1. Kite-Designs Intersecting in Pairwise Disjoint Blocks
Chin-Mei Fu
Department of Mathematics
Tamkang University,
Tamsui, Taipei Shien, Taiwan.
Wen-Chung Huang
Soochow University
Taipei, Taiwan.

2. Denote the kite by (a,b,c;d)

3. 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-disjoint decomposition of Kn into copies of kites.
If kite-design of order n exists then
Then n≡0,1 (mod 8)

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

5. n = 9
(1, 2, 5; 7) (mod 9) 1 3 2
(1, 2, 5; 7) 4
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) }

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

7. (4,5,7;) and (6,7,2;) has a common vertex 7
Example: KD(8);
B1={(1,2,4;),(2,3,5;),(3,4,6;), (4,5,7;),(5,6,1;),(6,7,2;),(7,1,3;)}
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

8. {1},{0,1},{0,1,3} for v=3,7,9, respectively
In 2004, Yeow Meng Chee give:
Two STS (X, A) and (X, B) are said to intersect in m pairwise disjoint blocks if |AB|=m and all blocks in AB 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 v13.
{1},{0,1},{0,1,3} for v=3,7,9, respectively

9. Kite-design Intersecting in Pairwise Disjoint Blocks
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.

10. Let Jd(n) = {0,1,…,[n/4]}
Id(n) = {k| there exist 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}
Lemma (Billington)
{0, 1}Id(n), for all n 0 or 1 (mod 8).

11. 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). ⋄

12. Ex: 0Id(K2,2,2) 1 1 1 1’ 1’ 1’ 2 2 2 3’ 3’ 3’ 2’ 2’ 2’ 3 3 3

13. 1 1 1’ 1’ 2 2 3’ 3’ 3’ 2’ 2’ 2’ 3 3 1 1’ 2 3

14. 1 1 1’ 2 2 3’ 3’ 2’ 2’ 3 1 1’ 1’ 2 3’ 3 2’ 3

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

16. 1 1 1 1’ 1’ 1’ 2 2 2 3’ 3’ 3’ 2’ 2’ 2’ 3 3 3

17. 1 1 1’ 1’ 2 2 2 3’ 3’ 2’ 2’ 3 3 3 1 1’ 3’ 2’

18. 1 1 1’ 1’ 2 2 2 3’ 3’ 2’ 2’ 3 3 3 1 1’ 3’ 2’

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

20. KD1(K2,2,2)={(1,3,2;1’),(3,2’,1’;3’), (1,2’,3’;2)}
0Id(K2,2,2)

21. Ex: 0Id(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 0Id(K2,2,2), we have 0Id(K2n,2n,2n)

22. 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*})

23. n=8k+1  2k

24. 2k0 or 2 (mod 6) 

25.  K9 K9 K9 K9

26. 
K4,4,4

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

28. 2k4 (mod 6) 

29. 
K17 K9 K9 K9

30. 
K4,4,4

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

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

33. 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).
The remaining 5 kites must come from the edges of K4,4
and {{1,4},{2,4},{5,8}, {6,8}}

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

35. n=16k 2k

36. 2k0 or 2 (mod 6)

37. K16
K16
K16
K16

38. K8,8,8

39. From Id(16) = {0,1,2,3,4}, and 0Id(K8,8,8), 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 0Id(K8,8,8), we have Id(16k)=Jd(16k), for 2k  4 (mod 6), 2k  10.

40. n=16k+8

41. 2k0 or 2 (mod 6)

42. K24\K8
K24\K8
K24

43. K8,8,8

44. From Id(24) = {0,1,2,3,4,5,6}, Id(K24\K8)={0,1,2,3,4} and 0Id(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 0Id(K8,8,8), we have Id(16k+8)=Jd(16k+8), for 2k  4 (mod 6), 2k  10.

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