Finite Fields, Applications and Open ProblemsFinite FieldsCombinatorial ObjectsLatin Squares and SudokuCostas ArraysOAs, CAs and OOAs Finite Fields, Applications and Open Problems

Finite Fields Combinatorial Objects Latin Squares and Sudoku Costas Arrays OAs, CAs and OOAs

Finite Fields, Applications and Open Problems

Daniel PanarioSchool of Mathematics and Statistics

Carleton [email protected]

LAWCI School, Campinas, July 2018

Finite Fields, Applications and Open Problems Daniel Panario


Summary

Lecture 1: Applications in Combinatorics

Brief review of finite fields.

Introduction to combinatorics objects (designs, latin squares,several types of arrays).

Classical results (latin squares and Sudoku; Costas arrays).

Orthogonal arrays and their constructions based on finitefields.

Some applications in cryptography/coding theory (brief):

secret sharing and combinatorial designs;orthogonal arrays and codes.

Orthogonal array variants (covering arrays, ordered orthogonalarrays) and their constructions based on finite fields.



Summary (cont.)

Lecture 2: Applications in cryptography

Applications of finite fields (brief).

Differential map, differential uniformity, and differentialcryptanalysis (special functions and desired properties,nonlinearity and low differential uniformity).

Examples of S-box functions and their characteristics.

Perfect nonlinear (PN) and almost perfect nonlinear (APN)functions.

Permutation polynomials and their cycle decomposition.

Generating pseudorandom sequences: how random is asequence, requirements for sequences in cryptography.



History

Finite fields were originally treated, for some particular cases, byFermat (1601-1665), Euler (1707-1783), Lagrange (1736-1813),Legendre (1752-1833) and Gauss (1777-1855).

The crucial researcher for finite fields is Evariste Galois(1811-1832). His paper Sur la theorie des nombres marks thebeginning of finite fields, or as they are also called, Galois fields.

By the end of the 19th century all the structure of finite fields wasknown. The 20th century was the time for the applications offinite fields. Of course, the main reason for these applications toflourish was the appearance of computers.



Application Areas

Many projects undertaken in finite fields can be applied almostimmediately to “real-world” problems.

Finite fields are used extensively in areas such as:

coding theory (for the recovery of errors),

cryptography (for the secure transmission of data),

communications and electrical engineering,

computer science,

· · · .

The vast majority of these applications work on the finite field F2

that we introduce next.



Groups

Definition. A group (G, ∗) is a set G together with a binaryoperation ∗ such that

(a) for all a, b ∈ G, a ∗ b ∈ G;

(b) for all a, b, c ∈ G, a ∗ (b ∗ c) = (a ∗ b) ∗ c;(c) there exists an element e ∈ G such that a ∗ e = e ∗ a = a for

all a ∈ G;

(d) for all a ∈ G, there exists an element b ∈ G such thata ∗ b = b ∗ a = e.

The group G is abelian if G is a group and

(e) for all a, b ∈ G, a ∗ b = b ∗ a.

Examples: (Z,+), and (Q \ 0, ·).



Finite Fields

Definition. A field (F,+, ·) is a set F together with operations +and · such that:

(1) (F,+) is an abelian group;

(2) (F \ 0, ·) is an abelian group;

(3) Distributive laws hold, that is, for a, b, c ∈ F , we have

a · (b+ c) = a · b+ a · c,(b+ c) · a = b · a+ c · a.

If #F is finite, then we say that F is a finite field.

Example:

Z/pZ is a field if and only if p is a prime.

p = 2: (0, 1,+, ·) is the field F2 of two elements!



Background on Finite Fields I

(Existence and Uniqueness) Up to isomorphisms, there isexactly one finite field with q = pn elements, denotedFpn = Fq for all primes p and positive integers n.The characteristic of the finite field Fq is p.

In Fq, aq = a for all a ∈ Fq.(Freshman’s Dream) We have that for 0 < i < p

(p

i

)=p(p− 1) . . . (p− i+ 1)

i!≡ 0 (mod p).

Hence, if α, β ∈ Fp, we have (α+ β)p = αp + βp. Thisgeneralizes for powers pn.



Background on Finite Fields II

The multiplicative group of Fq is cyclic. The generators ofthis multiplicative group are primitive elements.

A polynomial f ∈ Fq[x] is irreducible over Fq if f has positivedegree and f = gh with g, h ∈ Fq[x] implies that g or h is aconstant. Otherwise, f is reducible. An irreducible polynomialwith primitive roots is a primitive polynomial.

(Subfield Criterion) Every subfield of Fqn is of the form Fqkfor k dividing n.

The extension field Fqn can be seen as a vector space ofdimension n over Fq. Important in practice are polynomialand normal bases.



Examples: 1 + 1 = 0, and 1 + 1 + 1 = 0

The tables for addition and multiplication in F2 are:

+ 0 1

0 0 11 1 0

· 0 1

0 0 01 0 1


+ 0 1 2

0 0 1 21 1 2 02 2 0 1

· 0 1 2

0 0 0 01 0 1 22 0 2 1

There are finite fields of order pn for prime p and integer n ≥ 1.



Examples: 1 + 1 = 0, and 1 + 1 + 1 = 0


+ 0 1

0 0 11 1 0

· 0 1

0 0 01 0 1


+ 0 1 2

0 0 1 21 1 2 02 2 0 1

· 0 1 2

0 0 0 01 0 1 22 0 2 1

There are finite fields of order pn for prime p and integer n ≥ 1.



Classical Combinatorial Objects



Combinatorial Designs

Combinatorial designs are combinatorial objects like arrays and setsystems with some type of balance property.

Examples: latin squares, Steiner triple systems, t-designs, blockdesigns, orthogonal and covering arrays, finite geometry, etc.



Exemplo 1: Latin Squares

1 2 3 4 5

2 4 1 5 3

3 5 4 2 1

4 3 5 1 2

5 1 2 3 4

Definition

A latin square of order n is a matrix n×n such that each symbol of1, 2, . . . , n appear exactaly once in each row and in each column.

experimental farm medication test Sudoku game

Sudoku Squares andChromatic PolynomialsAgnes M. Herzberg and M. Ram Murty

The Sudoku puzzle has become a very

popular puzzle that many newspapers

carry as a daily feature. The puzzle con-

sists of a 9×9 grid in which some of the

entries of the grid have a number from

1 to 9. One is then required to complete the grid

in such a way that every row, every column, and

every one of the nine 3× 3 sub-grids contain the

digits from 1 to 9 exactly once. The sub-grids are

shown in Figure 1.

Figure 1. A Sudoku grid.

Agnes M. Herzberg is professor emeritus of mathemat-

ics at Queen’s University, Canada. Her email address is

[email protected].

M. Ram Murty is professor of mathematics at Queen’s

University, Canada. His email address is murty@mast.

queensu.ca.

Research of both authors is partially supported by Natu-

ral Sciences and Engineering Research Council (NSERC)

grants.

Recall that a Latin square of rank n is an n× narray consisting of the numbers such that eachrow and column has all the numbers from 1 ton. In particular, every Sudoku square is a Latinsquare of rank 9, but not conversely because ofthe condition on the nine 3 × 3 sub-grids. Figure2 (taken from [6]) shows one such puzzle withseventeen entries given.

1

4

2

5 4 7

8 3

1 9

3 4 2

5 1

8 6

Figure 2. A Sudoku puzzle with 17 entries.

For anyone trying to solve a Sudoku puzzle,several questions arise naturally. For a given puz-zle, does a solution exist? If the solution exists, isit unique? If the solution is not unique, how manysolutions are there? Moreover, is there a system-atic way of determining all the solutions? Howmany puzzles are there with a unique solution?What is the minimum number of entries that canbe specified in a single puzzle in order to ensurea unique solution? For instance, Figure 2 showsthat the minimum is at most 17. (We leave it to

708 Notices of the AMS Volume 54, Number 6



Applications: Latin Squares

experimental designs (statistics)

mathematical puzzles: Sudoku

error correcting codes (orthogonal latin squares):

A =0 1 21 2 02 0 1

, B =0 1 22 0 11 2 0

Consider the following construction:

location sA sB(0, 0) 0 0(0, 1) 1 1(0, 2) 2 2(1, 0) 1 2(1, 1) 2 0(1, 2) 0 1(2, 0) 2 1(2, 1) 0 2(2, 2) 1 0

−→

codewords000001110222101211201201202121022210

código capaz de corrigir 2 erros1 erro



Example 2: Orthogonal and Covering Arrays

Definition

An orthogonal array OA(t, k, v) is a vt × k array with each entryfrom a set V of size v and satisfying the following property: for anyvt × t subarray each t-tuple of V t appears exactly once as a row.

OA(2, 4, 3) =

000001221220220220210211211011011012



Covering Arrays

Definition

A covering array CA(N ; t, k, v) is an N × k array with each entryfrom a set V of size v and satisfying the following property: for anyN × t subarray each t-tuple of V t appears at least once as a row.

CAN(t, k, v) = minN∈NN : ∃ CA(N ; t, k, v).

We comment on relations of orthogonal arrays to other objects likeMOLS (mutually orthogonal latin squares) and codes, as well as onconstructions of covering arrays based on finite fields.



Application of Covering Arrays to Software Testing

Test a system with k = 4 components, each one with g = 3 options:

ComponentWeb Browser Operating Connection Printer

System Type ConfigConfig: Netscape(0) Windows(0) LAN(0) Local (0)

IE(1) Macintosh(1) PPP(1) Networked(1)Other(2) Linux(2) ISDN(2) Screen(2)

Table 3: Four components, each with 3 configurations

Test Case Browser OS Connection Printer

1 NetScape Windows LAN Local2 NetScape Linux ISDN Networked3 NetScape Macintosh PPP Screen4 IE Windows ISDN Screen5 IE Macintosh LAN Networked6 IE Linux PPP Local7 Other Windows PPP Networked8 Other Linux LAN Screen9 Other Macintosh ISDN Local

Table 4: Test Suite to Cover all Pairs from Table 3

agriculture and manufacturing [64]. It has entered the software testing community, appear-ing in practitioner’s guidebooks [70, 89], and provided in simple spreadsheet formats [40, 41].The use of covering arrays in software testing was pioneered by Mandl [83] and Brownlie etal. [9, 102], and statistical foundations were explored in [46, 47, 48, 49, 82, 91]. Empiricalresults indicate that testing of all pairwise interactions in a software system indeed finds alarge percentage of existing faults [45, 78]. Indeed, Burr et al. [11] provide more empiricalresults to show that this type of test coverage leads to useful code coverage as well. Dalalet al. present empirical results to argue that the testing of all pairwise interactions in asoftware system finds a large percentage of the existing faults [45]. Dunietz et al. link thee!ectiveness of these methods to software code coverage. They show that high code blockcoverage is obtained when testing all two-way interactions, but higher subset sizes are neededfor good path coverage [54]. Kuhn et al. examined fault reports for three software systems.They show that 70% of faults can be discovered by testing all two-way interactions, while90% can be detected by testing all three way interactions. Six-way coverage was required inthese systems to detect 100% of the faults reported [78]. This study was followed by similarexperiments, such as one of 109 software-controlled medical devices that were recalled bythe U.S. Food and Drug Administration (FDA) [79]. These experiments found that 97% ofthe flaws in these 109 cases could be detected with pair-wise testing of parameter settings.Only three devices required coverage higher than two.

Williams et al. [132] quantify the coverage for a particular interaction strength. Forinstance, if we have four factors, any new test case can contribute at most

!42

", or 6 new

31

Exaustively testing all options requires 34 = 81 tests. In general, errors

are caused by the “interaction” of t components (t << k). A covering

array with t = 2, k = 4, g = 3 covers all possible pairs with just 9 tests.

ComponentWeb Browser Operating Connection Printer

System Type ConfigConfig: Netscape(0) Windows(0) LAN(0) Local (0)

IE(1) Macintosh(1) PPP(1) Networked(1)Other(2) Linux(2) ISDN(2) Screen(2)

Table 3: Four components, each with 3 configurations

Test Case Browser OS Connection Printer

1 NetScape Windows LAN Local2 NetScape Linux ISDN Networked3 NetScape Macintosh PPP Screen4 IE Windows ISDN Screen5 IE Macintosh LAN Networked6 IE Linux PPP Local7 Other Windows PPP Networked8 Other Linux LAN Screen9 Other Macintosh ISDN Local

Table 4: Test Suite to Cover all Pairs from Table 3

agriculture and manufacturing [64]. It has entered the software testing community, appear-ing in practitioner’s guidebooks [70, 89], and provided in simple spreadsheet formats [40, 41].The use of covering arrays in software testing was pioneered by Mandl [83] and Brownlie etal. [9, 102], and statistical foundations were explored in [46, 47, 48, 49, 82, 91]. Empiricalresults indicate that testing of all pairwise interactions in a software system indeed finds alarge percentage of existing faults [45, 78]. Indeed, Burr et al. [11] provide more empiricalresults to show that this type of test coverage leads to useful code coverage as well. Dalalet al. present empirical results to argue that the testing of all pairwise interactions in asoftware system finds a large percentage of the existing faults [45]. Dunietz et al. link thee!ectiveness of these methods to software code coverage. They show that high code blockcoverage is obtained when testing all two-way interactions, but higher subset sizes are neededfor good path coverage [54]. Kuhn et al. examined fault reports for three software systems.They show that 70% of faults can be discovered by testing all two-way interactions, while90% can be detected by testing all three way interactions. Six-way coverage was required inthese systems to detect 100% of the faults reported [78]. This study was followed by similarexperiments, such as one of 109 software-controlled medical devices that were recalled bythe U.S. Food and Drug Administration (FDA) [79]. These experiments found that 97% ofthe flaws in these 109 cases could be detected with pair-wise testing of parameter settings.Only three devices required coverage higher than two.

Williams et al. [132] quantify the coverage for a particular interaction strength. Forinstance, if we have four factors, any new test case can contribute at most

!42

", or 6 new

31

(Example due to Colbourn, 2004.)



Exemplo 3: Steiner Triple Systems

Definition

A Steiner triple systems STS(n) of order n is a set of triples,subsets of X = 1, 2, . . . , n, such that for each pair of elementsof X appears in exactely one of the triples.

STS(7) :1, 2, 4, 1, 3, 7, 1, 5, 6, 2, 3, 5, 2, 6, 7, 3, 4, 6, 4, 5, 7



Steiner Triple Systems

The next examples of STS are “resolvable” in “parallel classes”STS(9) has 9 points and 12 triples, 4 parallel classes

1 2 3

4 5 6

7 8 9

STS(15) has 15 points, 35 triples and 7 parallel classes

Kirkman problem “girl school parade” (1850)balanced scheduling working groups.

Kirkman's schoolgirl problemFrom Wikipedia, the free encyclopedia

Kirkman's schoolgirl problem is a problem in combinatorics proposed by Rev.Thomas Penyngton Kirkman in 1850 as Query VI in The Lady's and Gentleman's Diary(pg.48). The problem states:

Fifteen young ladies in a school walk out three abreast for seven days insuccession: it is required to arrange them daily so that no two shall walk twiceabreast.[1]

Contents1 Solution2 History3 Generalization4 Other applications5 Notes6 References

Solution

If the girls are numbered from 01 to 15, the following arrangement is one solution:[2]

Sun. Mon. Tues. Wed. Thurs. Fri. Sat.01, 06, 11 01, 02, 05 02, 03, 06 05, 06, 09 03, 05, 11 05, 07, 13 11, 13, 0402, 07, 12 03, 04, 07 04, 05, 08 07, 08, 11 04, 06, 12 06, 08, 14 12, 14, 0503, 08, 13 08, 09, 12 09, 10, 13 12, 13, 01 07, 09, 15 09, 11, 02 15, 02, 0804, 09, 14 10, 11, 14 11, 12, 15 14, 15, 03 08, 10, 01 10, 12, 03 01, 03, 0905, 10, 15 13, 15, 06 14, 01, 07 02, 04, 10 13, 14, 02 15, 01, 04 06, 07, 10

A solution to this problem is an example of a Kirkman triple system,[3] which is a



Kirkman Problem (“Girl School Parade”)

Kirkman (1850) problem (how to arrange a girl school parade)deals with balanced scheduling work groups:

Kirkman's schoolgirl problemFrom Wikipedia, the free encyclopedia

Kirkman's schoolgirl problem is a problem in combinatorics proposed by Rev.Thomas Penyngton Kirkman in 1850 as Query VI in The Lady's and Gentleman's Diary(pg.48). The problem states:

Fifteen young ladies in a school walk out three abreast for seven days insuccession: it is required to arrange them daily so that no two shall walk twiceabreast.[1]

Contents1 Solution2 History3 Generalization4 Other applications5 Notes6 References

Solution

If the girls are numbered from 01 to 15, the following arrangement is one solution:[2]

Sun. Mon. Tues. Wed. Thurs. Fri. Sat.01, 06, 11 01, 02, 05 02, 03, 06 05, 06, 09 03, 05, 11 05, 07, 13 11, 13, 0402, 07, 12 03, 04, 07 04, 05, 08 07, 08, 11 04, 06, 12 06, 08, 14 12, 14, 0503, 08, 13 08, 09, 12 09, 10, 13 12, 13, 01 07, 09, 15 09, 11, 02 15, 02, 0804, 09, 14 10, 11, 14 11, 12, 15 14, 15, 03 08, 10, 01 10, 12, 03 01, 03, 0905, 10, 15 13, 15, 06 14, 01, 07 02, 04, 10 13, 14, 02 15, 01, 04 06, 07, 10

A solution to this problem is an example of a Kirkman triple system,[3] which is a



Threshold Secret Sharing Scheme Application

group of 3 managers2 managers can open box

1 manager cannot have info

1 2 3

4 5 6

7 8 9

Example: key bM1 receives “5”M2 receives “2”M3 receives “8”

shadows keys(3 people) (secret)

1, 2, 3 a4, 5, 6 a7, 8, 9 a

1, 4, 7 b2, 5, 8 b3, 6, 9 b

1, 5, 9 c2, 6, 7 c3, 4, 8 c

1, 6, 8 d2, 4, 9 d3, 5, 7 d



Theorem

If q ≡ 1 (mod 6) is a prime power, then there exists a Kirkmantriple system of order 2q + 1.

Proof: Let q = 6t+ 1 and let α ∈ Fq a primitive element. Letθ = (αt + 1)2−1. We define X = (Fq × 1, 2) ∪ ∞. Let usconstruct the first parallel class with the following ste of blocks:

Π0 = ∞, (0, 1), (0, 2)∪ (αi, 1), (αi+t, 1), (θαi, 2) :

0 ≤ i ≤ t− 1, 2t ≤ i ≤ 3t− 1, 4t ≤ i ≤ 5t− 1∪ θαi+t, 2), (θαi+3t, 2), (θαi+5t, 2) : 0 ≤ i ≤ t− 1.

The other parallel classes are constructed developing through Fq(add 1 to the elements of Fq in each block, sucessively).



Example for q = 7: α = 3 is primitive in F7 andθ = (3 + 1)× 2−1 = 4× 4 = 2.Blocks of Π0:∞, (0, 1), (0, 2),(1, 1), (3, 1), (2, 2), (2, 1), (6, 1), (4, 2), (4, 1), (5, 1), (1, 2)(6, 2), (5, 2), (3, 2)Blocks of Π1:∞, (1, 1), (1, 2),(2, 1), (4, 1), (3, 2), (3, 1), (0, 1), (5, 2), (5, 1), (6, 1), (2, 2)(0, 2), (6, 2), (4, 2)...Blocks of Π6:∞, (6, 1), (6, 2),(0, 1), (2, 1), (1, 2), (1, 1), (5, 1), (3, 2), (3, 1), (4, 1), (0, 2)(5, 2), (4, 2), (2, 2)



Latin Squares and Sudoku



Sudoku History

1 Modern puzzle designed by Howard Garns (at age 74), andfirst published by Dell Magazines in 1979 with the namenumber place. (This was only rediscovered around 2005.)

2 In Japan, Nikoli, Inc. first published puzzles in the MonthlyNikolist in 1984.

3 Maki Kaji (Nikoli President) originally named named thepuzzle Suuji Wa Dokushin Ni Kagiru (”the numbers must besingle”), then abbreviated it to “Sudoku” (Su = number,Doku = single).

4 International hit by 2005.



Sudoku Definition

A Sudoku square is a 9× 9 array using the numbers 1, . . . , 9arranged so that

1 Each row has each number once.

2 Each column has each number once.

3 Each of the 9, 3× 3 subsquares has each number once.

We also use numbers 0, . . . , 8 for convenience.

There are innumerous generalizations of Sudoku including diagonalSudoku, even-odd Sudoku, colored Sudoku, geometry Sudoku(irregular regions), and many more.



A Sudoku square:

0 4 8 7 2 3 5 6 15 6 1 0 4 8 7 2 37 2 3 5 6 1 0 4 88 0 4 3 7 2 1 5 61 5 6 8 0 4 3 7 23 7 2 1 5 6 8 0 44 8 0 2 3 7 6 1 56 1 5 4 8 0 2 3 72 3 7 6 1 5 4 8 0



Here is a Sudoku puzzle from the above Sudoku square

2 3 157 2 3 5 0 48 0 3 2 61 6 8 0 77 1 58 3 7 6 1 5

6 0 21 5 8



Latin Squares

Let n be a positive integer. A Latin square of order n is an n× narray on n distinct symbols such that every symbol appears exactlyonce in every row and column. Here are two examples:

L1 =

0 1 2

1 2 0

2 0 1

L2 =

0 1 2

2 0 1

1 2 0

Two Latin squares are called orthogonal if when superimposedeach of the n2 pairs appear exactly once.

(L1, L2):

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

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

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



A set L1, . . . , Lt of Latin squares is mutually orthogonal(MOLS) if Li is orthogonal to Lj for all i 6= j.

Mutually orthogonal Latin squares were originally considered byEuler (1779) for military parade arrangements:

Six different regiments have six officers, each one holdinga different rank (of six different ranks altogether). Canthese 36 officers be arranged in a square formation sothat each row and column contains one officer of eachrank and one from each regiment?

The solution requires a pair of MOLS of order 6. The answer isnegative: we cannot have this arrangement for n = 6 (or n = 2).For n = 3 (3 regiments and 3 officers), see the previous slide!



Let N(n) be the maximum number of MOLS of orden n.

Theorem

Given n ≥ 2, there does not exist n MOLS(n), that is,N(n) ≤ n− 1.

Proof: Let s MOLS(n): L1, . . . , Ls, and assume without loss ofgenerality that the first row of each Li is [1, 2, . . . , n]. The valuesL1(2, 1), . . . , Ls(2, 1) are all distinct since ifLi(2, 1) = Lj(2, 1) = x the pair (x, x) would appear in positions(2, 1) and (1, x). Since L(1, 1) = 1, then Li(2, 1) 6= 1, for all1 ≤ i ≤ s. Since we have s distinct elements of 2, . . . , n, wehave s ≤ n− 1.

Check the case n = 3: why there can not be more than n− 1 = 2Latin squares?



Bose (1938) proved that if q is a prime power, N(q) = q − 1.

Idea: Let α ∈ F∗q and define the Latin square Lα(i, j) = i+ αj,where i, j ∈ Fq. The set of Latin squares Lα : α ∈ F∗q is a set ofq − 1 MOLS of order q.

We only know that this is true in the prime power case where wecan use finite fields.

Big open problem: (Prime Power Conjecture)There are n− 1 MOLS order n iff n is prime power.



Relations

Sudoku puzzles are a special case of Latin squares; any solution toa Sudoku puzzle is a Latin square.

Sudoku imposes the additional restriction that nine particular 3adjacent subsquares must also contain the digits 1 to 9.

One can construct some classes of Sudokus using ideas from Latinsquares like rotations. One can also use 3× 3 subsquares close to“magic” squares. . . like in our previous example!However these are easy Sudokus.

A magic square of order n has each of the numbers 1, . . . , n2

exactly once, and has every row, every column and every diagonalsumming to a constant value (magic sum) n(n2 + 1)/2.



Relations

Sudoku puzzles are a special case of Latin squares; any solution toa Sudoku puzzle is a Latin square.

Sudoku imposes the additional restriction that nine particular 3adjacent subsquares must also contain the digits 1 to 9.

One can construct some classes of Sudokus using ideas from Latinsquares like rotations. One can also use 3× 3 subsquares close to“magic” squares. . . like in our previous example!However these are easy Sudokus.

A magic square of order n has each of the numbers 1, . . . , n2

exactly once, and has every row, every column and every diagonalsumming to a constant value (magic sum) n(n2 + 1)/2.



Albrecht Durer ‘Melencolia’ (1514)



The Passion: Facade of the Sagrada Familia : 33

1 14 14 4

11 7 6 9

8 10 10 5

13 2 3 15

16 3 2 13

5 10 11 8

9 6 7 12

4 15 14 1



Magic Squares as Stamps

Macau 2014 stamps (The Guardian Science, November 3, 2014):



Some Sudoku Math Tidbits

1 L9 = # LSs order 9 is 9!8!377, 597, 570, 964, 258, 816

2 # Sudoku sqs. is 6,670,903,752,021,072,936,960 = L9828186

3 # “essentially different” Sudoku sqs. (rotations, reflections,permutations and relabellings) is 5,472,730,538.

4 Can have 77 of 81 cells filled, but no unique solution.



5 17 of 81 is minimum number of known cells filled for whichthe puzzle has unique solution; 49151 such puzzles known (asof today); here is one of them

142

5 4 78 31 9

3 4 25 1

8 6Finite Fields, Applications and Open Problems Daniel Panario


6 Problem. It was proved (through an exhaustive search) in2011 that there are no unique solution with 16 of the 81numbers given. It took a computational year; however, thereis no mathematical proof of this fact yet.

7 Problem. Given a Sudoku solution square, how does onedelete numbers so that the resulting Sudoku puzzle always hasa unique solution?

8 Problem Same thing for a given a Sudoku solution puzzle:what are the different numbers of cells that can be left filled,and still have unique solution. For example in our earlierexample, we had 35 clues given in the original puzzle.



Costas Arrays

Costas arrays were introduced by John Costas in 1965 for a sonarapplication. These arrays have low auto-ambiguity function, usedto counter-attack echo. This make them useful in applications insonar and radar communications, as well as CDMA (code-divisionmultiple access) fiber-optic local area networks.

A Costas array of order n is an n× n array of dots and blankswhich satisfies

n dots, n(n− 1) blanks, with exactly one dot in each row andcolumn; and

all segments between pairs of dots are different.



Example

n = 3:

··

·

··

·

··

·

···

Question: how many Costas arrays are there of order n = 4?

Shifted left-right in time and up-down in frequency, copies of thepattern can only agree with the original in one dot, no dots, or alln dots at once. This allows the recovery of the information.



Radar or Sonar Echo



Radar or Sonar Echo



Radar or Sonar Echo



Radar or Sonar Echo



Radar or Sonar Echo



Constructions

All known constructions of Costas arrays (Welch, Lempel andGolomb) are based on finite fields. Then, there are computationalexperiments. There are Costas arrays for infinitely many n, but notfor all n; the smallest not known size is n = 32.

Welch Construction: n = p− 1, α a primitive element in Fp.

The multiplicative group of Fq is cyclic. The generators of thismultiplicative group are called primitive elements.

Example: 2 is not primitive in F7 since

21 = 2, 22 = 4, 23 = 1, 24 = 2, 25 = 4, 26 = 1,

but 3 is primitive in F7 since

31 = 3, 32 = 2, 33 = 6, 34 = 4, 35 = 5, 36 = 1.



Constructions

All known constructions of Costas arrays (Welch, Lempel andGolomb) are based on finite fields. Then, there are computationalexperiments. There are Costas arrays for infinitely many n, but notfor all n; the smallest not known size is n = 32.

Welch Construction: n = p− 1, α a primitive element in Fp.

The multiplicative group of Fq is cyclic. The generators of thismultiplicative group are called primitive elements.

Example: 2 is not primitive in F7 since

21 = 2, 22 = 4, 23 = 1, 24 = 2, 25 = 4, 26 = 1,

but 3 is primitive in F7 since

31 = 3, 32 = 2, 33 = 6, 34 = 4, 35 = 5, 36 = 1.



Constructions (cont)

Example

Let p = 7, n = 6, α = 3; for 1 ≤ j ≤ 6, ai,j has a dot iff αj = i.

··

···

·3 2 6 4 5 1

We have αj+k −αj = αi+k −αi implies that either i = j or k = 0.



Orthogonal, Covering andOrdered Orthogonal Arrays



Orthogonal Arrays

Previously, we consider an orthogonal array OA(t, k, v) as a vt × karray with entries from a set V of size v satisfying that for anyvt × t subarray each t-tuple of V t appears exactly once as a row.

Let us consider t = 2, and index (number of repetitions) equal to 1.

Definition

Let us consider integers k ≥ 2 and n ≥ 1. An orthogonal arrayOA(k, n) is an array A with dimension n2 × k and entries from aset X of cardinality n, such that in any two columns every orderedpair of symbols from X appears exactly in 1 row of A.

Orthogonal arrays are related to various combinatorial objectsincluding MOLS and codes.



Orthogonal arrays and MOLS

Theorem

An OA(s+ 2, n) exists if and only if s MOLS(n) exist.

Idea of proof: Given L1, . . . , Ls, construct tuples(i, j, L1(i, j), . . . , Ls(i, j)) as rows of the OA for all 1 ≤ i, j ≤ n.

Each pair of symbols occurs in each pair of columns (a, b):

a = 1, b = 2 (by construction)

a = 1, b ≥ 3 (a row of Lb−2 is a permutation)

a = 2, b ≥ 3 (a column of Lb−2 is a permutation)

a, b ≥ 3 (La−2 and Lb−2 are orthogonal).



Example: OA(n+ 1, n) from n− 1 MOLS(n)

A =0 1 21 2 02 0 1

, B =0 1 22 0 11 2 0

Consider the following construction:

location sA sB(0, 0) 0 0(0, 1) 1 1(0, 2) 2 2(1, 0) 1 2(1, 1) 2 0(1, 2) 0 1(2, 0) 2 1(2, 1) 0 2(2, 2) 1 0

−→

codewords000001110222101211201201202121022210

código capaz de corrigir 2 erros1 erro

This gives an MDS (maximum distance separable) code: lengthn+ 1, minimum distance d = n, alphabet size n, and number forcodewords: M = n2.



OA Constructions via Finite Fields

Theorem

Let q be a prime power and 2 ≤ k ≤ q. Then, there exists anOA(k, q).

Proof: Let a1, . . . , ak distinct elements in Fq (they exist sincek ≤ q). Let us consider v1, v2 ∈ (Fq)k: v1 = (1, . . . , 1),v2 = (a1, . . . , ak), and define the rows of A, with indexes inFq × Fq, by

row (i, j) of A: iv1 + jv2.

To prove that A is an orthogonal array, pick any two columns cand d, 1 ≤ c < d ≤ k, and let x, y ∈ Fq. We need to show thatthere exist a unique row (i, j) of A such that A((i, j), c) = x andA((i, j), d) = y. This gives a system in the unknowns i and j:



i+ jac = x

i+ jad = y.

Subtracting we get j(ac − ad) = x− y. Since (ac − ad) 6= 0, thereexists a multiplicative inverse (ac − ad)−1 ∈ Fq.We conclude that j = (ac − ad)−1(x− y), and substituting wehave i = x− jac = x− ac(ac − ad)−1(x− y).

We can extend the above OA(q, q) to construct an OA(q + 1, q).

Theorem

Let q be a prime power. Then, there exists an OA(q + 1, q).

Prova: Construct an OA(q, q) as above. Include a column (q + 1)with A((i, j), q + 1) = j for all i, j. We get an OA(q + 1, q).



Applications of orthogonal arrays

OAs can be seen as maximum distance separable (MDS)codes (see, for example, Paterson and Stinson, 2014). Theseare codes that meet the Singleton bound (d+ k = n+ 1).

OAs can also be used for secret sharing, as we commented.An OA(t, k, n) is used to distribute n shares with threshold t,having nt possible keys.

The number of possible shares in a threshold scheme must begreater than or equal to the number of possible secrets. If thenumber of possible secrets in a threshold scheme equals thenumber of possible shares, the scheme is ideal. Ideal thresholdschemes are equivalent to combinatorial orthogonal arrays andmaximum distance separable (MDS) codes.



Orthogonal Arrays and Ramp Schemes

A (s, t, n)-ramp is a generalization of threshold schemes in whichthere are two thresholds: s is the lower threshold value and t is theupper threshold. In a ramp scheme, any t of the n players cancompute the secret, and no subset of s players can determine thesecret. A (t− 1, t, n)-ramp scheme is a (t, n) threshold scheme.

The relation between ramp schemes and combinatorial arrays isless clear; see the recent article by Stinson (Discrete Mathematics,2018) where he connects these schemes with augmentedorthogonal arrays.



Orthogonal Arrays, LFSRs and Primitive Polynomials

A polynomial f of degree m is called primitive if k = qm − 1is the smallest positive integer such that f divides xk − 1.

A shift-register sequence with characteristic polynomialf(x) = xm +

∑m−1i=0 cix

i is the sequence a = (a0, a1, . . .)defined by the recurrence relation

an+m = −m−1∑

i=0

ciai+n, for n ≥ 0.

If f is primitive over Fq, the sequence has period qm − 1.

A subset C of Fnq is an orthogonal array of strength t if for anyt-subset T = i1, i2, . . . , it of 1, 2, . . . , n and any t-tuple(b1, b2, . . . , bt) ∈ Ftq, there exists exactly |C|/qt elementsc = (c1, c2, . . . , cn) of C such that cij = bj for all 1 ≤ j ≤ t.



Let q = 2, n = 7, C ⊆ F72 with |C| = 8, and t = 2. Thus, we

want an orthogonal array with |C|/2t = 2 and any 2-tuple ofF22 = (0, 0), (0, 1), (1, 0), (1, 1) appearing exactly |C|/2t = 2

times:1 1 1 1 1 1 10 1 0 1 0 1 01 0 0 1 1 0 00 0 1 1 0 0 11 1 1 0 0 0 00 1 0 0 1 0 11 0 0 0 0 1 10 0 1 0 1 1 0

.

Theorem. Let C be a linear code over Fq. Then, C is anorthogonal array of maximal strength t if and only if C⊥, itsdual code, has minimum weight t+ 1.



Theorem (Munemasa). Let f be a primitive polynomial of degree

m over Fq and let 2 ≤ n ≤ qm − 1. Let Cfn be the set of allsubintervals of the shift-register sequence with length n generatedby f , together with the zero vector of length n. The dual code ofCfn is given by

(Cfn)⊥ = (b1, . . . , bn) :

n−1∑

i=0

bi+1xi is divisible by f.

Theorem (Munemasa). Let f(x) = xm + xl + 1 be a trinomialover F2 such that gcd(m, l) = 1. If g is a trinomial of degree atmost 2m that is divisible by f , then g(x) = xdeg g−mf(x),g(x) = f(x)2, or g(x) = x5 + x4 + 1 = (x2 + x+ 1)(x3 + x+ 1)or, its reciprocal, g(x) = x5 + x+ 1 = (x2 + x+ 1)(x3 + x2 + 1).

Cfn corresponds to an orthogonal array of strength 2 that has aproperty very close to being an orthogonal array of strength 3.



Example:Consider the orthogonal array constructed from the LFSR definedby the primitive polynomial f(x) = x3 + x+ 1 over F2:

x0 x1 x2 x3 x4 x5 x6

0 0 0 0 0 0 01 1 1 0 0 1 01 1 0 0 1 0 11 0 0 1 0 1 10 0 1 0 1 1 10 1 0 1 1 1 01 0 1 1 1 0 00 1 1 1 0 0 1

.

We have strength 3 for many columns, but we do not havestrength 3 for shifts of f(x) = x3 + x+ 1 and f(x)2 = x6 + x2 + 1.Check: x(x3 + x+ 1) = x4 + x2 + x and f(x)2 = x6 + x2 + 1.



Orthogonal Arrays

Definition

An orthogonal array OAλ(N ; t, k, v) is a N × k array with eachentry from a set V of size v and satisfying the following property:

For any N × t subarray each t-tuple of V t appears exactly λ = Nvt

times as a row.

λ: the index of the array.N : Number of rowst: Strengthk: Number of columnsv: Number of symbols



OAs were introduced by Rao (1946, 1947, 1949) for use indesign of experiments in Statistics (medicine, agriculture andmanufacturing).

OAs are used in computer science and cryptography.

Hedayat, Sloane and Stufken; Orthogonal Arrays: Theory andApplications. Springer, 1999.

Munemasa, A.: Orthogonal arrays, primitive trinomials, andshift-register sequences, Finite Fields Appl. 4, 252–260(1998).



Ordered Orthogonal Arrays

Let m and s be positive integers and Ω[m, s] be a set of size ms,partitioned into m blocks Bi of cardinality s, where

Bi = bis, . . . , b(i+1)s−1

for i = 0, . . . ,m− 1. Each block has the total ordering:

bis ≺ bis+1 ≺ . . . ≺ b(i+1)s−1.

The set Ω[m, s] has the structure of partially ordered set(poset): the union of m totally ordered sets with s elementseach.

An antiideal is a subset of Ω[m, s] closed under followers.



Ordered Orthogonal Arrays

Let m and s be positive integers and Ω[m, s] be a set of size ms,partitioned into m blocks Bi of cardinality s, where

Bi = bis, . . . , b(i+1)s−1

for i = 0, . . . ,m− 1. Each block has the total ordering:

bis ≺ bis+1 ≺ . . . ≺ b(i+1)s−1.

The set Ω[m, s] has the structure of partially ordered set(poset): the union of m totally ordered sets with s elementseach.

An antiideal is a subset of Ω[m, s] closed under followers.



Example: m = 3 and s = 3

Ω[3, 3] = 0, 1, 2, 3, 4, 5, 6, 7, 8B0 = 0, 1, 2, B1 = 3, 4, 5, B2 = 6, 7, 8

0

1

2

3

4

5

6

7

8

Antiideals of size 3

0, 1, 21, 2, 51, 2, 82, 4, 52, 5, 82, 7, 83, 4, 54, 5, 85, 7, 86, 7, 8




Ω[3, 3] = 0, 1, 2, 3, 4, 5, 6, 7, 8B0 = 0, 1, 2, B1 = 3, 4, 5, B2 = 6, 7, 8

0

1

2

3

4

5

6

7

8


0, 1, 21, 2, 51, 2, 82, 4, 52, 5, 82, 7, 83, 4, 54, 5, 85, 7, 86, 7, 8




Ω[3, 3] = 0, 1, 2, 3, 4, 5, 6, 7, 8B0 = 0, 1, 2, B1 = 3, 4, 5, B2 = 6, 7, 8

0

1

2

3

4

5

6

7

8


0, 1, 21, 2, 51, 2, 82, 4, 52, 5, 82, 7, 83, 4, 54, 5, 85, 7, 86, 7, 8




Ω[3, 3] = 0, 1, 2, 3, 4, 5, 6, 7, 8B0 = 0, 1, 2, B1 = 3, 4, 5, B2 = 6, 7, 8

0

1

2

3

4

5

6

7

8


0, 1, 21, 2, 51, 2, 82, 4, 52, 5, 82, 7, 83, 4, 54, 5, 85, 7, 86, 7, 8



Definition

Let t, m, s, v be positive integers such that s ≤ t ≤ ms. Anordered orthogonal array OOA(t,m, s, v) is a vt ×ms array Awith entries from a set V of size v, columns labeled by Ω[m, s],and satisfying the property:

For each antiideal I ⊂ Ω[m, s] of size t, each t-tuple of V t appearsexactly once in the t columns of A labeled by I.

When s = 1, OOA(t,m, 1, v) = OA(t,m, v).



Example

Ω[3, 3] = 0, 1, 2, 3, 4, 5, 6, 7, 8B0 = 0, 1, 2, B1 = 3, 4, 5, B2 = 6, 7, 8

OOA(t = 3,m = 3, s = 3, v = 2)

0 1 2 3 4 5 6 7 8

1 0 0 1 1 1 0 1 10 0 1 0 1 1 1 0 00 1 1 1 0 1 1 0 11 1 1 0 1 0 1 1 01 1 0 0 0 1 0 1 01 0 1 1 0 0 1 1 10 1 0 1 1 0 0 0 10 0 0 0 0 0 0 0 0


0, 1, 21, 2, 51, 2, 82, 4, 52, 5, 82, 7, 83, 4, 54, 5, 85, 7, 86, 7, 8



Example

Ω[3, 3] = 0, 1, 2, 3, 4, 5, 6, 7, 8B0 = 0, 1, 2, B1 = 3, 4, 5, B2 = 6, 7, 8

OOA(t = 3,m = 3, s = 3, v = 2)

0 1 2 3 4 5 6 7 8

1 0 0 1 1 1 0 1 10 0 1 0 1 1 1 0 00 1 1 1 0 1 1 0 11 1 1 0 1 0 1 1 01 1 0 0 0 1 0 1 01 0 1 1 0 0 1 1 10 1 0 1 1 0 0 0 10 0 0 0 0 0 0 0 0


0, 1, 21, 2, 51, 2, 82, 4, 52, 5, 82, 7, 83, 4, 54, 5, 85, 7, 86, 7, 8



Example

Ω[3, 3] = 0, 1, 2, 3, 4, 5, 6, 7, 8B0 = 0, 1, 2, B1 = 3, 4, 5, B2 = 6, 7, 8

OOA(t = 3,m = 3, s = 3, v = 2)

0 1 2 3 4 5 6 7 8

1 0 0 1 1 1 0 1 10 0 1 0 1 1 1 0 00 1 1 1 0 1 1 0 11 1 1 0 1 0 1 1 01 1 0 0 0 1 0 1 01 0 1 1 0 0 1 1 10 1 0 1 1 0 0 0 10 0 0 0 0 0 0 0 0


0, 1, 21, 2, 51, 2, 82, 4, 52, 5, 82, 7, 83, 4, 54, 5, 85, 7, 86, 7, 8



Example

Ω[3, 3] = 0, 1, 2, 3, 4, 5, 6, 7, 8B0 = 0, 1, 2, B1 = 3, 4, 5, B2 = 6, 7, 8

OOA(t = 3,m = 3, s = 3, v = 2)

0 1 2 3 4 5 6 7 8

1 0 0 1 1 1 0 1 10 0 1 0 1 1 1 0 00 1 1 1 0 1 1 0 11 1 1 0 1 0 1 1 01 1 0 0 0 1 0 1 01 0 1 1 0 0 1 1 10 1 0 1 1 0 0 0 10 0 0 0 0 0 0 0 0


0, 1, 21, 2, 51, 2, 82, 4, 52, 5, 82, 7, 83, 4, 54, 5, 85, 7, 86, 7, 8



Example

Ω[3, 3] = 0, 1, 2, 3, 4, 5, 6, 7, 8B0 = 0, 1, 2, B1 = 3, 4, 5, B2 = 6, 7, 8

OOA(t = 3,m = 3, s = 3, v = 2)

0 1 2 3 4 5 6 7 8

1 0 0 1 1 1 0 1 10 0 1 0 1 1 1 0 00 1 1 1 0 1 1 0 11 1 1 0 1 0 1 1 01 1 0 0 0 1 0 1 01 0 1 1 0 0 1 1 10 1 0 1 1 0 0 0 10 0 0 0 0 0 0 0 0


0, 1, 21, 2, 51, 2, 82, 4, 52, 5, 82, 7, 83, 4, 54, 5, 85, 7, 86, 7, 8



Motivation

Niederreiter (1987) introduced (t,m, s)-nets in base b; there areseveral applications of this object to numerical integration(quasi-Monte Carlo methods).

Niederreiter (1987) showed that a (t, t+ 2, s)-net in base b isequivalent to an OAbt(2, s, b).

Lawrence (1996) and Mullen and Schmid (1996) show that thereexists a (t,m, s)-net in base b if and only if there exists anOOAbt(m− t, s,m− t, b).

Ordered orthogonal arrays are a combinatorial characterization of(t,m, s)-nets.



Rosenbloom and Tsfasman (1997) and Skriganov (2002)constructed a class of maximum distance separable (MDS) codeswith respect to the NRT metric.

For q a prime power and s ≤ t they show that there exists an MDScode with respect to the NRT metric with length (q + 1)s,dimension t, and minimum distance (q + 1)s− t+ 1.

This class of MDS codes is known as Reed-Solomon m-codes;they are equivalent to an OOA(t, q + 1, t, q).



The existing OOA constructions prior to 2002 essentially repeatedcolumns of existing orthogonal arrays in clever ways so that theresulting arrays satisfied the required column coverage for theordered orthogonal array definition.

The work of Fuji-Hara and Miao (2002) for t = 3, 4 and the OOAconstruction of Castoldi et al (2017) for arbitrary t are the firstconstructions of OOAs which did not simply repeat columns.

Ordered orthogonal array construction using LFSR sequences,A. Castoldi, L. Moura, D. Panario and B. Stevens,IEEE Transactions on Information Theory, 63, 1336-1347, 2017.

A general construction of ordered orthogonal arrays using LFSRs,D. Panario, M. Saaltink, B. Stevens and D. Wevrick, submitted.



Theorem

For q a prime power and t ≥ 3, there exists an OOA(t, q + 1, t, q).

Let f(x) = c0 + c1x+ . . .+ ct−1xt−1 + xt be a degree-t

primitive polynomial over Fq = 0, β1, . . . , βq−1 and α ∈ Fqta root of f .

Label the columns of the subinterval array M(f) by

Zk = 0, 1, . . . , k − 1, where k = qt−1q−1 .

For each i = 1, . . . , q − 1, let kβi ∈ Zqt−1 such that

αkβi (α− βi) = 1.

Choose the columns of the subinterval array M(f) labeled bythe following indexes modulo k:



0

1

t− 2

t− 1

2t− 1

2t− 2

t+ 1

t

t+ tkβ1

t+ (t− 1)kβ1

t+ 2kβ1

t+ kβ1

t+ tkβq−1

t+ (t− 1)kβq−1

t+ 2kβq−1

t+ kβq−1



How many t-subsets of columns of an OOA(t, q + 1, t, q) have theproperty of cover all the t-tuples of Ftq?

Table: OOA(3, 4, 3, 3)

Primitive Polynomial 3-sets covered Percentage

1 + 2x2 + x3 163 0.740909

1 + x+ 2x2 + x3 156 0.709091

1 + 2x+ x3 156 0.709091

1 + 2x+ x2 + x3 162 0.736364

RT construction 120 0.545455

Number of 3-antiideals 20

Number of 3-subsets of a 12-set 220



Summary

In this lecture we revised several combinatorial objects where finitefields play a role in their construction.

We covered some historic results related to several types ofdesigns, latin squares and Costas arrays. We also showed anexample where finite fields did not produce interesting Sudokus.

Then we focused on combinatorial arrays such as orthogonalarrays, covering arrays and ordered orthogonal arrays. The finitefields constructions here are more sophisticated leading tocompetitive arrays.



Next: Covering Arrays


2 1 0 1

1 0 2 2

0 2 2 1

2 2 0 2

2 0 1 1

2 1 2 0

1 2 1 0

1 1 0 1

0 1 1 2

0 0 0 0

2 1 0 1

1 0 2 2

0 2 2 1

2 2 0 2

2 0 1 1

2 1 2 0

1 2 1 0

1 1 0 1

0 1 1 2

0 0 0 0

2 1 0 1

1 0 2 2

0 2 2 1

2 2 0 2

2 0 1 1

2 1 2 0

1 2 1 0

1 1 0 1

0 1 1 2

0 0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0

Example

0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2

0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0

1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0

0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1

1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0

2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1

1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2

1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1

2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1

0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2

1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0

1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1

1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1

0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1

0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0

2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0

0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2

2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0

1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2

2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1

2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2

1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2

0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1

2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0

2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2

2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Arrays from cyclic shifts of m-sequences

See survey by Moura, Mullen and Panario (DCC, 2016)

0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2

0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0

1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0

0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1

1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0

2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1

1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2

1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1

2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1

0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2

1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0

1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1

1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1

0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1

0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0

2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0

0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2

2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0

1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2

2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1

2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2

1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2

0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1

2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0

2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2

2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Arrays from cyclic shifts of m-sequences

Linearly dependent columns do not have the CA or OA property

0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2

0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0

1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0

0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1

1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0

2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1

1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2

1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1

2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1

0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2

1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0

1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1

1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1

0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1

0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0

2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0

0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2

2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0

1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2

2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1

2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2

1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2

0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1

2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0

2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2

2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2 + =

See survey by Moura, Mullen and Panario (DCC, 2016)

Arrays from cyclic shifts of m-sequences:Orthogonal arrays







Covering and orthogonal arrays

m-sequences

Orthogonal arrays from m-sequences

Covering arrays from m-sequences

0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2

0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0

1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0

0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1

1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0

2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1

1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2

1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1

2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1

0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2

1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0

1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1

1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1

0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1

0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0

2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0

0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2

2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0

1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2

2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1

2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2

1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2

0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1

2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0

2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2

2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2

0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0

1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0

0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1

1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0

2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1

1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2

1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1

2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1

0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2

1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0

1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1

1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1

0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1

0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0

2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0

0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2

2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0

1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2

2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1

2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2

1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2

0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1

2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0

2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2

2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2

0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0

1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0

0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1

1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0

2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1

1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2

1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1

2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1

0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2

1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0

1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1

1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1

0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1

0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0

2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0

0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2

2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0

1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2

2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1

2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2

1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2

0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1

2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0

2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2

2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2

0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0

1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0

0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1

1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0

2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1

1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2

1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1

2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1

0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2

1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0

1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1

1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1

0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1

0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0

2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0

0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2

2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0

1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2

2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1

2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2

1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2

0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1

2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0

2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2

2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2

0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0

1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0

0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1

1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0

2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1

1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2

1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1

2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1

0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2

1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0

1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1

1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1

0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1

0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0

2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0

0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2

2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0

1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2

2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1

2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2

1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2

0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1

2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0

2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2

2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0

0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0

1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0

0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1

1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0

0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1

1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0

1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1

0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1

0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0

1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0

1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1

1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1

0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1

0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0

0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0

0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0

0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0

1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0

0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1

0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0

1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0

0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1

0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0

0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0

0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2

0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0

1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0

0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1

1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0

2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1

1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2

1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1

2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1

0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2

1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0

1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1

1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1

0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1

0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0

2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0

0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2

2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0

1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2

2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1

2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2

1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2

0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1

2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0

2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2

2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2

0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0

1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0

0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1

1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0

2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1

1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2

1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1

2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1

0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2

1 1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0

1 1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1

1 0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1

0 0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1

0 2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0

2 0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0

0 2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2

2 1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0

1 2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2

2 2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1

2 1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2

1 0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2

0 2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1

2 2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0

2 2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2

2 0 0 1 0 1 2 1 1 2 0 1 1 1 0 0 2 0 2 1 2 2 1 0 2 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0

0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0

1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0

0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1

1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0

0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1

1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0

1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1

0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1

0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0

1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0

1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1

1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1

0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1

0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0

0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0

0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0

0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0

1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0

0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1

0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0

1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0

0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1

0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0

0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0

0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0

0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0

1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0

0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1

1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0

0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1

1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0

1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1

0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1

0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0

1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0

1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1

1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1

0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1

0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0

0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0

0 0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0

0 1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0

1 0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0

0 0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1

0 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0

1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0

0 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1

0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0

0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0

0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Linearly dependent columns do not have the CA or OA property

CA property

have better coverage

Covering arrays of strength t from m-sequences



Questions:


Questions:


Questions:


Questions:


Questions:


Corollaries

Questions:

Documents

Finite Fields, Applications and Open ProblemsFinite FieldsCombinatorial ObjectsLatin Squares and SudokuCostas ArraysOAs, CAs and OOAs Finite Fields, Applications and Open Problems