Chapter 10 Combinatorial Designs

BIBD

Example

(a,b,c) (a,b,d) (a,c,e) (a,d,f) (a,e,f)(b,c,f) (b,d,e) (b,e,f) (c,d,e) (c,d,f)

Here are 10 subsets of the 6 element set {a, b, c, d, e, f}.

BIBD

DefinitionA balanced incomplete block design is a collection of k-subsets,called blocks, of a v-set S, k < v, such that each pair of elements of Soccur together in exactly λ of the blocks.

Notation(v, k, λ)-BIBD or (v, k, λ) design. Alternately, we could say we have a(b, v, r, k, λ)-BIBD.

b is the number of blocks

r is the number of times each element appears in a block

Necessary Conditions

TheoremIn a (v, k, λ) design with b blocks, each element occurs in r blockssuch that

1 λ(v− 1) = r(k − 1)2 bk = vr

Example

ExistenceShow that no (11, 6, 2) design can exist.

Example

‘

ExistenceDoes a (43, 7, 1)-BIBD exist?

Example

Example

(7, 3, 1)-BIBD

Finite Projective Plane

Definition

A finite projective plane is a (n2 + n + 1, n + 1, 1) design.

Fisher’s Inequality

TheoremFisher’s Inequality: In any (v, k, λ) design, b ≥ v.

DefinitionThe incidence matrix of a (v, k, λ) design is a b× v matrix A = (aij)defined by

aij =

{1 if the ith block contains the jth element0 otherwise

Theorem We Need

TheoremIf A is a (v, k, λ) design, then

A′A = (r − λ)I + λJ

where A′ is the transpose of A, I is the v× v identity matrix and J isthe v× v matrix of all 1’s.

Proof of Fisher’s Inequality

Let A be the incidence matrix. We first show what A′A is nonsingularby showing that its determinant is non-zero. Now,

|A′A| =

∣∣∣∣∣∣∣∣∣∣∣∣

r λ · · · λλ r · · · λλ λ · · · λ...

......

...

λ λ... r

∣∣∣∣∣∣∣∣∣∣∣∣Subtract the first row from each of the other rows.

=

∣∣∣∣∣∣∣∣∣∣∣∣

r λ · · · λλ− r r − λ · · · 0λ− r 0 r − λ 0

......

......

λ− r 0... r − λ

∣∣∣∣∣∣∣∣∣∣∣∣

Proof of Fisher’s Inequality (cont.)

Now, add to the first column of the sum of all the other columns.

=

∣∣∣∣∣∣∣∣∣∣∣∣

r + (v− 1)λ λ · · · λ0 r − λ · · · 00 0 r − λ 0...

......

...

0 0... r − λ

∣∣∣∣∣∣∣∣∣∣∣∣= [r + λ(v− 1)](r − λ)v−1

= [r + r(k − 1)](r − λ)v−1

= rk(r − λ)v−1

But, k < v, so by property (1) r > λ, so |A′A| 6= 0, But, A′A is a v× vmatrix, so the rank ρ of A′A is ρ(A′A) = v. Finally, sinceρ(A′A) ≤ ρ(A) and since ρ(A) ≤ b (A has b rows), v ≤ ρ(A) ≤ b.

Existence

Example

Can a (16, 6, 1) design exist?

Complementary Designs

DefinitionLet D be a (b, v, r, k, λ) design on a set S of v elements. Then thecomplementary design D has as it’s blocks the complements S− B ofthe blocks B in D.

Complementary Designs

Theorem

Suppose that D is a (b, v, r, k, λ) design. Then D is a(b, v, b− r, v− k, b− 2r + λ) design provided that b− 2r + λ > 0.

Why must b− 2r + λ > 0?

Symmetric Designs

Corollary

If D is a symmetric (v, k, λ) design with v− 2k + λ > 0 then D is asymmetric (v, v− k, v− 2k + λ) design.

Residual Designs

DefinitionThe (v− 1, v− k, r, k − λ, λ) design obtained from a symmetric(v, k, λ) design by deleting all elements of one block is called aresidual design.

Example

The residual design created from a (7, 3, 1)-BIBD(1,2,4) (2,3,5) (3,4,6) (4,5,7) (5,6,1) (6,7,2) (7,1,3)

Affine Plane

Definition

A design with the parameters (n2, n, 1) is called an affine plane oforder n.

If we can arrange the blocks into groups so that each group containseach element exactly once, we say the design is resolvable.

Resolvability

DefinitionA BIBD is resolvable if the blocks can be arranged into v groups sothat the

(br

)=( v

k

)blocks of each group are disjoint and contain in

their union each element exactly once. The groups are calledresolution classes of parallel classes.

Kirkman(1850)

Fifteen young ladies in a school walk out three abreast for seven daysin succession; it is required to arrange them daily so that no two shallwalk abreast twice.

(4,2,1) Design

Suppose we wanted a league schedule for 4 teams where each teamplayed each other team one time.How many weeks do we need?How many total games?

The Turning Trick

A (2n, 2, 1) design exists for all integers n ≥ 1. But this would gettedious to develop for a large league unless we had a trick ...

If we wanted to construct a league schedule for 8 teams, what wouldthe parameters of the corresponding block design be?

The Turning Trick

u

uu u

u uu u

∞36

7 2

1

5 4

The Turning Trick

u

uu uu

uu u∞

3

6

7

2

1

5 4

The Turning Trick

u

uu uu

uu u∞

3

6

7

2

1

5 4

Back to the (4,2,1) Design

We can ‘go backwards’ from the residual design idea to build finiteprojective planes.These constructions show that affine planes of order n exist iff finiteprojective planes of order n exist. There is also a correspondence with...

Latin Squares

DefinitionA Latin square on n symbols is an n× n array such that each of the nsymbols occurs exactly once in each row and in each column. Thenumber n is called the order of the square.

Example

A B C DD A B CC D A BB C D A

Latin Squares and League Schedules

Suppose a league schedule has been arranged for 2n teams in 2n− 1rounds. Then, define a 2n× 2n array A = (aij) by

aii = n, aij = k i 6= j

where the ith and jth teams play in round k. Since each team playsprecisely one game per round, A is a Latin square.

Latin Squares and League Schedules

ExampleConstruct a Latin square of order 2n from a league schedule on 8teams.

u

uu uu

uu u∞

3

6

7

2

1

5 4

""

""""

TTTTTTT

TTTTTTTTTT

TTTTTTTTT

MOLS

So just how many Latin squares are there of order n, up to labeling? Isthere just one of each order?

[1 22 1

]Order 2

1 2 32 3 13 1 2

1 2 33 1 22 3 1

Order 3

MOLS

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

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

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

Order 4

MOLS

Definition

Join (A,B) is the n× n array where the i, jth entry is (aij, bij) whereaij ∈ A and bij ∈ B.

These two Latin squares are called mutually orthogonal. For short, wesay MOLS. For n ≥ 4, the MOLS are pairwise orthogonal.

MOLS

DefinitionA complete set of MOLS of order n consists of n− 1 pairwiseorthogonal Latin squares.

Notation: Number of MOLS of order n is given by N(n).

TheoremFor all numbers n ≥ 3, N(n) ≥ 2, except for N(6) = 1.

MOLS

DefinitionA complete set of MOLS of order n consists of n− 1 pairwiseorthogonal Latin squares.

Notation: Number of MOLS of order n is given by N(n).

TheoremFor all numbers n ≥ 3, N(n) ≥ 2, except for N(6) = 1.

MOLS

TheoremN(n) ≥ 2 whenever n is odd, n ≥ 3.

MOLS Example

ExampleConstruct 2 MOLS of order 3.

MOLS Example

ExampleConstruct 2 MOLS of order 5.

Moore-MacNeish

TheoremMoore-MacNeish: N(mn) ≥ min{N(m),N(n)}

Corollary

N(n) ≥ min{pαii }-1

Latin Squares and Finite Projective Planes

TheoremAn affine plane of order n exists iff a finite projective plane of order niff n− 1 MOLS of order n exists.

Construction

ExampleConstruct an affine plane of order 4 from the three MOLS of order 4.

Initial Designs

DefinitionA (cyclic) (v, k, λ) difference set (mod v) is a set D = {d1, d2, . . . , dk}of distinct elements of Zv such that each non-zero d ∈ Zv can beexpressed in the form d = di − dj in precisely λ ways.

Back to the (7,3,1) Design

Example

{1, 2, 4} is a (7, 3, 1) difference set.

Justification

Why is it that the design we will obtain will be balanced?

Translates

DefinitionIf D = {d1, d2, . . . , dk} is a (v, k, λ) difference set mod v) then the setD + A = {d1 + a, d2 + a, . . . , dk + A} is called a translate of D.

TheoremIf D = {d1, d2, . . . , dk} is a cyclic (v, k, λ) difference set then thetranslates D,D + 1, . . . ,D + (v− 1) are the blocks of a symmetric(v, k, λ) design.

Translate Example

Example

We will illustrate this with our {1, 2, 4} difference set.

Another Example

Example

Verify that {1, 2, 4, 10} is a (13,4,1) difference set in Z13.

Another Example

Example

Verify that {1, 3, 4, 5, 9}(mod 11) yields a (11,5,2) design.

Difference Sets in Groups Other Than Zv

DefinitionA (v, k, λ) difference set in an additive abelian group G of order v is aset D = {d1, d2, . . . , dk} of distinct elements of G such that eachnon-zero element g ∈ G has exactly λ representations as g = di − dj.

How The Example Gives a Design

We can obtain the translates in the same manner, but how does thisgive us a (16,6,2) design?

Difference Systems

DefinitionLet D1, . . . ,Dt be sets of size k in an additive abelian group G oforder v such that the differences arising from the Di give eachnon-zero element of G exactly λ times. The D1, . . . ,Dt are said toform a (v, k, λ) difference system in G.

Note: the Di need not be disjoint.

Difference System Example

Example

Show that {1, 2, 5},{1, 3, 9} form a (13,3,1) difference system in Z13.

Important: the differences are only taken within blocks.

Starters

DefinitionA starter in an abelian group G of order 2n− 1 is a set of n− 1unordered pairs {x1, y1}, . . . , {xn−1, yn−1} of elements of G such that

i. x1, y1, . . . , xn−1, yn−1 are precisely all the non-zero elements of G

ii. ±(x1 − y1), . . . ,±(xn−1 − yn−1) are precisely the non-zeroelements of G

Example of a Starter

Example

The pairs {1, 2}, {4, 8}, {5, 10}, {9, 7}, {3, 6} form a starter in Z11.

Whist Tournaments

Difference systems can also be used to construct whist tournaments.

DefinitionA whist tournament, denoted Wh(4n), on 4n players is a schedule ofgames involving two players against two others, such that:

(i) the games are arranged in 4n− 1 rounds, each of n games

(ii) each player plays in exactly one game each round

(iii) each player partners every other player exactly once

(iv) each player opposes every other player exactly twice

Example

Construct a Wh(4).

Example of Whist Tournament

ExampleVerify that

∞,0 v 4,5 1,10 v 2,8 3,7 v 6,9

is the initial round of a cyclic Wh(12).