jcalm 2017 introduction to designs and steiner triple systems · 2/124 outline introduction to...

OutlineIntroduction to Designs: Definitions and Examples

Steiner Triple Systems: Definition and ExistenceOther t-designs, Affine and Projective planes, and Conclusion

JCALM 2017Introduction to Designs and Steiner Triple Systems

Fionn Mc Inerney

Universite Cote d’Azur, Inria, CNRS, I3S, France

May 3, 2017

Fionn Mc Inerney JCALM 2017 Introduction to Designs and Steiner Triple Systems

JCALM 2017Decomposition of Complete Graphs

Fionn Mc Inerney

Universite Cote d’Azur, Inria, CNRS, I3S, France

May 3, 2017

Fionn Mc Inerney JCALM 2017 Decomposition of Complete Graphs

1 Introduction to Designs: Definitions and Examples

2 Steiner Triple Systems: Definition and Existence

3 Other t-designs, Affine and Projective planes, and Conclusion

(7,3,1) design: Decompose K7 into K3

Decomposition: Partition of edges.

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K4 and K5 don’t decompose into K3

K4 doesn’t decompose into K3:for all v ∈ V (K4) and u ∈ V (K3), deg(v) = 3 anddeg(u) = 2 but 2 - 3.

K5 doesn’t decompose into K3:|E (K5)| = 10 and |E (K3)| = 3 but 3 - 10.

Main question: when do designs exist?

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t-designs

Definition

A t − (n, k, λ) design is a set of subsets of size k (blocks) of a setV of n vertices such that all subsets of size t belong to exactly λblocks.

Abbreviated to (n, k , λ) design when t = 2.

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t-designs

Definition

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2-designs (t = 2) with λ = 1: Decompose Kn into Kk

Definition

An (n, k, 1) design is a set of subsets of size k (blocks) of a set Vof n vertices such that any pair of elements belongs to exactly 1block.

Definition

An (n, k , 1) design is a decomposition of Kn (edge partition) intoKk .

Each block of size k : Kk .

Any pair (t = 2) of elements in exactly one (λ = 1) block:each edge of Kn in exactly one Kk .

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Definition

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Definition

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Definition

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(n, 2, 1) designs (k = 2) trivially exist (decomposing Kn into K2).

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Decomposition obtained by rotation of K3.Bi = {{i , i + 2, i + 3} | i ∈ [7]} (indices (mod 7)).

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Decomposition obtained by rotation of two K3.Bi = {{i , i + 2, i + 7} | i ∈ [13]} (indices (mod 13)).Bj = {{j , j + 1, j + 4} | j ∈ [13]} (indices (mod 13)).

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Decomposition obtained by rotation of K4.Bi = {{i , i + 4, i + 5, i + 7} | i ∈ [13]} (indices (mod 13)).

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Blue blocks partition vertices.

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Red blocks partition vertices.

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Green blocks partition vertices.

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Gray blocks partition vertices.

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Resolvable designs

Definition

A design is resolvable if the blocks (Kk) can be grouped into setsthat partition the vertices.

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Resolvable (4,2,1) design:Resolvable decomposition of K4 into K2

Definition

A design is resolvable if theblocks (Kk) can be grouped intosets that partition the vertices.

For n even, Kn decomposes into n − 1 perfect matchings.

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Definition

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Definition

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Definition

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Kirkman’s Schoolgirl Problem:Resolvable decomposition of K15 into K3

Kirkman’s Schoolgirl Problem(Reverend Kirkman, 1850)

15 young ladies in a school walk out 3abreast for 7 days in succession. Is itpossible to arrange them daily, so that no2 walk twice abreast?

Seeked actual solution for resolvable(15,3,1) design.

Same can be applied to Coati team.

1110 12

1413 15

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1110 12

1413 15

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1110 12

1413 15

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Coati Schoolgirl Problem - Day 1

Jeanne-ClaudiaMichelle Christine

Frederica V.Nicole N. Patricia

AndreaFiona Stephanie

Nicole H.Juliana Joanna

Frederica G.Christelle Valentina

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Nicole N.Michelle Christelle

JulianaFiona Frederica V.

Nicole H.Andrea Patricia

ChristineFrederica G. Joanna

ValentinaStephanie Jeanne-Claudia

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Nicole N.Michelle Christelle

JulianaFiona Frederica V.

Nicole H.Andrea Patricia

ChristineFrederica G. Joanna

ValentinaStephanie Jeanne-Claudia

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FionaNicole N. Jeanne-Claudia

ChristelleJuliana Andrea

Frederica G.Nicole H. Stephanie

PatriciaChristine Valentina

MichelleJoanna Frederica V.

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FionaNicole N. Jeanne-Claudia

ChristelleJuliana Andrea

Frederica G.Nicole H. Stephanie

PatriciaChristine Valentina

MichelleJoanna Frederica V.

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Jeanne-ClaudiaChristelle Nicole H.

AndreaFrederica V. Christine

StephaniePatricia Michelle

ValentinaJoanna Fiona

JulianaNicole N. Frederica G.

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Jeanne-ClaudiaChristelle Nicole H.

AndreaFrederica V. Christine

StephaniePatricia Michelle

ValentinaJoanna Fiona

JulianaNicole N. Frederica G.

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ChristelleFiona Christine

Jeanne-ClaudiaJuliana Patricia

Nicole H.Frederica V. Valentina

Frederica G.Andrea Michelle

JoannaStephanie Nicole N.

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ChristelleFiona Christine

Jeanne-ClaudiaJuliana Patricia

Nicole H.Frederica V. Valentina

Frederica G.Andrea Michelle

JoannaStephanie Nicole N.

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Frederica V.Christelle Stephanie

AndreaJeanne-Claudia Joanna

ChristineNicole H. Nicole N.

PatriciaFrederica G. Fiona

MichelleValentina Juliana

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Frederica V.Christelle Stephanie

AndreaJeanne-Claudia Joanna

ChristineNicole H. Nicole N.

PatriciaFrederica G. Fiona

MichelleValentina Juliana

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StephanieChristine Juliana

JoannaPatricia Christelle

Nicole N.Valentina Andrea

FionaMichelle Nicole H.

Frederica V.Jeanne-Claudia Frederica G.

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StephanieChristine Juliana

JoannaPatricia Christelle

Nicole N.Valentina Andrea

FionaMichelle Nicole H.

Frederica V.Jeanne-Claudia Frederica G.

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Necessary condition (1) for an (n, k , 1) design:Decomposition of Kn into Kk

If an (n, k , 1) design exists, then n(n − 1) ≡ 0 (mod k(k − 1)).

Proof.

|E (Kn)| = n(n−1)2 must be a multiple of |E (Kk)| = k(k−1)

Necessary Condition (1)

n(n − 1) ≡ 0 (mod k(k − 1)).

K5 doesn’t decompose into K3.

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Proof.

n(n − 1) ≡ 0 (mod k(k − 1)).

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Proof.

n(n − 1) ≡ 0 (mod k(k − 1)).

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Proof.

n(n − 1) ≡ 0 (mod k(k − 1)).

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If an (n, k , 1) design exists, then n − 1 ≡ 0 (mod k − 1).

Proof.

For all v ∈ V (Kn) and u ∈ V (Kk), deg(v) = n − 1 must be amultiple of deg(u) = k − 1.

Necessary condition (2)

n − 1 ≡ 0 (mod k − 1).

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Proof.

n − 1 ≡ 0 (mod k − 1).

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Proof.

n − 1 ≡ 0 (mod k − 1).

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Proof.

n − 1 ≡ 0 (mod k − 1).

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Necessary Conditions are Sufficient in General

n(n − 1) ≡ 0 (mod k(k − 1)).

n − 1 ≡ 0 (mod k − 1).

In general, for n large enough, these 2 conditions are sufficient.

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Necessary Conditions are Sufficient in General

n(n − 1) ≡ 0 (mod k(k − 1)).

n − 1 ≡ 0 (mod k − 1).

In general, for n large enough, these 2 conditions are sufficient.

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Steiner Triple Systems: Decomposition of Kn into K3

Definition

A Steiner Triple System is an (n, 3, 1) design.

Steiner asked the question if they exist in 1853, withoutknowing Kirkman had already proved their existence in 1847.

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Steiner Triple Systems: Decomposition of Kn into K3

Definition

A Steiner Triple System is an (n, 3, 1) design.

Steiner asked the question if they exist in 1853, withoutknowing Kirkman had already proved their existence in 1847.

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Necessary conditions for Steiner Triple Systems:Decomposition of Kn into K3

If an (n, 3, 1) design exists, then n ≡ 1 or 3 (mod 6).

Proof.

NC(2) ⇒ n − 1 ≡ 0 (mod 2) ⇒ n odd ⇒ n ≡ 1, 3 or 5(mod 6).

NC(1) ⇒ n(n− 1) ≡ 0 (mod 6) ⇒ n ≡ 1 or 3 (mod 6) since

if n = 6k + 5, then n(n−1)6 = 6k+5(6k+4)

6 = 18k2+27k+103 /∈ Z.

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Proof.

if n = 6k + 5, then n(n−1)6 = 6k+5(6k+4)

6 = 18k2+27k+103 /∈ Z.

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Proof.

if n = 6k + 5, then n(n−1)6 = 6k+5(6k+4)

6 = 18k2+27k+103 /∈ Z.

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Necessary Conditions are Sufficient for Steiner TripleSystems: Decomposition of Kn into K3

Theorem (Kirkman, 1847)

An (n, 3, 1) design exists ⇔ n ≡ 1 or 3 (mod 6).

i.e. Kn can be decomposed into K3 ⇔ n ≡ 1 or 3 (mod 6).

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Lemma 1: K2m+1 dec. K3 ⇔ Km∗2 dec. K3

Lemma 1

K2m+1 can be decomposed into K3 ⇔K2,··· ,2 = Km∗2 can be decomposed into K3.

. . .2m

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Lemma 1

. . .2m

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Lemma 1

. . .2m

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Lemma 1

. . .2m

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Lemma 1

. . .2m

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Lemma 1

. . .2m

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Proof of Lemma 1: K2m+1 dec. K3 ⇔ Km∗2 dec. K3

Proof.

Let V (K2m+1) = {v1, · · · , v2m+1}, D be a decomposition ofK2m+1 into K3, and A be the set of all blocks containingv2m+1.

|A| = m so w.l.o.g. suppose A = {vi , vi+m, v2m+1 | i ∈ [m]}.Then, D \ A is a decomposition of H into K3 where H is thegraph K2m with the edges vivi+m, i ∈ [m] removed. H isisomorphic to Km∗2.

In the other direction, if D ′ is a decomposition of H into K3,then D ′ ∪ A is a decomposition of K2m+1 into K3.

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Proof.

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Proof.

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Theorem 1: K3∗q dec. K3

Theorem 1

Kq,q,q = K3∗q can be decomposed into K3.

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Theorem 1

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Theorem 1

Example q = 2:

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Theorem 1

Example q = 2:

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Theorem 1

Example q = 2:

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Theorem 1

Example q = 2:

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Theorem 1

Example q = 2:

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Theorem 1

Example q = 2:

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Proof of Theorem 1: K3∗q dec. K3

Proof.

Let (A,B,C ) be the tripartition of K3∗q with A = {ai | i ∈ [q]}and B and C defined analogously. {{ai , bj , ci+j} | i ∈ [q], j ∈ [q]}(indices (mod q)) is a decomposition of K3∗q into K3.

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Theorem 2: ∀q ≥ 3, q 6= 6, K4∗q dec. K4

Theorem 2

For all q ≥ 3, q 6= 6, Kq,q,q,q = K4∗q can be decomposed into K4.

Proof for q odd.

q q q q

ci−j

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Theorem 2

Proof for q odd.

q q q q

ci−j

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Theorem 2

Example q = 3:

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Theorem 2

Example q = 3:

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Theorem 2

Example q = 3:

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Theorem 2

Example q = 3:

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Theorem 2

Example q = 3:

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Theorem 2

Example q = 3:

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Theorem 2

Example q = 3:

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Theorem 2

Example q = 3:

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Theorem 2

Example q = 3:

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Theorem 2

Example q = 3:

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Theorem 2

Example q = 3:

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Proof of Theorem 2: ∀q ≥ 3, q 6= 6, K4∗q dec. K4

q q q q

ci−j

Proof. (q odd)

(i , j) and (i ′, j ′) don’t exist such that i + j = i ′ + j ′ (mod q) andi − j = i ′ − j ′ (mod q). Such a pair exists ⇒ 2i = 2i ′ (mod q)and 2j = 2j ′ (mod q) ⇒ i = i ′ and j = j ′ since q is odd.

Doesn’t work when q is even: (i , j) and (i + q2 , j + q

2 ).(0, 0, 0, 0) and (q2 ,

q2 , 0, 0).

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Proof of Theorem 2: ∀q ≥ 3, q 6= 6, K4∗q dec. K4

q q q q

ci−j

Proof. (q odd)

(i , j) and (i ′, j ′) don’t exist such that i + j = i ′ + j ′ (mod q) andi − j = i ′ − j ′ (mod q). Such a pair exists ⇒ 2i = 2i ′ (mod q)and 2j = 2j ′ (mod q) ⇒ i = i ′ and j = j ′ since q is odd.

Doesn’t work when q is even: (i , j) and (i + q2 , j + q

2 ).(0, 0, 0, 0) and (q2 ,

q2 , 0, 0).

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Theorem 3: For r ≡ 0 or 1 (mod 3), Kr dec. K3,K4,K6

Theorem 3

If r ≡ 0 or 1 (mod 3), then Kr can be decomposed into acombination of K3,K4, and K6.

Proof.

We prove by induction. The result is trivially true for r = 3, 4, 6and r = 7 we already saw. Suppose r = 9m + j with m ∈ Z+ andj ∈ {0, 1, 3, 4, 6, 7}.

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Theorem 3

If r ≡ 0 or 1 (mod 3), then Kr can be decomposed into acombination of K3,K4, and K6.

Proof.

We prove by induction. The result is trivially true for r = 3, 4, 6and r = 7 we already saw. Suppose r = 9m + j with m ∈ Z+ andj ∈ {0, 1, 3, 4, 6, 7}.

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Example: r = 9.

K9 = 3K3 ∪ K3,3,3

K3 K3 K3

Base Case: K3 dec. K3,K4,K6.

K3,3,3

Thm 1: K3∗q dec. K3.

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Example: r = 9.

K9 = 3K3 ∪ K3,3,3

K3 K3 K3

Base Case: K3 dec. K3,K4,K6.

K3,3,3

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r = 9m + j for m ∈ Z+ and j ∈ {0, 1, 3, 4, 6, 7}.

K9m = 3K3m ∪ K3m,3m,3m

K3m K3m K3m

I.H.: K3m dec. K3,K4,K6.

K3m,3m,3m

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r = 9m + j for m ∈ Z+ and j ∈ {0, 1, 3, 4, 6, 7}.

K9m = 3K3m ∪ K3m,3m,3m

K3m K3m K3m

I.H.: K3m dec. K3,K4,K6.

K3m,3m,3m

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r = 9m + j for m ∈ Z+ and j ∈ {0, 1, 3, 4, 6, 7}.

K9m+3 = 3K3m+1 ∪ K3m+1,3m+1,3m+1

K3m+1 K3m+1 K3m+1

I.H.: K3m+1 dec. K3,K4,K6.

K3m+1,3m+1,3m+1

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r = 9m + j for m ∈ Z+ and j ∈ {0, 1, 3, 4, 6, 7}.

K9m+3 = 3K3m+1 ∪ K3m+1,3m+1,3m+1

K3m+1 K3m+1 K3m+1

I.H.: K3m+1 dec. K3,K4,K6.

K3m+1,3m+1,3m+1

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Example: r = 10.

K10 = 3(K3 ∪ z) ∪ K3,3,3

3(K3 ∪ z)Base Case: K4 dec. K3,K4,K6.

K3,3,3

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Example: r = 10.

K10 = 3(K3 ∪ z) ∪ K3,3,3

3(K3 ∪ z)Base Case: K4 dec. K3,K4,K6.

K3,3,3

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r = 9m + j for m ∈ Z+ and j ∈ {0, 1, 3, 4, 6, 7}.

K9m+1 = 3(K3m ∪ z) ∪ K3m,3m,3m

3(K3m ∪ z)I.H.: K3m+1 dec. K3,K4,K6.

K3m,3m,3m

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r = 9m + j for m ∈ Z+ and j ∈ {0, 1, 3, 4, 6, 7}.

K9m+1 = 3(K3m ∪ z) ∪ K3m,3m,3m

3(K3m ∪ z)I.H.: K3m+1 dec. K3,K4,K6.

K3m,3m,3m

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r = 9m + j for m ∈ Z+ and j ∈ {0, 1, 3, 4, 6, 7}. h ∈ {1, 3, 4}.

K9m+3+h =

3K3m+1 ∪ Kh ∪ K3m+1,3m+1,3m+1,h

K3m+1 K3m+1 K3m+1 Kh

I.H.: K3m+1 dec. K3,K4,K6.

K3m+1,3m+1,3m+1,h

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r = 9m + j for m ∈ Z+ and j ∈ {0, 1, 3, 4, 6, 7}. h ∈ {1, 3, 4}.

K9m+3+h =

3K3m+1 ∪ Kh ∪ K3m+1,3m+1,3m+1,h

I.H.: K3m+1 dec. K3,K4,K6.

K3m+1,3m+1,3m+1,h

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r = 9m + j for m ∈ Z+ and j ∈ {0, 1, 3, 4, 6, 7}. h ∈ {1, 3, 4}.

K9m+3+h =

3K3m+1 ∪ Kh ∪ K3m+1,3m+1,3m+1,h

I.H.: K3m+1 dec. K3,K4,K6.

K3m+1,3m+1,3m+1,3m+1

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r = 9m + j for m ∈ Z+ and j ∈ {0, 1, 3, 4, 6, 7}. h ∈ {1, 3, 4}.

K9m+3+h =

3K3m+1 ∪ Kh ∪ K3m+1,3m+1,3m+1,h

I.H.: K3m+1 dec. K3,K4,K6.

K3m+1,3m+1,3m+1,h

K3m+1,3m+1,3m+1,h dec. K3,K4

(omit extra vertices of K4∗q dec. K4).

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Proof of Theorem 3: For r ≡ 0 or 1 (mod 3), Kr dec.K3,K4, and K6

Proof.

Suppose r = 9m. Partition V (K9m) into three sets A1,A2,A3

of size 3m.

Decompose K9m into 3 K3m induced by A1,A2,A3 and aK3∗3m with tripartition (A1,A2,A3).

By I.H., each K3m decomposes into a combo of K3,K4,K6.

K3∗3m decomposes into K3 by Thm 1.

Suppose r = 9m + 3. Partition V (K9m+3) into three sets ofsize 3m + 1.

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Proof.

of size 3m.

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Proof.

of size 3m.

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Proof.

of size 3m.

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Proof.

of size 3m.

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Proof.

Suppose r = 9m + 1. Partition V (K9m+1) into three sets ofsize 3m and a vertex z .

Decompose K9m+1 into 3 K3m+1 induced byA1 ∪ {z},A2 ∪ {z},A3 ∪ {z} and a K3∗3m with tripartition(A1,A2,A3).

By I.H., each K3m+1 decomposes into a combo of K3,K4,K6.

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Proof.

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Proof.

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Proof.

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Proof.

Suppose r = 9m + 4, 9m + 6 or 9m + 7. PartitionV (K9m+3+h) into three sets of size 3m + 1 and a set of hvertices (h ∈ {1, 3, 4}).

Decompose K9m+3+h into 3 K3m+1 induced by A1,A2,A3, aK3m+1,3m+1,3m+1,h, and a K3 or K4 if h = 3 or h = 4 resp.

K4∗3m+1 decomposes into K4 by Thm 2. This decomp.transforms into a decomp. of K3m+1,3m+1,3m+1,h into acombo of K3 and K4 by omitting the extra vertices (some K4

become K3).

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Proof.

become K3).

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Proof.

become K3).

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Proof.

become K3).

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Necessary conditions are Sufficient for Steiner TripleSystems: Decomposition of Kn into K3

i.e. Kn can be decomposed into K3 ⇔ n ≡ 1 or 3 (mod 6).

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Let r = n−12 . Then, r ≡ 0 or 1 (mod 3).

Thm 3: Kr dec. K3,K4,K6.

→1 1 11 1 1

Kr∗2Kr∗2 dec. K3∗2,K4∗2,K6∗2.

Lemma 1: K2m+1 dec. K3 ⇔ Km∗2 dec. K3.By Lemma 1: K7,K9,K13 dec. K3 ⇒ K3∗2,K4∗2,K6∗2 dec. K3 ⇒Kr∗2 dec. K3.K2r+1 = Kn dec. K3 by Lemma 1.

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→1 1 11 1 1

Kr∗2Kr∗2 dec. K3∗2,K4∗2,K6∗2.

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→1 1 11 1 1

Kr∗2Kr∗2 dec. K3∗2,K4∗2,K6∗2.

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→1 1 11 1 1

Kr∗2Kr∗2 dec. K3∗2,K4∗2,K6∗2.

Lemma 1: K2m+1 dec. K3 ⇔ Km∗2 dec. K3.

By Lemma 1: K7,K9,K13 dec. K3 ⇒ K3∗2,K4∗2,K6∗2 dec. K3 ⇒Kr∗2 dec. K3.K2r+1 = Kn dec. K3 by Lemma 1.

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→1 1 11 1 1

Kr∗2Kr∗2 dec. K3∗2,K4∗2,K6∗2.

Lemma 1: K2m+1 dec. K3 ⇔ Km∗2 dec. K3.By Lemma 1: K7,K9,K13 dec. K3 ⇒ K3∗2,K4∗2,K6∗2 dec. K3 ⇒Kr∗2 dec. K3.

K2r+1 = Kn dec. K3 by Lemma 1.

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→1 1 11 1 1

Kr∗2Kr∗2 dec. K3∗2,K4∗2,K6∗2.

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Proof for Steiner Triple Systems:Kn dec. K3 ⇔ n ≡ 1 or 3 (mod 6)

Proof.

Suppose r = n−12 . Then, r ≡ 0 or 1 (mod 3).

By Thm 3, there’s a decomposition D of Kr into a combo ofK3,K4,K6. Replace each vertex of Kr by an independent setof 2 vertices to obtain Kr∗2.

For i ∈ {3, 4, 6}, each Ki in D corresponds to a Ki∗2 in Kr∗2.

Already showed K7,K9 and K13 could be decomposed into K3

and so by Lemma 1, K3∗2,K4∗2, and K6∗2 can too. Hence,Kr∗2 can be decomposed into K3.

K2r+1 = Kn can be decomposed into K3 by Lemma 1.

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Proof.

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Proof.

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Proof.

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Proof.

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Recall t-designs

Definition

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Recall t-designs

Definition

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What else do we know?

Theorem (Kirkman, 1847 (k = 3) and Hanani, 1961 (k = 4), 1972(k = 5))

For λ = 1, t = 2, and k ∈ {3, 4, 5}, an (n, k , 1) design exists ⇔n(n − 1) ≡ 0 (mod k(k − 1)) and n − 1 ≡ 0 (mod k − 1).

For λ = 1, t = 2 and k = 6, no (36, 6, 1) design (Tarry, 1901).

36 officers problem (Euler, 1782): possible to arrange 6 regimentsconsisting of 6 officers each of different rank in a 6× 6 square sothat no rank or regiment will be repeated in any row or column.

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All designs exist . . . eventually

Theorem (Wilson, 1973 (t = 2) and Keevash, 2014 (all t))

For all k , λ, t, there exists n0 such that for all n ≥ n0 satisfying thenecessary conditions, there exists a t − (n, k , λ) design.

For k = 6, λ = 1, t = 2, n0 ≤ 801.

If n ∈ {16, 21, 36, 46} then there is no (n, 6, 1) design.

For 21 values of n we don’t know if an (n, 6, 1) design exists.

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For k = 6, λ = 1, t = 2, n0 ≤ 801.

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For k = 6, λ = 1, t = 2, n0 ≤ 801.

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For k = 6, λ = 1, t = 2, n0 ≤ 801.

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One of the few other cases fully solved

Theorem (Hanani, 1960)

A 3− (n, 4, 1) design exists ⇔ n ≡ 2 or 4 (mod 6).

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Affine plane ((4,2,1) design)

Definition

System of points and lines s.t.

Two distinct points belongto exactly one line.

Let p be a point and l aline not containing p.Exactly one line parallel tol intersects p.

There exist at least 3points that are notcollinear.

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Affine planes and (q2, q, 1) resolvable designs

If in a finite affine plane, a line contains q points, then:

All lines contain q points.

All points belong to q + 1 lines.

There are q2 points and q2 + q lines.

There are q + 1 sets of q parallel lines with each line in eachset passing through all the points in the plane.

An affine plane is equivalent to a (q2, q, 1) resolvable design.

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Affine plane to Projective plane((4,2,1) design to (7,3,1) design)

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∞g ∞b

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∞g ∞b

Fano Plane

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Projective plane

Definition

Two distinct points belong to exactly one line.

Two lines intersect in exactly one point.

There exist 4 points such that any 3 of them are not aligned.

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Projective plane

Definition

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Projective plane

Definition

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Projective plane

Definition

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Projective planes and (q2 + q + 1, q + 1, 1) designs

A projective plane is of order q if each line contains q + 1 points.It has the following properties:

All lines contain q + 1 points.

There are q2 + q + 1 points and q2 + q + 1 lines.

A projective plane of order q is equivalent to a(q2 + q + 1, q + 1, 1) design.

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Properties of Graphs obtained from Projective planes

The bipartite graph G (P) for a given projective plane P where onevertex class consists of the points of P and the other the lines of Phas the following properties:

Bipartite.

High regular degree: q + 1.

High girth: 6.

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Bipartite.

High girth: 6.

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Bipartite.

High girth: 6.

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Bipartite.

High girth: 6.

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Projective planes and (q2 + q + 1, q + 1, 1) designs theorem

Theorem

If q is a prime power, then Kq2+q+1 decomposes into Kq+1.

All the same problem:

Kq2+q+1 decomposes into Kq+1?

A (q2 + q + 1, q + 1, 1) design exists?

A projective plane of order q exists?

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Theorem

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Theorem

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Theorem

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Non-existence of some designs

Theorem (Bruck-Ryser, 1949)

If q is not a prime power, q is not the sum of two squares, andq ≡ 1 or 2 (mod 4), then there does not exist a (q2, q, 1) design.

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Existence of Projective Planes

Theorem (Lam, 1991)

No projective plane of order q = 10 exists.

Proven by heavy computer calculations.

q = 12 is the first case where we don’t know if it exists.

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Theorem (Lam, 1991)

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Theorem (Lam, 1991)

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Conclusion

We know about the existence of designs for large values of nthanks to Wilson and Keevash.

Still many values of n below the threshold of n0 where wedon’t know if the designs exist.

While we may know a lot of designs exist, we don’t know howto construct many of them.

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Conclusion

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Conclusion

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Thank you!

jcalm 2017 introduction to designs and steiner triple systems · 2/124 outline introduction to...

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