’s totient to count how
many distinct uniform cyclic neofields of order q we may construct.
Definition 3.9 If q is even, we define a function φN (q):
φN (q) = #{u|u / q}.
This function was first studied by Schemmel [20] in 1869; he was simply
generalizing the definiton of Euler’s totient function. To be clear, Schemmel
defined a class of totient functions:
Sb (n) = #{a | 1 ≤ a ≤ n − b, (a, n) = (a + 1, n) = · · · = (a + b − 1, n) = 1}.
The function we defined above has φn (q) = S2 (q − 1).
Prop. 3.8 implies that if q − 1 is prime φN (q) = q − 3. Props. 3.5 and
3.6 imply that if q ≥ 6, φN (q) ≥ 3 since at least we have the three suitable
characters 2, q/2, and q − 2. Prop. 3.6 also implies that φN (q) is always
odd. Schemmel’s work included a formula analogous to that for the ordinary
Euler totient function:
28
Proposition 3.10 If n is any positive integer,
Y
2
1−
,
S2 (n) = φN (n + 1) = n ·
p
p|n
We can prove this directly: notice that φN (n + 1) is a multiplicative
function, and that if n = pk , φN (pk + 1) = pk − 2pk−1 . This also holds when
n is even, since there are no uniform cyclic neofields of size n + 1 then; and
this function comes out 0 if 2|n.
In order to summarize what we can construct for small even q, we will
give a chart. (See Fig. 3.2)
q
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
List of 2 ≤ u ≤ q/2 for which u / q
2
2, 3
2, 3, 4
2, 5
any
any
2, 8
any
any
2, 5, 11
any
2, 3, 4, 7, 8, 9, 12, 13
2, 5, 8, 11, 14
any
any
2, 5, 8, 14, 17
2, 3, 4, 9, 12, 13, 17, 18
any
2, 5, 8, 11, 17, 20
φN (q)
1
3
5
3
9
11
3
15
17
5
21
15
9
27
29
9
15
35
11
Figure 3.2: The possible suitable characters for small q. Notice that for a
complete list, we must tack on the entries q − u in the middle column; i.e.
for q = 22, the complete list of five suitable characters is 2, 5, 11, 17, 20.
29
3.3
Latin Squares over Neofields
One of the main reasons we have been interested in finite fields is so that
we may use them to construct latin squares, and indeed sets of mutually
orthogonal latin squares. We now raise the question: to what extent may
(u)
we use these ideas to construct sets of latin squares using the neofields Nq
instead?
We have already noted that the addition table of any finite neofield forms
a latin square; this follows directly from the definition. Another way to say
this is: if we label the rows and columns of a square array with the neofield
elements, the “polynomial” x ⊕ y generates a latin square.
For finite fields, we looked at the set of polynomials {ax + y|a ∈ F∗q }.
(u)
In a similar way we will look at the analogous functions over Nq . We
cannot really call these polynomials over the neofield; since addition is not
associative or commutative the conventional definition of a polynomial is
not very useful. However, these are really only “linear combinations” and
these work fine in the context of neofields.
First we establish that each of these generates a latin square:
(u)
Proposition 3.11 If q is even, u / q, and a ∈ Nq
(u) 2
(u)
function fa : Nq
→ Nq defined by
is not zero, then the
fa (x, y) = ax ⊕ y,
gives a latin square of order q.
Proof. We need only check that fixing x or y and allowing the other
to vary, the function fa maps to all q elements of the neofield. This is clear
from the definition of a neofield (specifically, the part that guarantees left
or right addition by a fixed element is a bijection) and the fact that a has a
multiplicative inverse. (u)
It turns out that in this case, the squares given by the set {fa |a ∈ Nq }
are not mutually orthogonal; but superimposing them does give a regular
and predictable number of pairs. This fact was established by Keedwell and
Mullen in [10], but the proof given here is new.
(u)
Theorem 3.12 If a 6= b are non-zero elements of Nq , then the two latin
squares generated by fa and fb have, when superimposed, precisely 4q − 3
pairs occurring once each and q − 1 pairs occurring q − 3 times each.
30
Proof. We seek to classify the pairs (α, β) by how many solutions (x, y)
we can obtain to the simultaneous equations:
ax ⊕ y = α
(3.2)
bx ⊕ y = β
(3.3)
We shall take cases; essentially we need to exclude the possibilities that
any of the three terms in either equation are 0.
CASE I. Suppose α = 0. Then (3.2) implies that ax = y, and substituting into (3.3) we have (b ⊕ a)x = β, which means that in this case, we have
the (unique) solution:
x=
β
aβ
, y=
.
b⊕a
b⊕a
Putting this in the language of latin squares, for each of the q pairs for
which α = 0, there is exactly one location in the two squares at which the
pair (α, β) occurs.
CASE II. Suppose α 6= 0 and β = 0. In the same fashion as the previous
case, we determine a unique solution:
x=
α
bα
, y=
.
a⊕b
a⊕b
So for each of these q − 1 pairs, there is exactly one location in which
the pair occurs.
CASE III. Suppose α, β 6= 0 and α = β. Then we have, combining (3.2)
and (3.3), that ax ⊕ y = bx ⊕ y, implying that ax = bx and since a 6= b, we
have x = 0. Indeed, here we have the unique solution:
x = 0, y = α = β.
This case again encompasses q − 1 new pairs, each of which occurs once
only across the two squares.
CASE IV. Suppose α, β 6= 0 and αa = βb . Then, dividing (3.2) by a and
(3.3) by b and equating the two left-hand sides, we obtain ay = yb , implying
that y = 0. Here, then we have a unique solution:
x=
α
β
= , y = 0.
a
b
Again, we have q − 1 new pairs, each of which occurs exactly once. The
total for the first four cases gives 4q − 3 pairs, each occurring once.
31
CASE V. If (α, β) fall into none of the above cases, we must have that
x, y =
6 0; this implies that the additions are governed by (3.1).
We will then suppose that logθ x = X, logθ y = Y , logθ a = A, logθ b = B,
logθ α = P , logθ β = Q. Then using (3.1), we must solve for X and Y in the
equations:
(1 − u)(A + X) + uY ≡ P (mod q − 1) ,
(1 − u)(B + X) + uY ≡ Q (mod q − 1) .
These equations are, however, not independent; indeed for them to be
consistent we must have P + (u − 1)A ≡ Q + (u − 1)B, or in non-logarithmic
notation, au−1 α = bu−1 β. For any nonzero choice of α this determines a
unique β which does not fall into any of the first four cases; so for only
these q − 1 pairs (α, β) are there any solutions at all; for all other pairs the
equations are inconsistent.
In the case where there are solutions, let R ≡ P +(u−1)A ≡ Q+(u−1)B,
then we only need solve:
(1 − u)X + uY ≡ R (mod q − 1) .
(3.4)
Ostensibly, there ought to be q − 1 pairs (X, Y ) that solve this equation;
however, two of those pairs will have to be discarded. We cannot have
ax = y or bx = y or we will be in Case I or II from above. Pairing (3.4)
with each of A + X ≡ Y and B + X ≡ Y gives exactly one solution each;
these are the solutions we must discard, leaving q − 3 legitimate solutions of
(3.4).
So for each of the q − 1 pairs (α, β), this pair occurs in q − 3 locations
across the two squares.
This concludes the proof; indeed we can check that this works since we
got 4q − 3 pairs occurring once each in Cases I through IV, and in Case V
q − 1 pairs occurring q − 3 times each; this definitely takes care of all cases
since (4q − 3)(1) + (q − 1)(q − 3) = q 2 , or all the possible locations. We can now use this result to form sets of latin squares which, though
not mutually orthogonal, have the same orthogonality condition on each
pair of squares.
Corollary 3.13 If q ≥ 4 is even, we may form a set of q − 1 latin squares
of order q which are mutually (5q − 4)-orthogonal. We may do this in φN (q)
essentially different ways.
32
Proof. We need only select a u / q and take the set of squares given by
(u)
the functions {fa |a ∈ Nq \{0}}; each pair of these is taken care of by Thm.
3.12. By selecting different u, we get different sets of squares. 3.4
Aggregate Neofields
In the last section, we focused on sets of latin squares that are generated
over a single neofield. However, as we have seen, for q ≥ 6 there are always
at least three if not many more neofields to choose from, all of which produce
latin squares of the same order. Indeed, simply looking at addition tables for
some orders q produce latin squares far closer to being mutually orthogonal
than the squares generated over the same neofield in section 3.
(2)
(3)
For example, we examine the addition tables for N12 and N12 . Here we
have have one pair, namely (∞, ∞), which occurs twelve times down the
main diagonal; eleven more pairs of the type (a, a) which occur twice each
(once in the first row and once in the first column) but every other position
contains a distinct ordered pair; there are 110 of these. so these two squares
are 122-orthogonal. What is more, this same pattern occurs for any two
addition tables for q = 12, so we have nine mutually 122-orthogonal latin
squares. This is an improvement over the eleven mutually 56-orthogonal
latin squres given by Cor. 3.13.
(2)
(5)
However, if we look instead at the addition tables for N10 and N10 (see
Fig. 3.1) we get one pair occurring ten times, nine more occurring four times
and eighteen occring three times for a total of only 37 pairs, worse than the
46 we got from Cor. 3.13. So clearly this excellent behavior does not occur
for all even numbers q.
In order to dive deeper into this, we will define an object where we may
think about more than one different neofield addition in a single object.
Essentially, we will stitch together all the uniform cyclic neofields of one
order into a single aggregate object with one multiplication operation but
several different additions.
Definition 3.14 An aggregate neofield is a set N together with several
operations called additions as well as a single multiplication; the additions
are denoted ⊕u where u is allowed to range over a specified finite set of
symbols called A, and the multiplication is denoted · as usual. N has the
property that it is a neofield under ⊕u and · for each choice of u ∈ A.
Definition 3.15 Let q ≥ 4 be even. The aggreate uniform cyclic neofield
33
(u)
Nq is the aggregate of all the neofields Nq where A = {u|u / q}; that is it
(u)
features all the additions ⊕u inherited from the neofields Nq .
To illustrate this new notation, we now state Prop. 3.6 in a somewhat
more succinct way:
Corollary 3.16 Let q ≥ 4 be even and let u / q. Then for any α, β ∈ Nq ,
α ⊕u β = β ⊕q−u α.
3.5
Latin Squares over Aggregate Neofields
We will now explore what happens when we look at sets of latin squares
generated over aggregate neofields, using the advantage of a different addition for each square. We will begin by confining our attention to those q for
which q − 1 is a prime; in this way, we are allowed to use any u from 2 up
to q − 2 and can get a large number of squares with different u.
Theorem 3.17 Let q be an even number such that q−1 is prime. Let a, b be
nonzero elements of Nq and choose 2 ≤ u, v ≤ q−2 so that we are guaranteed
u, v / q. Then the two latin squares generated by the two functions:
f (x, y) = ax ⊕u y,
g(x, y) = bx ⊕v y,
have the following properties when superimposed:
(1) If u 6= v and a = b, we have one pair repeated q times, q − 1 pairs
repeated twice, and (q − 1)(q − 2) pairs occurring only once; so that the two
squares are (q 2 − 2q + 2)-orthogonal.
(2) If u 6= v and a 6= b, we have 2q − 2 pairs occurring twice and (q − 2)2
pairs occurring only once; so that the two sqaures are again (q 2 − 2q + 2)orthogonal.
(3) If u = v and a 6= b, we are in the scenario of Theorem 3.12 and we
have precisely 4q − 3 pairs occurring once each and q − 1 pairs occurring
q − 3 times each; so that the squares are (5q − 4)-orthogonal.
Proof.
(1) First we note that if a = b, we may as well take a = 1
since a is invertible in Nq and all the equations we are solving would have
equivalent behavior. We are therefore seeking to classify the pairs (α, β) by
the number of solutions (x, y) to the equations
x ⊕u y = α,
34
x ⊕v y = β.
We will take cases as in the proof of Thm. 3.12.
CASE I. α or β = 0. In either case, the equations imply x = y so that if
only one of α, β equal zero, there are no solutions while if both equal zero,
there are q solutions: all pairs (x, y) where x = y.
CASE II. α = β 6= 0. Here we have at least two solutions per pair,
namely (x, y) = (0, α) or (α, 0). We will see in the course of the next case
that there cannot be more.
CASE III. None of α, β, x, y are zero, and also α 6= β. We may legitimately claim this since all cases where x or y may equal zero have been
dealt with already.
As before, let logθ x = X, logθ y = Y , logθ α = P , logθ β = Q. Then we
have the equations:
(1 − u)X + uY ≡ P (mod q − 1) ,
(1 − v)X + vY ≡ Q (mod q − 1) .
Since v − u is invertible modulo q − 1, we have the unique solutions:
X ≡ (v − u)−1 (vP − uQ), Y ≡ (v − u)−1 ((v − 1)P − (u − 1)Q) .
Notice that here we cannot have X = Y since Y = X − (v − u)−1 (P − Q)
so X = Y if and only if P = Q. This is excluded by the assumptions of this
case; and also ensures that if we are back in case II, there are no additional
solutions.
This case had (q − 1)(q − 2) pairs (α, β); between the three cases, then
we have proved the whole of part (1).
(2) Here we are solving the simultaneous equations:
ax ⊕u y = α,
bx ⊕v y = β.
We take cases as before:
CASES I and II. (α = 0, β = 0) are identical with Thm. 3.12 so we omit
them here. As a summary, we get 2q − 1 pairs (α, β), each occurring once.
CASES III and IV. (α = β, α/a = β/b), but in these cases the listed
solutions (where x = 0 and y = 0 respectively) are not unique, they are
35
achieved once again for x, y 6= 0 in case V below. However, these do exhaust
all the options where x or y equals zero.
CASE V: α, β, x, y 6= 0.
We can let logθ x = X, logθ y = Y , logθ a = A, logθ b = B, logθ α = P ,
logθ β = Q. and we get linear equations again:
(1 − u)(A + X) + uY ≡ P (mod q − 1) ,
(1 − v)(B + X) + vY ≡ Q (mod q − 1) .
Here again we may solve directly for X and Y :
X ≡ (v − u)−1 (vP − uQ + (u − 1)vA − u(v − 1)B),
Y ≡ (v − u)−1 ((v − 1)P − (u − 1)Q + (u − 1)(v − 1)(A − B)).
However, we must exclude those solutions which would make ax = y
or bx = y, which would mean one of α, β = 0. Therefore, we cannot have
A + X ≡ Y or B + X ≡ Y .
Calculating from the above solutions for X and Y , we see that these
happen when:
Q = P + (v − 1)(A − B) or Q = P + (u − 1)(A − B).
These give 2q − 2 choices for P and Q (hence for α and β) which have no
solutions in this case. It is equally clear these did not fall into any earlier
case.
However, if P = Q or P − A = Q − B (as in Cases III or IV) we get a
second solution since these cases do not coincide with the exclusions above.
In summary, then we have 2q − 1 pairs (α, β) producing one solution
from Cases I and II, 2q − 2 pairs producing two solutions between Cases III
or IV plus V, 2q − 2 pairs producing no solutions in the exclusions of Case
V, and (q − 1)(q − 5) pairs producing a single solution in Case V. This gives
the totals in the theorem. Although this theorem was quite technical to prove, it allows us to construct sets of latin squares which are mutually close-to-orthogonal:
Corollary 3.18 If q − 1 is prime, we may form a set of q − 3 latin squares
of order q which are mutually (q 2 − 2q + 2)-orthogonal.
36
In fact, we may do this either with all pairs having pairwise criteria as
in cases (1) or (2) of Thm. 3.17; for case (1) use the functions:
fu (x, y) = x ⊕u y, 2 ≤ u ≤ q − 2.
That is, we just use the addition tables for all the uniform cyclic neofields
of order q. For case (2), we use:
gu (x, y) = ux ⊕u y, 2 ≤ u ≤ q − 2.
3.6
Future Directions
There are several directions in which to investigate further in this area.
First we should like to deal with those even q for which q−1 is composite.
In these cases, we have fewer suitable characters u, but also fewer pairs when
we superimpose additon tables from different suitable characters. It seems
appropriate to investigate and make these precise.
This will involve study of sub-neofields; if we have two even numbers
q1 , q2 for which (q1 − 1)|(q2 − 1), then there is a copy of some neofield of
(u)
order q1 inside any Nq2 . This can be made exact:
Proposition 3.19 If q1 , q2 are even and (q1 − 1)|(q2 − 1), and also u / q2 ,
(u)
and θ is a generator in Nq2 , let r = (q2 − 1)/(q1 − 1); then the elements
{0, 1, θr , θ2r , . . . , θ(q1 −2)r },
(v)
form a copy of the neofield Nq1 where v ≡ u (mod q1 − 1) and 2 ≤ v ≤ q1 −2.
That is, if v / q1 and u / q2 and v ≡ u (mod q1 − 1), then we have a
(v)
(u)
sub-neofield relation Nq1 ⊂ Nq2 .
Another future step would be to include in our study other neofields
based on cyclic multiplications but which are not uniform. Such neofields
certainly exist; for example, at q = 8 there are three other neofields; one
is, of course, the finite field F8 , the other two are non-commutative, nonuniform neofields. (See the paper [2] or the treatise [8] for a list of all small
neofields, and for other constructions related to orthogonal latin squares.)
Also, it is as well to note that cyclic neofields can have odd order. These
are substantially different in that they do not have characteristic 2 and
cannot be uniform (strictly speaking). There are some of these which are
nearly uniform, however, and which possess some of the same properties as
37
their even-order cousins. Further study is needed to see whether these are
useful in the study of latin squares.
We would also like to extend these results to cubes and hypercubes generated by neofields. The trouble here is that, since addition is not associative,
it is not clear whether the functions
f (x, y, z) = (ax ⊕ by) ⊕ z
or
f (x, y, z) = ax ⊕ (by ⊕ z)
will be the best to study, or whether there is another similar variety. It is
also unclear how to apply the idea of different neofield additions inside an
aggregate neofield to these ideas.
38
Chapter 4
Class-r Hypercubes.
When we first defined an orthogonality condition for latin hypercubes, we
said that each pair of symbols occurs equally often, rather than exactly
once each. This definition loses the ‘uniqueness’ condition that beautifully
characterized the one for latin squares. In order to regain some form of
uniqueness we need to increase the possible number of ordered pairs, and
we shall accomplish this by increasing the size of the alphabet.
We will then explore the construction over finite fields that produces
these latin objects with a larger alphabet, which are called class-r hypercubes; and in particular study how to create large sets of mutually orthogonal ones. It will turn out that the question boils down to some interesting
linear algebra over finite fields. We will have cause to study a new class of
matrices which has strong properties of non-singularity that connect to an
old problem of coding theory.
This chapter is based on several open problems from the paper [7].
4.1
Definition
We introduce a new kind of latin hypercube that has a very similar definition
to the standard one; the only change is that the size of the alphabet is now
a power of the length of each side.
Definition 4.1 Let 1 ≤ r ≤ d and 0 ≤ j ≤ d − r be integers. A dimensiond, class-r, type j hypercube of order n is a d-dimensional hypercubical array
on nr symbols such that fixing any j coordinates, in this d − j-dimensional
subarray each symbol occurs exactly nd−r−j times.
39
When r = 1, this is the same definition as for ordinary latin hypercubes.
When j = d − r, each symbol must occur once in each r-dimensional ‘slice’.
We can now extend the definition orthogonality as well. Because the
alphabet is now of size nr , there are now n2r possible ordered pairs. We
say therefore that if d ≥ 2r, two dimension-d, class-r, type-j hypercubes of
order n are called orthogonal when, superimposing them, we obtain each of
the n2r possible order pairs exactly nd−2r times. In the case where d = 2r,
we recover a version of orthogonality which has each possible pair occurring
exactly once.
As an example, here are two orthogonal dimension-4, class-2, type-2
hypercubes
of order

  3.
 

0
  5

  7

3

 
8


  1

6

  2
4


0
  4

  8

3

 
7


  2

6

  1
5

4
6
2
7
0
5
1
3
8
8
1
3
2
4
6
5
7
0
7
2
3
1
5
6
4
8
0
5
6
1
8
0
4
2
3
7











1
 3
 8
4
 6
 2
7
 0
5

1
 5
 6
4
 8
 0
7
 2
3
5
7
0
8
1
3
2
4
6
6
2
4
0
5
7
3
8
1
8
0
4
2
3
7
5
6
1
3
7
2
6
1
5
0
4
8











2
 4
 6
5
 7
 0
8
 1
3

2
 3
 7
5
 6
 1
8
 0
4
3
8
1
6
2
4
0
5
7
7
0
5
1
3
8
4
6
2

 

 


 
,

 


 
6
1
5
0
4
8
3
7
2
4
8
0
7
2
3
1
5
6
 
 

 


 
.

 


 
These come from a process which will be described below, and there are
actually two further hypercubes mutually orthogonal to both of these.
4.2
Basic Orthogonality Results
We recall that the most important question in orthogonality of latin squares
is: given an order n, what is the largest set of mutually orthogonal latin
squares (usually shortened to MOLS) of order n? The size of such a set is
denoted N (n), and by Prop. 1.3 we know that N (n) ≤ n − 1. This bound
is always achievable if n is a prime power.
We use the finite field Fq . Label the rows and columns of a q × q array
with the elements of Fq , and for each a ∈ F∗q , construct a square by filling
the cell (x, y) with the field element x + ay.∗ We saw above in Thm. 1.6
∗
This notation is backwards from the way latin squares’ polynomials are conventionally
40
that these form a set of q − 1 MOLS of order q.
Let us say this a different way: let f, g : F2q → Fq be two functions
representing latin squares, the two latin squares are orthogonal if and only
if the concatenation, (f, g) : F2q → F2q , (f, g)(x, y) = (f (x, y), g(x, y)), is a
bijection. By taking two latin squareshgenerated
i by x + ay, x + by, this says
1 a
that the linear function with matrix 1 b is bijective; this can be seen
easily since the matrix is non-singular when a 6= b.
This same language can be applied to class-r hypercubes. Suppose d =
2r and j = r so that we have a dimension-2r, class-r, type-r hypercube
of order q, where the alphabet we choose is r-tuples of elements of Fq .
r
Such an object may be represented by a function f : F2r
q → Fq where
fixing any r of the inputs gives a bijection (this is the latin condition).
Now two such hypercubes are orthogonal when the concatenation function
r 2 ∼ 2r
(f, g) : F2r
q → (Fq ) = Fq is a bijection.
In analogy to the above, we denote the maximum size of a set of mutually
orthogonal dimension-2r, class-r, type-r hypercubes of order n by N (r; n).
In the notation from above, N (n) = N (1; n). It can be shown (see [7]) that:
Proposition 4.2 For any r ≥ 1, n ≥ 2, we have N (r; n) ≤ (n − 1)r .
Now, we can form linear functions to generate latin hypercubes as follows: let A be an r × r matrix over Fq . Then the linear function described
by the r × 2r matrix [I A] describes a latin hypercube. For this hypercube
to be of type r, it is necessary that fixing any r of the 2r columns, we have
a non-singular matrix. This is equivalent to the condition that A is nonsingular and that all square submatrices of A of any size (including 1 × 1)
are also non-singular. We will call such a matrix strongly non-singular or
SNS.
We note that in the case r = 2, this says only that A is non-singular and
has nonzero entries. This is the case treated in the original paper [7], and
the theorem stated there uses this terminology. (We restate this below, in
Thm. 4.5, with our new terminology.) However, if r ≥ 3, it is important to
assume that A has no singular submatrices.
For example, if we take the matrix (over F7 ):


1 2 4
A =  5 1 2 ,
6 5 1
written (usually ax + y rather than x + ay) in order to make this chapter’s notation a bit
cleaner and more consistent, and also due to the connections with coding theory, which
are easier to see with this notation.
41
we have that A is nonsingular, and has no zero entries, but it does have a
singular 2 × 2 submatrix. (Actually, it has three.) This means that the map
F67 → F37 given by the matrix [I A] does not give a dimension-6, class-3,
type-3 hypercube. Fixing the first, second, and fourth coordinates equal to
zero, the “face” given by the other three is generated by the linear map:



0 2 4
x3
 0 1 2   x5  ,
1 5 1
x6
which is singular and therefore produces only a subset of the 73 required
symbols in this “face”; here, we recall, the “symbols” are actually ordered
triples over F7 . We should point out also that although this hypercube is
not type-3, it is type-2.
Suppose we are given an r × r matrix A and asked to determine what is
the largest j for which the hypercube generated by [I A] can be said to be
of type j. We know that in this case, fixing j of the coordinates means that
in the “hyper-face” determined by varying the remaining coordinates, each
symbol appears equally often. In terms of the linear algebra, this means
that selecting any 2r − j columns of [I A] yields an r × (2r − j) matrix of
rank r. If we wish to reduce this to a condition on the matrix A, we have:
Proposition 4.3 Suppose that A is an r ×r matrix over Fq . The map [I A]
generates a hypercube of type j, 1 ≤ j ≤ r, if and only if for any 1 ≤ t ≤ j,
each submatrix of A of dimensions t × (t + r − j) has rank t.
We note that if A generates a type-j hypercube, it may also be said to
generate a type-j1 hypercube for any 0 ≤ j1 ≤ j. The maximum j for which
this holds we will call the inherent type of A. If the condition fails at j = 1,
we say its inherent type is 0.
Also notice that if j = r, the condition in Proposition 4.3 reduces to that
of A being SNS; so that saying A is SNS is equivalent to saying its inherent
type is r.
We now need to make a characterization of orthogonality based on these
matrices. Suppose A and B are two r×r matrices; then thelatin hypercubes
I A
generated are orthogonal if and only if the matrix
is non-singular;
I B
by row-reduction, the determinant of this matrix is det(B −A). We therefore
make the following definition to streamline our terminology:
42
Definition 4.4 A set S consisting of r × r matrices over Fq is called a
pairwise-difference non-singular set (shortened as PDNS set) if for any
A, B ∈ S where A 6= B, A − B is non-singular.
The set will be called PDNS(j) if, in addition, all the matrices in S have
inherent type at least j. When the type is best possible, j = r, a PDNS(r)
set will be shortened PDNS*.
Theorem 4.5 Suppose S is a PDNS(j) set of r × r matrices over Fq . Then
the linear maps [I A] for A ∈ S generate mutually orthogonal class-r hypercubes which are dimension-2r and type-j.
This brings us to the point where we can ask the central question: given a
prime power q and a class r, can we construct a set of mutually orthogonal
class-r hypercubes? To do this, by the above theorem, it is sufficient to
construct a PDNS* set of r × r matrices over Fq .
The largest possible size of such a set is (q − 1)r , but this ideal may not
be possible to attain; this bound may not be sharp. In our work below, we
achieve sets which come very close to this bound. It is possible that future
combinatorial work will sharpen the bound and the construction we have is
best possible.
4.3
Basics of PDNS Sets
It is worth noting the following result, which is useful in shortening future
explanations.
Proposition 4.6 Suppose S is a PDNS set of r×r matrices over Fq . Then:
(1) |S| ≤ q r − 1.
(2) If we remove any elements of S that are not SNS, the remainder is
a PDNS* set. We shall denote this set S ∗ .
(3) |S ∗ | ≤ (q − 1)r .
Proof. (1) Consider the top rows of all the elements of S. None of these
rows can consist of all zeros, for this would mean that matrix is singular.
Also, no two rows can be the same, since then the difference of those two
matrices would have a zero row and be singular. So since there are only q
possibilities for each entry in a row of length r, not all of which are zero, we
have a maximum of q r − 1 matrices.
(2) This is immediate.
43
(3) We know that at least there cannot be any zero entries in a matrix
which is SNS. Therefore, the top rows of the matrices in S ∗ have no zeros
and are all distinct, meaning there are only (q − 1)r possibilities. It is not clear whether the bound in (3) is best possible for r ≥ 3. The
constructions we give later, plus numerical evidence, suggest that the actual
bound is tighter but of the same order of magnitude, O(q r ).
The following remark, although almost trivial, will also speed up the
proofs of later results.
Proposition 4.7 Suppose S is a set of r × r non-singular matrices over Fq
such that for any A, B ∈ S, A 6= B, A − B ∈ S. Then S is a PDNS set.
4.4
PDNS Sets at r = 1, 2
When r = 1, we only need 1 × 1 matrices, and here the solution is simple:
take S = {[a]|a ∈ F∗q }. These produce ordinary latin squares.
The case r = 2 was addressed in detail in the paper [7]. We outline the
results here, using our new terminology.
Proposition 4.8 Suppose q is a prime power.
(1a) If q is odd, choose ω ∈ Fq to be a non-square. Then the set
a b
|a, b ∈ Fq , (a, b) 6= (0, 0) ,
S=
ωb a
is a PDNS set of q 2 − 1 2 × 2 matrices; and (1b) S ∗ is a PDNS* set of
(q − 1)2 matrices.
(2a) If q = 22k for integer k, let ω be a non-cube in Fq . Then the set
a
b
|a, b ∈ Fq ,
S=
ωb2 a2
is a PDNS* set of q 2 − 1 2 × 2 matrices; and (2b) S ∗ is a PDNS* set of
(q − 1)2 matrices.
Proof. (1a) Note that this set is closed under subtraction as in Prop.
4.7, so we have only to
h show ithat each element is non-singular. Suppose
a
b
that for some a, b, det ωb a = 0; then we have a2 − ωb2 = 0, so that if
2
b 6= 0, ω = ab , contradicting that ω was assumed to be a non-square; and
44
if b = 0, then a = 0 and this was specifically excluded. Therefore, all the
elements of S are non-singular and S is a PDNS set.
(1b) This follows immediately from Prop. 4.6, part (2), and noting that
a b
∗
∗
S =
|a, b ∈ Fq ,
ωb a
which has (q − 1)2 elements.
(2a) Because these fields have characteristic 2, for any x, y ∈ F22k , x2 −
2
y = (x − y)2 . This means that the set S is closed under subtraction as
in Prop. 4.7 so that again we only need tohcheck that
i each matrix is nona
b
singular. Suppose that for some a, b, det ωb2 a2 = 0; then we have
3
a3 − ωb3 = 0; so that if b 6= 0, we have ω = ab , contradicting that ω was
assumed to be a non-cube; and if b = 0, then a = 0, another contradiction.
Therefore, S is a PDNS set.
(2b) As above, this follows from Prop. 4.6, part (2); noting that
a
b
∗
∗
S =
|a, b ∈ Fq ,
ωb2 a2
which again has (q − 1)2 elements.
The corollary to the above is that for prime powers q which are either
odd or of the form 22k , we can construct a complete set of (q−1)2 dimension4, class-2, type-2 latin hypercubes of order q by using the explicit PDNS*
set in this proposition. This immediately raises the question: what about
2k+1
q =
h 2 i? When q = 2, the only 2 × 2 matrix with non-zero entries
1 1
is 1 1 , which is singular, so here there is no hope; but what about
q = 8, 32, 128, etc.? This question was listed as an open problem in the paper
[7]. We have not been able to achieve a PDNS* set of (q − 1)2 elements, but
we can get close:
Proposition 4.9 Let q = 22k+1 . Let ω ∈ Fq be chosen so that x2 + x + ω
is irreducible. Then the set
a
b
S=
|a, b ∈ Fq , (a, b) 6= (0, 0) ,
ωb a + b
is a PDNS set of q 2 − 1 2 × 2 matrices over Fq ; and then S ∗ is a PDNS*
set of (q − 1)(q − 2) elements.
45
Proof. We use the same techniques as above; clearly S is closed under
subtraction so we only need toh check that
i each element is non-singular.
a
b
Suppose that for some a, b, det ωb a + b = 0; then we have:
2
If b 6= 0, a2 + ab + ωb2 = 0; ie. ab + ab + ω = 0. But this would mean
that the field element ab is a root of the polynomial x2 + x + ω, which was
assumed irreducible, a contradiction.
If instead b = 0, a = 0, which is also excluded. Therefore each element
of S is non-singular and S is a PDNS set.
We note that
a
b
∗
∗
S =
|a, b ∈ Fq , a 6= b ,
ωb a + b
which has (q − 1)(q − 2) elements. Note that we must exclude a = b here
since we need a + b 6= 0, and we are in characteristic 2. We conjecture that it is actually impossible to improve this size of a
PDNS* set for r = 2, q = 22k+1 . Even this would not necessarily imply
that a larger set of mutually orthogonal hypercubes does not exist, since
they may be generated in a way other than by linear functions over finite
fields; however, we have no precedent for such a situation since in all known
cases the best possible set of latin objects (squares, ordinary hypercubes,
frequency squares) can be obtained by linear functions over finite fields.
4.5
PDNS Sets for General r: First Case
We shall now develop two cases which extend the result of Prop. 4.8 to any
positive integer r; that is, we will be able to construct complete PDNS* sets,
and hence complete sets of class-r hypercubes, for arbitrary r. The first case
is a generalization of case (1) of this proposition.
We begin by making the following notational conveniences. Fix r and
q, and let a0 , a1 , . . . , ar−1 ∈ Fq be not all zero; we will use a to denote the
r-tuple (a0 , . . . ar−1 ). We define f (x) = a0 + a1 x + · · · + ar−1 xr−1 , and we
define a matrix as follows:


a0
a1
· · · ar−2 ar−1
 ar−1 ω
a0
· · · ar−3 ar−2 


 ar−2 ω ar−1 ω · · · ar−4 ar−3 
M1 (a, ω) = 
.

..
..
..
.. 
.
.

.
.
.
.
. 
a1 ω
a2 ω
46
···
ar−1 ω
a0
We will eventually prove that for certain r and q and a suitable choice of
ω, these matrices will form a PDNS set. These matrices are a special type
of Toeplitz matrices; these have been studied previously in contexts that
are quite separate from our subject, and therefore the limited known results
about them are not really applicable to our discussion here.
First we establish the following determinant lemma.
Lemma 4.10 Let ζ be a primitive r-th root of unity over Fq . Then using
the above definition of M1 (ω) and f (x), we have
det M1 (a, xr ) =
r−1
Y
f (ζ k x).
k=0
Proof. We have two proofs of this result.
METHOD I. We define a Vandermonde matrix:

1
1
1
···
1
2x
r−1 x
 x
ζx
ζ
·
·
·
ζ

2

(ζx)2
(ζ 2 x)2 · · ·
(ζ r−1 x)2
V (x) =  x
 ..
..
..
..
 .
.
.
.
xr−1 (ζx)r−1 (ζ 2 x)r−1 · · · (ζ r−1 x)r−1
Then we multiply:



M1 (a, xr )V (x) = 

f (x)
xf (x)
..
.
f (ζx)
ζxf (ζx)
..
.
···
···
xr−1 f (x) (ζx)r−1 f (ζx) · · ·




.


f (ζ r−1 x)
ζ r−1 xf (ζ r−1 x)
..
.
(ζ r−1 x)r−1 f (ζ r−1 x)



.

Taking determinants, we note that the matrix on the right has a common
factor of the form f (ζ k x) in each column, which can be factored out of the
determinant; the matrix left is simply V (x) again so that
!
r−1
Y
k
r
f (ζ x) det V (x),
det M1 (x ) det V (x) =
k=0
and since det V (x) 6= 0 we have the desired conclusion.
METHOD II. The Vandermonde matrix in the proof above essentially
uses column-reduction to compute the determinant of M1 (xr ). We can instead independently use row-reduction as follows: add xr−1 times the first
47
row, xr−2 times the second row, etc., down to x times the second-to-last
row, all to the last row. This produces a factor of f (x) in every entry of the
resulting last row, so we know that f (x)| det M1 (xr ).
We now notice that M1 (xr ) is invariant under the replacement x 7→ ζx,
so weQknow that the factorization of M1 (xr ) must be also. This means
that
f (ζ k x)|M1 (xr ), and equality follows by comparison of the leading
coefficients. Now we use this fact to show:
Proposition 4.11 Let r be a positive integer, and q a prime power such
that q ≡ 1 (mod r). Let ω be primitive in Fq . Then the set
S = M1 (a, ω)| a ∈ Frq \{0} ,
is a complete PDNS set of r × r matrices over Fq ; and S ∗ is a PDNS* set.
Proof.
Recalling Prop. 4.7, since S is closed under subtraction, we
need only show all its elements to be non-singular to complete this proof.
Suppose for some choice of a that det M1 (a, ω) = 0. Then by Lemma
4.10, ω = xr0 for some x0 a root of f (x). However, since f (x) is of degree
r − 1, x0 lives in a field extension of Fq of degree at most r − 1; while ω is
primitive and any r-th roots of it must live in a field extension of degree r.
This is a contradiction so the matrices M1 (a, ω) are all non-singular. We note that here, |S| = q r − 1, but the size of S ∗ is not clear for general
r. We will explore this question in some detail below.
4.6
Counting the Size of PDNS* Sets for r = 3, 4
In Prop. 4.11, we have complete PDNS sets for any prime power q ≡
1 (mod r). Each set of matrices S produces a set of mutually orthogonal class-r, dimension-2r hypercubes, of which there are q r − 1. However,
we have not specified the type. In fact, in this set we will have in general
all types from 0 up to r. In principle, we could examine each set on a caseby-case basis, determining for each A ∈ S the inherent type of A. We could
remove from S all the matrices we found with fewer than a given inherent
type to give a PDNS(j) set; and removing all those which are not SNS gives
a PDNS* set.
The question is: just how many of the matrices of the form M1 (a, ω) are
SNS? The answer is unfortunately not going to be (q − 1)r for all r, but we
will investigate for the first few r exactly what this is.
48
Specifically: set Sr (q) to be the set defined in Prop. 4.11 with a given
r and q; let Sr∗ (q) be the same set with all non-SNS matrices removed. We
seek to count the size |Sr∗ (q)| wherever possible.
We begin with r = 3. For the sake of convenience, we will use the letters
c, b, and a in place of a0 , a1 , and a2 respectively; and for convenience we
take ω primitive in Fq . Using this:


c
b a
M1 (a, b, c, ω) =  aω c b  .
bω aω c
Each M1 (a, b, c, ω) is itself non-singular unless a = b = c = 0 (we established this in Prop. 4.11). M1 (a, b, c, ω) has no zero entries if each of a, b, c is
nonzero; we have (q − 1)3 possibilities. We now need only to exclude those
choices of (a, b, c) which produce singular 2 × 2 submatrices.
The reason this problem is easier than it might be is that despite the fact
that we have nine possible 2 × 2 submatrices, we only have three essentially
different ones. For example,
the matrix we get from the upper left (rows
c b
, is exactly the same as that from the bottom right
1,2, cols 1,2),
aω c
c a
,
(rows 2,3, cols 2,3); and even the one from (rows 1,3, cols 1,3) is
bω c
which has the same determinant. If one of these is non-singular, all are; this
occurs when c2 6= abω.
Similarly, we have two
more cases:
c a
.
(rows 1,2, cols 1,3):
aω b
aω c
.
(rows 2,3, cols 1,2):
bω aω
b a
(rows 1,3, cols 2,3):
.
aω c
All three of these are non-singular if bc 6= a2 ω.
The last three submatrices are non-singular if b2 6= ac.
We record this in a lemma:
Lemma 4.12 If q ≡ 1 (mod 3) and ω is primitive in Fq , the matrix M1 (a, b, c, ω)
is SNS iff a, b, c 6= 0 and all the following are false:
(1) a2 ω = bc, (2) b2 = ac, (3) c2 = abω.
Proposition 4.13 If q ≡ 1 (mod 3) is a prime power, the set S3∗ (q) has
precisely (q − 1)2 (q − 4) elements.
49
Proof.
Beginning with Lemma 4.12 all that remains is to count how
many triples (a, b, c) drawn from Fq satisfy the conditions given above.
There are (q − 1)3 triples which have all three elements nonzero. In
addition, each condition (1), (2), and (3) exclude precisely (q − 1)2 triples.
For instance, in condition (1) fixing any nonzero a and b, there is precisely
one nonzero c for which equality holds. The same argument works for (2);
in equation (3) we choose any b and c and determine a.
We now claim that it is not possible for only two of these three conditions
to fail simultaneously. Suppose, for example, (1) and (2) are both true; that
is, a2 ω = bc and b2 = ac. Then:
bc2 = a2 cω = ab2 ω,
so now (3) holds as well. Similar computations show that the failure of any
two implies the failure of the third.
Lastly we claim that it is also impossible for all three to fail. Since
we would have a2 ω = bc and ac = b2 we would also have a3 cω = b3 c or
ω = (b/a)3 . This is a contradiction since ω was assumed primitive in Fq
where q ≡ 1 (mod 3), so that primitive elements have no cube roots in the
field. (See Prop. 1.5, part 10).
Therefore, we have (q − 1)3 possibilities and 3(q − 1)2 exceptions, none
repeated; so we have (q − 1)3 − 3(q − 1)2 = (q − 1)2 (q − 4) total triples (a, b, c)
that give SNS matrices. When we move up to r = 4, the situation becomes much more complex
because we must deal with 2 × 2 and 3 × 3 submatrices.
Here we make the notation as above:


d
c
b a
 aω d
c b 

M1 (a, b, c, d, ω) = 
 bω aω d c  .
cω bω aω d
Here there are 16 possible 3 × 3 submatrices and 36 possible 2 × 2. Due
to the symmetry in the construction of the matrix, however, this produces
only four different 3 × 3 determinants and eight 2 × 2 determinants. In fact,
the 2 × 2 matrices can be condensed even further; this is best explained by
an example:
d c
at (rows 1,2 cols 1,2):
; determinant d2 − acω.
aω d
d b
at (rows 1,3 cols 1,3):
; determinant d2 − b2 ω.
bω d
50
b a
at (rows 1,2, cols 3,4):
; determinant b2 − ac.
c b
Requiring that these three matrices are all nonsingular can be worded
in the following way: the three field elements d2 , acω, b2 ω, are all different.
A second group of three determinants reduces in this way; the other two
2 × 2 determinants are separate conditions.
We will summarize the results in the following lemma; this may verified
by direct computation.
Lemma 4.14 If q ≡ 1 (mod 4) and ω is primitive in Fq , the matrix M1 (a, b, c, d, ω)
is SNS if and only if a, b, c, d 6= 0 and all the following are false:
(A1) a3 ω 2 + cd2 + b2 cω − 2abdω − ac2 ω = 0;
(A2) b3 ω + a2 dω + c2 d − 2abcω − bd2 = 0;
(A3) c3 + ab2 ω + ad2 − 2bcd − a2 cω = 0;
(A4) d3 + bc2 ω + a2 bω 2 − 2acdω − b2 dω = 0;
(B1) The following are not all different: a2 ω, c2 , bd;
(B2) The following are not all different: b2 ω, d2 , acω;
(B3) ad = bc;
(B4) abω = cd.
We have numbered the conditions in this way to better organize them;
the (A) conditions come from the 3 × 3 submatrices and the (B) conditions
from the 2 × 2 submatrices.
Notice that these conditions are very symmetrical: if we send a 7→ b 7→
c 7→ d 7→ aω in any condition, we get another one (or a multiple by ω).
Indeed, this sends the (A) conditions on a loop, and cycles (B1) to (B2) and
back, and (B3) to (B4) and back. This mapping comes straight from the
matrix construction itself: if we have any size submatrix, and then form a
second submatrix by shifting down one place (cycling from bottom row to
top row if necessary), we pass through this reassignment (losing a common
factor of ω if cycling bottom to top).
Theorem 4.15 If q ≡ 1 (mod 4) is a prime power, the set S4∗ (q) has precisely (q − 1)2 (q − 5)(q − 7) elements.
Proof.
As above, we will now count how many quadruples (a, b, c, d)
from Fq satisfy the conditions of Lemma 4.14.
In the last proof, the counting was relatively painless since no two of
the three conditions could be fulfilled simultaneously; this is no longer true.
Here, we need a careful argument of an inclusion-exclusion variety.
51
To keep this concise, we will omit most of the direct calculations. We
organize this into a sequence of intermediate results.
STEP I. The conditions (A1) - (A4) are met (individually) by precisely
(q − 1)2 (q − 2) quadruples.
STEP II. The conditions (B1) and (B2) are met (individually) by precisely 2(q − 1)3 quadruples.
STEP III. The conditions (B3) and (B4) are met (individually) by precisely (q − 1)3 quadruples.
STEP IV. The pairs (A1),(B1); (A2),(B2); (A3),(B1); (A4),(B2) can be
met simultaneously, in 3(q − 1)2 ways each.
STEP V. No other pair of conditions may be met without others being
met as well.
STEP VI. The next smallest set of simultaneous conditions met occurs
when five conditions fail; there are four sets of conditions: (A1), (A2), (B1),
(B2), and (B4) are the basic set; rotating as described above give the other
three sets. Each of these sets can fail in (q − 1)2 ways each.
STEP VII. There are no other ways in which these conditions can be
met simultaneously.
COLLECTION: So we can have, working backwards through inclusionexclusion:
Five conditions (as in Step VI) simultaneously: (q − 1)2 each for four
combinations.
Two conditions (as in Step IV) but not more simultaneously in (q − 1)2
ways each for four combinations.
A single (A) condition (but no more) is met in (q − 1)2 (q − 5) ways each;
there are four of these.
A single (B1) or (B2) (but no more) is met in 2(q − 1)2 (q − 4) ways each.
A single (B3) or (B4) (but no more) is met in (q − 1)2 (q − 3) ways each.
Therefore the number of ways that at least one condition is met is in
total:
4(q−1)2 +4(q−1)2 +4(q−1)2 (q−5)+2(2(q−1)2 (q−4))+2(q−1)2 (q−3) = (q−1)2 (10q−34)
This leaves:
(q − 1)4 − (q − 1)2 (10q − 34) = (q − 1)2 (q 2 − 12q + 35) = (q − 1)2 (q − 5)(q − 7),
52
quadruples for which no condition causes a failure.
In principle, we could now tackle r = 5; however the difficulty here has
increased quite a bit from r = 4 and is rather intractable.
4.7
PDNS Sets for General r: Second Case (Frobenius)
We now seek to generalize the second case of Prop. 4.8, which dealt with a
field where q = 22k .
The first thing to notice is that the characteristic being 2 can hbe general-i
a
b
ized; if q = p2 (p being, here, another prime power) the matrices bp ω ap
form a PDNS set when ω is chosen to be a non-1 + p-th power.
Following the same outline as above, we make some notational conveniences. Here, again we fix r and q = pr prime powers, and assume
a0 , a1 , . . . , ar−1 ∈ Fq are not all zero. We again denote this r-tuple of coefficients by a. We set
2
g(x) = a0 + a1 x + a2 x1+p + a3 x1+p+p · · · + ar−1 x1+p+···+p
and we define a matrix as follows:

a0
a1
···
p
 ap ω
a
···
 r−1
0
 p2
p2
p
a ω ar−1 ω · · ·
Mp (a, ω) = 
 r−2
..
..
..

.
.
.

ap1
r−1
ω
ap2
r−1
ω
···
ar−2
apr−3
2
apr−4
..
.
r−1
apr−1 ω p
ar−1
apr−2
2
apr−3
..
.
r−2
ap0
r−1
r−2
,




.



Once again, these matrices will form a PDNS set for certain r and q. We
need an analogous determinant lemma:
Lemma 4.16 Let ζ be a primitive (1 + p + p2 + · · · + pr−1 )-th root of unity
over Fq . Then, with the above definition of Mp (ω) and g(x), we have:
det Mp (x
1+p+p2 +···+pr−1
)=
r−1
p+p2 +···+p
Y
g(ζ k x).
k=0
The proof follows the same reasoning as Method II of the proof of Lemma
4.10; the dimensions are a mismatch for Method I.
53
Proposition 4.17 Let r be a positive integer, and q a prime power such
that q = pkr for p prime and k an integer. Let ω be a primitive in Fq . Then
the set
S = Mp (a, ω)| a ∈ Frq \{0} ,
is a complete PDNS set of r × r matrices over Fq ; and S ∗ is a PDNS* set.
To count the size of the PDNS* set associated to this is rather infeasible;
although the symmetries of the submatrices are similar, the symmetry is not
perfect since ω comes raised to different powers in each column. The best
conjecture right now, based on computational evidence for q = 27, 64 and
125, is that here the size of the PDNS* set for r = 3 is (q − 1)2 (q − 7).
4.8
PDNS Sets for General r: Third Case.
We can extend Proposition 4.9 to get one more general class of PDNS set;
here, though we will need xr − x + ω to be an irreducible polynomial, so we
will have to require r|q. This means that for r which are not primes powers,
there is no set of this type. We will again choose coefficients in the r-tuple
a = (a0 , a1 , . . . , ar−1 ) and will use the same f (x) from section 5:
f (x) = a0 + a1 x + a2 x2 + · · · ar−1 xr−1 .
We will define the matrix:

a0
a1
a2
···
 ar−1 ω
a
+
a
a
···
0
r−1
1

 ar−2 ω ar−1 ω + ar−2 a0 + ar−1 · · ·
Mr (a, ω) = 

..
..
..
..

.
.
.
.
a1 ω
a2 ω + a1
a3 ω + a2 · · ·
ar−2
ar−3
ar−4
..
.
ar−1
ar−2
ar−3
..
.
ar−1 ω + ar−2 a0 + ar−1
We then have a determinant lemma:
Lemma 4.18 Wtih Mr (a, ω) and f (x) defined as above, we have:
Y
det Mr (a, xr − x) =
f (x − α).
α∈Fr
This lemma may be proved using a Vandermonde matrix as in Method
I of the proof of Lemma 4.10; Here, we need the entries in the second row
of V (x) to be x − α where α runs over the elements of Fr .
54







Proposition 4.19 Suppose r ≥ 1 is a prime power, and q is a prime power
with r|q. Let ω be a primitive element of Fq which is not in the image of the
mapping x 7→ xr − x. Then the set:
S = Mr (a, ω)| a ∈ Frq \{0} ,
is a complete PDNS set of r × r matrices, and S ∗ is a PDNS* set.
Computations to count the size of the PDNS* set in this proposition for
r = 3 do not reveal a polynomial dependence. For each valid q, which here
are powers of 3, the size of S ∗ is divisible by (q − 1) but not (q − 1)2 . The
orders of magnitude do go up on the order of q 3 , but no further conjecture
can be made here.
4.9
Directions for Further Study
Although we now have several satisfactory constructions of PDNS sets for
certain choices of r and q, there are still some embarrassing holes. For
instance, r = 3 and q = 11 is not covered.
Each of our three cases rely on the irreducibility of a polynomial with a
primitive ω ∈ Fq as a coefficient:
• The polynomial xr − ω is irreducible if q ≡ 1 (mod r).
• The polynomial x1+p+p
2 +···+pr−1
− ω is irreducible if q = pr .
• The polynomial xr − x − ω is irreducible if r|q.
One direction of future research may be to find other matrices associated
to other irreducible polynomials, perhaps which work for other fields.
Another exciting direction is in the area of coding theory. Matrices which
have no square singular submatrices are need in the construction of MDS
codes. In these cases, the matrices needed are rectangular, but we could
just use the first few rows of M1 (a, ω) if it is indeed SNS, or any of the other
matrices above, which would work just fine. However, the catch is that the
open part of the problem comes when q = r + 1, and for r = 3, 4 we found
that there were no SNS matrices for so small a field.
It is not actually necessary that the matrix be SNS, though, so long as
the reason it fails to be SNS lies only in the rows we omit. It is conceivable
that for large enough r and a small number of rows, we may have matrices
of this type which have no singular submatrices. More computational work
is needed before we can answer this question definitely, but as far as we can
tell this is a new construction, so there is some promise here.
55
Chapter 5
Blocking Sets in Partial
Latin Cubes
For our last chapter, we give an introduction to some results achieved in a
somewhat different context. Here our results are not algebraic in nature, but
are purely combinatorial. We will be looking at extensions of the famous
Evans’ conjecture on the completability of partially filled latin squares. This
chapter can be thought of as appendix to the main thesis above, since most
of the results here represent only partial progress toward the solution of the
problems raised. This material is included since the methods discussed in
sections 3 and 4 below are improvements over previously published methods.
5.1
Partial Latin Squares
A partial latin square is a square array (n × n) where some of the cells
are filled with symbols from an alphabet of n symbols and where no row
or column contains two filled cells with the same symbol. A partial latin
square is called completable if the unfilled cells may be filled in such a way
as to form a latin square. Sometimes we also are concerned with the concept
of uniquely completable, but this notion does not enter into any of our new
results.
The problem of characterizing conditions that guarantee or forbid completability features the names of some famous mathematicians. Here we
state some of the results; for the proofs of these statements, see [5].
Theorem 5.1 (M. Hall) If the filled cells of an order-n partial latin square
consist of an r ×n rectangle for any 1 ≤ r ≤ n−1, the square is completable.
56
Example: If we are given a
given, as:
0
3
5
6 × 6 latin square with the first three rows
1 2 3 4 5
5 4 0 2 1
2 1 4 0 3
we can certainly finish it:
0
3
5
2
1
4
1
5
2
4
0
3
2
4
1
5
3
0
3
0
4
1
2
5
4
2
0
3
5
1
5
1
3
0
4
2
though this completion is not unique.
Theorem 5.2 (Evans-Smetaniuk) Suppose a partial latin square of order n has no more than n − 1 filled cells. Then the square is completable.
The proof of this theorem took quite a number of years from when it
was first conjectured by Evans in 1960; Smetaniuk provided a complicated
but fascinating inductive argument in 1981.
The number n − 1 for the number of filled cells is the largest possible
number we can give here; if there are n filled cells it is possible for the square
to be incompletable. Here are two examples with n = 4:
0
∗
∗
∗
1
∗
∗
∗
∗
0 ∗
3
∗ 0
or
∗
∗ ∗
∗
∗ ∗
2
∗
∗
∗
∗
∗
0
∗
∗
∗
∗
1
As an example of Evans’ conjecture working, here is a partial latin square
of order 6 with only 5 symbols given:
0
1
∗
∗
∗
∗
2
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
2
∗
∗
∗
∗
57
∗
∗
∗
∗
∗
∗
∗
∗
3
∗
∗
∗
We can complete this:
0
1
2
3
4
5
2
3
4
0
5
1
3
5
0
4
1
2
4
2
5
1
0
3
5
0
1
2
3
4
1
4
3
5
2
0
The following theorem first appeared in [19], but its proof and several
extensions may also be found in [5].
Theorem 5.3 (Ryser) Suppose that a partial latin square of order n has
filled cells consisting only of a single completely filled r × s rectangle for
1 ≤ r, s ≤ n. This is completable is and only if each of the n symbols occurs
at least r + s − n times in these filled cells.
For example, here is a partial order-6 latin square whose filled symbols
fill a 4 × 4 rectangle, and in which each symbol occurs at least twice. (Twice
because 2 = 4 + 4 − 6.)
0
4
5
3
∗
∗
1
0
3
2
∗
∗
2
5
4
0
∗
∗
3
1
0
4
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
0
4
5
3
1
2
1
0
3
2
5
4
2
5
4
0
3
1
3
1
0
4
2
5
4
2
1
5
0
3
5
3
2
1
4
0
This is completable:
On the other hand, if we have a symbol that does not occur twice (1
occurs only once in this example) we get an incompletable square:
58
0
4
5
3
∗
∗
1
0
3
2
∗
∗
2
5
4
0
∗
∗
3
2
0
4
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
Ryser’s theorem does have one consequence which makes it in essence
an extension of the M. Hall theorem:
Corollary 5.4 Any partial latin square of order n whose filled cells fit into
a rectangle of size r × s where r + s < n is completable.
These results are the cleanest theorems we have at this time about partial latin squares. A survey of the literature shows that much active research
is underway further characterizing completable and incompletable squares.
Most of these results have lengthy combinatorial proofs which rely on computer assistance. It is no wonder the problem is so difficult; in fact it is now
known that determining whether a given partial latin square is completable
is actually an NP-complete problem (see [3]).
5.2
Blocking Sets
We will now turn our attention to larger-dimensional latin objects. We may
define a partial latin hypercube of dimension d, type j, and order n as an
array with some cells filled that meets the criteria given in Defintion 1.8.
We will be mostly concerned in this work with d = 3, and j = 1 or 2.
When d = 3, j = 2, the partial latin cube has no repeated symbols in any
of the “rows” in all three directions. When d = 3, j = 1, the condition is
a good deal weaker; here we require only that in any two-dimensional slice
(fixing one of the three coordinates) among the filled cells, no symbol occurs
more than n times.
All of the above results on partial latin squares are now considered standard; however, as we attempt to extend these results to the realm of partial
latin cubes and hypercubes, we find that little progress has been made.
This is probably due to the fact that the most natural analouges of
the above statements appear to be false for latin cubes. For instance, no
interesting version of the M. Hall theorem (Thm. 5.1) works for either j = 1
or j = 2 when d = 3. Since this result is an important lemma in the proofs
of almost all the other results, we are left with very little to work with.
59
Here, we make some preliminary progress on an extension of Evans’
conjecture (Thm. 5.2) to d = 3, j = 1 or 2. Specifically, we shall construct
configurations of a small number of given cells which are not completable.
We call an incompletable configuration a blocking set, and we denote by
B(d, j; n) is the size of the smallest blocking set in a dimension-d, type-j,
order-n hypercube.
For latin squares, Evans’ conjecture (Thm. 5.2) says B(2, 1; n) = n.
We note the following immediate results:
Lemma 5.5 If 1 ≤ j ≤ d − 1, B(d + 1, j + 1; n) ≤ B(d, j; n).
Proof. Every hypercube of dimension d + 1, type j + 1 has the property
that each dimension-d sub-array is a latin hyperpcube of type j. Therefore,
given a blocking set at (d, j) we may construct the same set inside any ddimensional subarray of a dimension-(d + 1) hypercube to block it as well.
Lemma 5.6 If 1 ≤ j ≤ d − 1, B(d + 1, j, ; n) ≤ B(d, j; n) · n.
Proof. Given a blocking set for a dimension-d, type-j cube, construct
a partial dimension-(d + 1) hypercube by stacking n identical copies of this
blocking set in the new direction. This is now blocked from beimg completed
as a type-j hypercube. Putting these together:
Theorem 5.7 B(d, j; n) ≤ nd−j .
Proof. By induction on d and j using the two lemmas, beginning with
the base-case of Evans’ conjecture, B(2, 1; n) = n. One conjecture that we could make at this stage is that the inequality in
Lemma 5.5 should actually be equality, but in no case is a proof of this result
completely finished. What this would mean if true is that all we would have
to do is determine the behavior of B(d, 1; n).
5.3
Blocking Sets for d = 3, j = 1.
Above, in Thm. 5.7 we established that B(3, 1; n) ≤ n2 . In this section, we
will sharpen this bound. Since this was based on a sort of “brute-force”
construction, we should do better by using a bit more finesse.
60
For this section, when giving examples of latin cubes (partial or full)
we give each level as a square array separately; these are then meant to be
stacked atop one another, so that the upper-left corners of each square form
one vertical column. We will give the coordinates of the cells as a triple;
the first two coordinates denote the location within a square and the third
which square in the list. All coordinates will run from 0 to n − 1.
For instance, in the following blocked type-1 cube of order 4,
0
0
0
0
1
1
1
1
2
2
2
2
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
3
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
The cell with the symbol 3 has coordinates (0, 3, 1): row 0 and column 3
within a square, and square number 1 (square 0 being the leftmost square).
As a side note, this example shows that B(3, 1; 4) ≤ 13.
But we can improve this: we can use the following construction to bound
B(3, 1; 4) ≤ 11:
0
1
2
∗
0
1
2
∗
0
∗
∗
∗
∗
∗
∗
3
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
3
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
3
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
3
This configuration is not completable since, in the bottom-most square
(the square at the far left, number 0) we must have four instances of the
symbol 3, but the last row and column of this slice may not contain any more
3s due to the four 3s already present in the perpendicular faces. Therefore,
there are only two free cells in which we are required to place three more 3s,
impossible.
This construction may be generalized: form a blocking set in a latin cube
of type 1 and order n by (1) placing the symbol n − 1 in the n cells with
coordinates (n − 1, n − 1, i) for 0 ≤ i ≤ n, and (2) placing any of the symbols
0 through n − 2 in the cells with coordinates (j, k, 0) where 0 ≤ j, k ≤ n − 2
leaving precisely n − 2 of these cells blank.
Proposition 5.8 If n ≥ 4, we have B(3, 1; n) ≤ n2 − 2n + 3 using the above
construction.
Proof. The fact that this is blocking set follow the same argument as
for the example above; we count the number of filled cells as n filled in step
61
(1) and (n − 1)2 − (n − 2) = n2 − 3n + 3 cells from step (2), given the total
in the statement. However, this construction is not best possible, since we can also show
B(3, 1; 4) ≤ 9 with the following:
0
0
∗
∗
0
0
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
0
0
∗
∗
∗
∗
∗
∗
0
0
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
1
Generalizing this:
Proposition 5.9 If n is a perfect square, B(3, 1; n) ≤ n2 − n3/2 + 1.
√
Proof. This is achieved by placing cubes of side length n filled with
√
0s up the main diagonal in the first n− n rows and columns (specifically, in
√
√
√
all cells (i, j, k) for each 0 ≤ i, j, k, ≤ n−1 and also n ≤ i, j, k ≤ 2 n−1,
√
√
etc. up to n − 2 n ≤ i, j, k ≤ n − n − 1), then placing a lone 1 at the
extreme corner, (n − 1, n − 1, n − 1). However, in general even this is not best possible. Consider the following
construction at n = 9, where the last result gave a blocking set of size 55,
but here we do better:
0
∗
∗
∗
(0)
∗
∗
∗
∗
(0)
0
∗
∗
∗
∗
∗
∗
∗
∗
(0)
0
∗
∗
∗
∗
∗
∗
∗
∗
(0)
0
∗
∗
∗
∗
∗
∗
∗
∗
(0)
0
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
(x8)
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
1
5
∗
∗
∗
∗
∗
∗
∗
2
6
∗
∗
∗
∗
∗
∗
∗
3
7
∗
∗
∗
∗
∗
∗
∗
4
8
Explaining further: we have the first square as shown copied eight times,
with the cells given parenthetically only used in one of the eight faces (which
one is not important). The ninth and “top” square is then given. Here, then
we have 45 given 0s and 8 given symbols in the top square for a total of only
53 symbols. In this top face, we need to place nine 0s, but there is not
enough room in the bottom-right square to do so.
This procedure in generalized as follows: let k 2 > n. Then fill the first
n − k rows and columns with 0s without using the top square, which will
take n2 − kn 0s. Now we need n 0s to be placed in the top square, but only
62
a k × k square in which to place them. If we put k 2 − n + 1 other symbols
in this part of the top square, the configuration is blocked. So our bound is
n2 − kn + k 2 − n + 1.
If we optimize this to give the smallest size set among the choices of k
√
for n < k < n, we get k = n/2 if n is even; and if n is odd both k = n−1
2
and k = n+1
achieve
the
same
minimum.
Substituting
these
in,
we
have:
2
3
B(3, 1; n) ≤ n2 − n + 1. (n even.)
4
3
5
B(3, 1; n) ≤ n2 − n + . (n odd.)
4
4
We combine these results to achieve the following:
Theorem 5.10 If n ≥ 5,
3 2
B(3, 1; n) ≤
n −n+1 .
4
This is the best general bound we have been able to achieve; it is a bit
unexpected that it is asymptotically smaller than n2 .
Although we could tentatively conjecture that this bound is best possible,
it seems unlikely that a proof of this conjecture is at all attainable with
current methods.
Conjecture
3 2
5.11 If a partial latin cube of type 1 contains no more than
n
−
n
symbols, then the cube is completable.
4
5.4
Extending Evans to d = 3, j = 2.
We now examine the problem of attempting to extend Evans’ conjecture to
latin cubes of type 2. This problem has been more heavily studied than the
type-1 problem since classically, the term ‘latin cube’ was most often used
to signify what we call type-2 objects. Although it is easy (indeed nearly
trivial) to see that B(3, 2; n) ≤ n, proving an equality is apparently quite
a challenging problem. Note that due to the fact that the proof outlined
here in incomplete, our discussion is somewhat more informal than previous
sections.
There does not appear to be any immediate way of using the fact that
no set of size n − 1 blocks a latin square to show that the same is true in a
type-2 latin cube. Also, Smetaniuk’s method of induction does not operate
63
for cubes. The technique which we explore here will use a technique called
exponentiation, the idea of which had previously appeared in the paper [11]
but which we extend slightly to handle more cases. (This paper offers an
excellent summary of the state of research into this problem at the time of
writing.) Indeed, we conjecture that the stated method is strong enough to
handle all cases, but the proof is not complete.
Definition 5.12 Let B and C denote two latin squares of order n. Suppose
that C0 is the square C, rearranged by column swapping so that the first
row contain the symbols in order 0 to n − 1. Let B(x, y), C0 (x, y) denote
the entries in the two squares at the x-th row and y-th column, where both
coordinates range from 0 to n − 1.
Then the latin cube B C is the cubical array whose (i, j, k)-th entry consists of the symbol C0 (B(i, j), k). This means that the base square (the
sub-array where k = 0) is just a copy of the square B; and each ‘vertical’
column (fixing i and j) of the cube is a copy of one of the columns of C.
As an example, take the order-5 squares
2
0
B= 4
1
3
4
3
0
2
1
0
2
1
3
4
1
4
3
0
2
0
2
C= 3
4
1
3
1
2
4
0
1
4
2
0
3
2
3
4
1
0
3
1
0
2
4
4
0
1
3
2
Then exponentiating:
C
B =
2
0
4
1
3
4
3
0
2
1
0
2
1
3
4
1
4
3
0
2
3
1
2
4
0
3
2
0
4
1
0
1
2
3
4
2
3
4
1
0
4
0
1
2
3
1
4
3
0
2
4
3
1
2
0
1
0
3
4
2
3
4
2
0
1
2
1
0
3
4
0
2
4
1
3
1
4
3
0
2
3
2
4
1
0
4
1
0
2
3
0
3
2
4
1
2
0
1
3
4
0
1
2
3
4
2
4
1
0
3
1
0
3
4
2
3
2
4
1
0
To illustrate, the vertical column formed by the upper-right entries at
each level is (3, 1, 0, 2, 4), a copy of the fourth column of C.
Our strategy for completing a given partial cube with n − 1 given cells
is to construct squares B and C for which B C agrees with the given cells.
The basic idea is illustrated by the following example, an order-6 cube
with five given symbols:
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
0
1
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
4
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
64
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
2
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
0
∗
...
...
...
...
...
...
4
3
0
2
1
All the given cells here lie in
square for C:
0
∗
2
C=
∗
∗
∗
just three columns so we have a partial
1
4
∗
∗
∗
∗
∗
∗
0
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
We can complete this (by the two-dimension Evans’ Theorem); for example:
0
3
2
C=
1
4
5
1
4
5
2
0
3
3
1
0
5
2
4
2
0
4
3
5
1
4
5
1
0
3
2
5
2
3
4
1
0
Then we form a partial base square B by placing the ‘seed’ of each
vertical column of the partial cube in its proper place; where the first symbol
is not given we take it from the completion of C:
∗
∗
∗
B=
∗
∗
∗
∗
∗
∗
∗
∗
∗
0
1
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
3
∗
2
3
4
5
0
1
0
1
2
3
4
5
3
4
5
0
1
2
4
5
0
1
2
3
5
0
1
2
3
4
This may be completed:
1
2
3
B=
4
5
0
65
Exponentiating:
1
2
3
BC =
4
5
0
2
3
4
5
0
1
0
1
2
3
4
5
3
4
5
0
1
2
4
5
0
1
2
3
5
0
1
2
3
4
4
0
1
5
2
3
0
1
5
2
3
4
3
4
0
1
5
2
1
5
2
3
4
0
5
2
3
4
0
1
2
3
4
0
1
5
5
4
0
1
3
2
4
0
1
3
2
5
2
5
4
0
1
3
0
1
3
2
5
4
1
3
2
5
4
0
3
2
5
4
0
1
...
...
...
...
...
...
This allows us to articulate the following result:
Theorem 5.13 Suppose a partial latin cube K of type 2 and order n with no
more than n − 1 symbols is given; let the collection of partially filled vertical
columns be called C. If the columns of C with duplicates deleted form the
columns of a partial latin square, then K is completablee.
Proof.
Given the partial square formed from C, since there are no
more than n − 1 symbols in it is must be completable; we shall denote th
completion C.
Now, form a partial square B under the following rule: if the cell (i, j, k)
is filled by symbol s in K, locate the column in C in which s occurs at the
k − th place, and let t be the symbol in the top cell of that column. Place
the symbol t in the (i, j)-th place in B. Repeat for all filled cells in K.
We claim that this B is indeed a partial latin square. It could only fail
in two ways; first, we could have tried to place two different symbols into
the same cell. Suppose the cell (i, j, k1 ) filled with s1 in K placed t1 at (i, j)
in B but that (i, j, k2 ) filled with s2 in K placed t2 at (i, j) in B. Then the
symbols s1 and s2 come from the same vertical column in K, so they are in
the same entry of C, so they are in the same column of C, so that t1 = t2 .
Second, we could have tried to place the same symbol twice into a row
or column. Without loss of generality, we assume it is a row: suppose that
(i, j1 , k1 ) was filled with s1 in K and placed t at (i, j1 ) in B; and that
(i, j2 , k2 ) was filled with s2 in K and placed t at (i, j2 ) in B. Now s1 and
s2 were from different vertical columns of K, but by assumption they are in
the same column of C (the one whose top entry is t) and hence the entries
of C must have been identical since otherwise they would have been in two
different columns of the partial latin square that formed C. However, this
would mean that s1 occurred at both (i, j1 , k1 ) and (i, j2 , k1 ), a contradiction.
Therefore, B is a partial latin square, and cannot have more than n − 1
entries. It is therefore completable. Then by construction, B C will be a
66
latin cube which agrees with K at all the filled cells, so K is completable.
We note a special case of the above, easier to articulate. This is the way
the result is stated in [11].
Corollary 5.14 Suppose a partial latin cube K of type 2 and order n with
no more than n − 1 symbols is given, and that all the filled cells of K occur
in different columns. Then K is completable.
Proof. In this case, the collection C of the vertical columns of K each
have only one entry in them; therefore, they can all be placed as separate
columns of a partial latin square unless two are identical, in which case once
duplicates are deleted we have no further problems. It is perfectly possible to construct partial cubes K for which the collection C has columns which are not identical but cannot form distinct columns
of a partial square. For example, if
0 0 ∗
∗ 1 ∗
2 ∗ 2
},
,
C={ ,
∗ ∗ ∗
∗ ∗ ∗
∗ ∗ ∗
these three columns cannot go separately into C; if we had only the first two
to deal with, we could simply start with one column beginning [0, 1, 2, ...]T ;
but in the actual case we might have:
0
∗
∗
K= ∗
∗
∗
∗
∗
0
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
1
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
and here we would be trying to have
0 ∗ ∗ ∗ ∗ ∗ ∗
B= ∗ 0 ∗ 0 ∗ ∗ ∗
..
..
..
.
.
.
67
2
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
2
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
...
...
...
...
...
...
...
A problem can also occur in the construction of C itself; we might have
0
∗
K= ∗
∗
∗
∗
2
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
1
∗
∗
∗
1
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
...
...
...
...
...
where we would be trying to start C as
0
1
C? = ∗
∗
∗
2
1
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
which is not a legitimate partial latin square.
These two issues mean that the method of Thm. 5.13 needs to be tweaked
before we can handle a partial cube which does not fulfill the conditions of
the theorem.
We achieve this adjustment by using intercalates. An intercalate is a 2×2
A B
subsquare of a latin square that takes the form
. The reason these
B A
are effective tools is that where present, they can be exchanged (switching
the two symbols A and B in these four cells only) and we have a new latin
square with only a few different cells.
In the last example, we could take
0
1
C= 3
∗
∗
2
3
1
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
∗
and now, by exchanging the bolded symbols, we could get either desired
column. This square can be completed (although there are now more than
four given symbols, so Evans’ Theorem does not apply):
0
1
C= 3
2
4
2
3
1
4
0
68
1
0
4
3
2
3
4
2
0
1
4
2
0
1
3
We could then introduce an intercalate into the base square B as well:
0
3
B= 4
2
1
3
2
0
1
4
1
0
2
4
3
4
1
3
0
2
2
4
1
3
0
and
0
3
C
B = 4
2
1
3
2
0
1
4
1
0
2
4
3
4
1
3
0
2
2
4
1
3
0
1
4
2
3
0
4
3
1
0
2
0
1
3
2
4
2
0
4
1
3
3
2
0
4
1
3
2
0
1
4
4
1
3
4
0
2
3
1
0
2
0
4
2
3
1
1
0
4
2
3
...
...
...
...
...
Here, three of the four desired cells in K are met; although there is a
3 where there should be a 1 at position (1, 1, 1); by exchanging the cubical
intercalate of the symbols 1 and 3, we get an altered latin cube which has a
1 in the correct place and still has the other three symbols correct as well.
For larger partial cubes, multiple intercalates may need to be introduced.
We therefore have two things to show before a proof encompassing all possible partial cubes is complete: first we must limit the number of intercalates
needed given the order n of the cube; second we must show that the partial
latin squares where more than n − 1 symbols are used can be completed
provided that some of the given cells form intercalates.
We conjecture as follows:
Conjecture 5.15 Given any partial latin cube K, the set C of its vertical
columns can be adapted to a suitable latin square C with the use of no more
than n/2 − 1 intercalates.
Conjecture 5.16 If a partial latin square consists of k intercalates and t
other symbols, then if k < n/2 − 1 and 2k + t < n, the square is completable.
If both conjectures were true, we would have established that any ordern partial latin cube with n − 1 given symbols could be completed and also
would have established an algorithm for doing so.
We also should touch briefly on a notion which, in practice, usually
simplifies the problem of completing a given partial latin cube. If the conditions of Thm. 5.13 are not met, usually because there is a clash between two
columns, we can often resolve the problem by simply rotating the partial
69
cube, treating a different direction as vertical. In the last example, treating
either of the other two directions as vertical would mean that all symbols
lay in different columns so the cube is completable by Cor. 5.14. The smallest example where any of the three directions would require an intercalate
would have n = 9, and would have the given entries all in one corner, looking
like:
0 1 ··· 1 0 ··· ···
K = 1 0 ··· 0 2 ··· ···
.. ..
.. ..
. .
. .
In this special case, an extension of the corollary to Ryser’s Theorem
(Cor. 5.4) to cubes would handle this case without any intercalates at all.
This idea opens up another avenue of rescuing the former line of proof if
either of the above conjectures is false: if we use the freedom to choose the
direction of ‘vertical,’ how far can the number of required intercalates be
reduced (meaning that a stronger Conj. 5.15 is true), and is this enough to
mean a weaker version of Conj. 5.16 will finish the proof? If even this fails
sometimes, can a extension of Ryser’s theorem be used to take care of these
cases?
In summary, the method of introducing intercalates is a promising but
not yet rigorous method for finishing the proof of Evans’ conjecture for latin
cubes. We cannot construct any particular examples for which this method
fails, although we also cannot yet prove that such examples do not exist. In
future work, we shall hope to prove the conjectures 5.15 and 5.16 to close
the door on this problem.
70
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71
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72
VITA
Daniel R. Droz was born in Lancaster, PA in 1990; he graduated from
Veritas Academy there in 2006. He attended Franklin and Marshall College
and completed a B.A. summa cum laude in Mathematics and Physics with
a minor in Music, graduating in 2010. He then enrolled in the Mathematics
program at the Pennsylvania State University, where with this dissertation
he completed a Ph.D. in Mathematics in 2016. Daniel has received numerous
awards for his scholarly and pedagogical work, including induction into Phi
Beta Kappa at F&M and the Charles H. Hoover Memorial Award in teaching
at Penn State. Daniel’s diverse outside interests include opera, Victorian
literature, golf, and contract bridge.