Relations among partitions. IV: Adjusting for more than one - - PowerPoint PPT Presentation

Relations among partitions. IV: Adjusting for more than one partition

• R. A. Bailey

University of St Andrews Combinatorics Seminar, Shanghai Jiao Tong University, November 2017

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Abstract

Adjusted orthogonality and adjusted uniformity both generalize to adjusting for more than one partition. This leads to the notion of universal balance, in which every partition has adjusted uniformity after adjusting for any subset of the others. Multi-stage Youden rectangles and multi-layered Youden rectangles combine this idea with orthogonality.

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Outline

◮ Questions and answers. ◮ More general versions of adjusted orthogonality and

adjusted uniformity.

◮ Universal balance. ◮ Multi-stage Youden rectangles and multi-layered Youden

rectangles.

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Outline

◮ Questions and answers. ◮ More general versions of adjusted orthogonality and

adjusted uniformity.

◮ Universal balance. ◮ Multi-stage Youden rectangles and multi-layered Youden

rectangles.

Bailey Relations among partitions 3/28

Question 1

Question

Can we have three partitions, with each pair having adjusted orthogonality with respect to the third one?

Bailey Relations among partitions 4/28

Question 1

Question

Can we have three partitions, with each pair having adjusted orthogonality with respect to the third one?

Answer

• Yes. Here are two examples.

Bailey Relations among partitions 4/28

Question 1

Question

Can we have three partitions, with each pair having adjusted orthogonality with respect to the third one?

Answer

• Yes. Here are two examples.
• 1. F⊥G but F ≺ G and G ≺ F, so that F ∨ G is not F or G.

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Question 1

Question

Can we have three partitions, with each pair having adjusted orthogonality with respect to the third one?

Answer

• Yes. Here are two examples.
• 1. F⊥G but F ≺ G and G ≺ F, so that F ∨ G is not F or G.

Then X⊤

F (I − PF∨G)XG = 0 (by definition of F ⊥ G)

Bailey Relations among partitions 4/28

Question 1

Question

Can we have three partitions, with each pair having adjusted orthogonality with respect to the third one?

Answer

• Yes. Here are two examples.
• 1. F⊥G but F ≺ G and G ≺ F, so that F ∨ G is not F or G.

Then X⊤

F (I − PF∨G)XG = 0 (by definition of F ⊥ G)

and X⊤

F∨G(I − PF) = X⊤ F∨G(I − PG) = 0

because VF∨G ≤ VF ∩ VG.

Bailey Relations among partitions 4/28

Question 1

Question

Can we have three partitions, with each pair having adjusted orthogonality with respect to the third one?

Answer

• Yes. Here are two examples.
• 1. F⊥G but F ≺ G and G ≺ F, so that F ∨ G is not F or G.

Then X⊤

F (I − PF∨G)XG = 0 (by definition of F ⊥ G)

and X⊤

F∨G(I − PF) = X⊤ F∨G(I − PG) = 0

because VF∨G ≤ VF ∩ VG.

• 2. Suppose that R, C and L are pairwise strictly orthogonal

(e.g. the three 2-dimensional slicings of a cube,

• r the rows, columns and letters of a Latin square).

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Question 1

Question

Can we have three partitions, with each pair having adjusted orthogonality with respect to the third one?

Answer

• Yes. Here are two examples.
• 1. F⊥G but F ≺ G and G ≺ F, so that F ∨ G is not F or G.

Then X⊤

F (I − PF∨G)XG = 0 (by definition of F ⊥ G)

and X⊤

F∨G(I − PF) = X⊤ F∨G(I − PG) = 0

because VF∨G ≤ VF ∩ VG.

• 2. Suppose that R, C and L are pairwise strictly orthogonal

(e.g. the three 2-dimensional slicings of a cube,

• r the rows, columns and letters of a Latin square). L⊥C ⇒

PL(VC) = P0(VC) ⇒ X⊤

R (I − PL)XC = X⊤ R (I − P0)XC = 0.

Bailey Relations among partitions 4/28

Question 1

Question

Can we have three partitions, with each pair having adjusted orthogonality with respect to the third one?

Answer

• Yes. Here are two examples.
• 1. F⊥G but F ≺ G and G ≺ F, so that F ∨ G is not F or G.

Then X⊤

F (I − PF∨G)XG = 0 (by definition of F ⊥ G)

and X⊤

F∨G(I − PF) = X⊤ F∨G(I − PG) = 0

because VF∨G ≤ VF ∩ VG.

• 2. Suppose that R, C and L are pairwise strictly orthogonal

(e.g. the three 2-dimensional slicings of a cube,

• r the rows, columns and letters of a Latin square). L⊥C ⇒

PL(VC) = P0(VC) ⇒ X⊤

R (I − PL)XC = X⊤ R (I − P0)XC = 0.

In fact, so long as Fi ⊥ Fj and Fk Fi ∨ Fj for {i, j, k} = {1, 2, 3}, this is true.

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Question 1 continued

Question

Are there any other ways of achieving three-way adjusted

• rthgonality?

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Question 1 continued

Question

Are there any other ways of achieving three-way adjusted

• rthgonality?

Answer

I do not know the answer today, but it may be “obvious”.

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Question 2

Question

Any Hadamard matrix of order 4n defines a SBIBD with nB = nL = 4n − 1 and kB = kL = 2n − 1. Can we construct a triple array from every such SBIBD?

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Question 2

Question

Any Hadamard matrix of order 4n defines a SBIBD with nB = nL = 4n − 1 and kB = kL = 2n − 1. Can we construct a triple array from every such SBIBD?

Answer

• 1. No if n = 2.
• 2. Yes if n ≥ 3 and 2n − 1 is a prime power.
• 3. Otherwise, I do not think that the answer is known.

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Question 3

Question

What do you mean by “describe the system” given by a collection of partitions of a set?

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Question 3

Question

What do you mean by “describe the system” given by a collection of partitions of a set?

Answer (First part)

I do not mean “we know the structure up to isomorphism”. Example 1 If nB = nL = 16 and kB = kL = 6 and B ⊲ ⊳ L then there are three isomorphism classes of SBIBDS.

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Question 3

Question

What do you mean by “describe the system” given by a collection of partitions of a set?

Answer (First part)

I do not mean “we know the structure up to isomorphism”. Example 1 If nB = nL = 16 and kB = kL = 6 and B ⊲ ⊳ L then there are three isomorphism classes of SBIBDS. Example 2 If nR = nC = nL = n and R⊥C, L⊥R, L⊥C and R ∧ C = R ∧ L = C ∧ L = E then we have a Latin square of order n. If n ≥ 4 then there is more than

• ne isomorphism class of Latin squares of order n.

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Question 3

Question

What do you mean by “describe the system” given by a collection of partitions of a set?

Answer (First part)

I do not mean “we know the structure up to isomorphism”. Example 1 If nB = nL = 16 and kB = kL = 6 and B ⊲ ⊳ L then there are three isomorphism classes of SBIBDS. Example 2 If nR = nC = nL = n and R⊥C, L⊥R, L⊥C and R ∧ C = R ∧ L = C ∧ L = E then we have a Latin square of order n. If n ≥ 4 then there is more than

• ne isomorphism class of Latin squares of order n.

Aside

Fisher and Yates always said that a Latin square for an experiment should be chosen at random from among all Latin squares of that size. (I happen to disagree.) They never said that a BIBD should be chosen at random from among all BIBDS with those parameters.

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Question 3: my motivation

Xu Guangqi experimented with sweet potatoes to see if they could be grown successfully here. Responses in agronomy are more variable than those in physics, so experiments need to be designed carefully.

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Question 3 continued

Example

Suppose that F = {R, C, L} = a family of partitions of Ω. I measure Yω on each element ω of Ω. I assume that if ω is in row i, column j and letter k then E(Yω) = αi + βj + γk. In order to estimate the γ-parameters, I have to project the data vector Y onto the orthogonal complement of VR + VC. Let P{R,C} be the matrix of orthogonal projection onto VR + VC.

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Question 3 continued

Example

Suppose that F = {R, C, L} = a family of partitions of Ω. I measure Yω on each element ω of Ω. I assume that if ω is in row i, column j and letter k then E(Yω) = αi + βj + γk. In order to estimate the γ-parameters, I have to project the data vector Y onto the orthogonal complement of VR + VC. Let P{R,C} be the matrix of orthogonal projection onto VR + VC. The information matrix for L is X⊤

L (I − P{R,C})XL,

which is a generalized inverse of the variance-covariance matrix for the estimators of γ1, γ2, . . . , γnL.

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Question 3 continued

Example

Suppose that F = {R, C, L} = a family of partitions of Ω. I measure Yω on each element ω of Ω. I assume that if ω is in row i, column j and letter k then E(Yω) = αi + βj + γk. In order to estimate the γ-parameters, I have to project the data vector Y onto the orthogonal complement of VR + VC. Let P{R,C} be the matrix of orthogonal projection onto VR + VC. The information matrix for L is X⊤

L (I − P{R,C})XL,

which is a generalized inverse of the variance-covariance matrix for the estimators of γ1, γ2, . . . , γnL. I want to know this (or, at least, its eigenvalues and their multiplicities). So I probably need to know the formula for P{R,C}.

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An attempt to answer Question 3

More generally, let F be a family of partitions on Ω. For each G in F, put PF\{G} = matrix of orthogonal projection onto

∑

F∈F\{G}

VF. The information matrix for G is X⊤

G

• I − PF\{G}
• XG.

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An attempt to answer Question 3

More generally, let F be a family of partitions on Ω. For each G in F, put PF\{G} = matrix of orthogonal projection onto

∑

F∈F\{G}

VF. The information matrix for G is X⊤

G

• I − PF\{G}
• XG.

Definition

A description of the system is a list showing, for each G in F,

◮ a name or symbol for G; ◮ nG; ◮ if G is not uniform, a list of its part-sizes,

with multiplicities;

◮ the spectrum of the information matrix for G.

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Outline

◮ Questions and answers. ◮ More general versions of adjusted orthogonality and

adjusted uniformity.

◮ Universal balance. ◮ Multi-stage Youden rectangles and multi-layered Youden

rectangles.

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Outline

◮ Questions and answers. ◮ More general versions of adjusted orthogonality and

adjusted uniformity.

◮ Universal balance. ◮ Multi-stage Youden rectangles and multi-layered Youden

rectangles.

Bailey Relations among partitions 11/28

A more general version of adjusted orthogonality

Eccleston and Russell (1975) actually proposed this more general definition.

Definition

Let L be a set of partitions of Ω. Put VL = ∑

L∈L

VL and let PL be the matrix of orthogonal projection onto VL. Then R and C have adjusted orthogonality with respect to L if X⊤

R (I − PL)XC = 0.

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A more general version of adjusted orthogonality

Eccleston and Russell (1975) actually proposed this more general definition.

Definition

Let L be a set of partitions of Ω. Put VL = ∑

L∈L

VL and let PL be the matrix of orthogonal projection onto VL. Then R and C have adjusted orthogonality with respect to L if X⊤

R (I − PL)XC = 0.

I think that this implies that X⊤

R (I − PL∪{C})XR = X⊤ R (I − PL)XR.

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A more general version of balance: first version

Definition

Let G be a set of partitions of Ω. Put VG = ∑

G∈G

VG and let PG be the matrix of orthogonal projection onto VG. Then L is balanced with respect to G if X⊤

L (I − PG)XL is completely symmetric but not zero.

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Revisit some types of orthogonality

Put QF = PF − P0.

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Revisit some types of orthogonality

Put QF = PF − P0. G F means that G is finer than F, in that every G-part is contained in an F-part. This is equivalent to all of the following. (i) NFGNGF is a multiple of I. (ii) QFQGQF = QF. (iii) VF ≤ VG so if v ∈ VF ∩ V⊤

0 then PGv = v.

(iv) P{F,G} = PG.

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Revisit some types of orthogonality

Put QF = PF − P0. G F means that G is finer than F, in that every G-part is contained in an F-part. This is equivalent to all of the following. (i) NFGNGF is a multiple of I. (ii) QFQGQF = QF. (iii) VF ≤ VG so if v ∈ VF ∩ V⊤

0 then PGv = v.

(iv) P{F,G} = PG. F⊥G means that F is strictly orthogonal to G, in the sense that PFPG = PGPF = P0. This is equivalent to all of the following. (i) NFGNGF is a multiple of J. (ii) QFQGQF = 0. (iii) If v ∈ VF ∩ V⊤

0 then PGv = 0.

(iv) P{F,G} = PF + PG − P0.

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Revisit balance

F ◮ G means that F is balanced with respect to G but not strictly orthogonal to G.

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Revisit balance

F ◮ G means that F is balanced with respect to G but not strictly orthogonal to G.

Theorem (James (1957); James and Wilkinson (1971))

F ◮ G is equivalent to all of the following. (i) NFGNGF = αI + βJ for non-zero scalars α and β. (ii) QFQGQF = µQF for some scalar µ with 0 < µ < 1. (iii) If v ∈ VF ∩ V⊤

0 then the angle between v and PGv is θ,

where cos2 θ = µ. (iv) P{F,G} = P0 + QG +

1 1−µ(QF − QGQF − QFQG + QGQFQG),

which simplifies to P0 + 1 1 − µ(QF + QG − QGQF − QFQG) if nF = nG.

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Revisit balance

F ◮ G means that F is balanced with respect to G but not strictly orthogonal to G.

Theorem (James (1957); James and Wilkinson (1971))

F ◮ G is equivalent to all of the following. (i) NFGNGF = αI + βJ for non-zero scalars α and β. (ii) QFQGQF = µQF for some scalar µ with 0 < µ < 1. (iii) If v ∈ VF ∩ V⊤

0 then the angle between v and PGv is θ,

where cos2 θ = µ. (iv) P{F,G} = P0 + QG +

1 1−µ(QF − QGQF − QFQG + QGQFQG),

which simplifies to P0 + 1 1 − µ(QF + QG − QGQF − QFQG) if nF = nG. Refinement and strict orthogonality are the extreme cases of balance as well as the extreme cases of general orthogonality.

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Balance, with both extremes

If G is a set of partitions of Ω, put QG = PG − P0.

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Balance, with both extremes

If G is a set of partitions of Ω, put QG = PG − P0. If VF ≤ VG then QFQGQF = QF, every vector in VF ∩ V⊤

0 is in VG,

and QG∪{F} = QG.

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Balance, with both extremes

If G is a set of partitions of Ω, put QG = PG − P0. If VF ≤ VG then QFQGQF = QF, every vector in VF ∩ V⊤

0 is in VG,

and QG∪{F} = QG. If (VF ∩ V⊤

0 ) ⊥ VG then QFQGQF = 0,

every vector in VF ∩ V⊤

0 is orthogonal to VG,

and QG∪{F} = QG + QF.

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Balance, with both extremes

If G is a set of partitions of Ω, put QG = PG − P0. If VF ≤ VG then QFQGQF = QF, every vector in VF ∩ V⊤

0 is in VG,

and QG∪{F} = QG. If (VF ∩ V⊤

0 ) ⊥ VG then QFQGQF = 0,

every vector in VF ∩ V⊤

0 is orthogonal to VG,

and QG∪{F} = QG + QF. Let us say that F is strictly balanced with respect to G if there is a scalar µ with 0 < µ < 1 such that QFQGQF = µQF. In this case, every vector in VF ∩ V⊤

0 makes angle θ with VG,

where cos2 θ = µ, and QG∪{F} = QG +

1 1−µ(QF − QGQF − QFQG + QGQFQG).

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Balance, with both extremes

If G is a set of partitions of Ω, put QG = PG − P0. If VF ≤ VG then QFQGQF = QF, every vector in VF ∩ V⊤

0 is in VG,

and QG∪{F} = QG. If (VF ∩ V⊤

0 ) ⊥ VG then QFQGQF = 0,

every vector in VF ∩ V⊤

0 is orthogonal to VG,

and QG∪{F} = QG + QF. Let us say that F is strictly balanced with respect to G if there is a scalar µ with 0 < µ < 1 such that QFQGQF = µQF. In this case, every vector in VF ∩ V⊤

0 makes angle θ with VG,

where cos2 θ = µ, and QG∪{F} = QG +

1 1−µ(QF − QGQF − QFQG + QGQFQG).

(Statements on the remaining slides may not always be clear about excluding the extremes.)

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Outline

◮ Questions and answers. ◮ More general versions of adjusted orthogonality and

adjusted uniformity.

◮ Universal balance. ◮ Multi-stage Youden rectangles and multi-layered Youden

rectangles.

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Outline

◮ Questions and answers. ◮ More general versions of adjusted orthogonality and

adjusted uniformity.

◮ Universal balance. ◮ Multi-stage Youden rectangles and multi-layered Youden

rectangles.

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Exactly three partitions

Suppose that partitions F, G and H each have n parts of size k, and that each pair are balanced (both ways).

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Exactly three partitions

Suppose that partitions F, G and H each have n parts of size k, and that each pair are balanced (both ways). Then F is balanced with respect to {G, H} if and only if NFGNGHNHF + NFHNHGNGF is completely symmetric. Equivalently, QF(QGQH + QHQG)QF is a non-zero multiple of QF.

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Exactly three partitions

Suppose that partitions F, G and H each have n parts of size k, and that each pair are balanced (both ways). Then F is balanced with respect to {G, H} if and only if NFGNGHNHF + NFHNHGNGF is completely symmetric. Equivalently, QF(QGQH + QHQG)QF is a non-zero multiple of QF. The above is implied by this stronger condition: NFGNGH is a linear combination of NFH and J.

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My attempt at a general definition

A set F of uniform partitions of Ω, all with n parts, has universal balance if whenever F ∈ F and L ⊆ F \ {F} then F is strictly balanced with respect to L.

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My attempt at a general definition

A set F of uniform partitions of Ω, all with n parts, has universal balance if whenever F ∈ F and L ⊆ F \ {F} then F is strictly balanced with respect to L. Equivalently, whenever F and L are as above, then there is a scalar µ with 0 < µ < 1 such that QFQLQF = µQF.

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Matrix conditions for universal balance

Theorem

If F has universal balance and L ⊆ F then QL is a linear combination of products of the matrices QL for L in L.

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Matrix conditions for universal balance

Theorem

If F has universal balance and L ⊆ F then QL is a linear combination of products of the matrices QL for L in L.

Corollary

If F has universal balance and L ⊂ F and F ∈ F \ L then X⊥

F QLXF is a sum of matrices of the form

NFL1NL1L2 · · · NLrF (1) where (L1, L2, . . . , Lr) is a sequence of partitions in L, possibly having repeated entries (I hope not). (The coefficient of this and of NFLr · · · NL2L1NL1F must be the same.)

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Matrix conditions for universal balance

Theorem

If F has universal balance and L ⊆ F then QL is a linear combination of products of the matrices QL for L in L.

Corollary

If F has universal balance and L ⊂ F and F ∈ F \ L then X⊥

F QLXF is a sum of matrices of the form

NFL1NL1L2 · · · NLrF (1) where (L1, L2, . . . , Lr) is a sequence of partitions in L, possibly having repeated entries (I hope not). (The coefficient of this and of NFLr · · · NL2L1NL1F must be the same.) So, if we can ensure that, whenever K is a product like (1) then K + K⊥ is completely symmetric, then we have universal balance.

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Known families, for n parts of size k

NFGNGHNHF + NFHNHGNGF is completely symmetric,

• r its generalization.

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Known families, for n parts of size k

NFGNGHNHF + NFHNHGNGF is completely symmetric,

• r its generalization.

◮ k = n − 1: remove a common transversal from a set of

mutually orthogonal n × n Latin squares, so that every N is J − I. (Done by many people.)

Bailey Relations among partitions 21/28

Known families, for n parts of size k

NFGNGHNHF + NFHNHGNGF is completely symmetric,

• r its generalization.

◮ k = n − 1: remove a common transversal from a set of

mutually orthogonal n × n Latin squares, so that every N is J − I. (Done by many people.)

◮ n ≡ 3 (mod 4) and k = (n + 1)/2 or k = (n − 1)/2:

if there is a doubly-regular tournament of size n, its adjacency matrix A satisfies I + A + A⊤ = J and A2 ∈ I, A, J; then ensure that each N is either I + A or I + A⊤ (or A or A⊤). (Done by many people, usually without using the words doubly regular tournament.)

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Known families, for n parts of size k, continued

NFGNGHNHF + NFHNHGNGF is completely symmetric,

• r its generalization.

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Known families, for n parts of size k, continued

NFGNGHNHF + NFHNHGNGF is completely symmetric,

• r its generalization.

◮ n = 22m and k = 22m−1 + 2m−1 or k = 22m−1 − 2m−1:

Cameron and Seidel (1973) have constructions of incidence matrices from quadratic forms, and the strong form of the condition is satisfied.

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Known families, for n parts of size k, continued

NFGNGHNHF + NFHNHGNGF is completely symmetric,

• r its generalization.

◮ n = 22m and k = 22m−1 + 2m−1 or k = 22m−1 − 2m−1:

Cameron and Seidel (1973) have constructions of incidence matrices from quadratic forms, and the strong form of the condition is satisfied.

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Known families, for n parts of size k, continued

NFGNGHNHF + NFHNHGNGF is completely symmetric,

• r its generalization.

◮ n = 22m and k = 22m−1 + 2m−1 or k = 22m−1 − 2m−1:

Cameron and Seidel (1973) have constructions of incidence matrices from quadratic forms, and the strong form of the condition is satisfied. (For n = 16 and k = 6 they get incidence matrices between 8 “things”, but these cannot be all realised as partitions of a single set. Four can be realised as partitions of a single set

• f size 96; this involves compatible Clebsch graphs which

form an amorphic association scheme. Their adjacency matrices A1, A2 and A3 satisfy A1 + A2 + A3 = J − I, A2

i = 5I + 2(Aj + Ak) and AiAj = 2J − 2I − Ak.)

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Problem: is this all?

Your task

◮ Find all possible sets of three or more incidence matrices

NFG satisfying the conditions.

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Problem: is this all?

Your task

◮ Find all possible sets of three or more incidence matrices

NFG satisfying the conditions.

◮ For each such set, realise them as incidence matrices of a

set of partitions with n parts of size k.

Bailey Relations among partitions 23/28

Problem: is this all?

Your task

◮ Find all possible sets of three or more incidence matrices

NFG satisfying the conditions.

◮ For each such set, realise them as incidence matrices of a

set of partitions with n parts of size k.

◮ For each such realisation, find another partition with

k parts of size n that is orthogonal to all the rest (surprisingly, this often makes the previous part easier).

Bailey Relations among partitions 23/28

Problem: is this all?

Your task

◮ Find all possible sets of three or more incidence matrices

NFG satisfying the conditions.

◮ For each such set, realise them as incidence matrices of a

set of partitions with n parts of size k.

◮ For each such realisation, find another partition with

k parts of size n that is orthogonal to all the rest (surprisingly, this often makes the previous part easier).

◮ What about two such sets, one with n parts of size k, the

• ther with k parts of size n, and every partition in one set
• rthogonal to every partition in the other set?

(If each set has two partitions, this is a double Youden rectangle, so I only require one of the sets to have at least three partitions.)

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Problem: is this all?

Your task

◮ Find all possible sets of three or more incidence matrices

NFG satisfying the conditions.

◮ For each such set, realise them as incidence matrices of a

set of partitions with n parts of size k.

◮ For each such realisation, find another partition with

k parts of size n that is orthogonal to all the rest (surprisingly, this often makes the previous part easier).

◮ What about two such sets, one with n parts of size k, the

• ther with k parts of size n, and every partition in one set
• rthogonal to every partition in the other set?

(If each set has two partitions, this is a double Youden rectangle, so I only require one of the sets to have at least three partitions.)

◮ Or three or more?

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Outline

◮ Questions and answers. ◮ More general versions of adjusted orthogonality and

adjusted uniformity.

◮ Universal balance. ◮ Multi-stage Youden rectangles and multi-layered Youden

rectangles.

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Outline

◮ Questions and answers. ◮ More general versions of adjusted orthogonality and

adjusted uniformity.

◮ Universal balance. ◮ Multi-stage Youden rectangles and multi-layered Youden

rectangles.

Bailey Relations among partitions 24/28

At the Aberystwyth BCC in 1973

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At the Aberystwyth BCC in 1973

I want universal balance among some partitions with 16 parts of size 6

Bailey Relations among partitions 25/28

At the Aberystwyth BCC in 1973

Ooh!—I know some suitable incidence ma- trices for those numbers I want universal balance among some partitions with 16 parts of size 6

Bailey Relations among partitions 25/28

A B C D E F G H I J K L M N O P A B C D E F G H I J K L M N O P A B C D E F G H I J K L M N O P H G F E D C B A P O N M L K J I G H E F C D A B O P M N K L I J B A D C F E H G J I L K N M P O K L I J O P M N C D A B G H E F J I L K N M P O B A D C F E H G D C B A H G F E L K J I P O N M M N O P I J K L E F G H A B C D I J K L M N O P A B C D E F G H E F G H A B C D M N O P I J K L O P M N K L I J G H E F C D A B F E H G B A D C N M P O J I L K L K J I P O N M D C B A H G F E P O N M L K J I H G F E D C B A C D A B G H E F K L I J O P M N N M P O J I L K F E H G B A D C

Bailey Relations among partitions 26/28

Preceding slide, from Preece and Cameron (1975)

Underlying set has size 96. 16 columns of size 6. 16 top letters of size 6. 16 middle letters of size 6. 16 bottom letters of size 6. Universal balance among the above, which are all strictly orthogonal to: 6 rows of size 16.

Bailey Relations among partitions 27/28

Preceding slide, from Preece and Cameron (1975)

Underlying set has size 96. 16 columns of size 6. 16 top letters of size 6. 16 middle letters of size 6. 16 bottom letters of size 6. Universal balance among the above, which are all strictly orthogonal to: 6 rows of size 16. Cameron says that he did not really understand this way of thinking about relations between partitions on a set until 25 years later, when he generalized this construction to arbitrary powers of 4 at the 2001 BCC in Sussex (Cameron, 2003).

Bailey Relations among partitions 27/28

Generalizing Youden rectangles (assume that n < m)

stages F1, . . . , Fr universal balance m parts of size n everything above is strictly

• rthogonal to

everything below layers G1 . . . , Gs universal balance n parts of size m

Bailey Relations among partitions 28/28

Generalizing Youden rectangles (assume that n < m)

stages F1, . . . , Fr universal balance m parts of size n everything above is strictly

• rthogonal to

everything below layers G1 . . . , Gs universal balance n parts of size m r = s = 1: rectangle

Bailey Relations among partitions 28/28

Generalizing Youden rectangles (assume that n < m)

stages F1, . . . , Fr universal balance m parts of size n everything above is strictly

• rthogonal to

everything below layers G1 . . . , Gs universal balance n parts of size m r = s = 1: rectangle r = 2, s = 1: Youden square

Bailey Relations among partitions 28/28

Generalizing Youden rectangles (assume that n < m)

stages F1, . . . , Fr universal balance m parts of size n everything above is strictly

• rthogonal to

everything below layers G1 . . . , Gs universal balance n parts of size m r = s = 1: rectangle r = 2, s = 1: Youden square r = s = 2: double Youden rectangle

Bailey Relations among partitions 28/28

Generalizing Youden rectangles (assume that n < m)

stages F1, . . . , Fr universal balance m parts of size n everything above is strictly

• rthogonal to

everything below layers G1 . . . , Gs universal balance n parts of size m r = s = 1: rectangle r = 2, s = 1: Youden square r = s = 2: double Youden rectangle r ≥ 3, s = 1: Freeman-Youden rectangle

• r

multi-letter Youden rectangle or multi-stage Youden rectangle

Bailey Relations among partitions 28/28

Generalizing Youden rectangles (assume that n < m)

stages F1, . . . , Fr universal balance m parts of size n everything above is strictly

• rthogonal to

everything below layers G1 . . . , Gs universal balance n parts of size m r = s = 1: rectangle r = 2, s = 1: Youden square r = s = 2: double Youden rectangle r ≥ 3, s = 1: Freeman-Youden rectangle

• r

multi-letter Youden rectangle or multi-stage Youden rectangle r = 2, s ≥ 3: multi-layered Youden rectangle

Bailey Relations among partitions 28/28

Generalizing Youden rectangles (assume that n < m)

stages F1, . . . , Fr universal balance m parts of size n everything above is strictly

• rthogonal to

everything below layers G1 . . . , Gs universal balance n parts of size m r = s = 1: rectangle r = 2, s = 1: Youden square r = s = 2: double Youden rectangle r ≥ 3, s = 1: Freeman-Youden rectangle

• r

multi-letter Youden rectangle or multi-stage Youden rectangle r = 2, s ≥ 3: multi-layered Youden rectangle Preece and Morgan (2017) introduced the last type. They gave some constructions and proved some results.

Bailey Relations among partitions 28/28

Generalizing Youden rectangles (assume that n < m)

stages F1, . . . , Fr universal balance m parts of size n everything above is strictly

• rthogonal to

everything below layers G1 . . . , Gs universal balance n parts of size m r = s = 1: rectangle r = 2, s = 1: Youden square r = s = 2: double Youden rectangle r ≥ 3, s = 1: Freeman-Youden rectangle

• r

multi-letter Youden rectangle or multi-stage Youden rectangle r = 2, s ≥ 3: multi-layered Youden rectangle Preece and Morgan (2017) introduced the last type. They gave some constructions and proved some results. Your task Keep going!

Bailey Relations among partitions 28/28