The Map, A New Method to Define Latin Square

2019 International Conference on Computation and Information Sciences (ICCIS 2019) ISBN: 978-1-60595-644-2

The Map, A New Method to Define

Latin Square

Zhaoqi Zhang

ABSTRACT

A Latin Square is generally defined as a matrix, in which each entry occurs exactly once in each row and exactly once in each column, and the range of the entries can be from 1 to the order of the matrix. In the past, the researches of Latin Squares were mainly focused on the 2-dimension case, because the definition as matrix limits the extension of the dimension. In this paper, the Latin Square is redefined by map and is renamed as Latin map. This new definition enables Latin map to describe high dimensional extended Latin square. By exploiting the properties of bijection, the propositions and theorems of Latin square are rebuilt to be algebraic results instead of the usual combinatoric results. The behaviors of Latin square are discussed when regarded as map such as the composition of Latin maps, which finally comes to the Extending Construction theorems.

1. INTRODUCTION

In combinatorics and in experimental design, a Latin square is an 𝑛 × 𝑛 array filled with 𝑛 different symbols, each occurring exactly once in each row and exactly once in each column. An example of a 3 × 3 Latin square is:

[𝐴 𝐵 𝐶𝐶 𝐴 𝐵 𝐵 𝐶 𝐴

]

One of the frequent research questions of Latin Squares is the orthogonal Latin square. The orthogonal Latin square discusses the behaviour of the conjunction of two Latin squares. Orthogonality here means that every pair of the entries in the conjuntion occurs exactly once. Here’s an example of orthogonal Latin squares:

[

(1,1) (2,2) (3,3) (4,4) (2,3) (1,4) (4,1) (3,2) (3,4) (4,3) (1,2) (2,1) (4,2) (3,1) (2,4) (1,3)

]

Orthogonal Latin squares were studied in detail by Leonhard Euler. He demonstrated methods for constructing orthogonal Latin squares where 𝑛 is odd or a multiple of 4. In April 1959, Parker, Bose, and Shrikhande showed that the orthogonal Latin squares exist for all orders 𝑛 ≥ 3 except 𝑛 = 6.[10]

This paper describes a new approach to use map to define Latin square, which results in a way to extend the dimension of Latin square. With this new definition of Latin square as map, questions related with Latin square about the normal extension and partial maps which are defined in this paper can be proved by the description of map. Two traditional questions, the orthogonality and tensor product1 is described by map in this paper, which comes to a concise expression of Euler’s theory. This paper in the end discusses a new question derived from the new definition. The question is the composition of Latin maps. The discussion comes to the Extending Construction theorem.

2. DEFINITION OF THE LATIN MAPS

2.1 Grid Space

To describe the Latin square by map, we start from the definition of the domain of the map. This domain is defined as grid space. The structure of a grid space determines the structure of a Latin map by a great means.

2.1.2 GRID SET AND GRID SPACE

Let 𝐺_{𝑛1𝑛2⋯𝑛𝑚} = {(𝑘₁, 𝑘₂, ⋯ , 𝑘_𝑚) ∈ ℤ𝑚|0 ≤ 𝑘_𝑖 ≤ 𝑛_𝑖− 1,1 ≤ 𝑖 ≤ 𝑚}, where ℤ𝑚

is the cartesian product of ℤ for 𝑚 times. Denote 𝐺_{𝑛1𝑛2⋯𝑛𝑚} by 𝐺_𝑛𝑚_{, if} 𝑛₁ = 𝑛₂ =

⋯ = 𝑛𝑚 = 𝑛. Also,set 𝐺𝑛 = 𝐺𝑛1. By definition, 𝐺𝑛1𝑛2⋯𝑛𝑚 is the direct product of the

𝐺𝑛𝑖s, i.e. 𝐺𝑛1𝑛2⋯𝑛𝑚= 𝐺𝑛1 × 𝐺𝑛2 × ⋯ × 𝐺𝑛𝑚.

Definition2.1 Call 𝐺_{𝑛1𝑛2⋯𝑛𝑚} = {(𝑘₁, 𝑘₂, ⋯ , 𝑘_𝑚) ∈ ℤ𝑚|0 ≤ 𝑘_𝑖 ≤ 𝑛_𝑖− 1,1 ≤ 𝑖 ≤ 𝑚} a grid set for 𝑛₁𝑛₂⋯ 𝑛_𝑚. Each element is called a grid.

The grid sets that is mainly discussed in this paper are the 𝐺_𝑛𝑚 _{type as defined}

1

above, where there is a natural definition for maps. Thus, I give a special definition to these kinds of sets.

Definition 2.2 If 𝑛 > 1, call 𝐺𝑛𝑚_an 𝑚_{-dimensional grid space of level} 𝑛_.

If the integer 𝑛 is a prime number 𝑝, 𝐺_𝑝𝑚 _{is isomorphic to the} 𝑚-dimensional

vector space over the field ℤ_𝑝.

In a grid space 𝐺𝑛𝑚_{, all grids are arranged elegantly as an} 𝑚-dimensional cube.

Larger cubes contain many smaller cubes. Some of them are lower dimensional, while some have shorter sides. Call these smaller cubes the subspace of 𝐺𝑛𝑚_{, no matter how}

they are embedded into the larger cubes.

To define the subspace more conveniently, I consider using the direct product of an original point set and a lower dimensional Grid space to represent a cube that hangs in the air. Thus, define 𝑂𝑛 = {(0,0, ⋯ ,0) ∈ ℤ𝑛}, i.e. an original point with 𝑛 coordinates that are all 0.

Definition 2.3 Let 𝐴 be a subset of the grid space 𝐺_𝑛𝑚 _{and choose a grid} 𝑥 in 𝐴.

Call 𝐴 a subspace of 𝐺_𝑛𝑚 _{if there is a grid space} 𝐺_𝑘𝑙, where 𝑘 ≤ 𝑛, 𝑙 ≤ 𝑚, such that

each grid in 𝐴 can be represented by the sum in the 𝑏𝑍𝑚 of the coordinate of 𝑥 with the coordinate of a point in 𝐺_𝑘𝑙 × 𝑂𝑚−𝑙. The point in 𝐺_𝑘𝑙 × 𝑂𝑚−𝑙 is called the origin.

Remark 2.4 In the product set 𝐺_𝑘𝑙 × 𝑂𝑚−𝑙, how the 𝑚 − 1 coordinates zeros of

𝑂𝑚−1 _{are inserted into the coordinates of the point in 𝐺}

𝑘𝑙 is not determined. For an

uncertain product, these zeros may be inserted in any place. For a certain set 𝐺_𝑘𝑙×

𝑂𝑚−𝑙_{, these zeros should be inserted in the same place compared with 𝐺}

𝑘𝑙. In this way,

the structure of 𝐺_𝑘𝑙× 𝑂𝑚−𝑙 is the same with the structure of 𝐺_𝑘𝑙.

2.1.2 UNIT

To describe the limit of Latin squares that each element in a certain part are different to each other, the concept of the unit is introduced.

Definition 2.5 In a grid space 𝐺_𝑛𝑚_{, the subspaces that have the same structure with}

𝐺_𝑛𝑘 (𝑘 ∈ 𝐺_(𝑚+1)) are called a 𝑘-unit. That is the set of 𝑚 − 𝑘 constant coordinates

while each of other coordinates take over all grids in 𝐺_𝑛.

By definition, each unit in a grid space is a subspace. Conversely, each subspace with level 𝑛 in 𝐺𝑛𝑚 _{is a unit. Thus we have the proposition:}

Proposition 2.6 In a unit in a grid space with level 𝑛, each coordinate either be a

constant, or takes over the 𝐺_𝑛.

The inverse proposition is also correct, proved by definition. Each unit is a grid space. So, it contains its units of all dimensions. As the level of all units are 𝑛, the unit of a unit is still a unit. Here call a unit of a unit a Subunit.

All trivial subspaces are units. The total space 𝐺_𝑛𝑚 _{is an} 𝑚-dimensional unit. The

single grid is a 0-dimensional unit. Here call the trivial subspaces as trivial units. Other units are non-trivial units.

Proposition 2.7 The intersection of two units in a grid space 𝐺_𝑚𝑛 is either an empty set or a unit.

respectively. The constant coordinates of 𝑈₁ are 𝑋_𝑖0 = 𝑥_𝑖0, 𝑋_𝑖1 = 𝑥_𝑖1, ⋯ , 𝑋_{𝑖𝑚−𝑘1−1} =

𝑥_{𝑖𝑚−𝑘1−1}. The constant coordinates of 𝑈₂ are 𝑋_𝑗0 = 𝑥_𝑗0, 𝑋_𝑗1 = 𝑥_𝑗1, ⋯ , 𝑋_{𝑗𝑚−𝑘2−1} =

𝑥_{𝑗𝑚−𝑘2−1}. Other coordinates takes over 𝐺_𝑛.

If {𝑖₀, 𝑖₁, ⋯ , 𝑖_{𝑚−𝑘1−1}} ∩ {𝑗₀, 𝑗₁, ⋯ , 𝑗_{𝑚−𝑘2−1}} = ∅, the 𝑈₁∩ 𝑈₂ is a subspace that has constant coordinates

{𝑖₀, 𝑖₁, ⋯ , 𝑖_{𝑚−𝑘1−1}} ∪ {𝑗₀, 𝑗₁, ⋯ , 𝑗_{𝑚−𝑘2−1}}.

Other coordinates take over 𝐺_𝑛. So it is a 𝑘₁+ 𝑘₂− 𝑚-dimensional unit.

If {𝑖0, 𝑖1, ⋯ , 𝑖𝑚−𝑘1−1} ∩ {𝑗0, 𝑗1, ⋯ , 𝑗𝑚−𝑘2−1} ≠ ∅ , let {𝑖0, 𝑖1, ⋯ , 𝑖𝑚−𝑘1−1} ∩

{𝑗₀, 𝑗₁, ⋯ , 𝑗_{𝑚−𝑘2−1}} = {𝑟₀, 𝑟₁, ⋯ , 𝑟_{𝑐(𝑟)−1}}, and the symmetric difference

{𝑖0, 𝑖1, ⋯ , 𝑖𝑚−𝑘1−1} △ {𝑗0, 𝑗1, ⋯ , 𝑗𝑚−𝑘2−1} = {𝑠0, 𝑠1, ⋯ , 𝑠𝑐(𝑠)−1}.

If ∃𝑟 ∈ {𝑟₀, 𝑟₁, ⋯ , 𝑟_{𝑐(𝑟)−1}} such that 𝑋_𝑟 ≡ 𝑎 in 𝑈₁ while 𝑋_𝑟 ≡ 𝑏 in 𝑈₂, where

𝑎 ≠ 𝑏. Then 𝑈1∩ 𝑈2 = ∅ due to the contradiction of the 𝑋𝑟 in the two units.

If ∀𝑟 ∈ {𝑟0, 𝑟1, ⋯ , 𝑟𝑐(𝑟)−1}, 𝑋𝑟 has same value in 𝑈1 and 𝑈2. Then, 𝑈1∩ 𝑈2

satisfies that the coordinates

{𝑟₀, 𝑟₁, ⋯ , 𝑟_{𝑐(𝑟)−1}} ∪ {𝑠₀, 𝑠₁, ⋯ , 𝑠_{𝑐(𝑠)−1}}

are constants. Other coordinates take over 𝐺_𝑛. So it is a 𝑚 − 𝑐(𝑟) − 𝑐(𝑠)-dimensional unit. 

The properties of units are similar to linear spaces. For example, consider the 2-dimensional units, their intersection may be a 2-dimensional or 1-dimensional unit or empty set. This corresponds to the fact that two plains in an Euclidian space maybe coincide, intersect or in different spaces. When regarded geometrically, each unit is a set represented by linear equations. Thus, each unit is the total subspace of a linear space. So the properties of linear spaces can be analogized into grid spaces.

Definition 2.8 Let 𝐴 be a subset of 𝐺𝑛𝑚_, 𝐺_𝑛𝑘 be a unit of 𝐺_𝑛𝑚_{. If} 𝐴 ⊆ 𝐺_𝑛𝑘 and for

any unit 𝐺_𝑛𝑙 of 𝐺_𝑛𝑚 _{that contains} 𝐴_{, we have} 𝐺_𝑛𝑘 ⊆ 𝐺_𝑛𝑙. Then, call 𝐺_𝑛𝑘 the

minimal generated unit of 𝐴.

By the partial ordering relation of containing, the minimal generated unit is unique. It is

𝐺_𝑛𝑘 = ⋂ 𝐺_𝑛𝑙

𝐴⊆𝐺_𝑛𝑙

.

For 𝐴 ⊆ 𝐺_𝑛𝑚, let 𝑥_𝑖 be the value of the 𝑖-th component that has the same value over all grids 𝑥 in 𝐴. Then, the set 𝐺(𝐴) = {𝑋 ∈ 𝐺_𝑛𝑚|𝑋_𝑖 = 𝑥_𝑖, ∀𝑥 ∈ 𝐴} is a unit of

𝐺_𝑛𝑚.

Proposition 2.9 For any 𝐴 ⊆ 𝐺_𝑛𝑚_, 𝐺(𝐴) _{is the minimal generated unit of} 𝐴_.

𝑥𝑖0, 𝑥𝑖1⋯ , 𝑥𝑖𝑚−𝑘−1 be their value.

𝐺(𝐴) = {(𝑋₀, 𝑋₁, ⋯ , 𝑋_𝑚−1)|𝑋_𝑖0 = 𝑥_𝑖0, 𝑋_𝑖1 = 𝑥_𝑖1⋯ , 𝑋_{𝑖𝑚−𝑘−1} = 𝑥_{𝑖𝑚−𝑘−1}}

By definition, 𝐺(𝐴) is a unit.∀𝑎 = (𝑎₀, 𝑎₁, 𝑎_𝑚−1) ∈ 𝐴, when 𝑖 ∈ {𝑖₀, 𝑖₁, ⋯ , 𝑖_{𝑚−𝑘−1}}, 𝑎𝑖 = 𝑥𝑖. When 𝑖 ∉ {𝑖0, 𝑖1, ⋯ , 𝑖𝑚−𝑘−1}, 𝑎𝑖 ∈ 𝐺𝑛. Thus, 𝑎 ∈ 𝐺(𝐴), i.e. 𝐴 ⊆ 𝐺(𝐴).

Let 𝑈 be a unit that contains 𝐴. Then ∀𝑋 ∈ 𝐺(𝐴), let 𝑋_𝑖−1 be its 𝑖-th

component. If there are two different grids in 𝐴 that their 𝑖-th component are different. Then the 𝑖-th component of grids in 𝑈 takes over 𝐺_𝑛. If the 𝑖-th component values constant over 𝐴, let this constant be 𝑋_𝑖−1. Then the 𝑖-th component of elements over 𝑈 is either 𝑋𝑖−1 or takes over 𝐺𝑛. Whatever it is, 𝑋 ∈ 𝑈. So 𝐺(𝐴) ⊆ 𝑈. Then we have

𝐺(𝐴) is the minimal generated unit of 𝐴.

The question about generated set can be related to the question of matroid[6], which will not be discussed in this paper.

2.1.3 DIMENSIONAL DISTANCE

Let’s describe the units in a grid space by the dimensional distance.

Definition 2.10 ∀𝑋 = (𝑥₀, 𝑥₁, ⋯ , 𝑥_𝑚−1) ∈ 𝐺_𝑛𝑚 _and 𝑌 = (𝑦₀, 𝑦₁, ⋯ , 𝑦_𝑚−1) ∈ 𝐺_𝑛𝑚_,

call

𝐷(𝑋, 𝑌) = ∑ 𝑠

𝑚−1

𝑘=0

𝑔𝑛|𝑥_𝑘− 𝑦_𝑘|

the dimensional distance between 𝑋 and 𝑌.

There are some basic properties of the dimensional distance, which can be used to describe the traits of units.

Proposition 2.11 Metric∀𝑋, 𝑌, 𝑍 ∈ 𝐺_𝑛𝑚, (1) 𝐷(𝑋, 𝑌) ≥ 0, 𝐷(𝑋, 𝑌) = 0 ⇔ 𝑋 = 𝑌 (2) 𝐷(𝑋, 𝑌) = 𝐷(𝑌, 𝑋)

(3) 𝐷(𝑋, 𝑌) ≤ 𝐷(𝑋, 𝑍) + 𝐷(𝑍, 𝑌)

Proof. (1) ∀𝑋, 𝑌 ∈ 𝐺_𝑛𝑚_, |𝑥_𝑘− 𝑦_𝑘| ≥ 0. The equivalent comes if and only if

𝑥_𝑘 = 𝑦_𝑘. Thus, 𝐷(𝑋, 𝑌) ≥ 0, 𝐷(𝑋, 𝑌) = 0 ⇔ 𝑋 = 𝑌. (2) ∀𝑋, 𝑌 ∈ 𝐺𝑛𝑚_, |𝑥_𝑘− 𝑦_𝑘| = |𝑦_𝑘− 𝑥_𝑘|. So

𝐷(𝑋, 𝑌) = ∑ 𝑠

𝑚−1

𝑘=0

𝑔𝑛|𝑥𝑘− 𝑦𝑘| = ∑ 𝑠 𝑚−1

𝑘=0

𝑔𝑛|𝑦𝑘− 𝑥𝑘| = 𝐷(𝑌, 𝑋)

𝐷(𝑋, 𝑍) + 𝐷(𝑍, 𝑌) = ∑ 𝑠

𝑚−1

𝑘=0

𝑔𝑛|𝑋𝑘− 𝑍𝑘| + ∑ 𝑠 𝑚−1

𝑘=0

𝑔𝑛|𝑍𝑘− 𝑌𝑘|

≥ ∑ 𝑠

𝑚−1

𝑘=0

𝑔𝑛(|𝑋_𝑘− 𝑍_𝑘| + |𝑍_𝑘− 𝑌_𝑘|)

≥ ∑ 𝑠

𝑚−1

𝑘=0

𝑔𝑛|𝑋_𝑘− 𝑌_𝑘| = 𝐷(𝑋, 𝑌)

(2.11.1)

These properties show that the dimensional distance is a metric on 𝐺𝑛𝑚_{. This}

distance represents the dimension difference between any two points in 𝐺_𝑛𝑚_{. This}

metric induces a discrete topology on 𝐺𝑛𝑚_{. Obviously,} ∀𝑋, 𝑌 ∈ 𝐺_𝑛𝑚,0 ≤ 𝐷(𝑋, 𝑌) ≤

𝑚.

Proposition 2.12 Existence∀𝑘 ∈ 𝐺𝑚+1, ∃𝑋, 𝑌 ∈ 𝐺𝑛𝑚_{, such that} 𝐷(𝑋, 𝑌) = 𝑘.

Proof. Let 𝑋𝑖 = (𝑥0, 𝑥1, ⋯ , 𝑥𝑚−1). If 𝑗 < 𝑖, 𝑥𝑗 = 0. If 𝑗 ≥ 𝑖, 𝑥𝑗 = 1. Then

∀𝑘 ∈ 𝐺_𝑚+1, 𝐷(𝑋₀, 𝑋_𝑘) = 𝑘. 

Corollary 2.13 ∀𝑡 ∈ 𝐺_𝑘+1, ∃𝑋, 𝑌 ∈ 𝐺_𝑛𝑚_{, and} 𝑋 _and 𝑌 _{are in a same} 𝑘_{-unit, such}

that 𝐷(𝑋, 𝑌) = 𝑡.

This corollary can be deduced by the property of "equivalent to unit" (will be discussed later) due to the existence of dimensional distance.

Proposition 2.14 Equivalent to Unit∀𝑋, 𝑌 ∈ 𝐺_𝑛𝑚, 𝐷(𝑋, 𝑌) ≤ 𝑘 ⟺ 𝑋, 𝑌is in the same

𝑘-unit.

Proof.𝐷(𝑋, 𝑌) ≤ 𝑘 ⇐ 𝑋, 𝑌 in a same 𝑘-unit:

Let 𝑋 and 𝑌 in a same 𝑘-unit. There are at least 𝑚 − 𝑘 components are the same between them. So, among all 𝑠𝑔𝑛|𝑥_𝑖 − 𝑦_𝑖|, there are at least 𝑚 − 𝑘 terms are 0. Thus, the terms of 𝑠𝑔𝑛|𝑥_𝑖 − 𝑦_𝑖| that value 1 are no more than 𝑘. This means

𝐷(𝑋, 𝑌) = ∑ 𝑠

𝑚−1

𝑖=0

𝑔𝑛|𝑥_𝑖 − 𝑦_𝑖| ≤ 𝑘

𝐷(𝑋, 𝑌) ≤ 𝑘 ⇒ 𝑋, 𝑌 is in a same 𝑘-unit:

If 𝐷(𝑋, 𝑌) ≤ 𝑘, then there are at most 𝑘 terms over 𝑠𝑔𝑛|𝑥_𝑖 − 𝑦_𝑖| are 1, at least 𝑚 − 𝑘 terms are 0. Choose the minimal generated unit 𝐺(𝑋, 𝑌), then there are at least 𝑚 − 𝑘 terms values the constant 𝑥_𝑖 over 𝐺(𝑋, 𝑌), so this is a unit that has dimension no more than 𝑘. 𝐺(𝑋, 𝑌) is naturally included in some 𝑘-unit, thus 𝑋 and 𝑌 is in a same 𝑘-unit.

This proposition shows that the concept of dimensional distance is equivalent to units. So it is only needed to count the dimensional distance between two points when trying to judge if they are in a same 𝑘-unit.

Corollary 2.15 ∀𝑋, 𝑌 ∈ 𝐺𝑛𝑚_, 𝐷(𝑋, 𝑌) = 𝑘 ⟺ 𝑋, 𝑌 _{in a same} 𝑘_{-unit and} 𝑋, 𝑌 _are

not in any same 𝑘 − 1-unit.

2.2 Latin Map

2.2.1 THE MAP REPRESENTATION OF LATIN SQUARES

Definition 2.16 An 𝑚-dimensional Latin Square of level 𝑛 with rank 𝑘 is a map 𝑓: 𝐺_𝑛𝑚 → 𝐺_𝑛𝑘, 𝑘 ≤ 𝑚, such that the limitation of 𝑓 on each 𝑘-unit is injective. This

map is called the Latin Map. If 𝑘 = 𝑚 or 𝑘 = 0, call this map a trivial Latin map. If 𝑘 = 1 and 𝑚 = 2, call this map the minimal non-trivial Latin map.

The Latin squares that used to be discussed are the type of 2-dimension with rank 1. These are the most common Latin maps. Most common questions are based on these kind of squares, such as the discussion of orthogonal Latin squares.

There are many Latin maps that is more than 2 dimension. For example, the map 𝑓(𝑥, 𝑦, 𝑧) ≡ 𝑥 + 𝑦 + 𝑧 mod 3. In any 1-unit, there are two of 𝑥, 𝑦 and 𝑧 that value constant, so the image is bijective to the component that takes over 𝐺₃. In themodule class, different grids value different in the unit, thus 𝑓 is injective on 1-units. So 𝑓 is a 3-dimensional Latin map.

The following is a basic proposition:

Proposition 2.17 An 𝑚-dimensional Latin map of level 𝑛 with rank 𝑘 is bijective on each 𝑘-unit.

The number of grids in each 𝑘-unit is equal to the 𝐺_𝑛𝑘’s. According to the

cardinality theorem of finite set, the injections are surjections.

Proposition 2.18 Let 𝑓 be an 𝑚-dimensional Latin map of level 𝑛 with rank 𝑘. Then

∀𝑡 ∈ 𝐺_(𝑚+1) and 𝑡 ≥ 𝑘, the limitation of 𝑓 on any 𝑡-unit of 𝐺_𝑛𝑚 is a Latin map with

rank 𝑘.

The change of domain does not change the bijectivity. So the limitation of 𝑓 on each 𝑘-unit is still bijective.

Definition 2.19 Let 𝑓 be an 𝑚-dimensional Latin map of level 𝑛 with rank 𝑘,

𝑡 ∈ 𝐺(𝑚+1) and 𝑡 ≥ 𝑘. Call the limitation of 𝑓 on a 𝑡-unit a submap of 𝑓. If 𝑡 = 𝑚

or 𝑡 = 𝑘, this submap is called a trivial submap.

Latin maps that have more than 2 dimension possess properties that the minimal non-trivial Latin maps do not possess. One of the most important property is that they might have non-trivial submaps. Non-trivial submaps imply the existence of partial properties on high-dimensional Latin maps.

2.2.2 REPRESENTATION MAP

To make the Latin maps visible, the concept of representation map is introduced.

Definition 2.20 Call any injection 𝑖: 𝐺_{𝑛1𝑛2⋯𝑛𝑚} → 𝐺_{(𝑛1𝑛2⋯𝑛𝑚)}a representation map.

The following propositions are obvious.

Proposition 2.21 Representation maps are bijective.

The representation map is used for compositing on Latin maps. Let 𝑓 be an 𝑚-dimensional Latin map of level 𝑛 with rank 𝑘, 𝑖 ∈ 𝐼_𝑛𝑚_{. Then the composite map}

𝑖 ∘ 𝑓 is the operation of filling numbers from 𝐺_(𝑛𝑘₎ into the grid space 𝐺_𝑛𝑚_{. This is the}

frequently used operation while discussing the minimal non-trivial Latin maps.

The representation map has another effect. Consider the representation map from 𝐺𝑛

to 𝐺_𝑛 (i.e. the 𝐼_𝑛), it is a bijection from 𝐺_𝑛 to itself. So, this is a permutation. 𝐼_𝑛 ≅ 𝑆_𝑛. If compositing the permutation group to any representation map in 𝐼_𝑛𝑚 , the composition is still bijective. So the composite map is still a representation map inin 𝐼_𝑛𝑚_.

The composition of permutation group with representation map means that one can composite the permutation group when filling numbers into 𝐺_𝑛𝑚_{. In another word, it is}

only needed to care about whether two numbers filled in two grids are the same rather than the calculation properties of the numbers.

The involve of Representation Map is to make the Latin squares easier to study. However, the representations of Latin maps with high-dimensional are too complicated to present. To simplify the representation, the following method is used.

Definition 2.22 Let 𝑋 = (𝑥₀, 𝑥₁, ⋯ , 𝑥_2𝑚−1) ∈ 𝐺_𝑛(2𝑚) , 𝑌 = (𝑦₀, 𝑦₁, ⋯ 𝑦_𝑚−1) ∈

𝐺_(𝑛2₎𝑚, ∀𝑚, 𝑛 ∈ ℤ∗. Let 𝑦_𝑖 = 𝑛 × 𝑥_𝑖 + 𝑥_𝑖+𝑚. Define map 𝜎₂: 𝐺_𝑛(2𝑚) ⟶ 𝐺_(𝑛2₎𝑚

𝑋 ↦ 𝑌 (2.22.1)

Call this map the standard quadratic map.

By the properties of division algorithm, the standard quadratic map is surjective. The cardinality of its domain and range are the same and are finite. So the standard quadratic map is bijective. This is the significant fundamental trait to construct a lower dimensional representation.

Definition 2.23 Let 𝑓 be a 2𝑚-dimensional Latin map of level 𝑛 with rank 2𝑘. 𝑔 is an 𝑚-dimensional Latin map of level 𝑛2 with rank 𝑘. If ∀𝑋 ∈ 𝐺_𝑛(2𝑚), we have

𝑔(𝜎₂(𝑋)) = 𝜎₂(𝑓(𝑋)). Then call 𝑔a quadratic lower-dimensional representation of 𝑓.

We have the commutative diagram:

Proof.As 𝜎₂ is bijective, ∀𝑌 ∈ 𝐺_(𝑛2₎𝑚, there ∃₁𝑋 ∈ 𝐺_𝑛(2𝑚) such that 𝑌 = 𝜎₂(𝑋).

For any 2𝑚-dimensional Latin map 𝑓 of level 𝑛 with rank 2𝑘, we directly define map 𝑔 by

𝑔: 𝐺_(𝑛2₎𝑚 ⟶ 𝐺_(𝑛2₎𝑘

𝜎₂(𝑋) ↦ 𝜎₂(𝑓(𝑋)) (2.24.1) Then 𝑔 is a quadratic lower-dimensional representation of 𝑓.

For an even-dimensional Latin map 𝑓 with even rank, the quadratic lower-dimensional representation is 𝑔 = 𝜎₂∘ 𝑓 ∘ 𝜎₂−1.

An example is representing a 4-dimensional Latin map of level 3 with rank 2 by a 2-dimensional Latin map of level 9 with rank 1. This representation forms a Sudoku square, which is a common puzzle game. This representation makes the discussion of Sudoku square in much ease[8].

Example 2.25 Fill the coordinate of grids from 𝐺₃4 into 𝐺₉2. This gives out a standard quadratic map. I give the result as follow. Each coordinate in a grid of 𝐺₉2 is the preimage of the grid under 𝜎₂ (the number 𝑎𝑏𝑐𝑑 represents the coordinate (𝑎, 𝑏, 𝑐, 𝑑)):

Moreover, I fill grids from 𝐺₃2 into 𝐺₉2. This represents a Latin map from 𝐺₃4 to 𝐺₃2. To make it more convenient, I composite a natural representation map 𝐺₃2 → 𝐺₉

while filling the chart. I give the result as follow:

𝜎_𝑡: 𝐺_𝑛(𝑡𝑚) ⟶ 𝐺_(𝑛𝑡₎𝑚

𝑋 ↦ 𝑌 (2.25.1)

by

𝑌_𝑖 = ∑ 𝑛𝑗 𝑡−1

𝑗=0

𝑋_{𝑖+(𝑡−1−𝑗)𝑚}.

Then, this map is a representation map that reduce more dimensions. The reduction of dimension is dividing by 𝑡.

For any dimension, it is always available to use representation maps to reduce its dimension until 1. However, this does not make it more convenient. Most time, it is only considered about the quadratic lower-dimensional representation.

2.2.3 NORMAL EXTENSION

Filling Latin Squares is a question that is often discussed by mathematicians. There isn’t a formal definition for the process of filling yet. So normal extension is introduced to define the process formally and solve the related questions mathematically.

Definition 2.26 Let 𝐴 ⊆ 𝐺_𝑛𝑚_{, a map} 𝑓: 𝐴 → 𝐺_𝑛𝑘. 𝑈 be any 𝑘-unit of 𝐺_𝑛𝑚_{. If the}

limitation of 𝑓 over any 𝐴 ∩ 𝑈 is injective, call 𝑓 is an (incomplete) Latin map over 𝐴 with rank 𝑘.

Definition 2.27 Let 𝐴 ⊂ 𝐵 ⊂ 𝐺_𝑛𝑚_{. If} 𝑓 is a Latin map over 𝐴 with rank 𝑘, 𝑔 is a

Latin map over 𝐵 with rank 𝑘 and 𝑔|_𝐴 = 𝑓, call 𝑔a normal extension of 𝑓, and call

𝑓a limitation of 𝑔.[7]

A larger Latin map is still a Latin map when taking limitation on any subset of its domain. But an extension of a Latin map might not still be a Latin map. Only the normal extension can preserve the Latin property of the map. The process of normal extensions is the process of filling in Latin squares to find a solution.

Definition 2.28 Let 𝐴 ⊆ 𝐺_𝑛𝑚_, 𝑓 is a Latin map over 𝐴 with rank 𝑘. If there is a

Latin map 𝑔 over 𝐺_𝑛𝑚 _{that is a normal extension of} 𝑓, call 𝑔 a _{solution of} 𝑓.

Moreover, if there does not exist a normal extension ℎ other than 𝑔, call 𝑔 the unique

solution of 𝑓.

Now, I’m going to use normal extension to describe some common theorems about filling Latin squares.

Theorem 2.29 (Extrinsic Linking Theorem).Let 𝑓 be a Latin map over𝐴 ⊆ 𝐺_𝑛𝑚with

rank 𝑘,𝑡 be a positive integer no greater than 𝑛𝑘, 𝑖₁, 𝑖₂, ⋯ , 𝑖_𝑡∈ 𝐺_𝑛𝑘 , and

𝑥₁, 𝑥₂, ⋯ , 𝑥_𝑡 ∈ 𝐺_𝑛𝑚 _{be in the same} 𝑘_-unit𝑈_{. If any normal extensions of} 𝑓 _satisfies

that 𝑖₁, ⋯ , 𝑖_𝑡 are not the images of the grids other than 𝑥₁, ⋯ , 𝑥_𝑡, then any solution of

𝑓 satisfies that the image of 𝑥1, ⋯ , 𝑥𝑡 can only be 𝑖1, ⋯ , 𝑖𝑡. i.e. the grids other than

𝑖₁, ⋯ , 𝑖_𝑡 cannot be the images of 𝑥₁, ⋯ , 𝑥_𝑡.

Proof. Let 𝑔 be a solution of 𝑓, it is a normal extension. Then, the preimages of 𝑖₁, ⋯ , 𝑖_𝑡 about 𝑔 in the unit 𝑈 are included in 𝑥₁, ⋯ , 𝑥_𝑡. As 𝑔 is bijective on 𝑈, the preimages of 𝑖₁, ⋯ , 𝑖_𝑡 in the unit 𝑈, i.e. the images about 𝑔−1 are 𝑥₁, ⋯ , 𝑥_𝑡. Conversely, the images of 𝑥1, ⋯ , 𝑥𝑡 about 𝑔 are exactly 𝑖1, ⋯ , 𝑖𝑡.

with rank 𝑘, 𝑡 be a positive integer no greater than 𝑛𝑘. 𝑖₁, 𝑖₂, ⋯ , 𝑖_𝑡 ∈ 𝐺_𝑛𝑘, and

𝑥₁, 𝑥₂, ⋯ , 𝑥_𝑡 ∈ 𝐺_𝑛𝑚 _{are in a single} 𝑘_-unit 𝑈_{. If any normal extension of} 𝑓 _satisfies

thatthe grids other than 𝑖1, ⋯ , 𝑖𝑡 are not the images of 𝑥1, ⋯ , 𝑥𝑡, i.e. the images of

𝑥₁, ⋯ , 𝑥_𝑡 are included in 𝑖₁, ⋯ , 𝑖_𝑡, then any solution of 𝑓 satisfies that 𝑖₁, ⋯ , 𝑖_𝑡

cannot be the images of grids in 𝑈 other than 𝑥₁, ⋯ , 𝑥_𝑡.

The intrinsic linking theorem is in fact the inverse theorem of the extrinsic linking theorem. They are separated for practical usage.

3. Questions Related to the Latin Map

3.1 Partial Maps and the Operation of Latin Maps

3.1.1 PARTIAL MAPS AND PROJECTION

As a submap of a Latin map is still a Latin map, it is eligible to find out the relationship between a Latin map with its submap. Noticing that their image sets are exactly the same, the main difference are between their domains. Thus, we need to discuss the relationship between their domains. Here, we induce the partial maps of a grid space.

Similar to projection, we can project a whole grid space onto a subunit to define the partial maps.

Definition 3.1 Let 𝐺_𝑛𝑚 _{be a grid space,} 𝑈 be one of its 𝑘-unit and the constant

components of 𝑈 are 𝑥𝑖0, 𝑥𝑖1, ⋯ , 𝑥𝑖𝑚−𝑘−1 . Denote the grids in 𝐺𝑛𝑚 by 𝑋 =

(𝑋₀, ⋯ , 𝑋_𝑚−1). We map:

𝑃𝑈: 𝐺𝑛𝑚 ⟶ 𝑈

𝑋𝑖 ↦ {𝑋_𝑥𝑖 𝑖 ∉ {𝑖1, ⋯ , 𝑖𝑚−𝑘} 𝑖 𝑖 ∈ {𝑖1, ⋯ , 𝑖𝑚−𝑘}

(3.1.1)

Call 𝑃_𝑈 the partial map from 𝐺_𝑛𝑚 to 𝑈.

Specially, if 𝑥_𝑖0 = 𝑥_𝑖1 = ⋯ = 𝑥_{𝑖𝑚−𝑘−1} = 0, call 𝑃_𝑈 a standard partial map, the

unit 𝑈 is called the standard partial of 𝐺𝑛𝑚_.

A partial map is projection regarding its algebraic structure. It satisfies the idempotent property (𝑃_𝑈2 _{= 𝑃}

𝑈). We have its orthogonal projection 𝑄𝑈 = 𝑖𝑑 − 𝑃𝑈, then

𝑃_𝑈 ∘ 𝑄_𝑈 = 𝑃_𝑈− 𝑃_𝑈2 _{= 0. Here, 𝑖𝑑 is the identity map, 0 is the constant map whose}

image is the intersect grid of 𝑃_𝑈 and 𝑄_𝑈 (which might be any point in the image of 𝑃𝑈). The additivity is calculated by the linear additivity via the original point as the

image of 0. Here call 𝑄_𝑈 the Orthogonal Partial Map of the partialmap 𝑃_𝑈, denotes by 𝑃𝑈 ⊥ 𝑄𝑈.

Remark 3.2 We are able to construct the orthogonal partial map to prove its existence. Let 𝑃_𝑈 be a partial map from 𝐺_𝑛𝑚 to its unit 𝑈, which maps the 𝑖-th component to 𝑥𝑖−1. The components of 𝑈 that take over 𝐺𝑛 are denoted by 𝑋𝑗−1 (the

value 𝑥𝑗−1 ∈ 𝐺𝑛, other components take over 𝐺𝑛. Then 𝑈 ∩ 𝑉 = {(𝑥0, 𝑥1, ⋯ , 𝑥𝑚−1)}.

Let 𝑄_𝑈 = 𝑃_𝑉 , then 𝑖𝑚(𝑃_𝑈∘ 𝑄_𝑈) = 𝑈 ∩ 𝑉 = {(𝑥₀, 𝑥₁, ⋯ , 𝑥_𝑚−1)}. Suppose 𝑋 = (𝑥0, 𝑥1, ⋯ , 𝑥𝑚−1), define the additivity between maps: ∀𝑌 ∈ 𝐺𝑛𝑚_, (𝑃_𝑈 + 𝑄_𝑈)(𝑌) −

𝑋 = (𝑃_𝑈(𝑌) − 𝑋) + (𝑄_𝑈(𝑌) − 𝑋) . Then (𝑃_𝑈+ 𝑄_𝑈)(𝑌) = (𝑥₀, 𝑥₁, ⋯ , 𝑥_𝑚−1) + (0, ⋯ , 𝑦_𝑖−1− 𝑥_𝑖−1, ⋯ ,0) + (0, ⋯ , 𝑦_𝑗−1− 𝑥_𝑗−1, ⋯ ,0) = (𝑦₀, 𝑦₁, ⋯ , 𝑦_𝑚−1) = 𝑌. Thus, 𝑄𝑈 is the orthogonal partial map of 𝑃𝑈 which satisfies 𝑃𝑈+ 𝑄𝑈 = 𝑖𝑑 and 𝑃𝑈∘ 𝑄𝑈 =

0 (0(𝐺_𝑛𝑚) = {𝑋}).

Let’s turn to its map structure. There is the subspace-invariant of partial maps:

Proposition 3.3 Let 𝑃_𝑈 be a partial map from 𝐺_𝑛𝑚 _{to its unit} 𝑈_, 𝐴 _{be a subspace of}

𝐺𝑛𝑚. Then 𝐴 ∩ 𝑈 = ∅ or 𝐴 ∩ 𝑈 = 𝑃_𝑈(𝐴).

Proof.If 𝐴 ∩ 𝑈 ≠ ∅, then ∀𝑎 ∈ 𝐴 ∩ 𝑈, 𝑃𝑈(𝑎) = 𝑎. So 𝐴 ∩ 𝑈 ⊆ 𝑃𝑈(𝐴). Obviously,

𝑃_𝑈(𝐴) ⊆ 𝑈, so it is only needed to prove that 𝑃_𝑈(𝐴) ⊆ 𝐴.

∀𝑎 ∈ 𝐴, let 𝑎 ∉ 𝑈, 𝑖 be a component of 𝑈 that values a constant 𝑢𝑖−1, 𝑎𝑗−1 be

the 𝑗-th component of 𝑎. Then 𝑃_𝑈 maps the 𝑖-th components of 𝑎 to 𝑢_𝑖−1 and maps the components other than 𝑖-th to 𝑎_𝑗−1 themselves.

Let 𝑋 = (𝑥0, ⋯ , 𝑥𝑚−1) be the original of the subspace, 𝐴 = 𝑋 + 𝐺𝑘𝑙. Then 𝑎_𝑗−1

has a unique division 𝑥_𝑗−1+ 𝑦_𝑗−1 as a component of a subspace, where 𝑦_𝑗−1 = 0 or 𝑦𝑗−1 ∈ 𝐺𝑘. As 𝐴 ∩ 𝑈 ≠ ∅, each grid in 𝐴 ∩ 𝑈 has a unique division. So 𝑢𝑖−1 has the

unique division 𝑥_𝑖−1+ 𝑦_𝑖−1, where 𝑦_𝑖−1= 0 or 𝑦_𝑖−1∈ 𝐺_𝑘. Then any component of 𝑃𝑈(𝑎) can be divided into the sum of the corresponding component of 𝑋 and {0} or

some number in 𝐺_𝑘. Thus, 𝑃_𝑈(𝑎) ∈ 𝐴, i.e. 𝑃_𝑈(𝐴) ⊆ 𝐴.

Corollary 3.4 Partial map maps a unit to a unit.

Proof.Suppose 𝑈, 𝑉 are units of 𝐺𝑛𝑚_{. It is needed to prove that} 𝑃_𝑈(𝑉) is a unit of

𝐺_𝑛𝑚_.

If 𝑈 ∩ 𝑉 ≠ ∅, 𝑉 is a subspace of 𝐺_𝑛𝑚, so 𝑃_𝑈(𝑉) = 𝑈 ∩ 𝑉 is a unit of 𝐺_𝑛𝑚. When 𝑈 ∩ 𝑉 = ∅, let 𝑖 be the component that takes over 𝐺_𝑛 in 𝑉, 𝑗 be the component that values constant in 𝑉. Then, 𝑃_𝑈 maps the 𝑖-th component to constant or takes over 𝐺𝑛 and maps the 𝑗-th component to another constant.

Thus 𝑃_𝑈(𝑉) is still a unit of 𝐺_𝑛𝑚_.

Partial map can also reflect the relationship between two grids, see the following proposition.

Proposition 3.5 Let 𝑃_𝑈 be the partial map from 𝐺_𝑛𝑚 to its unit 𝑈, 𝑄_𝑈 be its orthogonal partial map. Then ∀𝑋, 𝑌 ∈ 𝐺𝑛𝑚 _, 𝐷(𝑋, 𝑌) = 𝐷(𝑃_𝑈(𝑋), 𝑃_𝑈(𝑌)) +

𝐷(𝑄_𝑈(𝑋), 𝑄_𝑈(𝑌)).

Proof.Let 𝑖𝑚(𝑃_𝑈 ∘ 𝑄_𝑈) = {(𝑧₀, 𝑧₁, ⋯ , 𝑧_𝑚−1)} , 𝑋 = (𝑥₀, 𝑥₁, ⋯ , 𝑥_𝑚−1) , 𝑌 = (𝑦₀, 𝑦₁, ⋯ , 𝑦_𝑚−1).

It’s eligible to suppose the first 𝑖 components of 𝑈 are constant, and other components take over 𝐺𝑛.

Then, 𝑃_𝑈(𝑋) = (𝑧₀, 𝑧₁, ⋯ , 𝑧_𝑖−1, 𝑥_𝑖, 𝑥_𝑖+1, ⋯ , 𝑥_𝑚−1), 𝑄𝑈(𝑋) = (𝑥0, 𝑥1, ⋯ , 𝑥𝑖−1, 𝑧𝑖, 𝑧𝑖+1, ⋯ , 𝑧𝑚−1).

𝐷(𝑋, 𝑌) = ∑ 𝑠

𝑚−1

𝑘=0

𝑔𝑛|𝑥𝑘− 𝑦𝑘| = ∑ 𝑠 𝑖−1

𝑘=0

𝑔𝑛|𝑥𝑘− 𝑦𝑘| + ∑ 𝑠 𝑚−1

𝑘=𝑖

𝑔𝑛|𝑥𝑘− 𝑦𝑘|

= 𝐷(𝑃_𝑈(𝑋), 𝑃_𝑈(𝑌)) + 𝐷(𝑄_𝑈(𝑋), 𝑄_𝑈(𝑌)).

Corollary 3.6 𝐷(𝑃𝑈(𝑋), 𝑃𝑈(𝑌)) ≤ 𝐷(𝑋, 𝑌)

3.1.2 ORTHOGONAL LATIN MAPS

The orthogonal Latin squares is a very hot field in the discussion of Latin squares. Many mathematicians, represented by Euler, have proved that the minimal non-trivial orthogonal Latin squares exist when the level is not 2 or 6. This is a perfect result for the minimal non-trivial Latin maps. Moreover, I extend the discussion of the orthogonality to high-dimensional Latin maps. Here is the definition:

Definition 3.7 Let 𝑓 be an 𝑚-dimensional Latin map of level 𝑛 with rank 𝑘₁, 𝑔 be an 𝑚-dimensional Latin map of level 𝑛 with rank 𝑘₂. Define the Direct Sum map:

𝑓 ⊕ 𝑔: 𝐺𝑛𝑚 ⟶ 𝐺_{𝑛(𝑘1+𝑘2)}

𝑋 ↦ (𝑓(𝑋), 𝐺(𝑋)) (1)

If 𝑘1+ 𝑘2 ≤ 𝑚 and 𝑓 ⊕ 𝑔 is also a Latin map, then call the Latin map 𝑓 is

orthogonal to the Latin map 𝑔.

Call 𝑓 and 𝑔 are orthogonal Latin maps, denoted by 𝑓 ⊥ 𝑔.

When 𝑘₁ = 𝑘₂ = 1, 𝑚 = 2, noticing that 𝑓 ⊕ 𝑔 is the bijection from 𝐺_𝑛2 to itself, so the orthogonal Latin maps 𝑓 and 𝑔 are the usual discussed orthogonal Latin squares.

There are some direct relationships between 𝑓 ⊕ 𝑔 with 𝑓 and 𝑔. Let 𝑃_𝑈 be the standard partial map from 𝐺_𝑛(𝑘1+𝑘2) to 𝐺𝑛𝑘1, and 𝑄𝑈 be the standard partial map from

𝐺_𝑛(𝑘1+𝑘2) to 𝐺_𝑛𝑘2. Here is the proposition:

Proposition 3.8 𝑃_𝑈 ⊥ 𝑄_𝑈, and 𝑓 = 𝑃_𝑈 ∘ (𝑓 ⊕ 𝑔), 𝑔 = 𝑄_𝑈 ∘ (𝑓 ⊕ 𝑔).

Proof.The image of 𝑓 is 𝐺_𝑛𝑘1, this is a standard partial of 𝐺_𝑛(𝑘1+𝑘2) which is also a

unit. Let this unit be 𝑈, then 𝑃_𝑈 is the partial map from 𝐺_{𝑛(𝑘1+𝑘2)} to 𝑈. The image of 𝑔 is 𝐺_𝑛𝑘2, this is a standard partial of 𝐺𝑛(𝑘1+𝑘2) which is also a unit. Let this unit be 𝑉,

then 𝑃_𝑉 is the partial map from 𝐺_𝑛(𝑘1+𝑘2) to 𝑉. 𝐺_𝑛(𝑘1+𝑘2) = 𝑈 × 𝑉. By the definition

of product space, the components that 𝑈 and 𝑉 keeps constant do not intersect. Their components that takes over 𝐺𝑛 also do not intersect. 𝑈 ∩ 𝑉 = {(0,0, ⋯ ,0)}. So,

𝑖𝑚(𝑃_𝑈∘ 𝑄_𝑈) = 0.

At the same time, ∀𝑋 ∈ 𝐺_𝑛(𝑘1+𝑘2) , (𝑃𝑈 + 𝑄𝑈)(𝑋) − (0,0, ⋯ ,0) = (𝑃𝑈(𝑋) −

(0,0, ⋯ ,0)) + (𝑄_𝑈(𝑋) − (0,0, ⋯ ,0)) =

(𝑥₀, 𝑥₁, ⋯ , 𝑥_𝑘1−1, 0,0, ⋯ ,0) + (0,0, ⋯ ,0, 𝑥_𝑘1, 𝑥_𝑘1+1, ⋯ , 𝑥_{𝑘1+𝑘2−1}) = 𝑋 . So, 𝑃_𝑈 + 𝑄𝑈 = 𝑖𝑑. Thus, 𝑃𝑈 ⊥ 𝑄𝑈.

Moreover, ∀𝑋 ∈ 𝐺_𝑛𝑚_, 𝑃_𝑈((𝑓 ⊕ 𝑔)(𝑋)) = 𝑃_𝑈(𝑓(𝑋), 𝑔(𝑋)) = 𝑓(𝑋)_, 𝑄_𝑈((𝑓 ⊕

𝑔)(𝑋)) = 𝑄𝑈(𝑓(𝑋), 𝑔(𝑋)) = 𝑔(𝑋). So, 𝑓 = 𝑃𝑈∘ (𝑓 ⊕ 𝑔), 𝑔 = 𝑄𝑈∘ (𝑓 ⊕ 𝑔)

Theorem 3.9 (Euler’s First Theorem)The minimal non-trivial orthogonal Latin maps of odd prime level always exist.

To prove this theorem, I use the Latin map to describe a common method (the Euler’s traditional method).

Definition 3.10 Let 𝑓 be a surjection from 𝐺_𝑛2 to 𝐺_𝑛. We regard the domain 𝐺_𝑛2 as the linear space of ℤ𝑛. If there is a vector 𝑥 ∈ 𝐺𝑛2 such that ∀𝑦 ∈ 𝐺_𝑛2, 𝑓(𝑥 + 𝑦) = 𝑓(𝑦), call 𝑓 an 𝑥-translation map. If 𝑥 = (𝑚, 1), then call 𝑓 an 𝑚-translation map. If the 𝑚-translation map is a minimal non-trivial Latin map, call 𝑓 a 𝑚-translation Latin map.

Proposition 3.11 An 𝑚-translation map is a minimal non-trivial Latin map of level

𝑛 if and only if(𝑚, 𝑛) = 1.

Proof.First, let 𝑓 be an 𝑚-translation map of level 𝑛, 𝑥 = (𝑚, 1) ∈ 𝐺_𝑛2. Necessity:

Assuming 𝑓 to be a minimal non-trivial Latin map, then its rank is 1. If (𝑚, 𝑛) =

𝑑 > 1 , then ∀𝑦 ∈ 𝐺_𝑛2 , 𝑓(𝑛_𝑑𝑥 + 𝑦) = 𝑓(𝑦) . But 𝑛_𝑑𝑚 =𝑚_𝑑 𝑛 ≡ 0 mod 𝑛 ,

𝑛

𝑑 ≢ 0 mod 𝑛. So the first components between 𝑛

𝑑𝑥 + 𝑦 and 𝑦 are the same

(under module 𝑛), while the second components are not the same (under module 𝑛). Then, 𝑓 is not injective on some 1-unit. This contradict to that 𝑓 is a Latin map with rank 1. So (𝑚, 𝑛) = 1.

Sufficiency:

When (𝑚, 𝑛) = 1, the set {𝑘𝑚|𝑘 ∈ ℤ𝑛} takes over all elements in ℤ𝑛. Obviously,

𝑛𝑥 + 𝑦 = 𝑦, ∀𝑦 ∈ 𝐺_𝑛2. ∀𝑦 ∈ 𝐺_𝑛2, the preimage of 𝑓(𝑦) at least contains{𝑘𝑥 + 𝑦|𝑘 ∈ ℤ𝑛}. As the second component of 𝑘𝑥 + 𝑦 are different, the preimage of 𝑓(𝑦) has at

least 𝑛 elements. 𝑓 is surjective, so 𝑓(𝑦) has 𝑛 different values. Also, the preimage of 𝐺_𝑛 has only 𝑛2 elements. Thus, for each 𝑓(𝑦), its preimage has exact 𝑛 elements. They are exactly {𝑘𝑥 + 𝑦|𝑘 ∈ ℤ𝑛}. Their first component takes over {𝑘𝑚|𝑘 ∈ ℤ𝑛},

while the second component takes over ℤ_𝑛. So 𝑓(𝑦₁) = 𝑓(𝑦₂) and 𝑦₁ ≠ 𝑦₂ can deduce thatboth components of 𝑦₁ and 𝑦₂ are different respectively. Thus, 𝑓 is injective on each 1-unit, i.e. 𝑓 is a minimal non-trivial Latin map.

Proposition 3.12 The 𝑚-translation Latin map and 𝑛-translation Latin map of level 𝑝

are orthogonal if and only if (𝑛 − 𝑚, 𝑝) = 1.

Proof.Let 𝑡 = (𝑚, 0) ∈ 𝐺_𝑝2, ∀𝑦 ∈ 𝐺_𝑝2, 𝑇(𝑦) = 𝑦₂𝑡 + 𝑦, where 𝑦₂ is the second

component of 𝑦. Suppose 𝑓 is an 𝑚-translation Latin map, 𝑔 is an 𝑛-translation

limited on the unit that first component keeps constant, ⇔ 𝑔 ∘ 𝑇 is injective limited on the unit that first component keeps constant. At the same time, 𝑇 is bijective limited on the unit that second component keeps constant. So 𝑔 ∘ 𝑇 must be injective limited on the unit that second component keeps constant. So 𝑔 ∘ 𝑇 is injective limited on the unit that first component keeps constant, 𝑖𝑔ℎ𝑡𝑎𝑟𝑟𝑜𝑤𝑔 ∘ 𝑇 is injective on each 1-unit, ⇔ 𝑔 ∘ 𝑇 is a Latin map with rank 1, i.e. a minimal non-trivial Latin map. Noticing that 𝑔 ∘ 𝑇 is a (𝑥′− 𝑡)-translation Latin map, so 𝑔 ∘ 𝑇 is a minimal non-trivial Latin map ⇔ it is an (𝑛 − 𝑚)- translation map. By proposition 3.11, 𝑔 ∘ 𝑇 is an (𝑛 − 𝑚)- translation map ⇔ (𝑛 − 𝑚, 𝑝) = 1.

Corollary 3.13 An 𝑚-translation map and 𝑛-translation map of prime level 𝑝 are

orthogonal Latin maps if and only if 𝑚, 𝑛 and 𝑚 − 𝑛 are not the multiple of 𝑝.

An integer 𝑛 is coprime with prime number 𝑝 if and only if it is not a multiple of 𝑝. By the two propositions above, we get this corollary immediately. When 𝑝 > 2, there always exist 0 < 𝑚 < 𝑛 < 𝑝 satisfies the requirements. Thus, the minimal non-trivial orthogonal Latin maps of odd prime level always exist. Theorem 3.9 has been proved.

3.1.3 TENSOR PRODUCT OF LATIN MAPS

When talking about the orthogonality of Latin maps, a common method to prove the Euler theorem is by the so called tensor product (matrix sense). Here, the tensor product is regarding the minimal non-trivial Latin maps as matrix and use operation similar to the tensor product of matrix to construct a new Latin map. During this operation, the orthogonality is invariant, so it proves part of conclusions of the Euler theorem.

Here, I use similar method to extend the definition of tensor product. Then it is able to discuss the high-dimension case.

Definition 3.14 Let 𝑓 be an 𝑚-dimensional Latin map of level 𝑛₁ with rank 𝑘, 𝑔

be an 𝑚-dimensional Latin map of level 𝑛₂ with rank 𝑘. ∀𝑋 ∈ 𝐺_𝑛1𝑚, 𝑌 ∈ 𝐺_𝑛 2

𝑚, define map:

𝑓 ⊗ 𝑔: 𝐺_(𝑛1𝑛2)𝑚 ⟶ 𝐺_(𝑛 1𝑛2)𝑘

𝑛₂𝑋 + 𝑌 ↦ 𝑛₂𝑓(𝑋) + 𝑔(𝑌). (1)

Call 𝑓 ⊗ 𝑔 the tensor product of 𝑓 and 𝑔.

The definition of tensor product does make sense. We have following property:

Proposition 3.15 The tensor product of Latin maps is still a Latin map.

Proof. Let 𝑍₁ = 𝑛₂𝑋₁+ 𝑌₁ and 𝑍₂ = 𝑛₂𝑋₂+ 𝑌₂ are in same 𝑘-unit of 𝐺_(𝑛1𝑛2)𝑚. For any grid 𝑋, we denote its 𝑖 + 1-th component as𝑋𝑖. As 𝑌1, 𝑌2 ∈ 𝐺𝑛2𝑚, its each

component values in 𝐺_𝑛2. So 𝑍_1𝑖 = 𝑍_2𝑖 ⇔ 𝑋_1𝑖 = 𝑋_2𝑖 and 𝑌_1𝑖 = 𝑌_2𝑖. As 𝐷(𝑍₁, 𝑍₂) ≤ 𝑘, there are at least 𝑚 − 𝑘 terms among 𝑠𝑔𝑛|𝑍1𝑖− 𝑍2𝑖| take 0. So we have at least

Now, 𝑍1 ≠ 𝑍2 if and only if 𝑋1 ≠ 𝑋2 or 𝑌1 ≠ 𝑌2. When 𝑋1 ≠ 𝑋2, as 𝑋1 and 𝑋2

are in same 𝑘-unit of 𝐺_𝑛1𝑚_{, we have} 𝑓(𝑋₁) ≠ 𝑓(𝑋₂). So there exist 𝑖 ∈ 𝐺_{𝑘, such that}

𝑓(𝑋1)𝑖 ≠ 𝑓(𝑋2)𝑖 . Suppose 𝑓(𝑋1)𝑖 < 𝑓(𝑋2)𝑖 . Then (𝑓 ⊗ 𝑔)(𝑍1)𝑖 = 𝑛2𝑓(𝑋1)𝑖 +

𝑔(𝑌₁)_𝑖 < 𝑛₂𝑓(𝑋₁)_𝑖+ 𝑛₂ = 𝑛₂(𝑓(𝑋₁)_𝑖 + 1) ≤ 𝑛₂𝑓(𝑋₂)_𝑖 ≤ 𝑛₂𝑓(𝑋₂)_𝑖 + 𝑔(𝑌₂)_𝑖 = (𝑓 ⊗ 𝑔)(𝑍1)𝑖. Thus (𝑓 ⊗ 𝑔)(𝑍1) ≠ (𝑓 ⊗ 𝑔)(𝑍2). When 𝑋1 = 𝑋2 and 𝑌1≠ 𝑌2, as

𝑌1 and 𝑌2 are in same 𝑘-unit of 𝐺𝑛2𝑚, we have 𝑔(𝑌1) ≠ 𝑔(𝑌2). Thus, (𝑓 ⊗

𝑔)(𝑍₁) = 𝑛₂𝑓(𝑋₁) + 𝑔(𝑌₁) ≠ 𝑛₂𝑓(𝑋₂) + 𝑔(𝑌₂) = (𝑓 ⊗ 𝑔)(𝑍₂). In all, (𝑓 ⊗ 𝑔) is injective on each 𝑘-unit, i.e. (𝑓 ⊗ 𝑔) is a Latin map with rank 𝑘.

Similarly, we need the orthogonal-invariant property of tensor product:

Theorem 3.16 (Tensor product is orthogonal-invariant).Let 𝑓1 and 𝑓2 be

𝑚-dimensional Latin maps with rank 𝑘1, the level of 𝑓1 is 𝑛1, while the level of 𝑓2 is

𝑛₂. Let 𝑔₁ and 𝑔₂ be 𝑚-dimensional Latin maps with rank 𝑘₂, the level of 𝑔₁ also

is 𝑛1, while the level of 𝑔2 also is 𝑛2.

If 𝑓₁ ⊥ 𝑔₁ and 𝑓₂ ⊥ 𝑔₂, then 𝑓₁⊗ 𝑓₂ ⊥ 𝑔₁⊗ 𝑔₂.

Proof.∀𝑋 ∈ 𝐺_𝑛1𝑚, we have (𝑓₁⊕ 𝑔₁)(𝑋) = (𝑓₁(𝑋), 𝑔₁(𝑋)). ∀𝑌 ∈ 𝐺_𝑛2𝑚, we have (𝑓2⊕ 𝑔2)(𝑌) = (𝑓2(𝑌), 𝑔2(𝑌)) . ∀𝑍 ∈ 𝐺(𝑛1𝑛2)𝑚 , by the division algorithm,

∃₁𝑋 ∈ 𝐺_𝑛1𝑚, 𝑌 ∈ 𝐺_𝑛

2𝑚 such that 𝑍 = 𝑛2𝑋 + 𝑌. So ∀𝑍 ∈ 𝐺(𝑛1𝑛2)𝑚,

((𝑓₁⊕ 𝑔₁) ⊗ (𝑓₂ ⊕ 𝑔₂))(𝑍) = ((𝑓₁⊕ 𝑔₁) ⊗ (𝑓₂⊕ 𝑔₂))(𝑛₂𝑋 + 𝑌) = 𝑛2(𝑓1⊕ 𝑔1)(𝑋) + (𝑓2⊕ 𝑔2)(𝑌)

= 𝑛2(𝑓1(𝑋), 𝑔1(𝑋)) + (𝑓2(𝑌), 𝑔2(𝑌))

= (𝑛₂𝑓₁(𝑋) + 𝑓₂(𝑌), 𝑛₂𝑔₁(𝑋) + 𝑔₂(𝑌))

= ((𝑓1⊗ 𝑓2)(𝑛2𝑋 + 𝑌), (𝑔1⊗ 𝑔2)(𝑛2𝑋 + 𝑌))

= ((𝑓1⊗ 𝑓2)(𝑍), (𝑔1⊗ 𝑔2)(𝑍))

= ((𝑓₁⊗ 𝑓₂) ⊕ (𝑔₁⊗ 𝑔₂))(𝑍)

(3.16.1)

So (𝑓₁⊕ 𝑔₁) ⊗ (𝑓₂⊕ 𝑔₂) = (𝑓₁⊗ 𝑓₂) ⊕ (𝑔₁⊗ 𝑔₂). By the proposition ?, equality to the left is a Latin map, so equality to the right is also a Latin map. Thus, (𝑓1⊗ 𝑓2) ⊥ (𝑔1⊗ 𝑔2).

Here is a stronger proposition: (𝑓₁⊕ 𝑔₁) ⊗ (𝑓₂⊕ 𝑔₂) = (𝑓₁⊗ 𝑓₂) ⊕ (𝑔₁⊗ 𝑔2). i.e. The tensor product of direct sums is equal to the direct sum of tensor products. Corollary 3.17 The minimal non-trivial orthogonal Latin maps of odd level always exist.

Proof. By the Fundamental Theorem of Arithmetic, each odd number larger than 1 is the product of some odd prime numbers.

Each minimal non-trivial orthogonal Latin maps of odd prime level exists. After the tensor product operation, the minimal non-trivial orthogonal Latin maps of their product level also exists.

So the minimal

non-The use of Latin map can describe the invariancy of orthogonality under tensor product concisely as demonstrated in theorem 1.

product can be defined by a higher-dimensional Latin map and using a lower-dimensional representation to preserve its dimension. However, this is difficult, because the domain of tensor product is not an exact cubic grid space. It is rather a cuboid grid set. For some rank 𝑘, this is hard to define a uniform image space. This possible definition should be discussed in the future.

3.2 Composition of Latin Maps

3.2.1 EXISTENCE OF COMPOUND LATIN MAPS

Two Latin maps that can composite and form another Latin map are not easy to construct. Here, is a rare example:

Example 3.18 A 3-dimensional Latin map of level 3 with rank 2:

composites with a 2-dimensional Latin map of level 3 with rank 1:

The result is a 3-dimensional Latin map of level 3 with rank 1:

There exists the case that two Latin maps compound together and still form a Latin map.

Proposition 3.19 Let 𝑓 be the minimal non-trivial Latin map of level 2, then there

does not exist a non-trivial Latin map 𝑔 of level 2 with rank 2 such that 𝑓 ∘ 𝑔 is still

a Latin map.

Proof.If 𝑔 exist, we first notice that there are only two possible map for 𝑓. Moreover, these two maps can be transferred by a representation map. So suppose 𝑓 to be:

[0 1_{1 0}]

Let 𝑋 be a grid in the domain of 𝑔 and 𝑓(𝑔(𝑋)) = 0. Then for any grid 𝑌 that in the same 1 unit with 𝑋, 𝑓(𝑔(𝑌)) = 1. So 𝑔(𝑌) can only be (0,1) or (1,0). As 𝑔 is non-trivial, the dimension of 𝑔 is larger than 2. So there are more than2 grids that are in a same 1 unit with 𝑋. By the pigeonhole principle, there exist 𝑌1 ≠ 𝑌2,

𝐷(𝑋, 𝑌₁) = 𝐷(𝑋, 𝑌₂) = 1, such that 𝑔(𝑌₁) = 𝑔(𝑌₂). So 𝐷(𝑌₁, 𝑌₂) ≤ 𝐷(𝑋, 𝑌₁) + 𝐷(𝑋, 𝑌2) = 2, which means 𝑌1 and 𝑌2 are in a same 2-unit. This iscontradict to that

𝑔 is injective on each 2-unit. So 𝑔 does not exist.

To construct compoundable Latin maps, a natural way is to construct unit-preserving Latin maps. However, we find that most Latin maps do not preserve unit. Here, we have the following proposition:

Proposition 3.20 If a non-trivial Latin map with rank more than 1, then it cannot map all units to units.

Proof.Let 𝑓 be a non-trivial Latin map with rank 𝑘 > 1, its domain is 𝐺_𝑛𝑚. Suppose 𝑓 maps all units to units. As 𝑓 is bijective on each 𝑘 unit, it is also bijective on each 1-unit. So 𝑓 maps 1-unit to 1-unit. ∀𝑋 ∈ 𝐺_𝑛𝑚, there are totally 𝑚1-units that contains 𝑋. In 𝐺_𝑛𝑘, there are totally 𝑘1-units that contains 𝑓(𝑋). Because 𝑓 is

non-trivial and 𝑚 > 𝑘, by the pigeonhole principle, there exists two different 1-units 𝑈 and 𝑉 that contains 𝑋, such that 𝑓(𝑈) = 𝑓(𝑉). But the rank of 𝑓 is 𝑘 > 1. So the minimal generated unit 𝐺(𝑈 ∪ 𝑉) is included in some 𝑘-unit. Thus 𝑈 and 𝑉 are both in some 𝑘-unit 𝑊, where 𝑓 is bijective on 𝑊. This is contradict to that 𝑓(𝑈) = 𝑓(𝑉). So 𝑓 cannot map all units to units.

After enumerate by computer, I find that any 4-dimensional Latin map of level 3 with rank 2 is not a Latin map any more after composing with some minimal non-trivial Latin map of level 3. This quite common Latin map does not exist any example to properly composite. Again it proves that compoundable Latin maps are really rare.

To construct two non-trivial Latin maps such that their composition is still a Latin map, we need to develop an algorithm. Here, we have the significant Extending Construction Theorem.

3.2.2 THE EXTENDING CONSTRUCTION THEOREM

Definition 3.21 Let 𝑓 be an 𝑚-dimensional Latin map of level 𝑛 with rank 𝑘, 𝑔 be an 𝑙-dimensional Latin map of level 𝑘 with rank 1. If there is a non-trivial 𝑚-dimensional Latin map 𝑔′ of level 𝑛 with rank 𝑙, such that 𝑓 = 𝑔 ∘ 𝑔′, call 𝑔 an textitextending map, 𝑔′ is called the construction of 𝑓 about the extending 𝑔.

an 𝑚-dimensional Latin map 𝑓 of level 𝑛 with rank 𝑘 about an extending 𝑔, then

for any extending map ℎ of same dimension, there is a construction of 𝑓 about ℎ.

Proof.Let the dimension of the extending 𝑔 be 𝑙, where 𝑙 > 𝑘, 𝑓 = 𝑔 ∘ 𝑔′. Then we have the diagram:

Suppose ℎ be another 𝑙-dimensional extending map. Now, ∀𝑋 ∈ 𝐺_𝑛𝑘, let’s count the

amount of elements in 𝑔−1(𝑋). Let 𝑈 be the unit of 𝐺_𝑛𝑙 whose first 𝑘 component

take over 𝐺_𝑛, where other components keep constant. Then 𝑔 is bijective on 𝑈. So there are exactly 1 elements in 𝑈 ∩ 𝑔−1(𝑋). For any unit that parallel to 𝑈, i.e. the components that take over 𝐺_𝑛 same to 𝑈 and other components value different constant to 𝑈, they have same properties with 𝑈. There are 𝑛𝑙−𝑘 this kind of units. So there are 𝑛𝑙−𝑘 elements in 𝑔−1(𝑋). Similarly, there are 𝑛𝑙−𝑘 elements in ℎ−1(𝑋). So, there exist a bijection 𝑡𝑋 between 𝑔−1(𝑋) and ℎ−1(𝑋). As 𝑔 and ℎ are both

surjection from 𝐺_𝑛𝑙 to 𝐺_𝑛𝑘, so

⋃ 𝑔−1 𝑋∈𝐺_𝑛𝑘

(𝑋) = 𝐺_𝑛𝑙 = ⋃ ℎ−1

𝑋∈𝐺_𝑛𝑘

(𝑋).

Thus we can construct a bijection 𝑡 from 𝐺_𝑛𝑙 to itself such that the limitation of 𝑡 on

each 𝑔−1(𝑋) is 𝑡_𝑋. Then, we have the commutable diagram:

Now, define ℎ′ = 𝑡 ∘ 𝑔′, we have 𝑓 = 𝑔 ∘ 𝑔′= ℎ ∘ 𝑡 ∘ 𝑔′= ℎ ∘ ℎ′. Let 𝑉 be any

𝑙-unit in 𝐺𝑛𝑚, then 𝑔′ is bijective on 𝑉. As 𝑡 is bijective on 𝐺_𝑛𝑙, we have ℎ′= 𝑡 ∘ 𝑔′

is bijective from 𝑉 to 𝐺_𝑛𝑙. By definition, ℎ′ is a Latin map. So ℎ′ is a construction of

𝑓 about the extending ℎ.

When constructing a Latin map, the method is first to build up a Latin map on the subset of its domain, then use normal extension to complete the construction on the whole domain.

Example 3.23 Let 𝑓 be a 3-dimensional Latin map of level 3 with rank 1, 𝑔 be a minimal non-trivial Latin map of level 3. They have the following representation: 𝑓: 𝑔: _(3.23.1)

To construct the construction of 𝑓 of the extending 𝑔, the first step is to find out the preimage of 0,1,2 under 𝑔. For example:

𝑔−1_{(0) = {(0,0), (1,2), (2,1)} , 𝑔}−1_{(1) = {(0,1), (1,0), (2,2)} , 𝑔}−1_{(2) =}

{(0,2), (1,1), (2,0)}.

So, 𝑔−1(0) = {0,5,7}, 𝑔−1(1) = {1,3,8}, 𝑔−1(2) = {2,4,6}. Now, fill 0 to 8 directly into the first 2-unit of 𝐺₃3, which forms this:

The second step is to find out all grids in 𝐺₃3 in the preimage of 𝑔 that needed to fill in 0 via the representation of 𝑓. Mark them out:

Here, it is impossible to fill 0 or 5 into the place marked by @, so the only choice is 7. It is impossible to fill 0 or 7 into the place marked by #, so the only choice is 5. Then, the first row is filled out:

Use same method to fill out the first 3 rows, and get:

Finally, fill the preimage of other numbers into 𝐺₃3, and get a valid construction:

its preimage under 𝑔, and use normal extension to deduce other grids. The key of the algorithm is to make sure that the preimage of the entries of the same image should be different in a same unit. For the entries of different image, it does not need to be discussed since their preimage are naturally different. The First Extending Construction Theorem guarantees that any other extending can have same choice if there isa choice of extending.

After searching by computers, there are 8! × 24 different kinds of compoundable Latin maps of type 𝐺₃3 → 𝐺₃2 → 𝐺₃1. Here, 8! is the number of all possible map in the first 2-unit of 𝐺₃3, the only difference between them is just a representation map. So, the actual number of independent compoundable Latin maps of level 3 is only 24.

Theorem 3.24 (The Second Extending Construction Theorem).Let 𝑓 be an

𝑚-dimensional Latin map of level 𝑛 with rank 𝑘, 𝑔 be an 𝑙-dimensional extending

map. If there is a construction of 𝑓 about extending 𝑔, then any submap of 𝑓 that has

dimension more than 𝑙 has its construction about extending 𝑔.

Proof.Let 𝑔′ be the construction of 𝑓 about extending 𝑔, ℎ be an 𝑙-dimensional submap of 𝑓. Let the domain of ℎ be 𝑈, then the limitation of 𝑔′ on 𝑈 is a sub map of 𝑔′, denote it by ℎ′. So ℎ = 𝑓|_𝑈 = (𝑔 ∘ 𝑔′)|_𝑈 = 𝑔 ∘ (𝑔′|_𝑈) = 𝑔 ∘ ℎ′,i.e. ℎ′ is a construction about ℎ

The main value of the Second Extending Construction Theorem is its contra-positive theorem. For an extending map of same dimension, if a compoundable Latin map of a lower dimension does not exist, the higher dimension case does not exist either. A 4-dimensional Latin map of level 3 with rank 2 cannot compound with a minimal non-trivial Latin map of level 3, therefore higher dimension Latin maps obviously cannot compound with the minimal non-trivial Latin map of level 3 either. This means, the only Latin maps that can compound with the minimal non-trivial Latin map of level 3 are the 24 kinds of 𝐺₃3 → 𝐺₃2.

4. CONCLUSION

The Latin map extends the dimension of the traditional Latin square from 2-dimension to 3-dimension or above. This extension not only preserves all properties of Latin square in 2-dimension, but also builds up a correlative structure between Latin maps of different dimensions by the partial map. The composition questions derived from this definition has come to a completed result. The extending construction theorem has provided the algorithm to fill out the construction of a Latin map about an extending map.

REFERENCES

1. SuyunGeng. "Discrete Mathematics".

2. Leonhard Euler. 1779. 1782. "Recherches sur une nouvelle espece de quarres magiques".

4. Jia-yu Shao, Wan-di Wei. 1992."A Formula for the Number of Latin Squares," Discrete Mathematics.

110 (1992). 293-296.

5. Bhaskar Bagchi. Sept, 2012."General Introduction To Latin Squares".

6. Henning Bruhn, Reinhard Diestel, Matthias Kriesell, Rudi Pendavingh, Paul Wollan. "Axioms for

Infinite Matroids".

7. Douglas S Stones. 2013."Symmetries of partial Latin squares," European Journal of Combinatorics.

34 (2013). 1092-1107.

8. Xuyan Shao, Yong Zhang, Chengmin Wang. 2015. "Existence of a family of strongly symmetric

self-orthogonal diagonal Sudoku squares", Applied Mathematics - A Journal of Chinese Universities.

2015, 30(4):469-475.

9. JL Alperin, RB Bell. "Groups and Representations".