The Fine Triangle Intersection Problem for Minimum Kite Coverings

Transcription

1 42Æ 5 Vol.42, No ADVANCES IN MATHEMATICS Oct., 2013 The Fine Triangle Intersection Problem for Minimum Kite Coverings ZHANG Guizhi 1,2, CHANG Yanxun 1,, FENG Tao 1 (1. Institute of Mathematics, Beijing Jiaotong University, Beijing, , P. R. China; 2. School of Mathematical Sciences, Hulunbuir College, Hulunbuir, Inner Mongolia, , P. R. China) Abstract: In this article the fine triangle intersection problem for a pair of minimum kite coverings is investigated. Let Fin(v) = {(s,t) : a pair of minimum kite coverings of order v intersectinginsblocks ands+ttriangles}. Let Adm(v) = {(s,t) : s+t b v,s,tare nonnegative integers} and b v = v(v 1). It is established that Fin(v) = Adm(v)\{(b 8 v 1,0),(b v 1,1)} for any integer v 0,1(mod 8) and v 8; Fin(v) = Adm(v) for any integer v 2,3,,7(mod 8) and v 4. Key words: kite covering; triangle intersection; fine triangle intersection MR(2000) Subject Classification: 05B30 / CLC number: O157.2 Document code: A Article ID: (2013) Introduction The intersection problem was first introduced by Kramer and Mesner [22]. They asked that for which values of s it is possible to find two Steiner systems S(t,k,v) intersecting in s blocks, where both objects must be based on the same element set. This initial work was later extended to cover many other combinatorial structures, such as Latin squares (see [1 2, 13, 17 18, 23]) and G-designs. This paper is closely related to the intersection problem for G-designs. Let K v be the complete graph with v vertices and λk v denote the graph K v with each of its edges replicated λ times. Given a family G of graphs each of which is simple and connected, a λ-fold G-design of order v, denoted by a (λk v,g)-design, is a pair (X,B) where X is the vertex set of K v and B is a collection of subgraphs (called blocks) of λk v, such that each block is isomorphic to a graph in G, and each edge of λk v belongs to exactly λ blocks of B. If in the definition of G-designs we replace the term exactly with at most (or at least ), we have a (λk v,g)-packing (or covering). When λ = 1, a (K v,g)-packing (or covering) is called a G-packing (or covering) of order v. When G contains a single graph G, i.e., G = {G}, a (λk v,{g})-design (packing or covering) is simply written as a (λk v,g)-design (packing or covering). If G is the complete graph K k, a (K v,k k )-design is called a Steiner system S(2,k,v). Two (K v,g)-packings (or coverings) (X,B 1 ) and (X,B 2 ) are said to intersect in s blocks provided B 1 B 2 = s. If s = 0, (X,B 1 ) and (X,B 2 ) are said to be disjoint. The intersection problem for (K v,g)-packings (or coverings) is the determination of all integral pairs (v,s) such that there exists a pair of (K v,g)-packings (or coverings) intersecting in s blocks. Received date: Revised date: Foundation item: Supported by the Fundamental Research Funds for the Central Universities (No. 2011JBZ012, No. 2011JBM298) and NSFC (No , No ). yxchang@bjtu.edu.cn

2 5,, : The Fine Triangle Intersection Problem for Minimum Kite Coverings 677 A complete solution to the intersection problem for S(2,3,v)s was made by Lindner and Rosa [24]. Colbournetal. [15] solvedtheintersectionproblemforapairofsteinersystems(2,4,v)s, leaving a handful of possible exceptions when v = 25,28 and 37. Gionfriddo and Lindner [19] first examined the intersection problem for S(3, 4, v)s; this problem was finally dealt with by Hartman and Yehudai [20], apart from two undecided values of v = 14 and 26. Billington and Kreher [5] discussed the intersection problem for all connected simple graphs G with at most four vertices or at most four edges. Billington et al. [4] completely determined the intersection numbers of K 4 e designs. For more information on intersection problems, the interested reader may also refer to [3, 12, 14, 21, 26 28]. Let B be a simple graph. Denote by T(B) the set of all triangles of the graph B. For example, if B is the graph with vertices a, b, c, d and edges ab, ac, bc, cd (such a graph is called a kite), then T(B) = {{a,b,c}}. Two (K v,g)-packings (or coverings) (X,B 1 ) and (X,B 2 ) are said to intersect in t triangles provided that T(B 1 ) T(B 2 ) = t, where T(B i ) = B B i T(B), i = 1, 2. The triangle intersection problem for (K v,g)-packings (or coverings) is the determination of all integer pairs (v,t) such that there exists a pair of (K v,g)-packings (or coverings) intersecting in t triangles. Lindner and Yazici [25] first consideredthe triangleintersectionproblem, and gavea complete solution to the triangle intersection problem for kite systems. Billington et al. [6] solved the triangle intersection problem for (K 4 e)-designs. Chang et al. [7 8] investigated the triangle intersection problems for S(2, 4, v)s. Every block B in a (K v,g)-packing (or covering) contributes T(B) = T(G) triangles. If two (K v,g)-packings (or coverings) (X,B 1 ) and (X,B 2 ) intersect in s blocks, then they intersect in at least s T(G) triangles. It is natural to ask how about the triangle intersection problem for a pair of (K v,g)-packings (or coverings) intersecting in s blocks. Thus the fine triangle intersection problem was introduced in [10]. Define Fin G (v) = {(s,t) : a pair of (K v,g)- packings (or coverings) intersecting in s blocks and t + s T(G) triangles}. The fine triangle intersection problem for (K v,g)-packings (or coverings) is to determine Fin G (v). Chang et al. has completely solved the fine triangle intersection problems for kite systems [10] and (K 4 e)- designs [9 11]. Recently, the authors investigated the fine triangle intersection problem for maximum kite packings [29]. In this paper we shall examine the fine triangle intersection problem for minimum kite coverings. A (K v,g)-covering (or packing) (X,B) is called minimum (or maximum) if there does not exist any (K v,g)-covering (or packing) (X,B ) with B < B (or B < B ). When G is a kite, a minimum (K v,g)-covering (or maximum (K v,g)-packing) is said to be a minimum kite covering (or maximum kite packing). In a kite covering, every block B contributes only one triangle, so we define Fin(v) = {(s,t) : a pair of minimum kite coverings of order v intersecting in s blocks and t+s triangles}. The fine triangle intersection problem for minimum kite coverings is to determine Fin(v). It is easy to see that the number of blocks in a minimum kite covering is v(v 1) 8 at least. Let Adm(v) = {(s,t) : s+t b v,s,t are nonnegative integers}, where b v = v(v 1) 8. The following result is straightforward. Lemma 0.1 Fin(v) Adm(v) for any integer v 2,3,,7(mod 8) and v 4.

3 678 «42Æ As the main result of the present paper, we are to prove the following theorem. Theorem 0.2 Fin(v) = Adm(v)\{(b v 1,0),(b v 1,1)} for any integer v 0,1(mod 8) and v 8; Fin(v) = Adm(v) for any integer v 2,3,,7(mod 8) and v 4. 1 Small Orders In this section, we shall examine Fin(v) for 4 v 23 and v 8,9,16,17 by ad hoc methods. In the following we always denote the copy of the kite with vertices a,b,c,d and edges ab,ac,bc,cd by [a,b,c d]. Lemma 1.1 Fin(4) = Adm(4). Proof Take the vertex set X = {0,1,2,3}. Let B = {[0,1,3 2],[2,1,0 3]}. Then (X,B) is a minimum kite covering of order 4. Consider the following permutations on X. π 0,0 = (0 3)(1 2), π 0,1 = ( ), π 0,2 = (2 3), π 1,0 = (1 2), π 1,1 = (0 1), π 2,0 = (1). We have that for each (s,t) Adm(4), π s,t B B = s and T(π s,t B \B) T(B\π s,t B) = t. Lemma 1.2 Fin(5) = Adm(5). Proof Take the vertex set X = {0,1,2,3,4}. Let B = {[0,4,3 1],[0,1,2 3],[4,2,1 3]}. Then (X,B) is a minimum kite covering of order 5. Consider the following permutations on X. π 0,0 = (0 4)(2 3), π 0,1 = (0 1)(3 4), π 0,2 = (0 4 3)(1 2), π 0,3 = (1 2), π 1,0 = ( ), π 1,1 = ( ), π 1,2 = (0 4), π 2,0 = (0 2)(1 3), π 2,1 = (0 4)(1 2), π 3,0 = (1). We have that for each (s,t) Adm(5), π s,t B B = s and T(π s,t B \B) T(B\π s,t B) = t. Lemma 1.3 Fin(6) = Adm(6). Proof Take the vertex set X = {0,1,2,3,4,5}. Let B 1 = {[1,2,3 0],[3,4,5 2],[5,0,1 4],[0,2,4 1]}, B 2 = (B 1 \ {[5,0,1 4],[0,2,4 1]}) {[1,5,0 2],[0,1,4 2]}, and B 3 = (B 2 \{[0,1,4 2]}) {[1,2,4 0]}. Then (X,B i ) is a minimum kite coveringof order 6, i = 1,2,3. Consider the following permutations on X. π 0,0 = ( ), π 0,1 = (0 4)(2 3), π 0,2 = (0 1 4)(2 5), π 0,3 = (0 2 5)(1 3 4), π 0,4 = (0 3)(2 5), π 1,0 = (0 3 5)(1 4), π 1,1 = (3 5), π 1,2 = (0 4 2)(1 5 3), π 1,3 = (0 4 5)(1 2 3), π 2,0 = (3 4), π 2,1 = (0 5 4)(1 3), π 2,2 = ( )(1 4), π 3,0 = (1), π 3,1 = (0 4 2)(1 5 3), π 4,0 = (1). Let M 1 = Adm(6)\{(0,3),(1,0),(1,2),(2,0),(2,1),(3,0)} and M 2 = {(0,3),(1,0),(1,2),(2,0), (2,1)}. We have that for each (s,t) M i, i = 1,2, π s,t B i B i = s and T(π s,t B i \B i ) T(B i \ π s,t B i ) = t. For (s,t) = (3,0), π s,t B 3 B 2 = s and T(π s,t B 3 \B 2 ) T(B 2 \π s,t B 3 ) = t. Lemma 1.4 Fin(7) = Adm(7). Proof TakethevertexsetX = {0,1,2,3,4,5,6}. LetB 1 = {[0,2,1 6],[2,4,3 1],[1,4,5 0],[2,6,5 3],[6,4,0 3],[2,6,3 5]}, B 2 = (B 1 \ {[2,6,3 5]}) {[3,6,4 1]}, B 3 = (B 1 \ {[1,4,5 0],[2,6,5 3]}) {[1,4,5 3],[2,6,5 0]}, B 4 = (B 1 \ {[6,4,0 3],[2,6,3 5]}) {[6,3,0 4],[6,5,4 3]}, and B 5 = (B 1 \{[2,4,3 1],[2,6,3 5]}) {[2,4,3 6],[1,3,4 5]}. Let B 6 = {[2,0,1 3],[2,4,3 0],[1,4,5 3],[2,6,5 0],[4,0,6 1],[2,6,3 1]}, and B 7 = (B 6 \{[2,6,3 1]}) {[2,6,3 5]}, Then (X,B i ) is a minimum kite covering of order 7 for each i = 1,2,,7. Consider the following permutations on X.

4 5,, : The Fine Triangle Intersection Problem for Minimum Kite Coverings 679 π 0,0 = (0 5)(1 3 4)(2 6), π 0,1 = (0 4)(2 6 3), π 0,2 = ( ), π 0,3 = (0 1 2)(3 4 6), π 0,4 = (3 5)(4 6), π 0,5 = (1 6)(2 4), π 1,0 = ( )(3 5 6), π 1,1 = ( ), π 1,2 = ( )(2 3 4), π 1,3 = (0 2 4)(1 3 6), π 1,4 = ( )(2 4), π 2,0 = (0 1)(2 6)(3 5), π 2,1 = ( )(2 4), π 2,2 = ( ), π 2,3 = (0 6)(1 5), π 3,0 = (1 4), π 3,1 = (3 5), π 3,2 = (0 5)(2 4), π 3,3 = (0 5)(2 4). Let N = {(0,6),(1,5),(2,4),(4,0),(4,1),(4,2),(5,0),(5,1),(6,0)}, and take the identity permutation π s,t = (1) if (s,t) N. For each i,j,s,t in Table 1, B i π s,t B j = s and T(B i \π s,t B j ) T(π s,t B j \B i ) = t. Table 1 Fine triangle intersections for minimum kite coverings of order 7 i j (s, t) i j (s, t) 1 1 (0,0),,(0,5),(1,0),,(1,4), 1 2 (3,3),(5,0) (2,0),(2,1),(2,2),(3,0),(3,1),(3,2),(6,0) 1 3 (4,2) 1 4 (4, 0) 1 5 (4, 1) 1 6 (0,6),(2,3) 1 7 (1,5) 6 3 (2, 4) 6 7 (5, 1) Lemma 1.5 Fin(10) = Adm(10). Proof Take the vertex set X = {0,1,,9}. Let B 1 : [2,0,1 9], [4,7,0 5], [1,4,8 2], [1,7,5 8], [2,4,3 1], [7,8,6 5], [2,9,5 3], [9,0,8 3], [0,3,6 1], [6,9,4 5], [9,3,7 2], [2,6,3 1]; B 2 : [2,0,1 3], [4,0,7 2], [2,3,4 5], [6,4,9 1], [9,7,3 5], [1,7,5 0], [2,9,5 8], [0,3,6 5], [7,8,6 1], [1,4,8 3], [9,0,8 2], [2,6,3 0]. Take A i and A i for 3 i 14 as follows: i A i A i 3 [2,6,3 1] [2,6,4 1] 4 [2,6,3 1] [2,6,3 0] 5 [7,8,6 5] [2,9,5 3] [9,3,7 2] [9,2,7 8] [9,5,3 7] [2,5,6 7] [2,6,3 1] [6,8,1 0] 6 [2,9,5 3] [9,3,7 2] [9,3,5 2] [9,2,7 3] 7 [1,4,8 2] [9,0,8 3] [1,4,8 3] [9,0,8 2] 8 [1,4,8 2] [9,0,8 3] [2,6,3 1] [1,4,8 3] [9,0,8 2] [2,6,3 0] 9 [1,4,8 2] [9,0,8 3] [7,8,6 5] [1,4,8 3] [9,0,8 2] [7,8,6 1] [0,3,6 1] [0,3,6 5] 10 [1,4,8 2] [1,7,5 8] [7,8,6 5] [1,4,8 3] [1,7,5 3] [7,8,6 1] [2,9,5 3] [9,0,8 3] [0,3,6 1] [2,9,5 8] [9,0,8 2] [0,3,6 5] 11 [7,8,6 5] [2,6,3 1] [7,8,6 2] [5,6,4 1] 12 [1,4,8 2] [9,0,8 3] [2,6,3 1] [1,4,8 3] [9,0,8 2] [2,6,4 1] 13 [2,0,1 9] [2,4,3 1] [9,0,8 3] [2,0,1 3] [2,4,3 8] [8,0,9 1] [2,6,3 1] [2,6,4 1] 14 [2,0,1 9] [2,4,3 1] [9,0,8 3] [2,0,1 3] [2,4,3 8] [8,0,9 1] [2,9,5 3] [1,7,5 8] [2,6,3 1] [2,9,5 8] [1,7,5 3] [2,6,4 1] Let B i = (B 1 \ A i ) A i, 3 i 14. Then (X,B i) is a minimum kite covering of order 10 for each 1 i 14. Consider the following permutations on X. π 0,0 = ( )(2 8)(5 7), π 0,1 = (0 8 4)( ), π 0,2 = (0 7 1)(2 4)( ), π 0,3 = ( )( ), π 0,4 = ( ), π 0,5 = ( )(2 8 7),

5 680 «42Æ π 0,6 = (0 8 7)(1 3)(2 5)(4 9 6), π 0,7 = ( )(6 7 8), π 0,8 = (0 9 7)(1 2)(3 4 8), π 0,9 = (0 7)(2 5)(3 8), π 0,10 = (0 9)(1 5)(3 4), π 1,0 = ( ), π 1,1 = ( )(1 3)(2 8 6), π 1,2 = (0 3 8)( ), π 1,3 = ( ), π 1,4 = ( )(3 6 7), π 1,5 = ( )(1 5)(4 9), π 1,6 = (0 8 9)( )(3 6 7), π 1,7 = ( ), π 1,8 = ( )(1 5 2)(3 8), π 1,9 = (0 9)(1 5)(3 4), π 1,10 = (0 9)(1 5)(3 4), π 2,0 = ( ), π 2,1 = ( ), π 2,2 = (0 4 1)(6 8), π 2,3 = ( )(2 7)(3 5), π 2,4 = (1 3 5)(4 9)(7 8), π 2,5 = ( ), π 2,6 = ( )(1 2 5)(3 4), π 2,7 = (1 5), π 2,8 = (2 3), π 3,0 = ( ), π 3,1 = (2 4 6), π 3,2 = (0 8 1), π 3,3 = (0 3)(7 8), π 3,4 = (4 9)(7 8), π 3,5 = (0 9)(6 7), π 3,6 = (0 2)(3 6)(4 7)(5 8), π 3,7 = (0 9)(1 5)(3 4), π 3,8 = (0 9)(1 5)(3 4), π 4,0 = (4 6 7), π 4,1 = (4 6)(5 8), π 4,2 = (0 5), π 4,3 = ( )(1 2), π 4,4 = (5 9), π 4,5 = (0 8)(4 6), π 4,6 = (0 9)(1 5)(3 4), π 4,7 = (0 9)(1 5)(3 4), π 5,0 = (1 6), π 5,1 = (2 4), π 5,2 = (1 4), π 5,3 = (5 9), π 5,4 = (5 7), π 5,5 = (5 7), π 6,0 = (0 9), π 6,1 = (8 9), π 6,2 = (5 7), π 6,3 = (2 3), π 6,4 = (2 3), π 7,0 = (7 8), π 7,1 = (4 6), π 7,2 = (5 7), π 7,3 = (5 7), π 8,1 = (2 3), π 8,2 = (5 7). Let S 1 = {(x,y) : x+y 12,x {8,9,10,11},y is a nonnegative integer}\{(8,1),(8,2)}, and S 2 = {(0,11),(0,12),(1,11),(2,9),(2,10),(3,9),(4,8),(5,6),(5,7),(6,5),(6,6),(7,4),(7,5), (12,0)}. For each (s,t) S 1 S 2, take the identity permutation π s,t = (1). For each i,j,s,t in Table 2, B i π s,t B j = s and T(B i \π s,t B j ) T(π s,t B j \B i ) = t. Table 2 Fine triangle intersections for minimum kite coverings of order 10 i j (s, t) i j (s, t) i j (s, t) 1 1 (0,0),,(0,10),(1,0),,(1,8), 1 2 (0,12), (3,7) 1 3 (11,0) (2,0),, (2,7), (3,0),, (3,6), 1 4 (11,1) 1 5 (8,0), (5,4) (4,0),, (4,5), (5,1), (5,2), 1 6 (10,0), (6,4), (8,2) 1 7 (10,2) (6,0), (6,2), (7,0), (7,1), (12,0) 1 9 (8,4) 1 10 (6,6) 1 11 (10, 1) 1 12 (9, 2) 1 13 (8, 3) 1 14 (6,5) 2 2 (1,9), (5,0), (5,3), (6,1) 2 3 (0,11), (3,8) 2 4 (1, 11) 2 6 (2, 8) 2 7 (2, 10), (4, 6) 2 8 (3, 9) 2 9 (4, 8) 2 10 (5, 7) 2 12 (2,9), (4,7) 3 6 (9,0), (6,3), (7,2) 3 9 (1,10), (7,4) 3 10 (5, 6) 4 7 (9, 3) 4 9 (7, 5) 6 4 (9, 1), (7, 3) 6 8 (5, 5) 6 12 (8, 1) For counting Fin(v) for 11 v 23 and v 16,17, we may search for a large number of instances of minimum kite coverings of order v as we have done in Lemma 1.5. However, to reduce the computation, when v = 19, 20, we shall first determine the fine triangle intersection numbers of a pair of minimum (K v \ K 9,G)-coverings with the same vertex set and the same subgraph K 9 removed, where G is a kite. When v = 22, in [29], we have determined the fine triangle intersection numbers of a pair of kite-gdds of type with the same group set. When v 19,20,22, in [29], we have determined the fine triangle intersection numbers of a pair of (K v \K hv,g)-designs with the same vertex set and the same subgraph K hv removed, where G is a kite and 6, if v = 11,14, 5, if v = 12,13, h v = 7, if v = 15, 10, if v = 18,23, 12, if v = 21.

6 5,, : The Fine Triangle Intersection Problem for Minimum Kite Coverings 681 For later use these results are listed in Appendix. Lemma 1.6 Let M 19 = ({(s,t) : s+t 34,s {0,8},tis a nonnegativeinteger}\{(0,29), (0,30),, (0,33),(8,21),,(8,26)}) {(5,29),(10,24),(15,0),(15,19),(16,1),,(16,5),(16, 11),(17,0),(19,15),(21,0),(21,5),(22,6),(24,10),(26,0),(29,5),(32,2),(33,0),(33,1),(34,0)}. Let G be a kite and (s,t) M 19. Then there is a pair of minimum (K 19 \ K 9,G)-coverings with the same vertex set and the same subgraph K 9 removed, which intersect in s blocks and t+s triangles. Proof Take the vertex set X = {0,1,,18}. Let A 1 : [11,16,4 18], [14,9,4 15], [15,9,5 17], [16,10,5 18], [4,10,13 8], [5,11,13 17], [5,12,14 18], [12,4,17 3]; A 2 : [11,16,5 18], [14,9,5 15], [15,9,4 17], [16,10,4 18], [5,10,13 8], [4,11,13 17], [4,12,14 18], [12,5,17 3]; A 3 : [15,10,6 17], [16,9,6 18], [0,12,11 7], [2,10,11 8], [1,9,11 17], [14,6,11 18], [1,10,12 8], [2,9,12 7], [3,9,13 7], [6,12,13 18], [0,13,14 7], [3,10,14 8], [1,13,15 7], [2,14,15 8], [3,11,15 17], [0,15,16 7], [1,14,16 8], [2,13,16 17], [3,12,16 18], [7,9,17 1], [8,10,17 2], [0,18,17 14], [15,12,18 1], [7,10,18 2], [8,9,18 3], [0,9,10 1]; A 4 : [11,16,4 15], [14,9,4 18], [15,9,5 18], [16,10,5 17], [15,10,6 18], [16,9,6 17], [0,12,11 8], [2,10,11 7], [1,9,11 18], [14,6,11 17], [1,10,12 7], [2,9,12 8], [3,9,13 8], [4,10,13 7], [5,11,13 18], [6,12,13 17], [0,13,14 18], [3,10,14 7], [5,12,14 8], [1,13,15 17], [2,14,15 7], [3,11,15 8], [0,15,16 8], [1,14,16 7], [2,13,16 18], [3,12,16 17], [7,9,17 2], [8,10,17 1], [12,4,17 14], [0,18,17 3], [15,12,18 3], [7,10,18 1], [8,9,18 2], [0,9,10 2]. Let B 1 = A 1 A 3, B 2 = (B 1 \{[11,16,4 18],[14,9,4 15]}) {[11,16,4 15],[14,9,4 18]}, B 3 = (B 2 \{[15,9,5 17],[16,10,5 18],[0,9,10 1]}) {[15,9,5 18],[16,10,5 17],[0,9,10 2]}, B 4 = (B 3 \{[15,10,6 17],[16,9,6 18],[0,13,14 7],[3,10,14 8],[5,12,14 18]}) {[15,10,6 18],[16,9,6 17],[0,13,14 18],[3,10,14 7],[5,12,14 8]}, and B 5 = (B 4 \ {[5,11,13 17],[6,12,13 18],[1,13,15 7],[2,14,15 8],[3,11,15 17]}) {[5,11,13 18],[6,12,13 17],[1,13,15 17],[2,14,15 7],[3,11,15 8]}. Let B 6 = A 2 A 3, B 7 = A 4, B 8 = (B 1 \ {[0,9,10 1]}) {[0,9,10 2]} and B 9 = (B 1 \{[1,10,12 8]}) {[8,10,12 1]}. Then (X,B i ) is a minimum (K 19 \ K 9,G)-covering for each 1 i 9, where the removed subgraph K 9 is constructed on Y = {0,1,,8}. Now take the identity permutation π s,t = (1) if (s,t) {(0,26),(0,34),(5,29),(10,24),(15,19),(19,15),(24,10),(26,0),(29,5),(32,2),(33,0),(33,1),(34, 0)}. And consider the following permutations on X. π 0,0 = ( ), π 0,1 = ( )(14 17), π 0,2 = ( )(11 17)(14 16), π 0,3 = ( ), π 0,4 = ( )(12 18)( ), π 0,5 = (9 13)(10 12)(15 18)(16 17), π 0,6 = ( )(11 13)( ), π 0,7 = ( ), π 0,8 = ( )( ), π 0,9 = ( )( )(13 14), π 0,10 = ( )(10 13), π 0,11 = ( )(12 18)(13 16)(14 15), π 0,12 = (9 13)( )( ), π 0,13 = ( )( ), π 0,14 = ( )(13 16)(14 15), π 0,15 = ( )(11 12)(13 14),

7 682 «42Æ π 0,16 = (9 15)(10 16)( ), π 0,17 = ( )(10 11)(13 16), π 0,18 = (9 17)(10 16)(11 13)(12 14)(15 18), π 0,19 = (9 12)(10 11)( )(14 15), π 0,20 = (9 10)( )(13 14)(15 16), π 0,21 = (9 12)(10 11)(13 16)( ), π 0,22 = (3 6)(4 5)(7 8)(9 12)(10 11)( ), π 0,23 = ( )(4 5)(9 12)(10 11)(17 18), π 0,24 = ( )(3 5)(9 11)(14 16)(17 18), π 0,25 = (0 2)(3 5)(4 7 8)(9 11)(14 16)(17 18), π 0,27 = (3 6)(4 5)(9 12)(10 11), π 0,28 = (2 8), π 8,0 = ( )( ), π 8,1 = ( )(1 3), π 8,2 = (0 7)(2 5 4)(15 17), π 8,3 = ( ), π 8,4 = ( ), π 8,5 = (0 7 5)(2 3 8), π 8,6 = (0 4)(2 5 7)(9 13), π 8,7 = ( )(14 16), π 8,8 = ( )(15 16), π 8,9 = ( )(11 12)(15 16)(17 18), π 8,10 = (0 1 4)(10 11)(14 15), π 8,11 = (1 2)(5 7)(11 12)(15 16)(17 18), π 8,12 = ( )(5 6)(11 12)(15 16)(17 18), π 8,13 = (3 7)(5 6)(9 10)(11 12), π 8,14 = (0 5)(1 2)(3 4)(7 8)(9 10)(15 16), π 8,15 = (3 8)(5 6)(9 10)(11 12)(17 18), π 8,16 = (0 7 8)(1 2)(5 6)(11 12)(15 16)(17 18), π 8,17 = (1 8 7)(3 4)(5 6)(9 10)(11 12), π 8,18 = ( )(5 6)(9 10)(11 12), π 8,19 = (1 2)(3 4 5)(7 8)(9 10)(15 16)(17 18), π 8,20 = (0 1)( )(7 8)( )(14 15), π 15,0 = (0 5)(11 13), π 16,1 = ( ), π 16,2 = (2 5)(4 6), π 16,3 = ( ), π 16,4 = ( ), π 16,5 = (0 6)(11 13), π 16,11 = (1 2)(5 6)(11 12)(15 16)(17 18), π 17,0 = ( ), π 21,0 = (0 6 5), π 21,5 = (1 2)(4 5), π 22,6 = (2 7). We have that for each row in Table 3, π s,t Y = Y, B i π s,t B j = s and T(B i \π s,t B j ) T(π s,t B j \ B i ) = t. Table 3 Fine triangle intersections for minimum (K 19 \ K 9,G)-coverings i j (s, t) i j (s, t) i j (s, t) 1 1 M 19 \{(0,26), 1 2 (32,2) 1 3 (29,5) (0, 27),(0, 28),(0, 34),(5, 29), 1 4 (24, 10) 1 5 (19, 15) (8,15),(8,19),(10,24),(15,19), 1 6 (26,0),(8,19),(21,5) 1 7 (0,27),(0,28),(0,34),(8,15) (19, 15),(21, 5),(24, 10),(26, 0), 1 8 (33, 1) 1 9 (33, 0) (29,5),(32,2),(33,0),(33,1)} 3 7 (5,29) 4 7 (10,24) 5 7 (15, 19) 6 7 (0, 26) Lemma 1.7 LetM 20 = ({(s,t) : s+t 39,s {0,39},tisanonnegativeinteger}\{(0,31), (0,35),(0,36),(0,37)}) {(7,32),(8,0),,(8,16),(8,21),(8,26),(13,26),(14,13),(14,20),(16,0),,(16,8),(18,12),(20,14),(21,8),(19,20),(20,19),(24,0),(24,3),(24,8),(26,6),(26,8),(26,13), (29,0),(32,7),(34,0),(38,0),(38,1)}. Let G be a kite and (s,t) M 20. Then there is a pair of minimum (K 20 \K 9,G)-coverings with the same vertex set and the same subgraph K 9 removed, which intersect in s blocks and t+s triangles. Proof Take the vertex set X = {0,1,,19}. Let B 1 : [2,13,9 6], [5,16,9 17], [1,13,10 7], [4,16,10 19], [0,9,11 6], [3,13,11 19], [0,10,12 6], [1,9,12 7], [2,16,12 19], [5,15,12 18], [1,11,14 5], [2,10,14 6], [3,9,14 7], [4,12,14 17], [0,13,15 19], [2,11,15 6], [3,10,15 7], [4,9,15 18], [0,14,16 7], [13,6,16 11], [1,15,17 2], [3,12,17 7], [4,11,17 13], [5,10,17 6], [1,16,18 0], [4,13,18 2], [5,11,18 7], [6,10,18 3], [0,17,19 1], [3,16,19 2], [5,13,19 4], [7,9,19 6], [18,14,19 8], [7,13,11 2], [8,9,18 17], [8,11,10 9], [8,13,12 11], [8,15,14 13], [8,17,16 15];

8 5,, : The Fine Triangle Intersection Problem for Minimum Kite Coverings 683 B 2 : [2,13,9 17], [5,16,9 6], [1,13,10 19], [4,16,10 9], [0,9,11 19], [3,13,11 6], [0,10,12 7], [1,9,12 6], [2,16,12 18], [5,15,12 11], [1,11,14 6], [2,10,14 5], [3,9,14 17], [4,12,14 13], [0,13,15 6], [2,11,15 19], [3,10,15 18], [4,9,15 7], [0,14,16 11], [13,6,16 15], [1,15,17 7], [3,12,17 2], [4,11,17 6], [5,10,17 13], [1,16,18 2], [4,13,18 0], [5,11,18 3], [6,10,18 17], [0,17,19 2], [3,16,19 1], [5,13,19 8], [7,9,19 4], [18,14,19 6], [7,13,11 3], [8,9,18 7], [8,11,10 7], [8,13,12 19], [8,15,14 7], [8,17,16 7]. Take A j and A j for 3 j 12 as follows: (l j,j) A j A j (1,3) [7,13,11 2] [1,13,7 11] (1,4) [7,13,11 2] [7,13,11 3] (1,5) [1,16,18 0] [4,13,18 2] [1,16,18 2] [4,13,18 0] [0,17,19 1] [3,16,19 2] [0,17,19 2] [3,16,19 1] [5,13,19 4] [7,9,19 6] [5,13,19 8] [7,9,19 4] [18,14,19 8] [18,14,19 6] (1,8) [1,13,10 7] [0,9,11 6] [0,13,10 7] [1,9,11 6] [0,10,12 6] [1,9,12 7] [1,10,12 6] [0,9,12 7] [1,11,14 5] [0,13,15 19] [0,11,14 5] [1,13,15 19] [0,14,16 7] [1,15,17 2] [1,14,16 7] [0,15,17 2] [1,16,18 0] [0,17,19 1] [0,16,18 1] [1,17,19 0] (1,10) [8,9,18 17] [8,11,10 9] [8,17,18 9] [8,9,10 11] [8,13,12 11] [8,15,14 13] [8,11,12 13] [8,13,14 15] [8,17,16 15] [8,15,16 17] (5,6) [3,10,15 7] [4,9,15 18] [3,10,15 18] [4,9,15 7] [1,15,17 2] [3,12,17 7] [1,15,17 7] [3,12,17 2] [4,11,17 13] [5,10,17 6] [4,11,17 6] [5,10,17 13] (6,7) [2,13,9 6] [5,16,9 17] [2,13,9 17] [5,16,9 6] [0,9,11 6] [3,13,11 19] [0,9,11 19] [3,13,11 6] [0,13,15 19] [2,11,15 6] [0,13,15 6] [2,11,15 19] (8,9) [2,13,9 6] [5,16,9 17] [2,13,9 17] [5,16,9 6] [2,16,12 19] [5,15,12 18] [2,16,12 18] [5,15,12 11] [8,13,12 11] [1,13,15 19] [8,13,12 19] [1,13,15 6] [2,11,15 6] [3,10,15 7] [2,11,15 19] [3,10,15 18] [4,9,15 18] [4,9,15 7] (10,11) [0,13,15 19] [2,11,15 6] [0,13,15 6] [2,11,15 19] [3,10,15 7] [4,9,15 18] [3,10,15 18] [4,9,15 7] [1,15,17 2] [3,12,17 7] [1,15,17 7] [3,12,17 2] [4,11,17 13] [5,10,17 6] [4,11,17 6] [5,10,17 13] (11,12) [0,17,19 1] [3,16,19 2] [0,17,19 2] [3,16,19 1] [5,13,19 4] [7,9,19 6] [5,13,19 8] [7,9,19 4] [18,14,19 8] [7,13,11 2] [18,14,19 6] [7,13,11 3] Let B j = (B lj \A j ) A j, where 3 j 12 and (l j,j) runs over each value in the first column of the above table. It is readily checked that (X,B i ) is a minimum (K 20 \K 9,G)-covering for each 1 i 12, where the removed subgraph K 9 is constructed on Y = {0,1,,8}. Now assume that

9 684 «42Æ π 0,0 = ( )( ), π 0,1 = ( )( ), π 0,2 = ( )(15 17)(18 19), π 0,3 = (9 11)( )(15 19)(17 18), π 0,4 = ( )( ), π 0,5 = ( )(11 13)( ), π 0,6 = ( )(15 19)( ), π 0,7 = ( )( )( ), π 0,8 = (9 11)(10 14)(12 13)( )(16 18), π 0,9 = ( )( ), π 0,10 = (9 11)(10 13)(12 14)( ), π 0,11 = (9 12)(10 11)(13 14)(15 16)(17 19), π 0,12 = ( )( )( ), π 0,13 = (9 11)(10 13)(15 17)(16 18), π 0,14 = (9 11)(10 13)(12 14)( )(16 18), π 0,15 = (9 11)(10 13)(12 14)(16 18)(17 19), π 0,16 = ( )( ), π 0,17 = (4 7 8)(5 6)(16 19)(17 18), π 0,18 = (4 5)(6 7 8)(16 17)(18 19), π 0,19 = ( )( ), π 0,20 = (4 5)(10 12), π 0,21 = (2 4)(9 12), π 0,22 = ( ), π 0,23 = ( ), π 0,24 = (1 5 4), π 0,25 = (12 13), π 0,26 = (9 14), π 0,27 = (0 1 6), π 0,28 = (2 5 6), π 0,29 = (1 4), π 0,30 = (2 4), π 0,32 = (1 6), π 0,33 = (2 6), π 8,0 = ( )(9 10), π 8,1 = (1 8 4)(3 5)(16 18), π 8,2 = (0 2)(1 4 3)( ), π 8,3 = (1 2 7)(3 4)(10 14), π 8,4 = (0 2)(1 7 3)(4 6), π 8,5 = (0 3)(1 2 6)(10 13), π 8,6 = (1 5 4)(3 8)(17 19), π 8,7 = (0 6)(3 5)(17 18), π 8,8 = ( )(12 13), π 8,9 = (0 4)(1 7 8)(10 14), π 8,10 = (5 6)(10 12)(16 19), π 8,11 = (3 7 8)( ), π 8,12 = (1 6 7)(16 18), π 8,13 = (3 6 7)(4 8), π 8,14 = (2 8 7)(18 19), π 8,15 = (5 6)(16 19)(17 18), π 8,16 = (4 7 6)(16 18), π 14,13 = (2 4)(10 12)(15 17), π 16,0 = ( ), π 16,1 = ( ), π 16,2 = (0 1)(16 18), π 16,3 = (1 4)(12 13), π 16,4 = ( ), π 16,5 = (2 5)(11 13), π 16,6 = (1 7)(12 14), π 16,7 = (2 6)(3 5), π 16,8 = ( ), π 18,12 = (5 6 7), π 24,0 = (15 16), π 24,3 = (0 1)(16 18), π 24,8 = (4 7), π 26,6 = (4 7). Take π s,t = (1) if (s,t) {(0,34),(0,38),(0,39),(7,32),(8,21),(8,26),(13,26),(14,20),(20,14), (21,8),(19,20),(20,19),(26,8),(26,13),(29,0),(32,7),(34,0),(38,0),(38,1),(39,0)}. It is readily checked that for each row in Table 4, π s,t Y = Y, B i π s,t B j = s and T(B i \ π s,t B j ) T(π s,t B j \B i ) = t. Table 4 Fine triangle intersections for minimum (K 20 \ K 9,G)-coverings i j (s, t) i j (s, t) i j (s, t) 1 1 M 20 \{(0,20),,(0,30), 1 3 (38,0) 1 4 (38,1) (0,32),(0,33),(0,34),(0,38),(0,39), 1 5 (32,7) 1 6 (26,13) (7,32),(8,21),(8,26),(13,26), 1 7 (20,19) 1 8 (24,3),(29,0) (14,20),(16,0),(16,4),(16,6), 1 9 (21,8) 1 10 (16,0),(34,0) (19, 20),(20, 14),(20, 19),(21, 8), 1 11 (26, 8) 1 12 (20, 14) (24,3),(26,6),(26,8),(26,13), 2 2 (16,4),(16,6),(26,6) 2 3 (0,38) (29,0),(32,7),(34,0),(38,0),(38,1)} 2 5 (7,32) 2 6 (13,26) 2 7 (19, 20) 2 9 (8, 21) 1 2 (0,20),,(0,30),(0,32), 2 10 (0,34) 2 11 (8,26) (0, 33),(0, 39) 2 12 (14, 20) Lemma 1.8 Fin(v) = Adm(v) for 11 v 23 and v 16,17,22. Proof By Lemma 0.1, Fin(v) Adm(v). We need to show that Adm(v) Fin(v). When v = 21, our proof will rely on the fact that Fin(12) = Adm(12). Thus one can first make use of the following procedure to obtain Fin(v) = Adm(v) for 11 v 23 and v 16,17,21,22. Then the same procedure guarantees Fin(21) = Adm(21). For each 11 v 23 and v 16,17,22, take the corresponding M v from Lemmas

10 5,, : The Fine Triangle Intersection Problem for Minimum Kite Coverings 685 and from Table 5 in Lemma 3.1 in Appendix. Let G be a kite and (α v,β v ) M v. Let 6, if v = 11,14, 5, if v = 12,13, 7, if v = 15, h v = 9, if v = 19,20, 10, if v = 18,23, 12, if v = 21. By Lemmas and Lemma 3.1 in Appendix, there is a pair of minimum (K v \ K hv,g)- coverings(or(k v \K hv,g)-designs)(x,b (v) 1 )and(x,b(v) 2 )withthesamesubgraphk h v removed, which intersect in α v blocks and β v + α v triangles. Here the subgraph K hv is constructed on Y X. Let (γ v,ω v ) Adm(h v ) when v 19,20, and (γ v,ω v ) Adm(9)\{(8,0),(8,1)} when v = 19,20. Since [10] shows that Fin(9) = Adm(9) \ {(8,0),(8,1)}, combining the results of Lemmas , we have a pair of minimum kite coverings of order h v (Y,B (v) 1 ) and (Y,B (v) 2 ), with γ v common blocks and γ v + ω v common triangles. It follows that (X,B (v) 1 B (v) 1 ) and (X,B (v) 2 B (v) 2 ) are both minimum kite coverings of order v with α v +γ v common blocks and α v +β v +γ v +ω v common triangles. Thus we have, when v 19,20, and when v = 19,20, Fin(v) {(α v +γ v,β v +ω v ) : (α v,β v ) M v,(γ v,ω v ) Adm(h v )}, Fin(v) {(α v +γ v,β v +ω v ) : (α v,β v ) M v,(γ v,ω v ) Adm(9)\{(8,0),(8,1)}}. It is readily checked that for any pair of integers (s,t) Adm(v), we have (s,t) Fin(v). Lemma 1.9 Fin(22) = Adm(22). Proof Take the same set M 22 as in Lemma 3.2in Appendix. Let (α,β) M 22. By Lemma 3.2inAppendix, thereisapairofkite-gddsoftype withthesamegroupset, whichintersect in α blocks and β +α triangles. Let (γ 1,ω 1 ) and (γ 2,ω 2 ) Adm(8)\{(6,0),(6,1)}. It is shown in [10] that there is a pair of kite systems of order 8 intersecting in γ i common blocks and ω i + γ i common triangles for i = 1,2. Let (γ 3,ω 3 ) Adm(6). By Lemma 1.3, there is a pair of minimum kite covering of order 6 with γ 3 common blocks and ω 3 +γ 3 common triangles. So we obtain a pair of minimum kite covering of order 22 with α + 3 γ i common blocks and β + 3 ω i +(α+ 3 γ i) common triangles. Thus we have {( ) 3 3 Fin(22) α+ γ i,β + ω i : (α,β) M 22,(γ 3,ω 3 ) Adm(6), (γ 1,ω 1 ),(γ 2,ω 2 ) Adm(8)\{(6,0),(6,1)} }. It is readily checked that for any pair of integers (s,t) Adm(22), we have (s,t) Fin(22). 2 Main Theorem In this section we shall establish the main theorem of this paper. First we need to introduce some auxiliary designs.

11 686 «42Æ Let K be a set of positive integers. A group divisible design (GDD) K-GDD is a triple (X,G,A) satisfying the following properties: (1) G is a partition of a finite set X into subsets (called groups); (2) A is a set of subsets of X (called blocks), each of cardinality from K, such that every 2-subset of X is either contained in exactly one block or in exactly one group, but not in both. If G contains u i groups of size g i for 1 i r, then we call g u1 1 gu2 2 gur r the group type (or type) of the GDD. If K = {k}, we write a {k}-gdd as a k-gdd. We quote the following result for later use. Lemma 2.1 [16] Let g, t and u be nonnegative integers. There exists a 3-GDD of type g t u 1 if and only if the following conditions are all satisfied: (1) if g > 0, then t 3, or t = 2 and u = g, or t = 1 and u = 0, or t = 0; (2) u g(t 1) or gt = 0; (3) g(t 1)+u 0(mod 2) or gt = 0; (4) gt 0(mod 2) or u = 0; (5) g2 t(t 1) 2 +gtu 0(mod 3). Let H = {H 1,H 2,,H m } be a partition of a finite set X into subsets (called holes), where H i = n i for 1 i m. Let K n1,n 2,,n m be the complete multipartite graph on X with the i-th part on H i, and G be a subgraph of K n1,n 2,,n m. A holey G-design is a triple (X,H,B) such that (X,B) is a (K n1,n 2,,n m,g)-design. The hole type (or type) of the holey G-design is {n 1,n 2,,n m }. We use an exponential notation to describe hole types: the hole type g u1 1 gu2 2 gur r denotes u i occurrences of g i for 1 i r. Obviously if G is the complete graph K k, a holey K k -design is just a k-gdd. When G is kite, a holey G-design is said to be a kite-gdd. A pair of holey G-designs (X,H,B 1 ) and (X,H,B 2 ) of the same type is said to intersect in s blocks if B 1 B 2 = s. A pair of holey G-designs (X,H,B 1 ) and (X,H,B 2 ) of the same type is said to intersect in t triangles if T(B 1 ) T(B 2 ) = t, where T(B i ) = B B i T(B), i = 1,2. Lemma 2.2 [10] Let (s,t) {(0,0),(0,1),(0,12),(12,0)}. There exists a pair of kite-gdds of type 4 3 with the same group set intersecting in s blocks and t+s triangles. Let (s 1,t 1 ) and (s 2,t 2 ) be two pairs of nonnegative integers. Define (s 1,t 1 ) + (s 2,t 2 ) = (s 1 +s 2,t 1 +t 2 ). If X and Y are two sets of pairs of nonnegative integers, then X +Y denotes the set {(s 1,t 1 )+(s 2,t 2 ) : (s 1,t 1 ) X,(s 2,t 2 ) Y}. If X is a set of pairs of nonnegative integers and h is some positive integer, then h X denotes the set of all pairs of nonnegative integers which can be obtained by adding any h elements of X together (repetitions of elements of X allowed). Lemma 2.3 For any integer v 2,3,,7,10,11,,15(mod 24) and v 4, Fin(v) = Adm(v). Proof The cases of v = 4,5,6,7,10,11,,15 follow from Lemmas and Lemma 1.8. Assume that v 24. By Lemma 0.1, we have Fin(v) Adm(v). We need to show that Adm(v) Fin(v). Let v = 8u+a with u 0,1(mod 3), u 3 and a {2,3,,7}. Start from a 3-GDD of type 2 u from Lemma 2.1. Give each point of the GDD weight 4. By Lemma 2.2, there is a pair of kite-gdds of type 4 3 with α common blocks and α+β common triangles, (α,β) {(0,0),(0,12),(12,0)}. Then there exists a pair of kite-gdds of type 8 u which intersect in x α i blocks and x (α i + β i ) triangles, where x = 2u(u 1) 3 is the number of

12 5,, : The Fine Triangle Intersection Problem for Minimum Kite Coverings 687 blocks of the 3-GDD of type 2 u and (α i,β i ) {(0,0),(0,12),(12,0)} for 1 i x. Define A(8+c) = A 8+c, where c {2,3,4} and A 8+c is taken from Table 6 of Lemma 3.3 in Appendix. When c {5,6,7}, define A(8+c) = M 8+c, where M 8+c is taken from Table 5 of Lemma 3.1 in Appendix. By Lemmas 3.1 and 3.3, there is a pair of (K 8+a \K a,g)-designs with the same vertex set and the same subgraph K a removed with γ j common blocks and γ j + ω j commontriangles, where(γ j,ω j ) A(8+a), 1 j u 1, anda {2,3,,7}. ByLemmas1.5 and1.8, thereisapairofminimum kitecoveringoforder8+awithγ u commonblocksandγ u +ω u common triangles, (γ u,ω u ) Adm(8+a), then there exists a pair of minimum kite covering of order v = 8u+a with x α i + u j=1 γ j common blocks and x (α i +β i )+ u j=1 (γ j +ω j ) common triangles. Thus we have {( x ) u x u Fin(v) α i + γ j, β i + ω j : (α i,β i ) {(0,0),(0,12),(12,0)},1 i x, j=1 j=1 (γ j,ω j ) A(8+a),1 j u 1,(γ u,ω u ) Adm(8+a) { x = (α i,β i )+ u (γ j,ω j ) : (α i,β i ) {(0,0),(0,12),(12,0)},1 i x, j=1 (γ j,ω j ) A(8+a),1 j u 1,(γ u,ω u ) Adm(8+a) =x {(0,0),(0,12),(12,0)}+(u 1) A(8+a)+Adm(8+a). Let n be the number of blocks of a (K 8+a \K a,g)-design. Then n = 7+2a. Let m be the number of blocks of a minimum kite covering of order 8+a. Define S(z) = {(s,t) : s+t z,s,t 0}. Then it is readily checked that (u 1) A(8+a)+Adm(8+a) = (u 1) A(8+a)+S(m) = (u 2) A(8+a)+[A(8+a)+S(m)] = (u 2) A(8+a)+S(m+n) = S((u 1)n+m). } } Note that {(0,0),(0,12),(12,0)}+S(z) = S(z +12) (1) for any integer z 22. Since (u 1)n+m 22, using Formula (1) inductively x times, we have Fin(v) (x 1) {(0,0),(0,12),(12,0)}+({(0,0),(0,12),(12,0)}+S((u 1)n+m)) = (x 1) {(0,0),(0,12),(12,0)}+S((u 1)n+m+12) = S((u 1)n+m+12x) = S(b v ) = Adm(v). This completes the proof. Lemma 2.4 For any integer v 18,19,,23(mod 24) and v 18, Fin(v) = Adm(v).

13 688 «42Æ Proof The cases of v = 18,19,,23 follow from Lemmas Assume that v 42. By Lemma 0.1, we have Fin(v) Adm(v). We need to show that Adm(v) Fin(v). Let v = 8u+a with u 2(mod 3), u 5 and a {2,3,,7}. Start from a 3-GDD of type 2 u from Lemma 2.1. Give each point of the GDD weight 4. By Lemma 2.2, there is a pair of kite-gdds of type 4 3 with α common blocks and α+β common triangles, (α,β) {(0,0),(0,12),(12,0)}. Then we obtain a pair of kite-gdds of type 8 u with x α i common blocks and x (α i + β i ) common triangles, where x = 2(u+1)(u 2) 3 is the number of blocks of the 3-GDD of type 2 u and (α i,β i ) {(0,0),(0,12),(12,0)} for 1 i x. Define A(8+c) = A 8+c, where c {2,3,4} and A 8+c is taken from Table 6 of Lemma 3.3 in Appendix. When c {5,6,7}, define A(8 + c) = M 8+c, where M 8+c is taken from Table 5 of Lemma 3.1 in Appendix. By Lemmas 3.1 and 3.3, there is a pair of (K 8+a \K a,g)-designs with the same vertex set and the same subgraph K a removed with γ j common blocks and γ j +ω j common triangles, where (γ j,ω j ) A(8+a), 1 j u 2, and a {2,3,,7}. By Lemmas , there is a pair of minimum kite coverings of order 16+a with γ u 1 common blocks and γ u 1 + ω u 1 common triangles, (γ u 1,ω u 1 ) Adm(16 + a). So we obtain a pair of minimum kite coverings of order v = 8u + a with x α i + u 1 j=1 γ j common blocks and x (α i +β i )+ u 1 j=1 (γ j +ω j ) common triangles. Thus we have Fin(v) {( x u 1 α i + γ j, j=1 x ) u 1 β i + ω j j=1 : (α i,β i ) {(0,0),(0,12),(12,0)},1 i x, (γ j,ω j ) A(8+a),1 j u 2,(γ u 1,ω u 1 ) Adm(16+a) { x u 1 = (α i,β i )+ (γ j,ω j ) : (α i,β i ) {(0,0),(0,12),(12,0)},1 i x, j=1 (γ j,ω j ) A(8+a),1 j u 2,(γ u 1,ω u 1 ) Adm(16+a) =x {(0,0),(0,12),(12,0)}+(u 2) A(8+a)+Adm(16+a). Next by similar arguments as those in Lemma 2.3, we have Adm(v) Fin(v). Proof of Theorem 0.2 When v 0, 1(mod 8) and v 8, it is shown in [10] that Fin(v) = Adm(v)\{(b v 1,0),(b v 1,1)}. When v 2,3,,7(mod 8) and v 4, combining the results of Lemmas , we have Fin(v) = Adm(v). 3 Appendix Lemma 3.1 [29] Let G be a kite and M v be defined in Table 5. Let (s,t) M v and 6, if v = 11,14, 5, if v = 12,13, h v = 7, if v = 15, 10, if v = 18,23, 12, if v = 21. } }

14 5,, : The Fine Triangle Intersection Problem for Minimum Kite Coverings 689 Then there is a pair of (K v \K hv,g)-designs with the same vertex set and the same subgraph K hv removed, which intersect in s blocks and t+s triangles. Table 5 Fine triangle intersections for (K v \ K hv,g)-designs v M v 11 {(s,t) : s+t 10,s {0,1,10},t is a nonnegative integer} {(4,6),(5,0),(5,2),(6,4),(7,0),(8,1)} 12 ({(s,t) : s+t 14,s {0,3,8,12,14},t is a nonnegative integer} \{(3,9),(3,10),(8,2),(8,4),(8,5),(12,1)}) {(4,10),(5,9),(6,0), (6,2),(6,5),(6,8),(7,4),(7,7),(9,5),(10,0),(10,2),(10,4)} 13 ({(s,t) : s+t 17,s {0,3,6,9,15,17},t is a nonnegative integer} \{(0,15),(3,12),(6,8),(6,9),(9,4),(9,5),(9,7),(15,1)}) {(2,15),(5,11), (7,7),(8,9),(10,5),(11,4),(11,6),(12,0),(12,3),(13,0),(13,2),(14,3)} 14 ({(s,t) : s+t 19,s {0,4,8,17,19},t is a nonnegative integer} \{(4,13),(8,8),(8,9),(8,10),(17,1)}) {(7,9),(7,11),(10,5),(10,9), (11,6),(12,0),(12,1),(12,2),(13,6),(14,0),(14,3),(15,2),(16,3)} 15 {(0,0),(0,1),,(0,16),(0,18),(0,21),(4,17),(6,0),(6,1),,(6,8),(6,12),(8,13), (9,8),(12,0),(12,4),(12,6),(12,9),(13,2),(13,8),(17,1),(17,4),(18,0),(21,0) 18 {(0,0),(0,1),,(0,17),(0,27),(27,0),(3,24),(6,21),(8,0),(9,18), (12,5),(12,15),(15,0),(15,12),(18,9),(21,0),(24,3)} 21 {(0,0),(0,1),,(0,19),(0,24),(0,36),(36,0),(6,30),(7,0),, (7,7),(12,5),(12,24),(13,11),(18,18),(24,0),(24,12),(30,6)} 23 {(0,0),(0,1),,(0,18),(7,0),(7,6),(7,7),(7,10),(10,6),(0,22), (0,26),(0,34),(0,42),(0,52),(3,49),(5,21),(5,29),(5,37),(6,46), (7,19),(9,43),(10,24),(10,32),(12,14),(12,40),(14,0),(14,12), (15,19),(15,27),(15,37),(18,8),(18,34),(19,15),(20,22),(21,5), (21,31),(22,0),(22,20),(24,10),(24,28),(26,0),(27,15),(27,25), (28,24),(29,5),(31,21),(32,10),(34,0),(34,18),(37,5),(37,15), (40,12),(42,0),(43,9),(46,6),(49,3),(52,0)} Lemma 3.2 [29] Let M 22 = {(0,0),(0,1),,(0,20),(10,0),(10,1),,(10,6),(13,7), (0,28),(0,40),(7,33),(11,17),(14,26),(17,11),(21,19),(26,14),(28,0),(33,7),(40,0)}. Let(s,t) M 22. Then there is a pair of kite-gdds of type with the same group set, which intersect in s blocks and t+s triangles. Lemma 3.3 [29] Let G be a kite and (v,a) {(10,2),(11,3),(12,4)}. Let A v be defined in Table 6 and (s,t) A v. Then there is a pair of (K v \K a,g)-designs with the same vertex set and the same subgraph K a removed, which intersect in s blocks and t+s triangles. Table 6 Fine triangle intersections for (K v \ K a,g)-designs (v,a) A v (10,2) {(s,t) : s+t 11,s {0,11},t is a nonnegative integer}\{(0,9)} (11,3) {(s,t) : s+t 13,s {0,13},t is a nonnegative integer}\{(0,10),(0,12)} (12,4) {(s,t) : s+t 15,s {0,15},t is a nonnegative integer}\{(0,12),(0,13)} References [1] Adams, P., Billington, E.J., Bryant, D.E. and Mahmoodian, E.S., The three-way intersection problem for Latin squares, Discrete Math., 2002, 243(1/2/3): [2] Baker, C., The intersection problem for Latin squares with holes of size 2 and 3, Ph.D. Thesis, Auburn: Auburn University, [3] Billington, E.J., The intersection problem for combinatorial designs, Congr. Numer., 1993, 92: [4] Billington, E.J., Gionfriddo, M. and Lindner, C.C., The intersection problem for K 4 e designs, J. Statist. Plann. Inference, 1997, 58(1): [5] Billington, E.J., and Kreher, D.L., The intersection problem for small G-designs, Australas. J. Combin., 1995, 12: [6] Billington, E.J., Yazici, E.S. and Lindner, C.C., The triangle intersection problem for K 4 e designs, Utilitas Math., 2007, 73: 3-21.

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