1. 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 Pairwise Disjoint Blocks Sangeetha Srinivasan

2. In 1995, Billington give the spectrum of the intersection problem for small G-design: Some of G as following. P4P4 S3S3 D C4C4 K 4 -e K4K4

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

4. Kite-design A kite-design of order n is a pair (X, Ҝ), where X is the vertex set of K n and Ҝ is an edge- disjoint decomposition of K n into copies of kites. If kite-design of order n exists then Then n≡0,1 (mod 8)

5.  n = 8 (1, 2, 4;  ) (mod 7) (1, 2, 4;  ) 12 3 KD(8)={(1, 2, 4;  ),(2,3,5;  ),(3,4,6;  ), (4,5,7;  ),(5,6,1;  ),(6,7,2;  ),(7,1,3;  ) }

6.  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

7. Find all possible values k such that there are two kite-designs, (V, B 1 ) and (V, B 2 ), of order n with |B 1  B 2 |=k. Intersection Problem for kite-designs:

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

9. 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

10. Find all possible values k such that there are two kite-designs of order n, (V, B 1 ) and (V, B 2 ), in which |B 1  B 2 |=k and all blocks in B 1  B 2 are pairwise disjoint.  Kite-design Intersecting in Pairwise Disjoint Blocks

11.  Let J d (n) = {0,1,…,[n/4]} Lemma (Billington) {0, 1}  I d (n), for all {0, 1}  I d (n), for all n  0 or 1 (mod 8).  I d (n) = {k| there exist two kite- designs of order n, (V, B 1 ) and (V, B 2 ), in which |B 1  B 2 |=k and all blocks in B 1  B 2 are pairwise disjoint}

12.  Lemma: I d (9)=J d (9), I d (17)=J d (17), I d (16)=J d (16), I d (32)=J d (32), I d (K 24 \K 8 )=J d (K 24 \K 8 ), I d (24)=J d (24), I d (K 40 \K 8 )=J d (K 40 \K 8 ), I d (40)=J d (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 d (9)=J d (9).  ⋄

13. 11’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3 Ex: 0  I d (K 2,2,2 )

14. 11’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3

15. 11’ 2 2’ 3’ 3 1 1’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3

16. 11’ 2 2’ 3’ 3 1 1’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3 KD 1 (K 2,2,2 )={(1,3,2;1’),(3,2’,1’;3’),(1,2’,3’;2)}

17. 11’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3

18. 11’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3

19. 11’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3

20. 11’ 2 2’ 3’ 3 1 1’ 2 2’ 3’ 3 11’ 2 2’ 3’ 3 KD 2 (K 2,2,2 )={(1’,3’,2’;1),(3’,2,1;3),(1’,2,3;2’)}

21.  KD 2 (K 2,2,2 )={(1’,3’,2’;1),(3’,2,1;3), (1’,2,3;2’)}  KD 1 (K 2,2,2 )={(1,3,2;1’),(3,2’,1’;3’), (1,2’,3’;2)} 0  I d (K 2,2,2 )

22. Ex: 0  I d (K 2n,2n,2n )  Take a Latin square of order n, it is C 3 - decomposition of K n,n,n  Replacing each C 3 by K 2,2,2, and from 0  I d (K 2,2,2 ), we have 0  I d (K 2n,2n,2n )

23.  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*})

24. n=8k+1  2k

25. 2k  0 or 2 (mod 6) 

26.  K9K9 K9K9 K9K9 K9K9

27. K 4,4,4 

28.  From I d (9) = {0,1,2} and 0  I d (K 4,4,4 ), we have I d (8k+1)=J d (8k+1), for 2k  0 or 2 (mod 6), 2k  6.

29. 2k  4 (mod 6) 

30.  K9K9 K9K9 K9K9 K 17

31. K 4,4,4 

32.  From I d (9) = {0,1,2}, I d (17)={0,1,2,3,4} and 0  I d (K 4,4,4 ), we have I d (8k+1)=J d (8k+1), for 2k  4 (mod 6), 2k  10.

33. I d (8)?=?{0,1,2}=J d (8)

34. Suppose 2 ∈ I d (8). There is a kite-design of K ₈ 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 K 4,4 and {{1,4},{2,4},{5,8}, {6,8}}

35.  Therefore, I d (8)={0,1}  {0,1,2}=J d (8)

36. n=16k 2k

37. 2k  0 or 2 (mod 6)

38. K 16

39. K 8,8,8

40.  From I d (16) = {0,1,2,3,4}, and 0  I d (K 8,8,8 ), we have I d (16k)=J d (16k), for 2k  0,2 (mod 6), 2k  6.  Similar, from I d (16) = {0,1,2,3,4}, I d (32)={0,1,2,3,4,5,6,7,8} and 0  I d (K 8,8,8 ), we have I d (16k)=J d (16k), for 2k  4 (mod 6), 2k  10.

41. n=16k+8

42. 2k  0 or 2 (mod 6)

43. K 24 K 24 \K 8

44. K 8,8,8

45.  From I d (24) = {0,1,2,3,4,5,6}, I d (K 24 \K 8 )={0,1,2,3,4} and 0  I d (K 8,8,8 ), we have I d (16k+8)=J d (16k+8), for 2k  0,2 (mod 6), 2k  6.  Similar, from I d (24) = {0,1,2,3,4,5,6}, I d (K 24 \K 8 )={0,1,2,3,4} I d (K 40 \K 8 )={0,1,2,3,4, 5,6,7,8} and 0  I d (K 8,8,8 ), we have I d (16k+8)=J d (16k+8), for 2k  4 (mod 6), 2k  10.

46.  Theorem I d (n)=J d (n), for n≡0,1 (mod 8), n≠8 and I d (8)={0,1}.