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(1)

The α-Acyclic Hypergraph Decompositions of Complete Uniform Hypergraphs

Xiande Zhang

Joint work with J.-C. Bermond, Y. M. Chee and N. Cohen

September 2011, Melbourn

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Introduction Decomposition I Decomposition II Concluding Remarks

REFERENCES

Y. M. Chee, L. Ji, A. Lim, and A. K. H. Tung, Arboricity: an

acyclic hypergraph decomposition problem motivated by

database theory, Discrete Applied Mathematics, to appear

J.-C. Bermond, Y. M. Chee, N. Cohen, X. Zhang, The

α-arboricity of complete uniform hypergraphs, SIAM Journal

on Discrete Mathematics, vol. 34, no. 2, pp. 600-610, 2011

(3)

Introduction Decomposition I Decomposition II Concluding Remarks

Outline

Introduction

Decomposition I

Decomposition II

Concluding Remarks

(4)

Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Outline

1

Introduction Motivations α-Acyclicity Extreme Cases

2

Decomposition I Steiner System Construction

3

Decomposition II The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

4

Concluding Remarks

(5)

Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Motivations

A natural bijection between database schemas and hypergraphs:

• an attribute of a database schema ↔ a vertex

• a relation of attributes ↔ an edge

α-Acyclicity is an important property of databases. Many NP-hard problems concerning databases can be solved in polynomial time when restricted to instances for which the corresponding hypergraphs are α-acyclic.

Examples of such problems: determining global consistency,

evaluating conjunctive queries, and computing joins or

projections of joins.

(6)

Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Motivations

A natural bijection between database schemas and hypergraphs:

• an attribute of a database schema ↔ a vertex

• a relation of attributes ↔ an edge

α-Acyclicity is an important property of databases. Many NP-hard problems concerning databases can be solved in polynomial time when restricted to instances for which the corresponding hypergraphs are α-acyclic.

Examples of such problems: determining global consistency,

evaluating conjunctive queries, and computing joins or

projections of joins.

(7)

Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Motivations

A natural bijection between database schemas and hypergraphs:

• an attribute of a database schema ↔ a vertex

• a relation of attributes ↔ an edge

α-Acyclicity is an important property of databases. Many NP-hard problems concerning databases can be solved in polynomial time when restricted to instances for which the corresponding hypergraphs are α-acyclic.

Examples of such problems: determining global consistency,

evaluating conjunctive queries, and computing joins or

projections of joins.

(8)

Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

GYO Reduction (Graham, Yu and Ozsoyoglu, 1979)

Given a hypergraph H = (X , A), the GYO reduction applies the following operations repeatedly to H until none can be applied:

• If a vertex x ∈ X has degree one, then delete x from the edge containing it.

• If A, B ∈ A are distinct edges such that A ⊆ B, then delete A from A.

A hypergraph H is α-acyclic if GYO reduction on H results in

an empty hypergraph.

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

GYO Reduction (Graham, Yu and Ozsoyoglu, 1979)

Given a hypergraph H = (X , A), the GYO reduction applies the following operations repeatedly to H until none can be applied:

• If a vertex x ∈ X has degree one, then delete x from the edge containing it.

• If A, B ∈ A are distinct edges such that A ⊆ B, then delete A from A.

A hypergraph H is α-acyclic if GYO reduction on H results in

an empty hypergraph.

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Examples

Example (1)

H

1

= (X , A) with X and A as follows is α-acyclic.

X = {1, 2, 3, 4, 5, 6}

A = {{1, 2, 3}, {3, 4, 5}, {1, 5, 6}, {1, 3, 5}}.

Example (2)

H

2

= (X , A) with X and A as follows is not α-acyclic.

X = {1, 2, 3, 4}

A = {{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}}.

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Examples

Example (1)

H

1

= (X , A) with X and A as follows is α-acyclic.

X = {1, 2, 3, 4, 5, 6}

A = {{1, 2, 3}, {3, 4, 5}, {1, 5, 6}, {1, 3, 5}}.

Example (2)

H

2

= (X , A) with X and A as follows is not α-acyclic.

X = {1, 2, 3, 4}

A = {{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}}.

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

α-Arboricity

An α-acyclic decomposition of a hypergraph H = (X , A) is a set of α-acyclic hypergraphs {(X , A

i

)}

ci =1

such that A

1

, . . . , A

c

partition A, that is, A = F

c

i =1

A

i

. The size of the α-acyclic decomposition is c.

Definition

The α-arboricity of a hypergraph H, denoted αarb(H), is the minimum size of an α-acyclic decomposition of H.

Example: Denote the complete k-uniform hypergraph (X ,

Xk

) of order n by K

n(k)

. αarb(K

n(1)

) = αarb(K

n(n)

) = 1, since both K

n(1)

and K

n(n)

are α-acyclic.

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Cyclomatic Number

Given a hypergraph H = (X , A),

Line graph of H: L(H) = (V , E ), where V = A and E = {{A, B} ⊆

V2

 : A ∩ B 6= ∅}.

L

0

(H): the weighted L(H) by assigning to every edge {A, B}

of L(H) the weight |A ∩ B|.

w (H): the maximum weight of a forest in L

0

(H).

Cyclomatic number of H:

µ(H) = P

A∈A

|A| −

S

A∈A

A

− w (H).

µ(H) = 0 ⇔ H is conformal and chordal. (Acharya et al., 1982)

H is α-acyclic ⇔ H is conformal and chordal. (Beeri et al.,

1983)

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Cyclomatic Number

Given a hypergraph H = (X , A),

Line graph of H: L(H) = (V , E ), where V = A and E = {{A, B} ⊆

V2

 : A ∩ B 6= ∅}.

L

0

(H): the weighted L(H) by assigning to every edge {A, B}

of L(H) the weight |A ∩ B|.

w (H): the maximum weight of a forest in L

0

(H).

Cyclomatic number of H:

µ(H) = P

A∈A

|A| −

S

A∈A

A

− w (H).

µ(H) = 0 ⇔ H is conformal and chordal. (Acharya et al., 1982)

H is α-acyclic ⇔ H is conformal and chordal. (Beeri et al.,

1983)

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Cyclomatic Number

Given a hypergraph H = (X , A),

Line graph of H: L(H) = (V , E ), where V = A and E = {{A, B} ⊆

V2

 : A ∩ B 6= ∅}.

L

0

(H): the weighted L(H) by assigning to every edge {A, B}

of L(H) the weight |A ∩ B|.

w (H): the maximum weight of a forest in L

0

(H).

Cyclomatic number of H:

µ(H) = P

A∈A

|A| −

S

A∈A

A

− w (H).

µ(H) = 0 ⇔ H is conformal and chordal. (Acharya et al., 1982)

H is α-acyclic ⇔ H is conformal and chordal. (Beeri et al.,

1983)

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Cyclomatic Number

Given a hypergraph H = (X , A),

Line graph of H: L(H) = (V , E ), where V = A and E = {{A, B} ⊆

V2

 : A ∩ B 6= ∅}.

L

0

(H): the weighted L(H) by assigning to every edge {A, B}

of L(H) the weight |A ∩ B|.

w (H): the maximum weight of a forest in L

0

(H).

Cyclomatic number of H:

µ(H) = P

A∈A

|A| −

S

A∈A

A

− w (H).

µ(H) = 0 ⇔ H is conformal and chordal. (Acharya et al., 1982)

H is α-acyclic ⇔ H is conformal and chordal. (Beeri et al.,

1983)

(17)

Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Cyclomatic Number

Given a hypergraph H = (X , A),

Line graph of H: L(H) = (V , E ), where V = A and E = {{A, B} ⊆

V2

 : A ∩ B 6= ∅}.

L

0

(H): the weighted L(H) by assigning to every edge {A, B}

of L(H) the weight |A ∩ B|.

w (H): the maximum weight of a forest in L

0

(H).

Cyclomatic number of H:

µ(H) = P

A∈A

|A| −

S

A∈A

A

− w (H).

µ(H) = 0 ⇔ H is conformal and chordal. (Acharya et al., 1982)

H is α-acyclic ⇔ H is conformal and chordal. (Beeri et al.,

1983)

(18)

Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Cyclomatic Number

Given a hypergraph H = (X , A),

Line graph of H: L(H) = (V , E ), where V = A and E = {{A, B} ⊆

V2

 : A ∩ B 6= ∅}.

L

0

(H): the weighted L(H) by assigning to every edge {A, B}

of L(H) the weight |A ∩ B|.

w (H): the maximum weight of a forest in L

0

(H).

Cyclomatic number of H:

µ(H) = P

A∈A

|A| −

S

A∈A

A

− w (H).

µ(H) = 0 ⇔ H is conformal and chordal. (Acharya et al., 1982)

H is α-acyclic ⇔ H is conformal and chordal. (Beeri et al.,

1983)

(19)

Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Equivalent Definition

Corollary

A hypergraph H is α-acyclic if and only if µ(H) = 0, i.e., P

A∈A

|A| −

S

A∈A

A

− w (H) = 0.

Suppose that there is an α-acyclic k-uniform hypergraph of order n and size m, then km − n = w (H) ≤ (k − 1)(m − 1), thus

m ≤ n − k + 1. Let L

0k−1

(H) denote the spanning subgraph of L

0

(H) containing only those edges of L

0

(H) of weight k − 1.

Corollary

A k-uniform hypergraph H = (X , A) of order n and size n − k + 1

is α-acyclic if and only if L

0

(H) contains a spanning tree, each

edge of which has weight k − 1 (i.e., L

0k−1

(H) is connected).

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

αarb(K n (2) ) = dn/2e

Let the vertex set of K

n(2)

be Z

n

. Note that αarb(K

n(2)

) ≥ dn/2e.

n even: K

n(2)

has a decompostion into n/2 Hamiltonian paths.

Take the path 0, n − 1, 1, n − 2, 2, . . . , n/2 and generate n/2 paths by adding i modulo n, i = 0, 1, . . . , n/2 − 1. Hence αarb(K

n(2)

) = n/2 in this case.

n odd: consider a decompostion of K

n−1(2)

into (n − 1)/2 Hamiltonian paths, P

i

, i = 0, 1, . . . , (n − 1)/2 − 1. Let T

i

= P

i

∪ {{i , n − 1}}, i = 0, 1, . . . , (n − 1)/2 − 1. Let S = {{i , n − 1} : i = (n + 1)/2, . . . , n − 2}. Then ∪

i

T

i

∪ S is an α-acyclic decomposition of K

n(2)

over Z

n

. Hence

αarb(K

n(2)

) = (n + 1)/2 in this case.

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

αarb(K n (2) ) = dn/2e

Let the vertex set of K

n(2)

be Z

n

. Note that αarb(K

n(2)

) ≥ dn/2e.

n even: K

n(2)

has a decompostion into n/2 Hamiltonian paths.

Take the path 0, n − 1, 1, n − 2, 2, . . . , n/2 and generate n/2 paths by adding i modulo n, i = 0, 1, . . . , n/2 − 1. Hence αarb(K

n(2)

) = n/2 in this case.

n odd: consider a decompostion of K

n−1(2)

into (n − 1)/2 Hamiltonian paths, P

i

, i = 0, 1, . . . , (n − 1)/2 − 1. Let T

i

= P

i

∪ {{i , n − 1}}, i = 0, 1, . . . , (n − 1)/2 − 1. Let S = {{i , n − 1} : i = (n + 1)/2, . . . , n − 2}. Then ∪

i

T

i

∪ S is an α-acyclic decomposition of K

n(2)

over Z

n

. Hence

αarb(K

n(2)

) = (n + 1)/2 in this case.

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

k = n − 1, n − 2 & Conjecture

αarb(K

n(n−1)

) = dn/2e since any set of at most two edges of K

n(n−1)

is α-acyclic.

αarb(K

n(n−2)

) = dn(n − 1)/6e. (Li, 1999) In general, Li (1999) showed that

 1 k

 n k − 1



≤ αarb(K

n(k)

) ≤ 1 2

n + 1 k − 1

 .

Conjecture (Wang, 2008): αarb(K

n(k)

) =  1 k

 n

k − 1



. Substituting n − (k − 1) for k, Conjecture can be restated as:

αarb(K

n(n−(k−1))

) =  1 k

 n

k − 1



.

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Introduction Decomposition I Decomposition II Concluding Remarks

Motivations α-Acyclicity Extreme Cases

Our Goal

Determine αarb(K

n(k)

)

(24)

Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

Outline

1

Introduction Motivations α-Acyclicity Extreme Cases

2

Decomposition I Steiner System Construction

3

Decomposition II The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

4

Concluding Remarks

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Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

Steiner System

Definition

A Steiner system S (t, k, n) is a k-uniform hypergraph (X , A) of order n such that every T ∈

Xt

 is contained in exactly one edge (called block) in A.

Theorem (Handbook of Combinatorial Designs, 2007)

There exists an S (k − 1, k, n) whenever any one of the following conditions holds:

(1) k = 2 and n ≡ 0 (mod 2) (perfect matchings), or (2) k = 3 and n ≡ 1, 3 (mod 6) (Steiner triple systems), or (3) k = 4 and n ≡ 2, 4 (mod 6) (Steiner quadruple systems), or (4) k = 5 and n ∈ {11, 23, 35, 47, 71, 83, 107, 131}, or

(5) k = 6 and n ∈ {12, 24, 36, 48, 72, 84, 108, 132}.

(26)

Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

Steiner System

Definition

A Steiner system S (t, k, n) is a k-uniform hypergraph (X , A) of order n such that every T ∈

Xt

 is contained in exactly one edge (called block) in A.

Theorem (Handbook of Combinatorial Designs, 2007)

There exists an S (k − 1, k, n) whenever any one of the following conditions holds:

(1) k = 2 and n ≡ 0 (mod 2) (perfect matchings), or (2) k = 3 and n ≡ 1, 3 (mod 6) (Steiner triple systems), or (3) k = 4 and n ≡ 2, 4 (mod 6) (Steiner quadruple systems), or (4) k = 5 and n ∈ {11, 23, 35, 47, 71, 83, 107, 131}, or

(5) k = 6 and n ∈ {12, 24, 36, 48, 72, 84, 108, 132}.

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Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

Example

αarb(K

7(5)

) ≥ 7. Consider an S (2, 3, 7) (Z

7

, A), with A = {{i , i + 1, i + 3} : i ∈ Z

7

}.

H

{0,1,3}

= {{2, 3, 4, 5, 6}, {1, 2, 4, 5, 6}, {0, 2, 4, 5, 6}};

H

{1,2,4}

= {{0, 3, 4, 5, 6}, {0, 2, 3, 5, 6}, {0, 1, 3, 5, 6}};

H

{2,3,5}

= {{0, 1, 4, 5, 6}, {0, 1, 3, 4, 6}, {0, 1, 2, 4, 6}};

H

{3,4,6}

= {{0, 1, 2, 5, 6}, {0, 1, 2, 4, 5}, {0, 1, 2, 3, 5}};

H

{4,5,0}

= {{0, 1, 2, 3, 6}, {1, 2, 3, 5, 6}, {1, 2, 3, 4, 6}};

H

{5,6,1}

= {{0, 1, 2, 3, 4}, {0, 2, 3, 4, 6}, {0, 2, 3, 4, 5}};

H

{6,0,2}

= {{1, 2, 3, 4, 5}, {0, 1, 3, 4, 5}, {1, 3, 4, 5, 6}};

Since ∪

A∈A

H

A

is a decomposition of K

7(5)

, αarb(K

7(5)

) = 7.

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Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

Construction

Given an S (k − 1, k, n) (X , A), |A| = 1 k

 n k − 1

 . Note that αarb(K

n(n−(k−1))

) ≥  1

k

 n

k − 1



. Note that ∀ A ⊂ X of size k,

H

A

= {X \ B : B ⊂ A, |B| = k − 1} is an α-acyclic (n − (k − 1))-uniform hypergraph of size k and order n.

It is easy to check that ∪

A∈A

H

A

is a decomposition of K

n(n−(k−1))

. Hence, αarb(K

n(n−(k−1))

) = 1

k

 n k − 1



in this

case.

(29)

Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

Construction

Given an S (k − 1, k, n) (X , A), |A| = 1 k

 n k − 1

 . Note that αarb(K

n(n−(k−1))

) ≥  1

k

 n

k − 1



. Note that ∀ A ⊂ X of size k,

H

A

= {X \ B : B ⊂ A, |B| = k − 1} is an α-acyclic (n − (k − 1))-uniform hypergraph of size k and order n.

It is easy to check that ∪

A∈A

H

A

is a decomposition of K

n(n−(k−1))

. Hence, αarb(K

n(n−(k−1))

) = 1

k

 n

k − 1



in this

case.

(30)

Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

Construction

Given an S (k − 1, k, n) (X , A), |A| = 1 k

 n k − 1

 . Note that αarb(K

n(n−(k−1))

) ≥  1

k

 n

k − 1



. Note that ∀ A ⊂ X of size k,

H

A

= {X \ B : B ⊂ A, |B| = k − 1} is an α-acyclic (n − (k − 1))-uniform hypergraph of size k and order n.

It is easy to check that ∪

A∈A

H

A

is a decomposition of K

n(n−(k−1))

. Hence, αarb(K

n(n−(k−1))

) = 1

k

 n

k − 1



in this

case.

(31)

Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

Construction

Theorem (Chee et al., 2009)

If there exists an S (k − 1, k, n), then αarb(K

n(n−(k−1))

) =

k1 k−1n

.

Corollary

We have αarb(K

n(n−(k−1))

) =

k1 k−1n

 whenever any one of the following conditions holds:

(1) k = 2 and n ≡ 0 (mod 2), or (2) k = 3 and n ≡ 1, 3 (mod 6), or (3) k = 4 and n ≡ 2, 4 (mod 6), or

(4) k = 5 and n ∈ {11, 23, 35, 47, 71, 83, 107, 131}, or

(5) k = 6 and n ∈ {12, 24, 36, 48, 72, 84, 108, 132}.

(32)

Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

Construction

Theorem (Chee et al., 2009)

If there exists an S (k − 1, k, n), then αarb(K

n(n−(k−1))

) =

k1 k−1n

.

Corollary

We have αarb(K

n(n−(k−1))

) =

k1 k−1n

 whenever any one of the following conditions holds:

(1) k = 2 and n ≡ 0 (mod 2), or (2) k = 3 and n ≡ 1, 3 (mod 6), or (3) k = 4 and n ≡ 2, 4 (mod 6), or

(4) k = 5 and n ∈ {11, 23, 35, 47, 71, 83, 107, 131}, or

(5) k = 6 and n ∈ {12, 24, 36, 48, 72, 84, 108, 132}.

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Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

k = n − 3

Theorem (Chee et al., 2009) αarb(K

n(n−3)

) =

14 n3

.

Theorem (Chee et al., 2009)

Let δ be a positive constant. Then for k = n − O(log

1−δ

n), we have

αarb(K

n(k)

) = (1 + o(1)) 1 k

 n k − 1



,

where the o(1) is in n.

(34)

Introduction Decomposition I Decomposition II Concluding Remarks

Steiner System Construction

k = n − 3

Theorem (Chee et al., 2009) αarb(K

n(n−3)

) =

14 n3

.

Theorem (Chee et al., 2009)

Let δ be a positive constant. Then for k = n − O(log

1−δ

n), we have

αarb(K

n(k)

) = (1 + o(1)) 1 k

 n k − 1



,

where the o(1) is in n.

(35)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Outline

1

Introduction Motivations α-Acyclicity Extreme Cases

2

Decomposition I Steiner System Construction

3

Decomposition II The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

4

Concluding Remarks

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Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

The Idea

Given an S (k − 1, k, n) (X , A), |A| = 1 k

 n k − 1

 . Note that αarb(K

n(k)

) ≥  1

k

 n k − 1



.

The idea of our construction consists in using the blocks of a S (k − 1, k, n) as centers of our partitions of K

n(k)

into

α-acyclic hypergraphs. Each α-acyclic hypergraph is based on a center C (in this case a block from the Steiner system) to which are added n − k edges, each of which intersect the center on k − 1 vertices (we name these hypergraphs star-shaped).

An example for S (2, 3, 7) (Z

7

, A), with

A = {{i , i + 1, i + 3} : i ∈ Z

7

}.

(37)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

The Idea

Given an S (k − 1, k, n) (X , A), |A| = 1 k

 n k − 1

 . Note that αarb(K

n(k)

) ≥  1

k

 n k − 1



.

The idea of our construction consists in using the blocks of a S (k − 1, k, n) as centers of our partitions of K

n(k)

into

α-acyclic hypergraphs. Each α-acyclic hypergraph is based on a center C (in this case a block from the Steiner system) to which are added n − k edges, each of which intersect the center on k − 1 vertices (we name these hypergraphs star-shaped).

An example for S (2, 3, 7) (Z

7

, A), with

A = {{i , i + 1, i + 3} : i ∈ Z

7

}.

(38)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

The Idea

Given an S (k − 1, k, n) (X , A), |A| = 1 k

 n k − 1

 . Note that αarb(K

n(k)

) ≥  1

k

 n k − 1



.

The idea of our construction consists in using the blocks of a S (k − 1, k, n) as centers of our partitions of K

n(k)

into

α-acyclic hypergraphs. Each α-acyclic hypergraph is based on a center C (in this case a block from the Steiner system) to which are added n − k edges, each of which intersect the center on k − 1 vertices (we name these hypergraphs star-shaped).

An example for S (2, 3, 7) (Z

7

, A), with

A = {{i , i + 1, i + 3} : i ∈ Z

7

}.

(39)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Case n = 7, k = 3

({1, 2, 4} , 0) ({1, 2, 4} , 3) ({1, 2, 4} , 5) ({1, 2, 4} , 6) ({1, 5, 6} , 0) ({1, 5, 6} , 2) ({1, 5, 6} , 3) ({1, 5, 6} , 4) ({2, 3, 5} , 0) ({2, 3, 5} , 1) ({2, 3, 5} , 4) ({2, 3, 5} , 6) ({0, 1, 3} , 2) ({0, 1, 3} , 4) ({0, 1, 3} , 5) ({0, 1, 3} , 6) ({0, 4, 5} , 1) ({0, 4, 5} , 2) ({0, 4, 5} , 3) ({0, 4, 5} , 6) ({0, 2, 6} , 1) ({0, 2, 6} , 3) ({0, 2, 6} , 4) ({0, 2, 6} , 5) ({3, 4, 6} , 0) ({3, 4, 6} , 1) ({3, 4, 6} , 2) ({3, 4, 6} , 5)

{1, 2, 3}

{0, 5, 6}

{0, 2, 5}

{0, 1, 4}

{0, 1, 2}

{0, 3, 4}

{0, 2, 4}

{2, 3, 4}

{3, 5, 6}

{1, 3, 6}

{3, 4, 5}

{2, 5, 6}

{0, 1, 5}

{1, 3, 5}

{1, 4, 5}

{1, 2, 6}

{0, 1, 6}

{0, 4, 6}

{1, 4, 6}

{1, 2, 5}

{2, 3, 6}

{0, 3, 5}

{2, 4, 6}

{4, 5, 6}

{0, 3, 6}

{2, 4, 5}

{1, 3, 4}

{0, 2, 3}

(40)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Case n = 7, k = 3

({1, 2, 4} , 0) ({1, 2, 4} , 3) ({1, 2, 4} , 5) ({1, 2, 4} , 6) ({1, 5, 6} , 0) ({1, 5, 6} , 2) ({1, 5, 6} , 3) ({1, 5, 6} , 4) ({2, 3, 5} , 0) ({2, 3, 5} , 1) ({2, 3, 5} , 4) ({2, 3, 5} , 6) ({0, 1, 3} , 2) ({0, 1, 3} , 4) ({0, 1, 3} , 5) ({0, 1, 3} , 6) ({0, 4, 5} , 1) ({0, 4, 5} , 2) ({0, 4, 5} , 3) ({0, 4, 5} , 6) ({0, 2, 6} , 1) ({0, 2, 6} , 3) ({0, 2, 6} , 4) ({0, 2, 6} , 5) ({3, 4, 6} , 0) ({3, 4, 6} , 1) ({3, 4, 6} , 2) ({3, 4, 6} , 5)

{1, 2, 3}

{0, 5, 6}

{0, 2, 5}

{0, 1, 4}

{0, 1, 2}

{0, 3, 4}

{0, 2, 4}

{2, 3, 4}

{3, 5, 6}

{1, 3, 6}

{3, 4, 5}

{2, 5, 6}

{0, 1, 5}

{1, 3, 5}

{1, 4, 5}

{1, 2, 6}

{0, 1, 6}

{0, 4, 6}

{1, 4, 6}

{1, 2, 5}

{2, 3, 6}

{0, 3, 5}

{2, 4, 6}

{4, 5, 6}

{0, 3, 6}

{2, 4, 5}

{1, 3, 4}

{0, 2, 3}

(41)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Case n = 7, k = 3

({1, 2, 4} , 0) ({1, 2, 4} , 3) ({1, 2, 4} , 5) ({1, 2, 4} , 6) ({1, 5, 6} , 0) ({1, 5, 6} , 2) ({1, 5, 6} , 3) ({1, 5, 6} , 4) ({2, 3, 5} , 0) ({2, 3, 5} , 1) ({2, 3, 5} , 4) ({2, 3, 5} , 6) ({0, 1, 3} , 2) ({0, 1, 3} , 4) ({0, 1, 3} , 5) ({0, 1, 3} , 6) ({0, 4, 5} , 1) ({0, 4, 5} , 2) ({0, 4, 5} , 3) ({0, 4, 5} , 6) ({0, 2, 6} , 1) ({0, 2, 6} , 3) ({0, 2, 6} , 4) ({0, 2, 6} , 5) ({3, 4, 6} , 0) ({3, 4, 6} , 1) ({3, 4, 6} , 2) ({3, 4, 6} , 5)

{1, 2, 3}

{0, 5, 6}

{0, 2, 5}

{0, 1, 4}

{0, 1, 2}

{0, 3, 4}

{0, 2, 4}

{2, 3, 4}

{3, 5, 6}

{1, 3, 6}

{3, 4, 5}

{2, 5, 6}

{0, 1, 5}

{1, 3, 5}

{1, 4, 5}

{1, 2, 6}

{0, 1, 6}

{0, 4, 6}

{1, 4, 6}

{1, 2, 5}

{2, 3, 6}

{0, 3, 5}

{2, 4, 6}

{4, 5, 6}

{0, 3, 6}

{2, 4, 5}

{1, 3, 4}

{0, 2, 3}

(42)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Case n = 7, k = 3

({1, 2, 4} , 0) ({1, 2, 4} , 3) ({1, 2, 4} , 5) ({1, 2, 4} , 6) ({1, 5, 6} , 0) ({1, 5, 6} , 2) ({1, 5, 6} , 3) ({1, 5, 6} , 4) ({2, 3, 5} , 0) ({2, 3, 5} , 1) ({2, 3, 5} , 4) ({2, 3, 5} , 6) ({0, 1, 3} , 2) ({0, 1, 3} , 4) ({0, 1, 3} , 5) ({0, 1, 3} , 6) ({0, 4, 5} , 1) ({0, 4, 5} , 2) ({0, 4, 5} , 3) ({0, 4, 5} , 6) ({0, 2, 6} , 1) ({0, 2, 6} , 3) ({0, 2, 6} , 4) ({0, 2, 6} , 5) ({3, 4, 6} , 0) ({3, 4, 6} , 1) ({3, 4, 6} , 2) ({3, 4, 6} , 5)

{1, 2, 3}

{0, 5, 6}

{0, 2, 5}

{0, 1, 4}

{0, 1, 2}

{0, 3, 4}

{0, 2, 4}

{2, 3, 4}

{3, 5, 6}

{1, 3, 6}

{3, 4, 5}

{2, 5, 6}

{0, 1, 5}

{1, 3, 5}

{1, 4, 5}

{1, 2, 6}

{0, 1, 6}

{0, 4, 6}

{1, 4, 6}

{1, 2, 5}

{2, 3, 6}

{0, 3, 5}

{2, 4, 6}

{4, 5, 6}

{0, 3, 6}

{2, 4, 5}

{1, 3, 4}

{0, 2, 3}

(43)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Case n = 7, k = 3

({1, 2, 4} , 0) ({1, 2, 4} , 3) ({1, 2, 4} , 5) ({1, 2, 4} , 6) ({1, 5, 6} , 0) ({1, 5, 6} , 2) ({1, 5, 6} , 3) ({1, 5, 6} , 4) ({2, 3, 5} , 0) ({2, 3, 5} , 1) ({2, 3, 5} , 4) ({2, 3, 5} , 6) ({0, 1, 3} , 2) ({0, 1, 3} , 4) ({0, 1, 3} , 5) ({0, 1, 3} , 6) ({0, 4, 5} , 1) ({0, 4, 5} , 2) ({0, 4, 5} , 3) ({0, 4, 5} , 6) ({0, 2, 6} , 1) ({0, 2, 6} , 3) ({0, 2, 6} , 4) ({0, 2, 6} , 5) ({3, 4, 6} , 0) ({3, 4, 6} , 1) ({3, 4, 6} , 2) ({3, 4, 6} , 5)

{1, 2, 3}

{0, 5, 6}

{0, 2, 5}

{0, 1, 4}

{0, 1, 2}

{0, 3, 4}

{0, 2, 4}

{2, 3, 4}

{3, 5, 6}

{1, 3, 6}

{3, 4, 5}

{2, 5, 6}

{0, 1, 5}

{1, 3, 5}

{1, 4, 5}

{1, 2, 6}

{0, 1, 6}

{0, 4, 6}

{1, 4, 6}

{1, 2, 5}

{2, 3, 6}

{0, 3, 5}

{2, 4, 6}

{4, 5, 6}

{0, 3, 6}

{2, 4, 5}

{1, 3, 4}

{0, 2, 3}

(44)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Case n = 7, k = 3

({1, 2, 4} , 0) ({1, 2, 4} , 3) ({1, 2, 4} , 5) ({1, 2, 4} , 6) ({1, 5, 6} , 0) ({1, 5, 6} , 2) ({1, 5, 6} , 3) ({1, 5, 6} , 4) ({2, 3, 5} , 0) ({2, 3, 5} , 1) ({2, 3, 5} , 4) ({2, 3, 5} , 6) ({0, 1, 3} , 2) ({0, 1, 3} , 4) ({0, 1, 3} , 5) ({0, 1, 3} , 6) ({0, 4, 5} , 1) ({0, 4, 5} , 2) ({0, 4, 5} , 3) ({0, 4, 5} , 6) ({0, 2, 6} , 1) ({0, 2, 6} , 3) ({0, 2, 6} , 4) ({0, 2, 6} , 5) ({3, 4, 6} , 0) ({3, 4, 6} , 1) ({3, 4, 6} , 2) ({3, 4, 6} , 5)

{1, 2, 3}

{0, 5, 6}

{0, 2, 5}

{0, 1, 4}

{0, 1, 2}

{0, 3, 4}

{0, 2, 4}

{2, 3, 4}

{3, 5, 6}

{1, 3, 6}

{3, 4, 5}

{2, 5, 6}

{0, 1, 5}

{1, 3, 5}

{1, 4, 5}

{1, 2, 6}

{0, 1, 6}

{0, 4, 6}

{1, 4, 6}

{1, 2, 5}

{2, 3, 6}

{0, 3, 5}

{2, 4, 6}

{4, 5, 6}

{0, 3, 6}

{2, 4, 5}

{1, 3, 4}

{0, 2, 3}

(45)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Case n = 7, k = 3

αarb(K

7(3)

) ≥ 7

B

{3,4,6}

= {{3, 4, 6}, {4, 5, 6}, {2, 3, 6}, {1, 3, 4}, {0, 3, 4}}}.

B

{0,2,6}

= {{0, 2, 6}, {0, 5, 6}, {0, 2, 4}, {0, 3, 6}, {0, 1, 6}}}.

B

{0,4,5}

= {{0, 4, 5}, {0, 4, 6}, {0, 3, 5}, {0, 2, 5}, {1, 4, 5}}}.

B

{0,1,3}

= {{0, 1, 3}, {1, 3, 6}, {1, 3, 5}, {0, 1, 4}, {1, 2, 3}}}.

B

{2,3,5}

= {{2, 3, 5}, {2, 5, 6}, {3, 4, 5}, {1, 2, 5}, {0, 2, 3}}}.

B

{1,5,6}

= {{1, 5, 6}, {1, 4, 6}, {3, 5, 6}, {1, 2, 6}, {0, 1, 5}}}.

B

{1,2,4}

= {{1, 2, 4}, {2, 4, 6}, {2, 4, 5}, {2, 3, 4}, {0, 1, 2}}}.

Since ∪

A∈A

B

A

is a decomposition of K

7(3)

, αarb(K

7(3)

) = 7.

(46)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Construction and Proof

Let (X , A) be an S(k − 1, k, n).

Step 1 Define a bipartite graph G with V (G ) = P t Q, where P = {(A, x ) : A ∈ A and x ∈ X \ A} and Q =

Xk

 \ A, so that vertex (A, x ) ∈ P is adjacent to vertex K ∈ Q if and only if K ⊂ A ∪ {x }.

Step 2 Prove that G is k-regular. Hence, G has a perfect matching M.

Step 3 For each A ∈ A, define the k-uniform hypergraph H

A

= (X , B

A

), where

B

A

= {A} ∪ {K ∈ Q : {(A, x ), K } ∈ M for some x ∈ X \ A}.

Step 4 Prove each H

A

is α-acyclic. Then

Xk

 = F

A∈A

B

A

is a

α-acyclic decomposition of size

1k k−1n

.

(47)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Construction and Proof

Let (X , A) be an S(k − 1, k, n).

Step 1 Define a bipartite graph G with V (G ) = P t Q, where P = {(A, x ) : A ∈ A and x ∈ X \ A} and Q =

Xk

 \ A, so that vertex (A, x ) ∈ P is adjacent to vertex K ∈ Q if and only if K ⊂ A ∪ {x }.

Step 2 Prove that G is k-regular. Hence, G has a perfect matching M.

Step 3 For each A ∈ A, define the k-uniform hypergraph H

A

= (X , B

A

), where

B

A

= {A} ∪ {K ∈ Q : {(A, x ), K } ∈ M for some x ∈ X \ A}.

Step 4 Prove each H

A

is α-acyclic. Then

Xk

 = F

A∈A

B

A

is a

α-acyclic decomposition of size

1k k−1n

.

(48)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Construction and Proof

Let (X , A) be an S(k − 1, k, n).

Step 1 Define a bipartite graph G with V (G ) = P t Q, where P = {(A, x ) : A ∈ A and x ∈ X \ A} and Q =

Xk

 \ A, so that vertex (A, x ) ∈ P is adjacent to vertex K ∈ Q if and only if K ⊂ A ∪ {x }.

Step 2 Prove that G is k-regular. Hence, G has a perfect matching M.

Step 3 For each A ∈ A, define the k-uniform hypergraph H

A

= (X , B

A

), where

B

A

= {A} ∪ {K ∈ Q : {(A, x ), K } ∈ M for some x ∈ X \ A}.

Step 4 Prove each H

A

is α-acyclic. Then

Xk

 = F

A∈A

B

A

is a

α-acyclic decomposition of size

1k k−1n

.

(49)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Construction and Proof

Let (X , A) be an S(k − 1, k, n).

Step 1 Define a bipartite graph G with V (G ) = P t Q, where P = {(A, x ) : A ∈ A and x ∈ X \ A} and Q =

Xk

 \ A, so that vertex (A, x ) ∈ P is adjacent to vertex K ∈ Q if and only if K ⊂ A ∪ {x }.

Step 2 Prove that G is k-regular. Hence, G has a perfect matching M.

Step 3 For each A ∈ A, define the k-uniform hypergraph H

A

= (X , B

A

), where

B

A

= {A} ∪ {K ∈ Q : {(A, x ), K } ∈ M for some x ∈ X \ A}.

Step 4 Prove each H

A

is α-acyclic. Then

Xk

 = F

A∈A

B

A

is a

α-acyclic decomposition of size

1k k−1n

.

(50)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Construction and Proof

Let (X , A) be an S(k − 1, k, n).

Step 1 Define a bipartite graph G with V (G ) = P t Q, where P = {(A, x ) : A ∈ A and x ∈ X \ A} and Q =

Xk

 \ A, so that vertex (A, x ) ∈ P is adjacent to vertex K ∈ Q if and only if K ⊂ A ∪ {x }.

Step 2 Prove that G is k-regular. Hence, G has a perfect matching M.

Step 3 For each A ∈ A, define the k-uniform hypergraph H

A

= (X , B

A

), where

B

A

= {A} ∪ {K ∈ Q : {(A, x ), K } ∈ M for some x ∈ X \ A}.

Step 4 Prove each H

A

is α-acyclic. Then

Xk

 = F

A∈A

B

A

is a

α-acyclic decomposition of size

1k k−1n

.

(51)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Results

Theorem

If there exists an S (k − 1, k, n), then αarb(K

n(k)

) =

1k k−1n

.

Corollary

We have αarb(K

n(k)

) =

1k k−1n

 whenever any one of the following conditions holds:

(1) k = 2 and n ≡ 0 (mod 2), or (2) k = 3 and n ≡ 1, 3 (mod 6), or (3) k = 4 and n ≡ 2, 4 (mod 6), or

(4) k = 5 and n ∈ {11, 23, 35, 47, 71, 83, 107, 131}, or

(5) k = 6 and n ∈ {12, 24, 36, 48, 72, 84, 108, 132}.

(52)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Results

Theorem

If there exists an S (k − 1, k, n), then αarb(K

n(k)

) =

1k k−1n

.

Corollary

We have αarb(K

n(k)

) =

1k k−1n

 whenever any one of the following conditions holds:

(1) k = 2 and n ≡ 0 (mod 2), or (2) k = 3 and n ≡ 1, 3 (mod 6), or (3) k = 4 and n ≡ 2, 4 (mod 6), or

(4) k = 5 and n ∈ {11, 23, 35, 47, 71, 83, 107, 131}, or

(5) k = 6 and n ∈ {12, 24, 36, 48, 72, 84, 108, 132}.

(53)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

k = 3 & n ≡ 0, 4 (mod 6)

Let X = Y t Z , where |Y | = n − 3 and Z = {∞

1

, ∞

2

, ∞

3

}. Let (Y , A) be an S (2, 3, n − 3).

two different kinds: star-shaped hypergraphs whose centers belong to a Steiner triple system on Y , but also classes whose centers are two triples {y , ∞

1

, ∞

2

} and {y , ∞

1

, ∞

3

}

(intersecting on y , ∞

1

), for all y ∈ Y .

The decomposition is completed by another star-shaped class containing the triples {y , ∞

2

, ∞

3

} where y ∈ X \{∞

2

, ∞

3

}.

Define the bipartite graph and prove it is regular.

(54)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

k = 3 & n ≡ 0, 4 (mod 6)

Let X = Y t Z , where |Y | = n − 3 and Z = {∞

1

, ∞

2

, ∞

3

}. Let (Y , A) be an S (2, 3, n − 3).

two different kinds: star-shaped hypergraphs whose centers belong to a Steiner triple system on Y , but also classes whose centers are two triples {y , ∞

1

, ∞

2

} and {y , ∞

1

, ∞

3

}

(intersecting on y , ∞

1

), for all y ∈ Y .

The decomposition is completed by another star-shaped class containing the triples {y , ∞

2

, ∞

3

} where y ∈ X \{∞

2

, ∞

3

}.

Define the bipartite graph and prove it is regular.

(55)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

k = 3 & n ≡ 0, 4 (mod 6)

Let X = Y t Z , where |Y | = n − 3 and Z = {∞

1

, ∞

2

, ∞

3

}. Let (Y , A) be an S (2, 3, n − 3).

two different kinds: star-shaped hypergraphs whose centers belong to a Steiner triple system on Y , but also classes whose centers are two triples {y , ∞

1

, ∞

2

} and {y , ∞

1

, ∞

3

}

(intersecting on y , ∞

1

), for all y ∈ Y .

The decomposition is completed by another star-shaped class containing the triples {y , ∞

2

, ∞

3

} where y ∈ X \{∞

2

, ∞

3

}.

Define the bipartite graph and prove it is regular.

(56)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Vertices of the graph

Define the bipartite graph Γ with bipartition V (Γ) = P t Q, where

P = [

A∈A

{(A, x) : x ∈ X \ A}

!

 [

y ∈Y

{({y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}, z) : z ∈ Y \ {y }}

 , Q = X

3



\

(A ∪ {{y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}, {y , ∞

2

, ∞

3

} : y ∈ Y } ∪ {Z }).

(57)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Edges and Regularity

Edges:

(1) Vertex ({a, b, c}, x ) ∈ P is adjacent to vertices {a, b, x}, {a, c, x}, {b, c, x} ∈ Q.

(2) Vertex ({y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}, z) ∈ P is adjacent to vertices {y , z, ∞

h

} ∈ Q, h ∈ [3].

Regularity: ∀u, v ∈ Y , let A

uv

∈ A containing both u and v . (1) Every vertex in P being of degree 3.

(2) {a, b, c} ∈ Q is adjacent to (A

ab

, c), (A

bc

, a), and (A

ac

, b).

(3) {a, b, ∞

h

} ∈ Q is adjacent to (A

ab

, ∞

h

), ({b, ∞

1

, ∞

2

}, {b, ∞

1

, ∞

3

}, a) and ({a, ∞

1

, ∞

2

}, {a, ∞

1

, ∞

3

}, b).

Hence Γ is 3-regular and has a perfect matching M.

(58)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Edges and Regularity

Edges:

(1) Vertex ({a, b, c}, x ) ∈ P is adjacent to vertices {a, b, x}, {a, c, x}, {b, c, x} ∈ Q.

(2) Vertex ({y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}, z) ∈ P is adjacent to vertices {y , z, ∞

h

} ∈ Q, h ∈ [3].

Regularity: ∀u, v ∈ Y , let A

uv

∈ A containing both u and v . (1) Every vertex in P being of degree 3.

(2) {a, b, c} ∈ Q is adjacent to (A

ab

, c), (A

bc

, a), and (A

ac

, b).

(3) {a, b, ∞

h

} ∈ Q is adjacent to (A

ab

, ∞

h

), ({b, ∞

1

, ∞

2

}, {b, ∞

1

, ∞

3

}, a) and ({a, ∞

1

, ∞

2

}, {a, ∞

1

, ∞

3

}, b).

Hence Γ is 3-regular and has a perfect matching M.

(59)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Edges and Regularity

Edges:

(1) Vertex ({a, b, c}, x ) ∈ P is adjacent to vertices {a, b, x}, {a, c, x}, {b, c, x} ∈ Q.

(2) Vertex ({y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}, z) ∈ P is adjacent to vertices {y , z, ∞

h

} ∈ Q, h ∈ [3].

Regularity: ∀u, v ∈ Y , let A

uv

∈ A containing both u and v . (1) Every vertex in P being of degree 3.

(2) {a, b, c} ∈ Q is adjacent to (A

ab

, c), (A

bc

, a), and (A

ac

, b).

(3) {a, b, ∞

h

} ∈ Q is adjacent to (A

ab

, ∞

h

), ({b, ∞

1

, ∞

2

}, {b, ∞

1

, ∞

3

}, a) and ({a, ∞

1

, ∞

2

}, {a, ∞

1

, ∞

3

}, b).

Hence Γ is 3-regular and has a perfect matching M.

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Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Decompostion

(1) ∀A ∈ A, define H

A

= (X , B

A

), where

B

A

= {A} ∪ {T ∈ Q : {(A, x ), T } ∈ M for some x ∈ X \ A}.

(2) ∀y ∈ Y , define H

y

= (X , B

y

), where B

y

={{y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}}∪

{T ∈ Q : {({y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}, z), T } ∈ M for some z ∈ Y \ {y }}.

(3) Define H = (X , B), where

B = {{y , ∞

2

, ∞

3

} : y ∈ X \{∞

2

, ∞

3

}}

(61)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Decompostion

(1) ∀A ∈ A, define H

A

= (X , B

A

), where

B

A

= {A} ∪ {T ∈ Q : {(A, x ), T } ∈ M for some x ∈ X \ A}.

(2) ∀y ∈ Y , define H

y

= (X , B

y

), where B

y

={{y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}}∪

{T ∈ Q : {({y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}, z), T } ∈ M for some z ∈ Y \ {y }}.

(3) Define H = (X , B), where

B = {{y , ∞

2

, ∞

3

} : y ∈ X \{∞

2

, ∞

3

}}

(62)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Decompostion

(1) ∀A ∈ A, define H

A

= (X , B

A

), where

B

A

= {A} ∪ {T ∈ Q : {(A, x ), T } ∈ M for some x ∈ X \ A}.

(2) ∀y ∈ Y , define H

y

= (X , B

y

), where B

y

={{y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}}∪

{T ∈ Q : {({y , ∞

1

, ∞

2

}, {y , ∞

1

, ∞

3

}, z), T } ∈ M for some z ∈ Y \ {y }}.

(3) Define H = (X , B), where

B = {{y , ∞

2

, ∞

3

} : y ∈ X \{∞

2

, ∞

3

}}

(63)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Decompostion

Now, we have

X 3



= G

A∈A

B

A

! t

 G

y ∈Y

B

y

 t B,

so that {H

A

}

A∈A

∪ {H

y

}

y ∈Y

∪ {H} is an α-acyclic decomposition of K

n(3)

. The size of this decomposition is

(n − 3)(n − 4)

6 + (n − 3) + 1 = n(n − 1)

6 .

(64)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Final Results

Finally, we get the following results separately.

Lemma

αarb(K

n(3)

) = n(n − 1)/6 for all n ≡ 0, 4 (mod 6).

Lemma

αarb(K

n(3)

) = dn(n − 1)/6e for all n ≡ 5 (mod 6).

αarb(K

n(3)

) = dn(n − 1)/6e for all n ≡ 2 (mod 6).

Theorem

αarb(K

n(3)

) = dn(n − 1)/6e for all n ≥ 3.

(65)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Final Results

Finally, we get the following results separately.

Lemma

αarb(K

n(3)

) = n(n − 1)/6 for all n ≡ 0, 4 (mod 6).

Lemma

αarb(K

n(3)

) = dn(n − 1)/6e for all n ≡ 5 (mod 6).

αarb(K

n(3)

) = dn(n − 1)/6e for all n ≡ 2 (mod 6).

Theorem

αarb(K

n(3)

) = dn(n − 1)/6e for all n ≥ 3.

(66)

Introduction Decomposition I Decomposition II Concluding Remarks

The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

Final Results

Finally, we get the following results separately.

Lemma

αarb(K

n(3)

) = n(n − 1)/6 for all n ≡ 0, 4 (mod 6).

Lemma

αarb(K

n(3)

) = dn(n − 1)/6e for all n ≡ 5 (mod 6).

αarb(K

n(3)

) = dn(n − 1)/6e for all n ≡ 2 (mod 6).

Theorem

αarb(K

n(3)

) = dn(n − 1)/6e for all n ≥ 3.

(67)

Introduction Decomposition I Decomposition II Concluding Remarks

Outline

1

Introduction Motivations α-Acyclicity Extreme Cases

2

Decomposition I Steiner System Construction

3

Decomposition II The Idea

Construction and Proof k = 3 & n ≡ 0, 4 (mod 6)

4

Concluding Remarks

(68)

Introduction Decomposition I Decomposition II Concluding Remarks

Concluding Remarks

αarb(K

n(k)

) =  1 k

 n k − 1



for k = 1, 2, 3, n − 3, n − 2, n − 1, n αarb(K

n(k)

) = αarb(K

n(n−(k−1))

) =  1

k

 n

k − 1



if there exists an S(k − 1, k, n).

for general k: k = 4, 5, . . . , n − 5 ???

for general hypergraphs: for exmaple, multipartite

hypergraphs. ???

(69)

Introduction Decomposition I Decomposition II Concluding Remarks

Concluding Remarks

αarb(K

n(k)

) =  1 k

 n k − 1



for k = 1, 2, 3, n − 3, n − 2, n − 1, n αarb(K

n(k)

) = αarb(K

n(n−(k−1))

) =  1

k

 n

k − 1



if there exists an S(k − 1, k, n).

for general k: k = 4, 5, . . . , n − 5 ???

for general hypergraphs: for exmaple, multipartite

hypergraphs. ???

(70)

Introduction Decomposition I Decomposition II Concluding Remarks

Concluding Remarks

αarb(K

n(k)

) =  1 k

 n k − 1



for k = 1, 2, 3, n − 3, n − 2, n − 1, n αarb(K

n(k)

) = αarb(K

n(n−(k−1))

) =  1

k

 n

k − 1



if there exists an S(k − 1, k, n).

for general k: k = 4, 5, . . . , n − 5 ???

for general hypergraphs: for exmaple, multipartite

hypergraphs. ???

References

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