latin squares & steiner triple systems

Latin Squares & Steiner Triple Systems

An Old ProblemArrange the 16 face cards of a deck of playing cards in a 4 x 4 array so that each denomination (Ace, King, Queen, Jack) and each suit (Clubs, Hearts, Diamonds, Spades) appears only once in each row and column.

♠A ♥K ♦Q ♣J ♦J ♣Q ♠K ♥A ♣K ♦A ♥J ♠Q ♥Q ♠J ♣A ♦K

An enumeration by type of the solutions to this problem was published in 1723.

Latin SquaresIf we seperate the denominations and the suits we obtain:

A K Q J J Q K A K A J Q Q J A K

Each of these is a 4×4 Latin square.

A Latin square is an n × n square matrix whose entries consist of n symbols such that each symbol appears exactly once in each row and each column.

♠ ♥ ♦ ♣ ♦ ♣ ♠ ♥ ♣ ♦ ♥ ♠ ♥ ♠ ♣ ♦

Latin Squares

Latin squares have a long history. The concept probably originated with problems concerning the movement and disposition of pieces on a chess board. However, the earliest written reference is the solutions of the card problem published in 1723. The Latin square concept certainly goes back further than this written document. In his famous etching Melencholia I, the 16th Century artist Albrecht Dürer portrays an order 4 magic square, a relative of Latin squares, in the background. Magic squares can also be found in the ancient Chinese literature.

Melencholia I

Latin Squares The systematic development of Latin squares started with Euler (1779) and was carried on by Cayley (1877-1890) who showed that the multiplication table of a group is an appropriately bordered special Latin square. In the 1930's the concept arose once again in the guise of multiplication tables when the theory of quasi-groups and loops began to be developed as a generalization of the group concept. Latin squares played an important role in the foundations of finite geometries, a subject which was also in development at this time. Also in the 1930's, a big application area for Latin squares was opened by R.A.Fisher who used them and other combinatorial structures in the design of statistical experiments.

Quasigroups A Quasigroup (S,⊗) is a set S together with a binary operation (⊗) such that:1. The operation is closed (i.e., a⊗b∈S, ∀ a,b∈S)2. Given a,b∈S the equations a⊗x = b and y⊗a = b have unique solutions for x and y.

A simple example of a finite quasigroup is given by the set {0,1,2} with the operation ⊗ defined by a⊗b = 2a+b+1 where the operations on the right are the usual multiplication and addition modulo 3. The multiplication table for this quasigroup is given below:

⊗ 0 1 20 1 2 01 0 1 22 2 0 1

Quasigroups/Loops

A Loop is a quasigroup (S,⊗) that has an identity element, that is a quasigroup satisfying: There exists an element e ∈ S such that e⊗x = x⊗e = x ∀x ∈ S.

Loosely speaking, a loop is a non-associative group. In a group setting, the equation ax = b can be solved for x and by using all of the group axioms, one can show that the unique solution is x = a-1b. In a loop you also know that this equation has a unique solution, but you don't know what it is or how to find it in general.

Latin Squares and Quasigroups The simple result which causes our interest in these algebraic forms is:

Theorem: The multiplication table of a quasigroup is a Latin square.

Proof: Let a1 ,a2 ,...,an be the elements of the quasigroup and construct its multiplication table where the entry ars which occurs in the rth row and sth column is the product ar⊗as of the elements ar and as. If the same entry occured twice in the rth row, say in the sth and tth columns so that ars = art = b say, we would have two solutions to the equation ar⊗x = b, in contradiction to the quasigroup axiom.

Latin Squares and Quasigroups

Theorem: The multiplication table of a quasigroup is a Latin square.

Proof: Similarly, if the same entry occurred twice in the sth column, we would have two solutions to the equation y⊗as = c for some c. We conclude that each element of the quasigroup occurs exactly once in each row and column, and so the unbordered multiplication table (which is an nxn array) is a latin square.

Steiner Triple Systems When k = 3 and λ = 1, the BIBD is called a Steiner Triple System since the blocks are triples and each pair of varieties appears in exactly one triple. We can calculate that r = (v-1)/2 and b = v(v-1)/6. Since r is an integer, v must be odd. Since b is an integer and v is odd, either 3 divides v or 6 divides v-1, so v ≡ 3 mod 6 or v ≡ 1 mod 6. This necessary condition is also sufficient for the existence of a Steiner Triple System.

Theorem - [Kirkman, 1847] There exists a Steiner Triple System of v varieties iff v ≥ 3 and v ≡ 1 or 3 mod 6.

Proof: We have already shown the necessity of these conditions. The sufficiency is proved by construction.

Commutative Idempotent Quasigroups

A quasigroup (or any other algebraic structure) is said to be idempotent if x⊗x = x for all x in the quasigroup. In terms of the Latin square which is the multiplication table of the quasigroup, idempotency is recognized by the existence of an ordering of the elements of the quasigroup so that the entries in the main diagonal are in this order when the rows and columns are in order.

A quasigroup (or any other algebraic structure) is said to be commutative if x⊗y = y⊗x for every pair x,y in the quasigroup. The Latin square of a properly ordered commutative quasigroup is symmetric.


For example, the following is a multiplication table of an idempotent commutative quasigroup of order 5:

⊗ 1 2 3 4 5

1 1 5 2 3 42 5 2 4 1 33 2 4 3 5 14 3 1 5 4 25 4 3 1 2 5

Theorem: Idempotent commutative quasigroups exist iff they have odd order.


Theorem: Idempotent commutative quasigroups exist iff they have odd order.

Pf: Consider an appropriately bordered Latin square which is the multiplication table of an idempotent commutative quasigroup. Since the square is symmetric, each element appears an even number of times off the main diagonal and once on the main diagonal, so an odd number of times in all. Thus, the quasigroup has odd order. Now suppose n = 2s+1 and consider the addition table of ℤn. The elements 2i on the main diagonal are all distinct. For each j, rename (in all occurrences) the element 2j with j. By bordering the table appropriately we obtain an idempotent commutative quasigroup.

ExampleFor n = 5 we have

ℤ5,+ 0 1 2 3 4

0 0 1 2 3 41 1 2 3 4 02 2 3 4 0 13 3 4 0 1 24 4 0 1 2 3

0 3 1 4 23 1 4 2 01 4 2 0 34 2 0 3 12 0 3 1 4

0 1 2 3 40 0 3 1 4 21 3 1 4 2 02 1 4 2 0 33 4 2 0 3 14 2 0 3 1 4

Bose Construction (v ≡ 3 mod 6)Let v = 6n + 3 and (Q,⊗) an idempotent commutative quasigroup on the elements {1, ..., 2n+1}. Let S = Q x ℤ3. We define triples of elements of S of two types: Type 1: {(i,0), (i,1), (i,2) | i ԑ Q} and Type 2: {(i,k), (j,k), (i⊗ , + ) j k 1 mod 3 | i < j ԑ Q}.

And it works!The number of blocks in a Steiner Triple System of order v is v(v-1)/6. If we are considering a set of v(v-1)/6 triples of elements of a v-set and know that each pair elements is contained in at least one triple, then each pair must be contained in exactly one triple and we have an STS(v). Therefore we need only count the number of triples and verify that each pair of elements is contained in a triple in order to show that a system is an STS.

The Bose construction gives 2n+1 Type 1 triples and there are (2n+1)(n) choices for i < j and each gives 3 triples of Type 2. Thus there are (2n+1) + 3(2n+1)n = (2n+1)(3n+1) = (v/3)(v-1)/2 = v(v-1)/6 triples.

and it works!So, we only need to show that each pair of elements appear in a triple.

Let (a,b) and (c,d) be a pair of elements. Case 1: If a = c then this pair appears in a Type 1 triple. Case 2: If b = d, then a ≠ c and the Type 2 triple, (a,b)(c,b), (a ⊗ c, b+1) contains the pair. Case 3: If a ≠ c and b ≠ d then, since Q is a quasigroup,a⊗ = . x c has a unique solution for x Since Q is

idempotent and a ≠ , c we have that x ≠ . = + a If d b 1 ( , ), ( , ), (then the triple a b x b a⊗ , + ) = ( , ) .x b 1 c d works

= + , ⊗ = , If b d 1 then with y a solution of y c a the ( , ), ( , ), ( ⊗ , + ) = ( , ) . triple y d c d y c d 1 a b works

( ). So the Bose construction gives an STS v

½ Idempotent Commutative Quasigroups

A Latin square L of order 2n is said to be half-idempotent if the cells (i,i) and (n+i,n+i) contain the symbol i for 1 ≤ i ≤ n. A quasigroup is half-idempotent if it has a multiplication table which is half-idempotent.

We can form half-idempotent commutative quasigroups by starting with (ℤ2n,+) and renaming the elements.

1 2 3 41 1 3 2 42 3 2 4 13 2 4 1 34 4 1 3 2

Skolem Construction (v ≡ 1 mod 6)Let v = 6n + 1 and (Q,⊗) an half idempotent commutative quasigroup on the elements {1, ..., 2n}. Let S = {∞} U Q x ℤ3. We define triples of elements of S of three types: Type 1: {(i,0), (i,1), (i,2) | i ԑ [n]}, Type 2: {∞, (n+i, k), (i, k+1 mod 3) | i ԑ [n]}, and Type 3: {(i,k), (j,k), (i⊗ , + ) j k 1 mod 3 | i < j ԑ Q}.

Steiner Triple Systems There are a number of other constructions of STS's, we present only one of them.

Theorem : If there is a Steiner Triple System on e varieties and a Steiner Triple System on f varieties then there is a Steiner Triple System on ef varieties.

Proof: Let A be the STS whose elements are a1 ,a2 ,...,ae and B the STS whose elements are b1 ,b2 ,...,bf . Let S consist of the ef elements cij, where i = 1, ...,e and j = 1,...,f. A triple {cir,cjs,ckt} is in the STS based on S if and only if one of the following conditions hold: 1. r = s = t and {ai, aj, ak } is in A. 2. i = j = k and {br, bs, bt } is in B. 3. {ai, aj, ak } is in A and {br, bs, bt } is in B.We leave it as an exercise to show that these triples on S form a STS.

Steiner Triple Systems

It has been shown that there is only one STS of orders v = 3, 7 or 9. There are two nonisomorphic designs for v = 13 and 80 distinct STS's with v = 15. This work was done byhand but the v = 15 case was verified by computer. In 2004, Kaski and Östergård showed that there were11,084,874,829 distinct STS's with v = 19. Some upper and lower bounds are known for other STS's but no other exact counts.

OrthogonalityThe property needed by the pair of Latin squares in the solution of the 16 card problem is called orthogonality.

Two Latin squares L1 = | a

ij| and L

2 = | b

ij| on n symbols

1,2,...,n are said to be orthogonal if every ordered pair of symbols occurs exactly once among the n2 pairs (a

ij ,b

ij), i =

1,2,...,n; j = 1,2,...,n.

For example, these squares are orthogonal:

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

Orthogonal Mates

We often refer to one of a pair of orthogonal squares as being an orthogonal mate of the other.

Euler was originally interested in such pairs and in his writings he would always use Latin letters for the first square and Greek letters for the second. Thus, when Cayley referred to the first of the squares in Euler's work he referred to the Latin square.

When referring to both of the orthogonal squares he used the term Graeco-Latin squares, which is the way orthogonal squares are referred to in all the earlier literature.

Transversals Given a pair of orthogonal Latin squares, consider the cells in the first square which contain one particular symbol. Of course, by “Latinness”, there is only one of these cells in each row and column. Now consider the cells in the orthogonal mate which correspond to these cells in the first square. By orthogonality, the entries in these cells must all be different and so these cells form a transversal in the orthogonal mate. This observation applies to the cells corresponding to any symbol of the first square, so we have:

Theorem - A given Latin square possesses an orthogonal mate if and only if it has n disjoint transversals.

Transversals

1 2 3 42 1 4 33 4 1 24 3 2 1

A transversal

1 2 3 4 1 2 3 4 3 4 1 2 4 3 2 1 4 3 2 1 2 1 4 3 2 1 4 3 3 4 1 2

4 disjoint transversals

Odd Order

If the Latin square in question is the multiplication table of a group of odd order n, then it can be shown that the existence of a single transversal implies the existence of n disjoint transversals, so we conclude that the square has an orthogonal mate.

Theorem - The multiplication table of any group of odd order forms a Latin square which possesses an orthogonal mate.

Corollary - There exist pairs of orthogonal Latin squares of every odd order.

Even Order The existence question for pairs of orthogonal Latin squares of even order is much more difficult to settle and has a long and famous history. To start with, there are only two Latin squares of order 2 and they are not orthogonal. We have given an example of a pair of orthogonal squares of order 4. The next case, that of order 6, is the problem that originally interested Euler in the subject. Called the Problem of the 36 Officers, Euler stated it as follows in 1779, "Arrange 36 officers, 6 from each of 6 regiments, of 6 different ranks, into a 6x6 square, so that each row and each file contains one officer of each rank and one officer of each regiment." Clearly, the problem can be solved, as with the card problem, by finding a pair of orthogonal latin squares of order 6.

Euler's ConjectureAs brilliant a mathematician as Euler was, he was unable to find such a pair of squares and unable to prove that they did not exist. Based on his experience with the problem and some other pieces of evidence (such as the Corollary, which he was aware of), Euler made a conjecture which included and went beyond this problem:

Euler's Conjecture: There does not exist an orthogonal mate for any Latin square whose order has the form n = 4k + 2 (oddly even integers as he put it).

Tarry's Result120 years after Euler first stated the problem, Tarry in 1900 settled the problem of the 36 officers in the negative. His method was straight-forward, he listed out all of the 812,851,200 Latin squares of order 6 and examined each pair for orthogonality and found none [ actually, by working with reduced squares he simplified the problem to checking only 9408 pairs – but of course this was all done by hand].

It was beginning to look like the old master had pulled off a coupe, but in 1960 Bose, Shrikhande and Parker shocked the mathematical community by proving that if n > 6 Euler'sconjecture is false. Their original method is long, complicated and involved, looking at a number of special cases, but has since been simplified.

MOLS A set of Latin squares of the same order, each of which is an orthogonal mate of each of the others is called a set of mutually orthogonal Latin squares. This mouthful is often shortened to its acronym MOLS. For example, a set of 3 MOLS of order 4 is given by:

1 0 3 2 2 3 0 1 3 2 1 0 0 1 2 3

2 3 0 1 3 2 1 0 1 0 3 2 0 1 2 3

3 2 1 0 1 0 3 2 2 3 0 1 0 1 2 3

Question: How large can a set of MOLS be?

Equivalent MOLSTwo sets of MOLS with the same number of squares are said to be equivalent sets of MOLS if one can be obtained from the other by any combination of simultaneously permuting the rows of all the squares, simultaneously permuting the columns of all the squares and renaming the elements of any square.

Lemma - Any set of MOLS is equivalent to a set where each square has the first row in natural order and one of the squares (usually the first) is reduced (i.e., it also has its first column in natural order). Pf: Given a set of MOLS, we can convert it to an equivalent set by renaming the elements in each square so that the first rows are all in natural order. Now take any square and simultaneously permute the rows of all the squares so that the first column of this square is in natural order (this will not affect the first row since it is in natural order and so starts with the smallest element). The result is an equivalent set with the required properties.

Number of MOLS A set of MOLS in the form described by this lemma is said to be in standard form, and the lemma merely says that any set of MOLS is equivalent to a set of MOLS in standard form.

Theorem - No more than n-1 MOLS of order n can exist. Pf: Any set of MOLS of order n is equivalent to a set in standard form, which has the same number of squares in it. Consider the entries in first column and second row of all of the squares in standard form. No two squares can have the same entry in this cell. Suppose two squares had an r, say, in this cell, then in the superimposed square the ordered pair (r,r) would appear in this cell and also in the r-th cell of the first row because both squares have the same first row, and so, the two squares can not be orthogonal →← Now, we can not have a 1 in this cell, since it appears in the first column of the first row. Thus, there are only n-1 possible entries for this cell and so there can be at most n-1 squares.

Complete Sets of MOLS A set of MOLS of order n containing n-1 squares is called a complete set. We now have an existence question, for which orders do complete sets of MOLS exist?

We know by examples that complete sets exist for orders 2 (only one square is needed), 3 and 4 and also that no complete set exists for orders 6 and 10. Using finite fields, complete sets of MOLS for any order which is a prime or power of a prime can be constructed. However, it is an open research question of long standing whether or not a complete set of MOLS exists for any composite order.

Instant fame will go to any mathematician who can settle this question.

So what is known?n max

known MOLS(n)

n max known

MOLS(n)

n max known

MOLS(n)3 2 9 8 15 4

4 3 10 2 (< 9) 16 15

5 4 11 10 17 16

6 1 12 5 18 3

7 6 13 12 19 18

8 7 14 3 (< 13) 20 4

Field Construction

Construction - Let b1 ,b

2 ,...,b

n be the elements of a

finite field GF(n) where n = pk .

Let b1 be the multiplicative identity ( = 1) and b

n be the

additive identity ( = 0) of GF(n).

For e = 1,2, ..., n-1 define the n x n array L(e) = (aij(e) ) by

taking a

ij(e) = (b

e x b

i ) + b

j ,

where the operations on the right are in the field. The resulting squares form a complete set of MOLS of order n.

Affine Planes and MOLSComplete sets of MOLS(n) are interesting because they are equivalent to affine planes of order n. The affine planes obtained from the field construction of MOLS(q) are the AG(2,q), but the equivalence is more general.

Affine Plane of order n → MOLS(n)

Since an affine plane is resolvable, there are n+1 parallel classes each containing n disjoint lines, and two lines from different parallel classes intersect in a unique point. Choose two parallel classes to play special roles, say PX and PY. Label (arbitrarily) the lines in each of these with the labels 1, ..., n. Now each point of the affine plane is on

Affine Planes and MOLSAffine Plane of order n → MOLS(n)

a unique line, say m, of PX and a unique line, say s, of PY. The point can thus be associated with the ordered pair (m,s). Arbitrarily label each line of each other parallel class with the labels 1,...,n. To each of these parallel classes we associate a square Li. The entry in the ms position of this square is the label of the unique line in this parallel class that the point associated with (m,s) lies on.

Each Li is a Latin square. Suppose two positions in a row, say ms1 and ms2 have the same entry, say r. Then line r in the parallel class

Affine Planes and MOLSAffine Plane of order n → MOLS(n) Each Li is a Latin square. contains the points associated with (m,s1) and (m,s2), but these points are also on the line m of parallel class PX. →←A similar contradiction arises if there are two equal entries in a single column. Thus Li is a Latin square.

Li and Lj are orthogonal if i ≠ j.If these squares are not orthogonal, then there exists a pair of points associated with the ordered pairs (m,s) and (p,r) which are on the same line of the parallel class associated with Li and also on a common line of the parallel class associated with Lj. But this line can not be in both parallel classes. → ←

Affine Planes and MOLS

This process is easily reversible giving the implication that MOLS(n) Affine Plane of order n. →

MacNeish's ResultsTheorem (MacNeish [1922]) Suppose that there exist r MOLS of order n and r MOLS of order m, then there exist r MOLS of order mn.

Proof: This proof is a notational nightmare, but the idea behind it is actually very simple.

Let A(1), A(2), ... , A(r) be the set of MOLS of order n and B(1), B(2), ... , B(r) be the set of MOLS of order m. For e = 1,2,...,r, let (a

ij(e), B(e)) represent the m x m matrix whose

h,k entry is the ordered pair (aij

(e),bhk

(e)). Let C(e) be the mn x mn matrix that can be represented schematically by:


〚 a11e , Be a12

e , Be ⋯ a1ne , Be

a21e , Be a22

e , Be ⋯ a2ne , Be

⋮ ⋮ ⋱ ⋮an1

e , Be an2e , Be ⋯ ann

e , Be〛We will show that C(1),C(2),...,C(r) is a set of MOLS of order mn.


To see that C(e) is a Latin square, note first that in a given row, two entries in different columns are given by (a

ij(e) ,b

uv(e))

and (aik

(e),buw

(e)), and so are distinct since A(e) and B(e) are Latin squares. In a given column two entries in different rows are distinct by the same reasoning. To see that C(e) and C(f) are orthogonal, suppose that ((a

ij(e) ,b

uv(e) ) , (a

ij(f) ,b

uv(f) )) = ((a

pq(e) ,b

st(e) ) , (a

pq(f) ,b

st(f) ))

Then it follows that, (a

ij(e) ,a

ij(f) ) = (a

pq(e) ,a

pq(f) )

so, by orthogonality of A(e) and A(f) , i = p and j = q. Similarly, orthogonality of B(e) and B(f) implies that u = s and v = t.

Theorem - Suppose that is the prime powerdecomposition of n, n > 1, and r is the smallest of the quantities Then there are at least r MOLS(n).

n= p1a p2

b⋯ pst

{ p1a−1, p2

b−1,⋯ , pst−1}

MacNeish's Results

Proof: For each prime power p* in the decomposition we know that there are p* - 1 MOLS of that order. Thus there are r MOLS for each p*, since r is the smallest of these values. Repeated use of the previous theorem yields r MOLS of order n.

Orthogonal Arrays

An orthogonal array OA(n,q,λ) is a λn2 x q array of n symbols, such that in any two columns of the array every one of the possible n2 pairs of symbols occurs in exactly λ rows.

An OA(n,q,1) is equivalent to a q-net or set of (q-2) mutually orthogonal Latin Squares of order n. Use two of the columns to determine positions and then each of the remaining columns to fill in the entries of a Latin Square.

ExampleTo illustrate this last point we will construct an OA(5,4,1) from 3 MOLS(4). A complete set of MOLS(4)

1 2 3 42 1 4 33 4 1 24 3 2 1

1 2 3 43 4 1 24 3 2 12 1 4 3

1 2 3 44 3 2 12 1 4 33 4 1 2

1 1 1 1 11 2 2 2 21 3 3 3 31 4 4 4 42 1 2 3 42 2 1 4 32 3 4 1 22 4 3 2 13 1 3 4 23 2 4 3 13 3 1 2 43 4 2 1 34 1 4 2 34 2 3 1 44 3 2 4 14 4 1 3 2

Exampleand 2 MOLS(3) from an OA(4,3,1):

1 1 1 11 2 2 21 3 3 32 1 2 32 2 3 12 3 1 23 1 3 23 2 1 33 3 2 1

↓ L1 L2 ↓

1 2 3 2 3 1 L1

3 1 2

1 2 3 3 1 2 L2

2 3 1

Another Construction of OA'sWe can construct OA's from complete sets of MOLs which in turn can be constructed from finite fields. Let's do this more directly.

Theorem: There exists an OA(k,q,1) for every prime power q and 2 ≤ k ≤ q.Pf: Consider two vectors in (GF(q))k, v1 = (1, ..., 1) and v2 = (a1, ..., ak)where the ai are distinct elements of GF(q). Define a matrix A whose rows are indexed by the ordered pairs (i,j) of Fq x Fq where the row is the k-tuple i v1 + j v2.

Another Construction of OA'sTheorem: There exists an OA(k,q,1) for every prime power q and 2 ≤ k ≤ q.

Pf: (cont.) We show that A is an OA(k,q,1). To do this we need to find, for each c and d with 1 ≤ c < d ≤ k and each pair x,y in GF(q) a row (i,j) of A so that A(i,j),c = x and A(i,j),d = y.This condition gives the system: i + jac = x, i + jad = y.Since ac ≠ ad we have j = (ac – ad)-1(x – y) and i = x – jac = x – ac(ac -ad)-1(x-y).

and Another Construction

Theorem: There exists an OA(q+1,q,1) for any q a prime power.

Pf: Construct an OA(q,q,1) as in the last theorem. Adjoin a new column (labeled q+1) and define A(i,j),q+1 = j for all rows.This extended array is an OA(q+1,q,1).

Transversal DesignsAnother structure that is equivalent to sets of MOLS is called a transversal design.

Let k ≥ 2 and n ≥ 1. A transversal design TD(k,n) is a triple (X,G,B) such that: a) X is a set of kn elements called points, b) G is a partition of X into k subsets of size n called groups (or as some would like to have it groops), c) B is a set of k-subsets of X called blocks, d) any group and any block contain exactly 1 common point, and e) every pair of points from distinct groups is contained in exactly one block.

Transversal DesignsA simple example of a transversal design TD(2,n) is given by the complete bipartite graphs Kn,n.

In Kn,n, the X is the set of vertices, |X| = 2n, and each partite set is a groop. The edges are blocks which are 2-subsets and the conditions are clearly satisfied.

The definition of a transversal design can be tweaked to permit any complete Km,n to be an example with k = 2, however the “tweak” disappears as soon as k > 2. That is to say, we can relax the definition so that the groops do not have to have the same size, but then prove that they do if k>2. There does not appear to be any advantage to doing this.

TDs and OAsTD(k,n)s and OA(k,n)s are equivalent.

Let A be an OA(k,n) on the symbol set [n], with columns labeled 1,...,k and rows labeled 1,...,n2. Define X to be the set of ordered pairs [n] x [k] (and note |X| = kn), and define the groops to be the sets {[n]x{i}} for each i ԑ [k], and the n2 blocks to be given by the rows of A, i.e., row r gives {(ar,i, i) | ar,i ԑ A, i ԑ [k]}. This construction is reversible. Given a TD(k,n) we can arbitrarily associate each element in each groop with the integers 1,...,n. Now, each point of the TD can be labeled with an ordered pair (i,j) where j is the label of the groop that the point is in and i the associated element label in that groop. The rows of an OA(k,n) can be formed from the blocks of the TD as above.

Example1 1 1 1 11 2 2 2 21 3 3 3 31 4 4 4 42 1 2 3 42 2 1 4 32 3 4 1 22 4 3 2 13 1 3 4 23 2 4 3 13 3 1 2 43 4 2 1 34 1 4 2 34 2 3 1 44 3 2 4 14 4 1 3 2

OA(5,4,1)

(1,1) (1,2) (1,3) (1,4) (1,5)(1,1) (2,2) (2,3) (2,4) (2,5)(1,1) (3,2) (3,3) (3,4) (3,5)(1,1) (4,2) (4,3) (4,4) (4,5)(2,1) (1,2) (2,3) (3,4) (4,5)(2,1) (2,2) (1,3) (4,4) (3,5)(2,1) (3,2) (4,3) (1,4) (2,5)(2,1) (4,2) (3,3) (2,4) (1,5)(3,1) (1,2) (3,3) (4,4) (2,5)(3,1) (2,2) (4,3) (3,4) (1,5)(3,1) (3,2) (1,3) (2,4) (4,5)(3,1) (4,2) (2,3) (1,4) (3,5)(4,1) (1,2) (4,3) (2,4) (3,5)(4,1) (2,2) (3,3) (1,4) (4,5)(4,1) (3,2) (2,3) (4,4) (1,5)(4,1) (4,2) (1,3) (3,4) (2,5)

TD(5,4)

Summary

Theorem: For n ≥ 2 and k ≥3, the existence of any one of the following implies the existence of the other two:

a) k – 2 MOLS(n), b) an OA(k,n), c) a TD(k,n).

latin squares & steiner triple systems

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