Chin-Mei Fu Department of Mathematics Tamkang University, Tamsui, Taipei Shien, Taiwan.

Wen-Chung Huang Department of Mathematics Soochow University Taipei, Taiwan.

Kite-Designs Intersecting in PaiKite-Designs Intersecting in Pairwise Disjoint Blocksrwise Disjoint Blocks

a

b

c d

• Denote the kite by (a,b,c;d)

Kite-design A kite-design of order n is a pair (X, Ҝ), where

X is the vertex set of Kn and Ҝ is an edge-disjoi

nt decomposition of Kn into copies of kites.

If kite-design of order n exists thenIf kite-design of order n exists then( 1)

8

n nN

Then n≡0,1 (mod 8)

n = 8

(1, 2, 4; ) (mod 7)

(1, 2, 4; )

1 2

3

KD(8)={(1, 2, 4;),(2,3,5;),(3,4,6;),(4,5,7;),(5,6,1;),(6,7,2;),(7,1,3;) }

n = 9

(1, 2, 5; 7) (mod 9)

KD(9)={(1,2,5;7),(2,3,6;8),(3,4,7;9),(4,5,8;1),(5,6,9;2),(6,7,1;3),(7,8,2;4),(8,9,3;5),(9,1,4;6) }

(1, 2, 5; 7)

1 3 2

4

Find all possible values k such that there are two kite-designs, (V, B1) and (V, B2), of order n with |B1 B2|=k.

Intersection Problem for kite-designs:

Example: KD(8);

B2={(1,2,4;3),(5,2,3;),(,4,6;3),(4,5,7;),(6,1,5;),(6,7,2;),(3,7,1;)}}B1 B2={(4,5,7;),(6,7,2;)}

|B1 B2|=2

(4,5,7;) and (6,7,2;) has a common vertex 7

B1={(1,2,4;),(2,3,5;),(3,4,6;), (4,5,7;),(5,6,1;),(6,7,2;),(7,1,3;)}

In 2004, Yeow Meng Chee give:

Two STS (X, A) and (X, B) are said to intersect in m pairwise disjoint blocks if |AB|=m and all blocks in AB are pairwise disjoint

He prove that the spectrum of the problem in order v is

{0,1,…,(v-1)/3} if v 1 (mod 6); {0,1,…,v/3} if v 3 (mod 6); for v13.

{1},{0,1},{0,1,3} for v=3,7,9, respectively

Find all possible values k such that there are two kite-designs of order n, (V, B1) and (V, B2), in which |B1 B2|=k and all blocks in B1 B2 are pairwise disjoint.

Kite-design Intersecting in Pairwise Disjoint Blocks

Let Jd(n) = {0,1,…,[n/4]}

Lemma (Billington)Lemma (Billington){0, 1}{0, 1}IIdd(n), for all (n), for all n 0 or 1 (mod 8).

Id(n) = {k| there exist two kite-designs of order n, (V, B 1) and (V, B2), in which |B1 B2|=k and all blocks in B1 B2 are pairwise disjoint}

Lemma: Id(9)=Jd(9), Id(17)=Jd(17), Id(16)=Jd(16), Id(32)=Jd(32), Id(K24\K8)=Jd(K24\K8), Id(24)=Jd(24), Id(K40\K8)=Jd(K40\K8), Id(40)=Jd(40).

Pf: n=9. Let K={(1,2,3;4),(5,6,7;8),(4,8,9;6),(2,5,8;1),(1,9,7;3),(3,8,6;1),(1, 5,4;6),(4,7,2;6),(3,5,9;2)}.

Let π=(1,2)(5,6). πK={(2,1,3;4),(6,5,7;8),(4,8,9;5),(1,6,8;2),(2,9,7;3),(3,8,5;2),(2, 6,4;5),(4,7,1;5),(3,6,9;1)}.

Then K∩πK={(1,2,3;4),(5,6,7;8)}. From the result and above Lemma, we obtain I(9)=J(9).

⋄

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

Ex: 0Id(K2,2,2)

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1

1’

2

2’

3’

3

1 1’

2

2’

3’3

1 1’

2

2’

3’

3

1

1’

2

2’

3’

3

1 1’

2

2’

3’3

KD1(K2,2,2)={(1,3,2;1’),(3,2’,1’;3’),(1,2’,3’;2)}

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1 1’

2

2’

3’

3

1

1’

2

2’

3’

3

1 1’

2

2’

3’ 3

KD2(K2,2,2)={(1’,3’,2’;1),(3’,2,1;3),(1’,2,3;2’)}

KD2(K2,2,2)={(1’,3’,2’;1),(3’,2,1;3), (1’,2,3;2’)}

KD1(K2,2,2)={(1,3,2;1’),(3,2’,1’;3’), (1,2’,3’;2)}

0Id(K2,2,2)

Ex: 0Id(K2n,2n,2n)

Take a Latin square of order n, it is C3-decomposition of Kn,n,n

Replacing each C3 by K2,2,2, and from 0Id(K2,2,2), we have 0Id(K2n,2n,2n)

If 2m 0 or 2 (mod 6), 2m 6, there exist GDD(2m, 3,2)

If 2m 4 (mod 6), 2m 10, there exist GDD(2m, 3,{2,4*})

n=8k+1

2k

•2k0 or 2 (mod

6)

K9 K9K9 K9

K4,4,4

From Id(9) = {0,1,2} and 0Id(K4,4,4), we have Id(8k+1)=Jd(8k+1), for 2k 0 or 2 (mod 6), 2k 6.

2k4 (mod 6)

K9 K9K9K17

K4,4,4

From Id(9) = {0,1,2}, Id(17)={0,1,2,3,4} and 0Id(K4,4,4), we have Id(8k+1)=Jd(8k+1), for 2k 4 (mod 6), 2k 10.

Id(8)?=?{0,1,2}=Jd(8)

Suppose 2 I(8). There is a kite-design of ∈K₈ containing two kites of the form (1,2,3;4),(5,6,7;8).

1

2

3

4

5

6

7

8

The remaining 5 kites must come from the edges of K4,4 and {{1,4},{2,4},{5,8}, {6,8}}

Therefore, Id(8)={0,1} {0,1,2}=Jd(8)

n=16k

2k

2k0 or 2 (mod 6)

K16 K16K16 K16

K8,8,8

From Id(16) = {0,1,2,3,4}, and 0Id(K

8,8,8), we have Id(16k)=Jd(16k), for 2k 0,2 (mod 6), 2k 6.

Similar, from Id(16) = {0,1,2,3,4}, Id(32)={0,1,2,3,4,5,6,7,8} and 0Id(K8,8,8), we have Id(16k)=Jd(16k), for 2k 4 (mod 6), 2k 10.

n=16k+8

2k0 or 2 (mod 6)

K24K24\K8

K24\K8

K8,8,8

From Id(24) = {0,1,2,3,4,5,6}, Id(K24\K

8)={0,1,2,3,4} and 0Id(K8,8,8), we have Id(16k+8)=Jd(16k+8), for 2k 0,2 (mod 6), 2k 6.

Similar, from Id(24) = {0,1,2,3,4,5,6}, Id(K24\K8)={0,1,2,3,4} Id(K40\K8)={0,1,2,3,4,

5,6,7,8} and 0Id(K8,8,8), we have Id(16k+8)=Jd(16k+8), for 2k 4 (mod 6), 2k 10.

Theorem

Id(n)=Jd(n), for n≡0,1 (mod 8), n≠8 and Id(8)={0,1}.