Ti設計的三角形交集問題

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(1)東吳大學數學系碩士班碩士論文. 指導教授:黃文中 教授. Ti 設計的三角形交集問題 The triangle intersection problem for Ti-design. 研究生:林穆佑 撰 中華民國一百零三年七月.

(2) 誌謝(Acknowledgment) 能夠完成這篇論文,首先要感謝我的家人與指導老師黃文 中教授,若沒有家人精神上的支持以及指導老師在學術上的指導 與關心,這份論文就沒有辦法順利完成。在此也要感謝蔡風順老 師與蔡漢彬老師在百忙之中撥空指導論文與口試。 在研究所的日子,我過得非常開心與充實,不論是課堂上 每位老師盡心盡力的教學,下課後參與系上的許多活動,都讓我 留下許多美好回憶。在此感謝盛吉助教、麗君助教、慧娟助教、 彥洲助教、育鋒學長,在我修課的時候時常請助教們幫忙解答課 業上的問題,才讓我的課程能夠順利過關,也感謝研究室的各位: 世揚、雅君、惠安、智凱、郁穎、育誌、仕寬、富翔、鈞元、鈺 麟、小萍老師、皓強、孟霖、承軒、思瑩、與祥、國祥、賢煜、 昱伶、和磬、保錡、韋廷以及志凱,還有系上的李秘書、任秘書 以及各位教授,非常感謝。 很開心能夠在東吳大學完成我的學士學位與碩士學位,不 管這個路程是辛苦的、是艱難的,但是這裡的環境與老師帶領著 我,讓我成功的取得學位,也讓我留下深刻的甜美回憶。 最後,我要將此篇論文獻給我的家人。.

(3) Abstract A Ti-design of G is a collection of subgraphs of G, those subgraph is isomorphic to Ti-graph, such that the collection of subgraphs from a partition of G. A Ti-design of order n is a Ti-design of Kn. A Ti-design of order n exists precisely when n ≡ 0, 1 (mod 5). Two Ti-designs (X, ℬ1) and (X, ℬ2) are said to intersect in m blocks if | ℬ1 ⋂ ℬ2|=m. Furthermore, Two Ti-design of order v, (X, ℬ1) and (X, ℬ2) are said to intersect in t triangles if | T(ℬ1) ⋂ T(ℬ2)|=t, where T(ℬi)=. ℬ. and. T(B) is the set of all triangles of the graph B.For i=1, 2, 3, let Ji(n)={0, 1, 2, ... ,. }, and Ii(n) be the set of integers k such. that there exists a pair of Ti-design of order n intersecting in k triangles. In this thesis, we prove that Ji(n)=Ii(n), for all n ≡ 0, 1 (mod 5)..

(4) 摘要 圖 G 的 Ti 設計是一些圖 G 的子圖所形成的集合,這些子圖與 Ti圖同構,使得這些子圖形成圖 G 的分割。一個 n 階 Ti-設計是一個完全 圖 Kn 的 Ti-設計。當 n≡0,1(mod 5),存在一個 n 階的 Ti-設計。兩個 Ti-設計(X, 1)和(X, 2)滿足|. 1. ∩. 2. |=m 時,我們說此兩設計交集 m. 個區塊。兩個 Ti-設計(X, 1)和(X, 2)滿足| T( 1) ∩T( 2)|=t,則我 們說此兩設計交集 t 個三角形,其中 T(ℬi)=. ,且 T(B)是圖. B 的三角形個數的一個集合。 對於 i=1、2、3 而言,令 Ji(n)={0, 1, 2, ... ,. },而. Ii(n)={k|存在一對 Ti-設計交集 k 個三角形},而在此篇論文裡,我們 要去證明在 n≡0,1(mod 5),Ii(n)= Ji(n)。.

(5) Contents 1 Introduction. 2. 1.1. Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2. 1.2. Designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 5. 2 Small Cases. 14. 2.1. Small order of T1 -design. . . . . . . . . . . . . . . . . . . . . . 15. 2.2. Small order of T2 -design. . . . . . . . . . . . . . . . . . . . . . 24. 2.3. Small order of T3 -design. . . . . . . . . . . . . . . . . . . . . . 34. 3 Main Theorem. 43. 1.

(6) 1. Introduction. 1.1. Graphs. Definition 1.1 A graph G is an ordered pair (V , E), where V is a set of points called vertices, denoted by V (G), and E is a collection of 2-element subsets of V called edges, denoted by E(G). If |V (G)| = p and |E(G)| = q, then we called G is a (p, q)-graph.. If e = {x, y} is an edge, denote by xy for convenience, then we say that the vertices x and y are adjacent and the edge e is incident with the vertex x. Furthermore, if e1 and e2 are distinct edges of G incident with a common vertex, then e1 and e2 are adjacent edges.. Example 1.2 Let V = {1, 2, 3, 4} and E = {{1, 4}, {1, 2}, {2, 3}, {1, 3}}. Then G = (V, E) is a graph (see Figure 1).. 1. t. t @ @ @ @. 2. 4. t. @t. 3. Figure 1: Graph of Example 1.1. 2.

(7) Example 1.3 T1 = ({a, b, c, d, e}, {ab, bc, ac, bd, ce}), denote this graph by (a, b, c, d, e)1 ; T2 = ({a, b, c, d, e}, {ab, bc, ac, cd, ce}), denote this graph by (a, b, c, d, e)2 ; T3 = ({a, b, c, d, e}, {ab, bc, ac, cd, de}), denote this graph by (a, b, c, d, e)3 . (see Figure 2). d y.    T1 : b y @ @ @ @ @y. ye d y ye JJ .  J.  J. y c. T2 : c Jy @ @ @ @ @y y. a a b Figure 2: Graphs T1 , T2 , and T3. e y d y T3 :    y. a. c y Q  Q. Q Q Qy. b. Definition 1.4 Let V = {u1 , u2 , . . . , un } and E = {{ui , uj } | 1 ≤ i < j ≤ n}, then (V, E) is a graph, called complete graph of order n and denoted by Kn .. 1. t Q  B Q   B Q  Q B   Qt t 2 l B , 3 D  , B ,  D ll  B , l D  B , D  l  D  ,,l B  l B  D , BBt l Dt,  l. 4. 5. Figure 3: Complete graph K5 3.

(8) Definition 1.5 Let V = {u1 , u2 , . . . , um } ∪ {v1 , v2 , . . . , vn }, and E = {{ui , vj } | 1 ≤ i ≤ m, 1 ≤ j ≤ n}, then (V, E) is a graph, called bipartite graph and denoted by Km,n . The set {u1 , u2 , . . . , um } and {v1 , v2 , . . . , vn } are called partite sets of the bipartite graph Km,n . u1. u2. u3. t t t B @ B  B@  B  B @  B  B @ B  B @ B  B  @ B  B t @B t B @B. v1. u1. u2. u3. v1. v2. v3. t t t Z  JZ. J. J Z J . J Z.  J.  J. J Z  Z. J Z J. J. ZJ  ZJ. J. t. Jt.  Z Jt. v2 K3,2. K3,3. Figure 4: Complete bipartite graph K3,2 and K3,3. Definition 1.6 Let G = (V, E) and G′ = (V ′ , E ′ ) be two graphs, then we say G′ is a subgraph of G if V ′ ⊆ V and E ′ ⊆ E.. s , l B ,  Bl s ls H ,  B  Q B Q    B B  Q  Q B  BB s QBs. s , ,  s,  B  B  BBs. s   s. s l B B lls s B  Q  Q Q B Q B s QBs. Figure 5: Two subgraph of K5. 4.

(9) 1.2. Designs. Definition 1.7 Let v, k, and λ be positive integers such that v > k ≥ 2. A (v, k, λ)-balanced incomplete block design (which we abbreviate to (v, k, λ)-BIBD) is an ordered pair (X, A), where X is a set of elements called points, and A is a collection of nonempty subsets of X called blocks such that the following properties are satisfied: 1. |X| = v. 2. eack block of A contains exactly k points of X, and 3. every pair of distinct points is contained in exactly λ blocks.. Theorem 1.8 In a (v, k, λ)-BIBD, every point occurs exactly in r blocks, where. r=. λ(v−1) . k−1. Theorem 1.9 A (v,k,λ)-BIBD has exactly b blocks, where. b=. vr k. =. λ(v 2 −v) . k2 −k. Example 1.10 X = {1, 2, 3, 4, 5, 6, 7}, and A = {123, 345, 156, 147, 257, 367, 246}, then (X, A) is a (7, 3, 1)-BIBD. (see Figure 6). 5.

(10) 1. t A  A A  A  A  2 t At 6 QQ 7 A Qt A  Q A  Q  Q  A  Q A  Q Q At  t t. 3. 4. 5. Figure 6: A (7, 3, 1)-BIBD. Definition 1.11 The Steiner triple system of order v is a (v, 3, 1)-design, denote by STS(v).. Example 1.12 Let X={1, 2, 3, 4, 5, 6, 7, 8, 9}, and A={123, 456, 789, 147, 258, 369, 159, 267, 348, 168, 249, 357}, then (X, A) is an STS(9).. Lemma 1.13 If there exists an STS(v), then v ≡ 1, 3(mod 6), v ≥ 7.. Proof: Since k = 3 and λ = 1, we have r = λ(v − 1)/(k − 1) = (v − 1)/2 . Hence v = 2r + 1; i.e. v is odd. From b = vr/k = v(v − 1)/6 and b ∈ Z, we have v(v − 1) ≡ 0(mod 6). The congruence is true if and only if v ≡ 0, 1, 3, 4(mod 6). Since v is odd, we see that v ≡ 1, 3(mod 6). Since v > k in a BIBD, we obtain that v ≥ 7. 6.

(11) Theorem 1.14 [7] If v ≡ 1, 3(mod 6), there exists an STS(v).. Definition 1.15 Suppose v ≥ 2, λ ≥ 1, and K ⊆ {n ∈ Z | n ≥ 2}. A (v, K, λ)-pairwise balanced design (which we abbreviate to (v, K, λ)-PBD) is an ordered pair (X, A) such that the following properties are satisfied: 1. |X| = v 2. |A| ∈ K for all A ∈ A, and 3. every pair of distinct points of X is contained in exactly λ blocks. A (v,K,1)-PBD is often denoted simply as a (v,K)-PBD.. Example 1.16 Let X = {1, 2, 3, 4, 5, 6} and A = {123, 145, 16, 24, 256, 346, 35}. Then (X, A) is a (6, {2, 3}, 1)-PBD.. Theorem 1.17 [7] If v ≡ 5(mod 6), there exists a (v, {3, 5∗ })-PBD (i.e. with one block of size 5 and the rest of size 3).. Definition 1.18 A group-divisible design (which we abbreviate to GDD) is a triple (X, G, A) such that the following properties are satisfied:. 1. X is a finite set of elements called points, 2. G is a partition of X called groups (Note that groups of size one are allowed),. 7.

(12) 3. A is a set of subsets of X called blocks such that |A| ≥ 2 for all A ∈ A, 4. a group and a block contain at most one common point, and 5. every pair of points from distinct groups is contained in exactly one block.. Remark 1.19 We denoted the design by GDD(v, M, N ), where |X| = v, M = {|G| | G ∈ G} and N = {|A| | A ∈ A}.. Example 1.20 Let X = {1, 2, . . . , 12}, G = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}} be a partition of X, and B be the collection of the following 4-element subset of X: {1, 4, 7, 10}, {1, 5, 8, 11}, {1, 6, 9, 12}, {2, 4, 8, 12}, {2, 5, 9, 10}, {2, 6, 7, 11}, {3, 4, 9, 11}, {3, 5, 7, 12}, {3, 6, 8, 10}. Then (X, G, B) is a GDD(12, 3, 4).. Theorem 1.21 If v ≡ 0, 2 (mod 6), then there exists a GDD (v, 2, 3).. Proof: Let (X, A) be an STS(v + 1), and x ∈ X. Set G = {A \ {x} | A ∈ A, x ∈ A} and B = {A | A ∈ A, x ̸∈ A}. Then (X \ {x}, G, B) is a GDD (v, 2, 3).. 8.

(13) Example 1.22 A (9, 3, 1)-BIBD, (X, A), was presented in Example 1.12: X = {1, 2, 3, 4, 5, 6, 7, 8, 9} and A = {123, 456, 789, 147, 258, 369, 159, 267, 348, 168, 249, 357}. Deleting the point 1, we have the GDD(Γ, G, B), where Γ = {2, 3, 4, 5, 6, 7, 8, 9}, G = {23, 47, 59, 68}, and B = {456, 789, 258, 369, 267, 348, 249, 357}. The GDD contains four groups of size 2 and eight blocks of size 3. Theorem 1.23 [7] If v ≡ 4 (mod 6), then there exists a GDD (v, {2, 4∗ }, 3), where the number of group size 4 is one.. Proof: Let (X, A) be a (v + 1, {3, 5∗ }, 1)-PBD, and x ∈ G, where G is the only group size 5. Set G = {A \ {x} | A ∈ A, x ∈ A} and B = {A | A ∈ A, x ̸∈ A}. Then (X \ {x}, G, B) is a GDD (v, {2, 4∗ }, 3). Example 1.24 Let (X, A) be a (11, {3, 5∗ }, 1)-PBD, where X = {1, 2, . . . , 11} and A = {{1, 2, 3, 4, 5}, {1, 6, 7}, {1, 8, 9}, {1, 10, 11}, {2, 6, 9}, {2, 7, 11}, {2, 8, 10}, {3, 6, 11}, {3, 7, 8}, {3, 9, 10}, {4, 6, 10}, {4, 7, 9}, {4, 8, 11}, {5, 6, 8}, {5, 7, 10}, {5, 9, 11}}. Deleting the point 1, we have the GDD(10, {2, 4∗ }, 3), (Γ, G, B), where Γ = {2, 3, . . . , 11}, G = {{2, 3, 4, 5}, {6, 7}, {8, 9}, {10, 11}}, and B = {{2, 6, 9}, {2, 7, 11}, {2, 8, 10}, {3, 6, 11}, {3, 7, 8}, {3, 9, 10}, {4, 6, 10}, {4, 7, 9}, {4, 8, 11}, {5, 6, 8}, {5, 7, 10}, {5, 9, 11}}.. Definition 1.25 Suppose that G is a simple graph and H1 , H2 , . . . , Hn are 9.

(14) subgraph of G. If. 1. E(H1 ) ∪ E(H2 ) ∪ . . . ∪ E(Hn ) = E(G), 2. For 1 ≤ i ̸= j ≤ n, E(Hi ) ∩ E(Hj ) = ∅, then G is decomposable into subgraphs {H1 , H2 , . . . , Hn }. If Hi ∼ = H, ∀i, then we say that G is H-decomposition, and denoted by H | G.. Definition 1.26 Let G be a graph. A Ti -design of G is a Ti -decomposition of G. A Ti -design of order n is a Ti -design of Kn .. Example 1.27 Let A = {(3, 2, 1, 4, 5)1 , (2, 5, 6, 4, 1)1 , (6, 4, 3, 1, 5)1 }. Then A is a T1 -design of order 6. (See Figure 7) 3.. 2.. 6.. 2. 1. 5. 6. 4. 3. 4. 5. 4. 1. 1. 5. Figure 7: T1 -design of K6. Example 1.28 Let A = {(9, 10, 1, 6, 8)2 , (2, 9, 7, 1, 10)2 , (10, 3, 8, 4, 7)2 , (4, 6, 7, 3, 5)2 , (5, 6, 8, 9, 2)2 , (2, 6, 10, 4, 5)2 , (3, 6, 9, 4, 5)2 }. Then A is a T2 -design of K10 \ K5 . (See Figure 8) 10.

(15) 6. 8. 1. 1.. 10. 9. 9. 2. 9. 2. 4. 5. 8.. 3. 6. 3. 5 7.. 10. 4. 6. 4. 5. 10. 5. 7 8.. 7.. 10. 6. 4. 9. 2. 6. 3. Figure 8: T2 -design of K10 \ K5. Example 1.29 Let A = {(11, 10, 6, 7, 4)3 , (10, 9, 5, 8, 6)3 , (9, 8, 4, 11, 1)3 , (11, 8, 3, 7, 1)3 , (11, 9, 2, 10, 4)3 , (10, 8, 1, 9, 3)3 , (5, 11, 7, 9, 6)3 , (2, 8, 7, 10, 3)3 }. Then A is a T3 -design of K11 \ K6 . (See Figure 9). 10. 4. 6. 1. 1. 7. 8. 11. 7. 6.. 5.. 4.. 3.. 11. 9. 10. 8. 11. 9. 8. 11.

(16) 9. 4. 3. 6. 3. 10. 9. 9. 10. 2.. 1.. 7.. 7.. 11. 8. 10. 11. 5. 8. 2. Figure 9: T3 -design of K11 \ K6. Theorem 1.30 If there exists a Ti -design of order v, then v ≡ 0, 1(mod 5).. Proof. If there exists a Ti -design of order n, we obtain n(n − 1) ≡ 0 (mod 10).. 1. n = 5k, n(n − 1) = 5k(5k − 1) ≡ 0 (mod 10) 2. n = 5k + 1, n(n − 1) = 5k(5k + 1) ≡ 0 (mod 10) 3. n = 5k +2, n(n−1) = (5k +2)(5k +1) = 25k 2 +15k +2 = 5k(5k +3)+2 ≡ 2 (mod 10)̸= 0 (mod 10) 4. n = 5k +3, n(n−1) = (5k +3)(5k +2) = 25k 2 +25k +6 = 5k(5k +5)+6 ≡ 6 (mod 10)̸= 0 (mod 10). 12.

(17) 5. n = 5k+4, n(n−1) = (5k+4)(5k+3) = 25k 2 +35k+12 = 5k(5k+7)+12 ≡ 2 (mod 10)̸= 0 (mod 10). From the proof, we obtain n ≡ 0, 1 (mod 5).. The purpose of the thesis is to solve the triangle intersection problem for Ti -designs where i=1,2,3. If | B1 ∩ B2 |= s, we say two Ti -design of order v, (X, B1 ) and (X, B2 ) are intersect in s blocks. Example 1.31 | B1 ∩ B2 |= 2, where B1 = {(3, 2, 1, 4, 5)1 , (2, 5, 6, 4, 1)1 , (6, 4, 3, 1, 5)1 } and B2 = {(3, 2, 1, 4, 6)1 , (2, 6, 5, 4, 1)1 , (5, 4, 3, 1, 6)1 }. Then two T1 -design B1 and B2 of order 6 are said to intersect in 2 blocks.. Let B be a simple graph. Denote by T (B) the set of all triangles of the graph B. For example, if B = (a, b, c, d, e)1 , then T (B) = {a, b, c}. Two Ti -design of order v (X, B1 ) and (X, B2 ) are said to intersect in t triangles if |T (B1 ) ∩ T (B2 )| = t, where T (Bi ) =. ∪ B∈Bi. T (B), i = 1, 2. The triangle. intersection problem for Ti -design is the determination of all integer pairs (v, t) such that there exists a pair of Ti -design of order v intersecting in t triangles.. In what follows, let Ii (v) denote the set of integer t for which there exists 13.

(18) two Ti -designs of order v, (X, B1 ) and (X, B2 ), such that |T (B1 ) ∩ T (B2 )| = t. For v ≡ 0, 1 (mod 5), let Ji (v) = {0, 1, 2, . . . , v(v − 1)/10}. In other words, Ji (v) denotes the triangle intersection numbers one expects to achieve with a Ti -design of order v. We also modify this notation slightly and let Ii (H) and Ji (H) denote respectively the achievable and expected triangle intersection numbers for Ti -design of the graph H. It is clear that Ii (H) ⊆ Ji (H); it is the reverse containment that we deal with here.. 2. Small Cases. In this section, we give the solutions for the triangles-intersection problems for Ti -design of small order followed by examples which is necessary for the construction in next section. For convenience, let V (Kn ) = {1, 2, . . . , n}; V (Kn \ Kv ) = {1, 2, . . . , n}, where V (Kv ) = {1, 2, . . . , v}; V (Ki,j,k ) = X1 ∪ X2 ∪ X3 , where X1 = {1, 2, . . . , i}, X2 = i + {1, 2, . . . , j} and X3 = (i + j) + {1, 2, . . . , k}. Let A and B be two sets and k be a positive integer, then A+B = {a+b | a ∈ A, b ∈ B}, k · A = A +A+ . . . + A}. {z | k. 14.

(19) Theorem 2.1 Ti -design of order 5 does not exist, for i = 2, 3.. Proof: Case 1: Suppose there exists a T2 -design (X, B) of order 5. Then there are two blocks in B, say B1 and B2 . If B1 is a T2 -graph, then B2 = K4 −e (See Figure 10). So, there does not exist a T2 -design of order 5.. s , l B ,  Bl , ls H s  B  Q B Q    B B  Q  Q B  BB s QBs. s , ,  s,  s  B   B   BBs s. s l B B lls s B Q  Q Q B  Q B s QBs. Figure 10: Graph decomposition of Case 1. Case 2: Suppose there exists a T3 -design (X, B) of order 5. Then there are two blocks in B, say B1 and B2 . If B1 is a T3 -graph, then B2 = C4 + e (See Figure 5). So, there does not exist a T3 -design of order 5.. 2.1. Small order of T1 -design.. Lemma 2.2 I1 (5) = J1 (5).. Proof: Let B1 , B2 and B3 be three T1 -designs of order 5, where B1 = {(2, 1, 3, 5, 4)1 , (2, 5, 4, 3, 1)1 }.B2 = {(2, 4, 3, 5, 1)1 , (2, 5, 1, 3, 4)1 }, B3 = {(1, 2, 15.

(20) 3, 5, 4)1 , (1, 5, 4, 3, 2)1 }. Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 1, |T (B1 ) ∩ T (B1 )| = 2. From these results, we have I1 (5) = J1 (5). ⋄. Lemma 2.3 I1 (6) = J1 (6).. Proof: Let B1 , B2 , B3 and B4 be T1 -designs of order 6, where B1 = {(3, 2, 1, 4, 5)1 , (2, 5, 6, 4, 1)1 , (6, 4, 3, 1, 5)1 }, B2 = {(3, 2, 6, 4, 5)1 , (2, 5, 1, 4, 6)1 , (1, 4, 3, 6, 5)1 }, B3 = {(2, 3, 1, 4, 5)1 , (3, 5, 6, 4, 1)1 , (6, 4, 2, 1, 5)1 }, B4 = {(3, 2, 1, 4, 6)1 , (2, 6, 5, 4, 1)1 , (5, 4, 3, 1, 6)1 }. Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 1, |T (B1 ) ∩ T (B4 )| = 2, |T (B1 ) ∩ T (B1 )| = 3. From those results, we have I1 (6) = J1 (6). ⋄. Lemma 2.4 I1 (K10 \ K5 ) = J1 (K10 \ K5 ), and I1 (10) = J1 (10).. Proof: Consider the following T1 -designs of K10 \ K5 . B1 = {(10, 1, 6, 7, 8)1 , (6, 2, 7, 8, 4)1 , (7, 5, 8, 6, 1)1 , (5, 9, 10, 8, 7)1 , (4, 8, 10, 3, 2)1 , (4, 6, 9, 3, 1)1 , (7, 3, 9, 10, 2)1 }. B2 = {(10, 1, 6, 7, 8)1 , (2, 6, 7, 3, 4)1 , (5, 7, 8, 3, 1)1 , (5, 9, 10, 1, 3)1 , (4, 8, 10, 3, 7)1 , (4, 6, 9, 5, 3)1 , (8, 2, 9, 10, 7)1 }. 16.

(21) B3 = {(1, 6, 10, 8, 3)1 , (2, 6, 7, 3, 4)1 , (7, 5, 8, 6, 1)1 , (5, 9, 10, 4, 7)1 , (10, 4, 8, 6, 3)1 , (8, 2, 9, 10, 3)1 , (1, 7, 9, 3, 6)1 }. B4 = {(1, 6, 10, 8, 3)1 , (7, 2, 6, 10, 3)1 , (7, 5, 8, 6, 3)1 , (5, 9, 10, 4, 8)1 , (2, 8, 9, 4, 6)1 , (7, 1, 9, 8, 3)1 , (10, 4, 7, 6, 3)1 }. B5 = {(1, 6, 10, 4, 9)1 , (2, 6, 7, 5, 3)1 , (7, 5, 8, 9, 4)1 , (2, 8, 9, 6, 3)1 , (7, 1, 9, 8, 6)1 , (7, 4, 10, 9, 5)1 , (8, 3, 10, 6, 2)1 }. B6 = {(1, 6, 10, 4, 9)1 , (6, 2, 7, 10, 3)1 , (2, 8, 9, 6, 3)1 , (9, 1, 7, 8, 5)1 , (10, 4, 7, 9, 8)1 , (3, 8, 10, 4, 5)1 , (9, 5, 6, 8, 3)1 }. B7 = {(1, 6, 10, 2, 9)1 , (8, 2, 9, 7, 3)1 , (9, 1, 7, 8, 6)1 , (4, 7, 10, 3, 2)1 , (3, 8, 10, 7, 5)1 , (6, 8, 9, 7, 4)1 , (4, 6, 8, 3, 5)1 }. B8 = {(8, 2, 9, 7, 10)1 , (7, 1, 9, 6, 10)1 , (4, 7, 10, 8, 1)1 , (10, 3, 8, 9, 5)1 , (6, 5, 9, 10, 4)1 , (4, 6, 8, 2, 1)1 , (3, 6, 7, 10, 5)1 }.. Intersection T (B1 ) ∩ T (B1 ) T (B1 ) ∩ T (B3 ) T (B1 ) ∩ T (B5 ) T (B1 ) ∩ T (B7 ). Table 1 Size Intersection Size 7 T (B1 ) ∩ T (B2 ) 6 5 T (B1 ) ∩ T (B4 ) 4 3 T (B1 ) ∩ T (B6 ) 2 1 T (B1 ) ∩ T (B8 ) 0. 17.

(22) From the Table 1, we have I1 (K10 \ K5 ) = J1 (K10 \ K5 ). The graph K10 can be regarded as a union of graphs K10 \ K5 and K5 . Therefore, we have I1 (10) ⊇ I1 (K10 \ K5 ) + I1 (K5 ) = J1 (K10 \ K5 ) + J1 (K5 ) = {0, 1, 2, . . . , 7} + {0, 1, 2} = {0, 1, 2, . . . , 9} = J1 (10).⋄ Lemma 2.5 I1 (K11 \ K6 ) = J1 (K11 \ K6 ), and I1 (11) = J1 (11).. Proof: Consider the following T1 -designs of K11 \ K6 . B1 = {(8, 1, 7, 11, 6)1 , (1, 9, 10, 6, 2)1 , (9, 2, 7, 8, 3)1 , (3, 8, 10, 9, 4)1 , (8, 4, 11, 9, 3)1 , (5, 9, 11, 3, 2)1 , (10, 5, 7, 8, 4)1 , (10, 6, 11, 8, 7)1 }. B2 = {(7, 1, 8, 11, 2)1 , (1, 9, 10, 6, 2)1 , (2, 7, 9, 3, 8)1 , (8, 3, 10, 11, 4)1 , (8, 4, 11, 9, 2)1 , (11, 5, 9, 8, 3)1 , (5, 7, 10, 4, 6)1 , (7, 6, 11, 8, 10)1 }. B3 = {(8, 1, 7, 11, 4)1 , (1, 9, 10, 6, 4)1 , (2, 7, 9, 3, 8)1 , (3, 8, 10, 5, 6)1 , (11, 4, 8, 9, 6)1 , (11, 5, 9, 7, 3)1 , (6, 7, 11, 10, 3)1 , (11, 2, 10, 8, 5)1 }. B4 = {(8, 1, 7, 11, 4)1 , (1, 9, 10, 6, 4)1 , (2, 7, 9, 3, 4)1 , (3, 8, 10, 6, 7)1 , (4, 8, 11, 2, 5)1 , (6, 7, 11, 5, 9)1 , (2, 10, 11, 6, 3)1 , (8, 5, 9, 10, 3)1 }. B5 = {(8, 1, 7, 11, 3)1 , (1, 9, 10, 3, 5)1 , (7, 2, 9, 8, 11)1 , (10, 3, 8, 11, 6)1 , (7, 18.

(23) 6, 11, 9, 4)1 , (2, 10, 11, 6, 8)1 , (9, 5, 8, 11, 4)1 , (10, 4, 7, 9, 5)1 }. B6 = {(8, 1, 7, 11, 3)1 , (1, 9, 10, 3, 6)1 , (7, 2, 9, 8, 6)1 , (7, 6, 11, 8, 3)1 , (2, 10, 11, 3, 8)1 , (9, 5, 8, 10, 3)1 , (4, 7, 10, 5, 8)1 , (9, 4, 11, 8, 5)1 }. B7 = {(8, 1, 7, 11, 5)1 , (1, 9, 10, 3, 6)1 , (11, 6, 7, 9, 2)1 , (11, 2, 10, 9, 3)1 , (9, 5, 8, 10, 2)1 , (4, 7, 10, 9, 8)1 , (9, 4, 11, 8, 5)1 , (11, 3, 8, 7, 6)1 }. B8 = {(8, 1, 7, 11, 5)1 , (11, 6, 7, 10, 2)1 , (11, 2, 10, 9, 1)1 , (5, 8, 9, 6, 1)1 , (4, 7, 10, 9, 8)1 , (11, 4, 9, 8, 6)1 , (3, 8, 11, 2, 5)1 , (9, 3, 10, 7, 5)1 }. B9 = {(11, 6, 7, 10, 1)1 , (2, 10, 11, 8, 1)1 , (9, 5, 8, 7, 1)1 , (4, 7, 10, 8, 1)1 , (11, 4, 9, 8, 1)1 , (3, 8, 11, 6, 5)1 , (3, 9, 10, 6, 5)1 , (9, 2, 7, 8, 3)1 }.. Intersection T (B1 ) ∩ T (B1 ) T (B1 ) ∩ T (B3 ) T (B1 ) ∩ T (B5 ) T (B1 ) ∩ T (B7 ) T (B1 ) ∩ T (B9 ). Table 2 Size Intersection Size 8 T (B1 ) ∩ T (B2 ) 7 6 T (B1 ) ∩ T (B4 ) 5 4 T (B1 ) ∩ T (B6 ) 3 2 T (B1 ) ∩ T (B8 ) 1 0. From the Table 2, we have I1 (K11 \ K6 ) = J1 (K11 \ K6 ). The graph K11 can be regarded as a union of graphs K11 \ K6 and K6 . Therefore, we have I1 (11) ⊇ I1 (K11 \ K6 ) + I1 (K6 ) = J1 (K11 \ K6 ) + J1 (K6 ) = 19.

(24) {0, 1, 2, . . . , 8} + {0, 1, 2, 3} = {0, 1, 2, . . . , 11} = J1 (11).⋄. Lemma 2.6 {0, 5, 10, 15} ⊆ I1 (K5,5,5 ).. Proof: Let B1 , B2 , B3 and B4 be four T1 -designs of K5,5,5 , where B1 = {(11, 1, 6, 8, 3)1 , (4, 9, 15, 1, 3)1 , (9, 2, 12, 14, 5)1 , (12, 4, 6, 8, 13)1 , (3, 9, 14, 13, 8)1 , (11, 2, 7, 8, 3)1 , (10, 5, 15, 7, 1)1 , (3, 10, 11, 13, 8)1 , (8, 5, 13, 14, 4)1 , (6, 2, 15, 10, 8)1 , (8, 3, 12, 13, 1)1 , (1, 7, 13, 15, 2)1 , (9, 5, 11, 6, 4)1 , (1, 10, 14, 12, 6)1 , (14, 4, 7, 10, 12 )1 }, B2 = {(8, 2, 14, 11, 6)1 , (1, 8, 15, 3, 4)1 , (3, 6, 13, 11, 8)1 , (14, 4, 10, 6, 5)1 , (7, 5, 14, 11, 3)1 , (4, 8, 11, 12, 10)1 , (9, 5, 13, 12, 4)1 , (15, 3, 7, 11, 13)1 , (13, 2, 10, 6, 3 )1 , (12, 1, 10, 13, 15)1 , (6, 5, 15, 8, 9)1 , (4, 9, 12, 2, 6)1 , (14, 1, 9, 6, 3)1 , (7, 2, 12, 15, 3)1 , (1, 7, 11, 4, 9)1 }, B3 = {(11, 2, 7, 8, 3)1 , (4, 9, 15, 2, 3)1 , (9, 5, 12, 14, 1)1 , (12, 4, 7, 8, 13)1 , (3, 9, 14, 13, 8)1 , (11, 5, 6, 8, 3)1 , (10, 1, 15, 6, 2)1 , (3, 10, 11, 13, 8)1 , (8, 1, 13, 14, 4)1 , (7, 5, 15, 10, 8)1 , (8, 3, 12, 13, 2)1 , (2, 6, 13, 15, 5)1 , (9, 1, 11, 7, 4)1 , (2, 10, 14, 12, 7)1 , (14, 4, 6, 10, 12)1 }, B4 = {(11, 1, 8, 6, 3)1 , (4, 9, 15, 1, 3)1 , (9, 2, 12, 14, 5)1 , (12, 4, 8, 6, 13)1 , (3, 9, 14, 13, 6)1 , (11, 2, 7, 6, 3)1 , (10, 5, 15, 7, 1)1 , (3, 10, 11, 13, 6)1 , (6, 5, 13, 14, 4)1 , (8, 2, 15, 10, 6)1 , (6, 3, 12, 13, 1)1 , (1, 7, 13, 15, 2)1 , (9, 5, 11, 8, 4)1 , (1, 10, 14, 12, 8)1 , (14, 4, 7, 10, 12)1 }. 20.

(25) Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 5, |T (B1 ) ∩ T (B4 )| = 10 and |T (B1 ) ∩ T (B1 )| = 15. From those results, we have {0, 5, 10, 15} ⊆ I1 (K5,5,5 ). ⋄. Lemma 2.7 {0, 5, 10, 15, 17} ⊆ I1 (K5,5,6 ).. Proof: Let B1 , B2 , B3 , B4 , B5 and B6 be five T1 -designs of K5,5,6 , where B1 = {(11, 1, 6, 8, 2)1 , (9, 4, 14, 15, 3)1 , (8, 2, 14, 11, 6)1 , (5, 9, 13, 15, 4)1 , (10, 2, 13, 15, 7)1 , (11, 5, 7, 8, 4)1 , (12, 2, 7, 9, 15)1 , (10, 5, 15, 16, 6)1 , (9, 3, 16, 7, 6)1 , (11, 4, 8, 12, 15)1 , (10, 3, 11, 15, 9)1 , (9, 1, 12, 15, 10)1 , (13, 3, 8, 6, 12)1 , (1, 7, 16, 14, 2)1 , (10, 4, 16, 6, 8)1 , (10, 1, 14, 13, 5)1 , (5, 6, 12, 13, 3)1 }, B2 = {(8, 1, 12, 6, 2)1 , (1, 9, 15, 14, 2)1 , (5, 7, 11, 1, 6)1 , (6, 3, 12, 13, 10)1 , (16, 2, 6, 8, 14)1 , (13, 1, 10, 14, 11)1 , (7, 4, 15, 11, 10)1 , (9, 5, 12, 14, 4)1 , (6, 4, 13, 10, 8)1 , (3, 7, 14, 12, 4)1 , (14, 2, 10, 9, 3)1 , (3, 9, 11, 13, 1)1 , (5, 8, 15, 11, 6)1 , (9, 4, 16, 8, 1)1 , (16, 3, 8, 15, 14)1 , (7, 2, 13, 11, 5)1 , (10, 5, 16, 6, 7)1 }, B3 = {(11, 2, 7, 8, 3)1 , (9, 4, 14, 15, 1)1 , (8, 3, 14, 11, 7)1 , (5, 9, 13, 15, 4)1 , (10, 3, 13, 15, 6)1 , (11, 5, 6, 8, 4)1 , (12, 3, 6, 9, 15)1 , (10, 5, 15, 16, 7)1 , (9, 1, 16, 6, 7)1 , (11, 4, 8, 12, 15)1 , (10, 1, 11, 15, 9)1 , (9, 2, 12, 15, 10)1 , (13, 1, 8, 7, 12)1 , (2, 6, 16, 14, 3)1 , (10, 4, 16, 7, 8)1 , (10, 2, 14, 13, 5)1 , (5, 7,. 21.

(26) 12, 13, 1)1 }, B4 = {(11, 4, 6, 8, 2)1 , (9, 1, 14, 15, 3)1 , (8, 2, 14, 11, 6)1 , (5, 9, 13, 15, 1)1 , (10, 2, 13, 15, 7)1 , (11, 5, 7, 8, 1)1 , (12, 2, 7, 9, 15)1 , (10, 5, 15, 16, 6)1 , (9, 3, 16, 7, 6)1 , (11, 1, 8, 12, 15)1 , (10, 3, 11, 15, 9)1 , (9, 4, 12, 15, 10)1 , (13, 3, 8, 6, 12)1 , (4, 7, 16, 14, 2)1 , (10, 1, 16, 6, 8)1 , (10, 4, 14, 13, 5)1 , (5, 6, 12, 13, 3)1 }, B5 = {(11, 1, 6, 15, 13)1 , (9, 4, 14, 6, 10)1 , (8, 2, 14, 15, 6)1 , (5, 9, 13, 15, 7)1 , (10, 2, 13, 6, 4)1 , (11, 5, 7, 14, 3)1 , (12, 2, 7, 11, 14)1 , (5, 10, 15, 12, 4)1 , (9, 3, 16, 15, 8)1 , (11, 4, 8, 12, 1)1 , (10, 3, 11, 6, 9)1 , (1, 9, 12, 2, 6)1 , (8, 3, 13, 14, 1)1 , (16, 1, 7, 14, 15)1 , (16, 4, 10, 7, 1)1 , (5, 6, 16, 15, 2)1 , (5, 8, 12, 15, 3)1 }. Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 5, |T (B1 ) ∩ T (B4 )| = 10, |T (B1 ) ∩ T (B5 )| = 15 and |T (B1 ) ∩ T (B1 )| = 17. From those results, we have {0, 5, 10, 15, 17} ⊆ I1 (K5,5,6 ). ⋄. Lemma 2.8 I1 (K15 \ K5 ) = J1 (K15 \ K5 ) and I1 (15) = J1 (15).. Proof: The graph K15 \ K5 can be regarded as a union of tripartite graph K5,5,5 , and 2 copies of K5 . Therefore, we have I1 (K15 \ K5 ) ⊇ I1 (K5,5,5 ) + 2 · I1 (K5 ) ⊇ {0, 5, 10, 15} + 2 · J1 (K5 ) = J1 (K15 \ K5 ). The graph K15 can be regarded as a union of graphs K15 \ K5 and K5 . Therefore, we have I1 (15) ⊇ I1 (K15 \ K5 ) + I1 (K5 ) = J1 (K15 \ K5 ) + J1 (K5 ) = {0, 1, 2, . . . , 19} + 22.

(27) {0, 1, 2} = {0, 1, 2, . . . , 21} = J1 (15). ⋄. Lemma 2.9 I1 (K16 \ K6 ) = J1 (K16 \ K6 ) and I1 (16) = J1 (16).. Proof: The graph (K16 \ K6 ) can be regarded as a union of graphs (K5,5,6 ) and K5 .. Therefore, we have I1 (K16 \ K6 ) ⊇ I1 (K5,5,6 ) + 2 · I1 (K5 ) ⊇. {0, 5, 10, 15, 17} + 2 · J1 (K5 ) = J1 (K16 \ K6 ). The graph (K16 ) can be regarded as a union of graphs (K16 \ K6 ) and K6 . Therefore, we have I1 (16) ⊇ I1 (K16 \ K6 ) + I1 (K6 ) = J1 (K16 \ K6 ) + J1 (K6 ) = {0, 1, 2, . . . , 21} + {0, 1, 2, 3} = {0, 1, 2, . . . , 24} = J1 (16). ⋄. Lemma 2.10 I1 (20) = J1 (20) and I1 (21) = J1 (21).. Proof: The graph K20 can be regarded as a union of graphs K5,5,5 , K10 \ K5 and I1 (K5 ). Therefore, we have I1 (20) ⊇ I1 (K5,5,5 )+3·I1 (K10 \K5 )+I1 (K5 ) ⊇ {0, 15} + 3 · J1 (K10 \ K5 ) + J1 (K5 ) = {0, 15} + 3 · {0, 1, 2, . . . , 7} + {0, 1, 2} = {0, 15} + {0, 1, 2, . . . , 21} + {0, 1, 2} = {0, 1, 2, . . . , 38} = J1 (20). The graph K21 can be regarded as a union of graphs K5,5,5 , K11 \ K6 and K6 . Therefore, we have I1 (21) ⊇ I1 (K5,5,5 ) + 3 · I1 (K11 \ K6 ) + I1 (K6 ) ⊇ 23.

(28) {0, 15} + 3 · J1 (K11 \ K6 ) + J1 (K6 ) = {0, 15} + 3 · {0, 1, 2, . . . , 8} + {0, 1, 2, 3} = {0, 15} + {0, 1, 2, . . . , 24} + {0, 1, 2, 3} = {0, 1, 2, . . . , 42} = J1 (21). ⋄. Lemma 2.11 I1 (25) = J1 (25) and I1 (26) = J1 (26).. Proof: The graph K25 can be regarded as a union of graphs K5,5,5 , K10 \ K5 and I1 (K5 ). Therefore, we have I1 (25) ⊇ 2 · I1 (K5,5,5 ) + 4 · I1 (K10 \ K5 ) + I1 (K5 ) ⊇ 2·{0, 15}+4·J1 (K10 \K5 )+J1 (K5 ) = 2·{0, 15}+4·{0, 1, 2, . . . , 7}+ {0, 1, 2} = {0, 15, 30}+{0, 1, 2, . . . , 28}+{0, 1, 2} = {0, 1, 2, . . . , 60} = J1 (25). The graph K26 can be regarded as a union of graphs K5,5,6 , K10 \ K5 and K6 . Therefore, we have I1 (26) ⊇ 2 · I1 (K5,5,6 ) + 4 · I1 (K10 \ K5 ) + I1 (K6 ) ⊇ 2 · {0, 17}+4·J1 (K10 \K5 )+J1 (K6 ) = 2·{0, 17}+4·{0, 1, 2, . . . , 7}+{0, 1, 2, 3} = {0, 17, 34} + {0, 1, 2, . . . , 28} + {0, 1, 2, 3} = {0, 1, 2, . . . , 65} = J1 (26). ⋄. 2.2. Small order of T2 -design.. Lemma 2.12 I2 (6) = J2 (6). Proof: Let B1 , B2 , B3 and B4 be T2 -designs of order 6, where B1 = {(2, 3, 1, 4, 5)2 , (2, 4, 5, 3, 6)2 , (3, 4, 6, 1, 2)2 }, B2 = {(1, 6, 4, 3, 5)2 , (5, 6, 2, 1, 4)2 , (1, 5, 3, 2, 6)2 }, B3 = {(1, 2, 3, 4, 6)2 , (1, 6, 4, 2, 5)2 , (2, 6, 5, 1, 3)2 }, B4 = {(1, 24.

(29) 2, 3, 4, 5)2 , (2, 4, 5, 1, 6)2 , (1, 4, 6, 3, 2)2 }. Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 1, |T (B1 ) ∩ T (B4 )| = 2, |T (B1 ) ∩ T (B1 )| = 3. From those results, we obtain I2 (6) = J2 (6). ⋄ Lemma 2.13 {0, 7} ⊆ I2 (K10 \ K5 ).. Proof: Let A and B be two T2 -designs of K10 \ K5 , where A = {(9, 10, 1, 6, 8)2 , (2, 9, 7, 1, 10)2 , (10, 3, 8, 4, 7)2 , (4, 6, 7, 3, 5)2 , (5, 6, 8, 9, 2)2 , (2, 6, 10, 4, 5)2 , (3, 6, 9, 4, 5)2 } and B = {(7, 8, 1, 9, 10)2 , (8, 9, 4, 7, 10)2 , (5, 10, 6, 1, 4)2 , (2, 6, 8, 5, 10)2 , (7, 10, 3, 6, 8)2 , (5, 9, 7, 2, 6)2 , (2, 10, 9, 3, 6)2 }. Then |T (A) ∩ T (B)| = 0 and |T (A) ∩ T (A)| = 7. From those results, we obtain {0, 7} ⊆ I2 (K10 \ K5 ). ⋄ Lemma 2.14 {0, 3, 6} ⊆ I2 (K10 \ K6 ).. Proof: Let B1 , B2 and B3 be three T2 -designs of I2 (K10 \ K6 ), where B1 = {(5, 8, 7, 6, 1)2 , (6, 9, 8, 1, 2)2 , (1, 10, 9, 2, 5)2 , (2, 7, 10, 5, 6)2 , (7, 9, 3, 8, 10)2 , (8, 10, 4, 7, 9)2 ), B2 = {(6, 8, 7, 1, 2)2 , (1, 9, 8, 2, 3)2 , (2, 10, 9, 3, 6)2 , (3, 7, 10, 6, 1)2 , (7, 9, 4, 8, 10)2 , (8, 10, 5, 7, 9)2 } and 25.

(30) B3 = {(6, 8, 7, 3, 1)2 , (3, 9, 8, 1, 2)2 , (1, 10, 9, 2, 6)2 , (2, 7, 10, 6, 3)2 , (7, 9, 5, 8, 10)2 , (8, 10, 4, 7, 9)2 }. Then |T (B1 ) ∩ T (B3 )| = 3, |T (B1 ) ∩ T (B2 )| = 0 and |T (B1 ) ∩ T (B1 )| = 6. From those results, we have {0, 3, 6} ⊆ I2 (K10 \ K6 ). ⋄. Lemma 2.15 I2 (10) = J2 (10).. Proof: The graph K10 can be regarded as a union of graphs K10 \ K6 and K6 . Therefore, we have I2 (10) ⊇ I2 (K10 \K6 )+I2 (K6 ) ⊇ {0, 3, 6}+J2 (K6 ) = {0, 3, 6} + {0, 1, 2, 3} = {0, 1, 2, . . . , 9} = J2 (10).⋄. Lemma 2.16 {0, 4, 8} ⊆ I2 (K11 \ K6 ), and I2 (11) = J2 (11).. Proof: Let B1 , B2 and B3 be three T2 -designs of K11 \ K6 , where B1 = {(4, 8, 9, 6, 7)2 , (11, 2, 9, 3, 1)2 , (11, 5, 7, 6, 4)2 , (8, 3, 11, 4, 1)2 , (10, 1, 8, 6, 5)2 , (2, 8, 7, 3, 1)2 , (11, 6, 10, 4, 2)2 , (9, 5, 10, 7, 3)2 }, B2 = {(4, 8, 9, 6, 7)2 , (11, 2, 9, 1, 5)2 , (5, 11, 7, 4, 6)2 , (3, 11, 8, 2, 7)2 , (3, 9, 10, 1, 6)2 , (4, 10, 11, 1, 6)2 , (2, 10, 7, 1, 3)2 , (5, 10, 8, 1, 6)2 } and. 26.

(31) B3 = {(3, 10, 9, 4, 5)2 , (4, 10, 11, 2, 9)2 , (2, 7, 10, 1, 6)2 , (5, 10, 8, 2, 6)2 , (1, 8, 11, 3, 5)2 , (6, 11, 7, 3, 5)2 , (1, 7, 9, 2, 6)2 , (4, 7, 8, 3, 9)2 }. Then |T (B1 ) ∩ T (B3 )| = 0, |T (B1 ) ∩ T (B2 )| = 4 and |T (B1 ) ∩ T (B1 )| = 8. From those results, we have {0, 4, 8} ⊆ I2 (K11 \ K6 ). The graph K11 can be regarded as a union of graphs K11 \ K6 and K6 . Therefore, we have I2 (11) ⊇ I2 (K11 \ K6 ) + I2 (K6 ) ⊇ {0, 4, 8} + J2 (K6 ) = {0, 4, 8} + {0, 1, 2, 3} = {0, 1, 2, . . . , 11} = J2 (11).. Lemma 2.17 {0, 5, 15} ⊆ I2 (K5,5,5 ).. Proof: Let B1 , B2 and B3 be three T2 -designs of K5,5,5 , where B1 = {(11, 6, 1, 12, 7), (11, 10, 2, 12, 6), (11, 9, 3, 12, 10), (11, 8, 4, 12, 9), (11, 7, 5, 12, 8), (13, 8, 1, 14, 9), (13, 7, 2, 14, 8), (13, 6, 3, 14, 7), (13, 10, 4, 14, 6), (13, 9, 5, 14, 10), (15, 1, 10, 12, 14), (15, 2, 9, 12, 14), (15, 3, 8, 12, 14), (15, 4, 7, 12, 14), (15, 5, 6, 12, 14)},. B2 = {(12, 7, 2, 13, 8), (12, 6, 3, 13, 7), (12, 10, 4, 13, 6), (12, 9, 5, 13, 10), (12, 8, 1, 13, 9), (14, 9, 2, 15, 10), (14, 8, 3, 15, 9), (14, 7, 4, 15, 8), (14, 6,. 27.

(32) 5, 15, 7), (14, 10, 1, 15, 6), (11, 2, 6, 13, 15), (11, 3, 10, 13, 15), (11, 4, 9, 13, 15), (11, 5, 8, 13, 15), (11, 1, 7, 13, 15)} and. B3 = {(11, 7, 2, 12, 6), (11, 10, 1, 12, 7), (11, 9, 3, 12, 10), (11, 8, 4, 12, 9), (11, 6, 5, 12, 8), (13, 8, 2, 14, 9), (13, 6, 1, 14, 8), (13, 7, 3, 14, 6), (13, 10, 4, 14, 7), (13, 9, 5, 14, 10), (15, 2, 10, 12, 14), (15, 1, 9, 12, 14), (15, 3, 8, 12, 14), (15, 4, 6, 12, 14), (15, 5, 7, 12, 14)}.. Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 5 and |T (B1 ) ∩ T (B1 )| = 15. From those results, we have {0, 5, 15} ⊆ I2 (K5,5,5 ). ⋄. Lemma 2.18 {0, 30} ⊆ I2 (K5,5,5,5 ).. Proof: Let A and B be two T2 -designs of K5,5,5,5 , where A = {(6, 11, 1, 12, 8), (10, 11, 2, 6, 18), (7, 13, 2, 14, 8), (8, 16, 4, 9, 20), (10, 18, 4, 6, 19), (9, 18, 5, 10, 19), (18, 8, 11, 4, 19), (19, 9, 12, 16, 5), (17, 10, 13, 4, 19), (14, 3, 16, 2, 10), (7, 4, 17, 5, 11), (16, 13, 9, 17, 3), (6, 17, 14, 4, 18), (7, 18, 15, 4, 19), (10, 20, 14, 5, 9), (6, 16, 15, 5, 20), (1, 15, 10, 12, 19), (3, 20, 7, 19, 14), (5, 20, 6, 12, 19), (3, 12, 18, 13, 6), (8, 17, 12, 4, 2), (11, 9, 20, 12, 8), (6, 13, 3, 11, 10), (15, 9, 2, 19, 20), (14, 19, 1, 9, 16), (3, 19, 8, 14, 15), (3, 15, 17,. 28.

(33) 1, 2), (13, 20, 1, 7, 18), (8, 13, 5, 11, 16), (11, 16, 7, 5, 12)} and B = {(8, 13, 2, 11, 9), (7, 13, 3, 8, 19), (6, 15, 3, 12, 9), (9, 20, 5, 10, 18), (7, 19, 5, 8, 17), (10, 19, 1, 7, 17), (19, 9, 13, 5, 17), (17, 10, 11, 20, 1), (16, 7, 15, 5, 17), (12, 4, 20, 3, 7), (6, 5, 16, 1, 13), (20, 15, 10, 16, 4), (8, 16, 12, 5, 19), (6, 19, 14, 5, 17), (7, 18, 12, 1, 10), (8, 20, 14, 1, 18), (2, 14, 7, 11, 17), (4, 18, 6, 17, 12), (1, 18, 8, 11, 17), (4, 11, 19, 15, 8), (9, 16, 11, 5, 3), (13, 10, 18, 11, 9), (8, 15, 4, 13, 7), (14, 10, 3, 17, 18), (12, 17, 2, 10, 20), (4, 17, 9, 12, 14), (4, 14, 16, 2, 3), (15, 18, 2, 6, 19), (9, 15, 1, 13, 20), (13, 20, 6, 1, 11)}. Then |T (A) ∩ T (B)| = 0 and |T (A) ∩ T (A)| = 30. From those results, we have {0, 30} ⊆ I2 (K5,5,5,5 ). ⋄. Lemma 2.19 {0, 19} ⊆ I2 (K15 \ K5 ).. Proof: Let A and B be two T2 -designs of I2 (K15 \ K5 ), where A = {(12, 1, 7, 2, 13)2 , (12, 5, 8, 1, 13)2 , (12, 4, 9, 5, 13)2 , (12, 3, 10, 6, 9)2 , (12, 2, 11, 1, 9)2 , (7, 3, 14, 15, 12)2 , (14, 2, 8, 15, 3)2 , (14, 1, 9, 15, 2)2 , (14, 5, 10, 11, 1)2 , (14, 4, 11, 15, 5)2 , (6, 7, 5, 13, 15)2 , (6, 8, 4, 13, 15)2 , (6, 9, 3, 13, 11)2 , (13, 10, 2, 6, 15)2 , (13, 14, 6, 1, 11)2 , (7, 9, 8, 11, 10)2 , (12, 6, 15, 10, 3)2 , (4, 10, 7, 11, 15)2 , (1, 15, 13, 11, 12)2 } and B = {(11, 1, 6, 2, 12)2 , (11, 5, 7, 1, 29.

(34) 12)2 , (11, 4, 8, 5, 12)2 , (11, 3, 9, 15, 8)2 , (11, 2, 10, 1, 8)2 , (6, 3, 13, 14, 11)2 , (13, 2, 7, 14, 3)2 , (13, 1, 8, 14, 2)2 , (13, 5, 9, 10, 1)2 , (13, 4, 10, 14, 5)2 , (15, 6, 5, 12, 14)2 , (15, 7, 4, 12, 14)2 , (15, 8, 3, 12, 10)2 , (12, 9, 2, 15, 14)2 , (12, 13, 15, 1, 10)2 , (6, 8, 7, 10, 9)2 , (11, 15, 14, 9, 3)2 , (4, 9, 6, 10, 14)2 , (1, 14, 12, 10, 11)2 }. Then |T (A) ∩ T (B)| = 0 and |T (A) ∩ T (A)| = 19. From those results, we have {0, 19} ⊆ I2 (K15 \ K5 ). ⋄. Lemma 2.20 {0, 4, 8, 12, 15, 18} ⊆ I2 (K15 \ K6 ) and I2 (15) = J2 (15).. Proof: Let B1 , B2 , . . ., B6 be six T2 -designs of K15 \ K6 , where B1 = {(11, 2, 10, 8, 6)2 , (11, 9, 3, 12, 10)2 , (11, 5, 7, 8, 6)2 , (13, 8, 1, 14, 9)2 , (13, 7, 2, 14, 8)2 , (13, 9, 5, 14, 10)2 , (15, 1, 10, 12, 13)2 , (15, 2, 9, 12, 14)2 , (3, 8, 15, 5, 14)2 , (15, 7, 4, 9, 10)2 , (7, 12, 14, 4, 10)2 , (8, 11, 4, 12, 13)2 , (6, 13, 12, 2, 5)2 , (6, 9, 8, 5, 12)2 , (3, 13, 14, 6, 8)2 , (6, 11, 15, 12, 13)2 , (1, 12, 11, 13, 14)2 , (9, 10, 7, 1, 3)2 }, B2 = {(12, 2, 14, 8, 5)2 , (12, 9, 3, 10, 14)2 , (12, 6, 7, 8, 5)2 , (11, 8, 1, 15, 9)2 , (11, 7, 2, 15, 8)2 , (11, 9, 6, 15, 14)2 , (13, 1, 14, 10, 11)2 , (13, 2, 9, 10, 15)2 , (3, 8, 13, 6, 15)2 , (13, 7, 4, 9, 14)2 , (7, 10, 15, 4, 14)2 , (8, 12, 4, 10, 11)2 , (5,. 30.

(35) 11, 10, 2, 6)2 , (5, 9, 8, 6, 10)2 , (3, 11, 15, 5, 8)2 , (5, 12, 13, 10, 11)2 , (1, 10, 12, 11, 15)2 , (9, 14, 7, 1, 3)2 }, B3 = {(11, 2, 14, 8, 6)2 , (11, 9, 3, 10, 14)2 , (11, 5, 7, 8, 6)2 , (12, 8, 1, 15, 9)2 , (12, 7, 2, 15, 8)2 , (12, 9, 5, 15, 14)2 , (13, 1, 14, 10, 12)2 , (13, 2, 9, 10, 15)2 , (3, 8, 13, 5, 15)2 , (13, 7, 4, 9, 14)2 , (7, 10, 15, 4, 14)2 , (8, 11, 4, 10, 12)2 , (6, 12, 10, 2, 5)2 , (6, 9, 8, 5, 10)2 , (3, 12, 15, 6, 8)2 , (6, 11, 13, 10, 12)2 , (1, 10, 11, 12, 15)2 , (9, 14, 7, 1, 3)2 }, B4 = {(11, 2, 14, 8, 6)2 , (11, 9, 3, 10, 14)2 , (11, 5, 7, 8, 6)2 , (12, 8, 1, 13, 9)2 , (12, 7, 2, 13, 8)2 , (12, 9, 5, 13, 14)2 , (15, 1, 14, 10, 12)2 , (15, 2, 9, 10, 13)2 , (3, 8, 15, 5, 13)2 , (15, 7, 4, 9, 14)2 , (7, 10, 13, 4, 14)2 , (8, 11, 4, 10, 12)2 , (6, 12, 10, 2, 5)2 , (6, 9, 8, 5, 10)2 , (3, 12, 13, 6, 8)2 , (6, 11, 15, 10, 12)2 , (1, 10, 11, 12, 13)2 , (9, 14, 7, 1, 3)2 }, B5 = {(11, 2, 10, 8, 6)2 , (11, 9, 3, 13, 10)2 , (11, 5, 7, 8, 6)2 , (12, 8, 1, 14, 9)2 , (12, 7, 2, 14, 8)2 , (12, 9, 5, 14, 10)2 , (15, 1, 10, 13, 12)2 , (15, 2, 9, 13, 14)2 , (3, 8, 15, 5, 14)2 , (15, 7, 4, 9, 10)2 , (7, 13, 14, 4, 10)2 , (8, 11, 4, 13, 12)2 , (6, 12, 13, 2, 5)2 , (6, 9, 8, 5, 13)2 , (3, 12, 14, 6, 8)2 , (6, 11, 15, 13, 12)2 , (1, 13, 11, 12, 14)2 , (9, 10, 7, 1, 3)2 } and. 31.

(36) B6 = {(11, 2, 10, 8, 6)2 , (11, 9, 3, 14, 10)2 , (11, 5, 7, 8, 6)2 , (13, 8, 1, 12, 9)2 , (13, 7, 2, 12, 8)2 , (13, 9, 5, 12, 10)2 , (15, 1, 10, 14, 13)2 , (15, 2, 9, 14, 12)2 , (3, 8, 15, 5, 12)2 , (15, 7, 4, 9, 10)2 , (7, 14, 12, 4, 10)2 , (8, 11, 4, 14, 13)2 , (6, 13, 14, 2, 5)2 , (6, 9, 8, 5, 14)2 , (3, 13, 12, 6, 8)2 , (6, 11, 15, 14, 13)2 , (1, 14, 11, 13, 12)2 , (9, 10, 7, 1, 3)2 }. Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 4, |T (B1 ) ∩ T (B4 )| = 8, |T (B1 ) ∩ T (B5 )| = 12, |T (B1 ) ∩ T (B6 )| = 15 and |T (B1 ) ∩ T (B1 )| = 18. From those results, we have {0, 4, 8, 12, 15, 18} ⊆ I2 (K15 \ K6 )2 . The graph K15 can be regarded as a union of graphs K15 \ K6 and K6 . Therefore, we have I2 (15) ⊇ I2 (K15 \ K6 ) + I2 (K6 ) = I2 (K15 \ K6 ) + J2 (K6 ) ⊇ {0, 4, 8, 12, 15, 18} + {0, 1, 2, 3} = {0, 1, 2, . . . , 21} = J2 (15). ⋄. Lemma 2.21 {0, 21} ⊆ I2 (K16 \ K6 ) and I2 (16) = J2 (16).. Proof: I2 (K16 \ K6 ) ⊇ I2 (K5,5,5 ) + 2 · I2 (K6 ) ⊇ {0, 15} + 2 · J2 (K6 ) = {0, 15} + 2 · {0, 1, 2, 3} = {0, 15} + {0, 1, 2, . . . , 6} ⊇ {0, 21}. I2 (16) ⊇ I2 (K5,5,5 ) + 3 · I2 (K6 ) ⊇ {0, 5, 15} + 3 · J2 (K6 ) = {0, 5, 15} + 3 · {0, 1, 2, 3} = {0, 5, 15} + {0, 1, 2, . . . , 9}) = {0, 1, 2, . . . , 24} = J2 (16). ⋄ 32.

(37) Lemma 2.22 I2 (20) = J2 (20) and I2 (21) = J2 (21).. Proof: I2 (20) ⊇ I2 (K5,5,5 ) + 2 · I2 (K10 \ K5 ) + I2 (K10 ) ⊇ {0, 5, 15} + 2 · I2 (K10 \ K5 ) + J2 (K10 ) = {0, 5, 15} + 2 · {0, 7} + {0, 1, 2, . . . , 9} = {0, 5, 15} + {0, 7, 14} + {0, 1, 2, . . . , 9} = {0, 1, 2, . . . , 38} = J2 (20). I2 (21) ⊇ I2 (K5,5,5 ) + 2 · I2 (K11 \ K6 ) + I2 (K11 ) = I2 (K5,5,5 ) + 2 · I2 (K11 \ K6 ) + J2 (K11 ) ⊇ {0, 5, 15} + 2 · {0, 4, 8} + {0, 1, 2, . . . , 11} = {0, 5, 15} + {0, 4, 8, 12, 16} + {0, 1, 2, . . . , 11} = {0, 1, 2, . . . , 42} = J2 (21). ⋄. Lemma 2.23 I2 (25) = J2 (25) and I2 (26) = J2 (26).. Proof: I2 (25) ⊇ I2 (K5,5,5,5 ) + 3 · I2 (K10 \ K5 ) + I2 (K10 ) ⊇ I2 (K5,5,5,5 ) + 3 · J2 (K10 \K5 )+J2 (K10 ) ⊇ {0, 30}+3·{0, 7}+{0, 1, 2} = {0, 30}+{0, 7, 14, 21}+ {0, 1, 2, . . . , 9} = {0, 1, 2, . . . , 60} = J2 (25). I2 (26) ⊇ I2 (K5,5,5,5 )+3·I2 (K11 \K6 )+I2 (K11 ) = I2 (K5,5,5,5 )+3·I2 (K11 \K6 )+ J2 (K11 ) ⊇ {0, 30}+3·{0, 4, 8}+{0, 1, 2, . . . , 11} = {0, 30}+{0, 4, 8, 12, 16, 20, 24}+ {0, 1, 2, . . . , 11} = {0, 1, 2, . . . , 65} = J2 (26). ⋄. 33.

(38) 2.3. Small order of T3 -design.. Lemma 2.24 I3 (6) = J3 (6). Proof: Let B1 , B2 , B3 and B4 be T3 -designs of order 6, where B1 = {(1, 3, 2, 4, 6)3 , (2, 6, 5, 3, 4)3 , (4, 5, 1, 6, 3)3 }, B2 = {(4, 5, 3, 1, 2)3 , (1, 6, 4, 2, 5)3 , (2, 3, 6, 5, 1)3 }, B3 = {(3, 2, 1, 5, 6)3 , (4, 5, 3, 6, 2)3 , (6, 1, 4, 2, 5)3 }, B4 = {(1, 2, 3, 6, 4)3 , (5, 6, 2, 4, 1)3 , (4, 3, 5, 1, 6)3 }. Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 1, |T (B1 ) ∩ T (B4 )| = 2, |T (B1 ) ∩ T (B1 )| = 3. From those results, we obtain I3 (6) = J3 (6). ⋄. Lemma 2.25 {0, 7} ⊆ I3 (K10 \ K5 ). Proof: Let A and B be two T3 -designs of K10 \ K5 , where A = {(1, 7, 6, 3, 10)3 , (2, 8, 7, 9, 1)3 , (3, 8, 9, 5, 7)3 , (4, 9, 10, 8, 5)3 , (5, 6, 10, 1, 8)3 , (6, 9, 2, 10, 7)3 , (8, 6, 4, 7, 3)3 } and B = {(8, 10, 1, 7, 6)3 , (5, 7, 9, 1, 6)3 , (2, 6, 10, 9, 3)3 , (4, 9, 6, 5, 8)3 , (6, 8, 3, 10, 7)3 , (7, 8, 4, 10, 5)3 , (8, 9, 2, 7, 3)3 }. Then |T (A) ∩ T (B)| = 0 and |T (A) ∩ T (A)| = 7. From those results, we obtain {0, 7} ⊆ I3 (K10 \ K5 ). ⋄. Lemma 2.26 {0, 3, 6} ⊆ I3 (K10 \ K6 ).. Proof: Let B1 , B2 and B3 be three T3 -designs of K10 \ K6 , where 34.

(39) B1 = {(5, 8, 7, 4, 9)3 , (6, 9, 8, 1, 7)3 , (1, 10, 9, 2, 8)3 , (2, 7, 10, 3, 8)3 , (3, 7, 9, 5, 10)3 , (4, 8, 10, 6, 7)3 }, B2 = {(6, 8, 7, 5, 9)3 , (1, 9, 8, 2, 7)3 , (2, 10, 9, 3, 8)3 , (3, 7, 10, 4, 8)3 , (4, 7, 9, 6, 10)3 , (5, 8, 10, 1, 7)3 } and B3 = {(5, 8, 7, 4, 9)3 , (6, 9, 8, 2, 7)3 , (2, 10, 9, 3, 8)3 , (3, 7, 10, 1, 8)3 , (1, 7, 9, 5, 10)3 , (4, 8, 10, 6, 7)3 }. Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 3 and |T (B1 ) ∩ T (B1 )| = 6. From those results, we obtain {0, 3, 6} ⊆ I3 (K10 \ K6 ). ⋄. Lemma 2.27 I3 (10) = J3 (10).. Proof: The graph K10 can be regarded as a union of graphs K10 \ K6 and K6 . Therefore, we have I3 (10) ⊇ I3 (K10 \K6 )+I3 (K6 ) ⊇ {0, 3, 6}+J3 (K6 ) = {0, 3, 6} + {0, 1, 2, 3} = {0, 1, 2, . . . , 9} = J2 (10).⋄. Lemma 2.28 {0, 4, 8} ⊆ I3 (K11 \ K6 ), and I3 (11) = J3 (11).. Proof: Let B1 , B2 and B3 be three T3 -designs of K11 \ K6 , where. 35.

(40) B1 = {(11, 10, 6, 7, 4)3 , (10, 9, 5, 8, 6)3 , (9, 8, 4, 11, 1)3 , (11, 8, 3, 7, 1)3 , (11, 9, 2, 10, 4)3 , (10, 8, 1, 9, 3)3 , (5, 11, 7, 9, 6)3 , (2, 8, 7, 10, 3)3 }, B2 = {(11, 10, 6, 9, 2)3 , (10, 9, 5, 8, 2)3 , (4, 8, 9, 11, 2)3 , (3, 11, 8, 10, 4)3 , (6, 7, 8, 1, 9)3 , (7, 11, 1, 10, 3)3 , (2, 10, 7, 5, 11)3 , (3, 9, 7, 4, 11)3 } and B3 = {(7, 8, 6, 9, 4)3 , (1, 7, 11, 6, 10)3 , (2, 7, 10, 5, 9)3 , (7, 9, 3, 8, 1)3 , (5, 8, 11, 3, 10)3 , (2, 8, 9, 11, 4)3 , (1, 9, 10, 11, 2)3 , (8, 10, 4, 7, 5)3 }. Then |T (B1 ) ∩ T (B3 )| = 0, |T (B1 ) ∩ T (B2 )| = 4 and |T (B1 ) ∩ T (B1 )| = 8. From those results, we have {0, 4, 8} ⊆ I3 (K11 \ K6 ). The graph K11 can be regarded as a union of graphs K11 \ K6 and K6 . Therefore, we have I3 (11) ⊇ I3 (K11 \ K6 ) + I3 (K6 ) ⊇ {0, 4, 8} + J3 (K6 ) = {0, 4, 8} + {0, 1, 2, 3} = {0, 1, 2, . . . , 11} = J3 (11). ⋄. Lemma 2.29 {0, 5, 15} ⊆ I3 (K5,5,5 ).. Proof: Let B1 , B2 and B3 be three T3 -designs of K5,5,5 , where B1 = {(11, 6, 1, 7, 12)3 , (11, 10, 2, 6, 12)3 , (11, 9, 3, 10, 12)3 , (11, 8, 4, 9, 12)3 , (11, 7, 5, 8, 12)3 , (13, 8, 1, 14, 9)3 , (13, 7, 2, 14, 8)3 , (13, 6, 3, 14, 7)3 , 36.

(41) (13, 10, 4, 14, 6)3 , (13, 9, 5, 14, 10)3 , (15, 1, 10, 5, 12)3 , (15, 2, 9, 1, 12)3 , (15, 3, 8, 2, 12)3 , (15, 4, 7, 3, 12)3 , (15, 5, 6, 4, 12)3 }, B2 = {(11, 6, 2, 7, 12)3 , (11, 10, 3, 6, 12)3 , (11, 9, 4, 10, 12)3 , (11, 8, 5, 9, 12)3 , (11, 7, 1, 8, 12)3 , (13, 8, 2, 14, 9)3 , (13, 7, 3, 14, 8)3 , (13, 6, 4, 14, 7)3 , (13, 10, 5, 14, 6)3 , (13, 9, 1, 14, 10)3 , (15, 2, 10, 1, 12)3 , (15, 3, 9, 2, 12)3 , (15, 4, 8, 3, 12)3 , (15, 5, 7, 4, 12)3 , (15, 1, 6, 5, 12)3 } and B3 = {(12, 10, 5, 7, 13)3 , (12, 6, 2, 10, 13)3 , (12, 9, 3, 6, 13)3 , (12, 8, 4, 9, 13)3 , (12, 7, 1, 8, 13)3 , (11, 8, 5, 14, 9)3 , (11, 7, 2, 14, 8)3 , (11, 10, 3, 14, 7)3 , (11, 6, 4, 14, 10)3 , (11, 9, 1, 14, 6)3 , (15, 5, 6, 1, 13)3 , (15, 2, 9, 5, 13)3 , (15, 3, 8, 2, 13)3 , (15, 4, 7, 3, 13)3 , (15, 1, 10, 4, 13)3 }. Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 5 and |T (B1 ) ∩ T (B1 )| = 15. From those results, we have {0, 5, 15} ⊆ I3 (K5,5,5 ). ⋄. Lemma 2.30 {0, 30} ⊆ I3 (K5,5,5,5 ).. Proof: Let A and B be two T3 -designs of K5,5,5,5 , where A = {(6, 11, 1, 7, 12)3 , (10, 11, 2, 17, 5)3 , (7, 13, 2, 14, 5)3 , (8, 16, 4, 11, 7)3 , (10, 18, 4, 20, 15)3 , (9, 5, 18, 6, 4)3 , (19, 9, 12, 2, 6)3 , (17, 10, 13, 4, 9)3 , (7, 4, 17, 1, 8)3 , 37.

(42) (16, 13, 9, 1, 12)3 , (14, 17, 6, 12, 4)3 , (7, 18, 15, 19, 4)3 , (10, 20, 14, 18, 1)3 , (6, 16, 15, 5, 11)3 , (1, 15, 10, 3, 9)3 , (3, 20, 7, 19, 2)3 , (5, 20, 6, 19, 13)3 , (3, 18, 12, 16, 10)3 , (12, 17, 8, 14, 4)3 , (11, 9, 20, 12, 10)3 , (3, 6, 13, 18, 2)3 , (15, 9, 2, 20, 8)3 , (1, 19, 14, 9, 17)3 , (3, 19, 8, 15, 4)3 , (3, 15, 17, 11, 3)3 , (13, 20, 1, 16, 7)3 , (8, 13, 5, 7, 14)3 , (5, 10, 19, 11, 16)3 , (18, 11, 8, 2, 16)3 , (3, 14, 16, 5, 12)3 } and B = {(7, 11, 2, 8, 12)3 , (6, 11, 3, 17, 1)3 , (8, 13, 3, 14, 1)3 , (9, 16, 5, 11, 8)3 , (6, 18, 5, 20, 15)3 , (10, 1, 18, 7, 5)3 , (19, 10, 12, 3, 7)3 , (17, 6, 13, 5, 10)3 , (8, 5, 17, 2, 9)3 , (16, 13, 10, 2, 12)3 , (14, 17, 7, 12, 5)3 , (8, 18, 15, 19, 5)3 , (6, 20, 14, 18, 2)3 , (7, 16, 15, 1, 11)3 , (2, 15, 6, 4, 10)3 , (4, 20, 8, 19, 3)3 , (1, 20, 7, 19, 13)3 , (4, 18, 12, 16, 6)3 , (12, 17, 9, 14, 5)3 , (11, 10, 20, 12, 6)3 , (4, 7, 13, 18, 3)3 , (15, 10, 3, 20, 9)3 , (2, 19, 14, 10, 17)3 , (4, 19, 9, 15, 5)3 , (4, 15, 17, 11, 4)3 , (13, 20, 2, 16, 8)3 , (9, 13, 1, 8, 14)3 , (1, 6, 19, 11, 16)3 , (18, 11, 9, 3, 16)3 , (4, 14, 16, 1, 12)3 }. Then |T (A) ∩ T (B)| = 0 and |T (A) ∩ T (A)| = 30. From those results, we have {0, 30} ⊆ I3 (K5,5,5,5 ). ⋄. Lemma 2.31 {0, 19} ⊆ I3 (K15 \ K5 ).. Proof: Let A and B be two T3 -designs of K15 \ K5 , where A = {(1, 6, 8, 13, 9)3 , (1, 7, 9, 4, 6)3 , (1, 10, 11, 8, 14)3 , (2, 6, 7, 5, 8)3 , (2, 8, 9, 11, 3)3 ,. 38.

(43) (3, 13, 14, 6, 12)3 , (8, 12, 10, 14, 11)3 , (11, 13, 6, 5, 9)3 , (3, 9, 15, 10, 7)3 , (4, 7, 11, 15, 6)3 , (5, 11, 12, 4, 14)3 , (2, 12, 15, 5, 13)3 , (1, 13, 12, 7, 3)3 , (1, 15, 14, 5, 10)3 , (7, 15, 13, 2, 11)3 , (9, 12, 14, 7, 8)3 , (4, 13, 10, 2, 14)3 , (6, 9, 10, 3, 12)3 , (4, 15, 8, 3, 6)3 } and B = {(2, 7, 9, 13, 10)3 , (2, 8, 10, 5, 7)3 , (2, 6, 11, 9, 14)3 , (3, 7, 8, 1, 9)3 , (3, 9, 10, 11, 4)3 , (4, 13, 14, 7, 12)3 , (9, 12, 6, 14, 11)3 , (11, 13, 7, 1, 10)3 , (4, 10, 15, 6, 8)3 , (5, 8, 11, 15, 7)3 , (1, 11, 12, 5, 14)3 , (3, 12, 15, 1, 13)3 , (2, 13, 12, 8, 4)3 , (2, 15, 14, 1, 6)3 , (8, 15, 13, 3, 11)3 , (10, 12, 14, 8, 9)3 , (5, 13, 6, 3, 14)3 , (7, 10, 6, 4, 12)3 , (5, 15, 9, 4, 7)3 }. Then |T (A) ∩ T (B)| = 0 and |T (A) ∩ T (A)| = 19. From those results, we have {0, 19} ⊆ I3 (K15 \ K5 ). ⋄. Lemma 2.32 {0, 4, 8, 12, 14, 18} ⊆ I3 (K15 \ K6 ) and I3 (15) = J3 (15).. Proof: Let B1 , B2 , . . ., B6 be six T3 -designs of K15 \ K6 , where B1 = {(1, 7, 8, 14, 10)3 , (3, 8, 11, 14, 15)3 , (2, 9, 14, 7, 12)3 , (1, 9, 15, 8, 12)3 , (5, 13, 14, 4, 12)3 , (6, 9, 12, 5, 7)3 , (2, 10, 13, 12, 1)3 , (4, 7, 15, 12, 10)3 , (3, 13, 15, 2, 7)3 , (11, 15, 5, 10, 7)3 , (3, 12, 14, 1, 10)3 , (3, 9, 10, 8, 2)3 , (1, 11, 13, 7, 3)3 , (10, 15, 6, 8, 5)3 , (2, 12, 11, 9, 5)3 , (4, 8, 13, 6, 14)3 , (6, 11, 7, 9, 8)3 , (10, 11, 4, 9, 13)3 }, 39.

(44) B2 = {(2, 7, 8, 14, 10)3 , (4, 8, 11, 14, 15)3 , (3, 9, 14, 7, 12)3 , (2, 9, 15, 8, 12)3 , (6, 13, 14, 5, 12)3 , (1, 9, 12, 6, 7)3 , (3, 10, 13, 12, 2)3 , (5, 7, 15, 12, 10)3 , (4, 13, 15, 3, 7)3 , (11, 15, 6, 10, 7)3 , (4, 12, 14, 2, 10)3 , (4, 9, 10, 8, 3)3 , (2, 11, 13, 7, 4)3 , (10, 15, 1, 8, 6)3 , (3, 12, 11, 9, 6)3 , (5, 8, 13, 1, 14)3 , (1, 11, 7, 9, 8)3 , (10, 11, 5, 9, 13)3 }, B3 = {(3, 8, 9, 14, 10)3 , (5, 9, 11, 14, 15)3 , (2, 7, 14, 8, 12)3 , (3, 7, 15, 9, 12)3 , (1, 13, 14, 4, 12)3 , (6, 7, 12, 1, 8)3 , (2, 10, 13, 12, 3)3 , (4, 8, 15, 12, 10)3 , (5, 13, 15, 2, 8)3 , (11, 15, 1, 10, 8)3 , (5, 12, 14, 3, 10)3 , (5, 7, 10, 9, 2)3 , (3, 11, 13, 8, 5)3 , (10, 15, 6, 9, 1)3 , (2, 12, 11, 7, 1)3 , (4, 9, 13, 6, 14)3 , (6, 11, 8, 7, 9)3 , (10, 11, 4, 7, 13)3 }, B4 = {(2, 7, 8, 14, 10)3 , (1, 8, 11, 14, 15)3 , (3, 9, 14, 7, 12)3 , (2, 9, 15, 8, 12)3 , (5, 13, 14, 4, 12)3 , (6, 9, 12, 5, 7)3 , (3, 10, 13, 12, 2)3 , (4, 7, 15, 12, 10)3 , (1, 13, 15, 3, 7)3 , (11, 15, 5, 10, 7)3 , (1, 12, 14, 2, 10)3 , (1, 9, 10, 8, 3)3 , (2, 11, 13, 7, 1)3 , (10, 15, 6, 8, 5)3 , (3, 12, 11, 9, 5)3 , (4, 8, 13, 6, 14)3 , (6, 11, 7, 9, 8)3 , (10, 11, 4, 9, 13)3 }, B5 = {(2, 7, 8, 14, 10)3 , (3, 8, 11, 14, 15)3 , (1, 9, 14, 7, 12)3 , (2, 9, 15, 8, 12)3 , (5, 13, 14, 4, 12)3 , (6, 9, 12, 5, 7)3 , (1, 10, 13, 12, 2)3 , (4, 7, 15, 12, 10)3 , (3,. 40.

(45) 13, 15, 1, 7)3 , (11, 15, 5, 10, 7)3 , (3, 12, 14, 2, 10)3 , (3, 9, 10, 8, 1)3 , (2, 11, 13, 7, 3)3 , (10, 15, 6, 8, 5)3 , (1, 12, 11, 9, 5)3 , (4, 8, 13, 6, 14)3 , (6, 11, 7, 9, 8)3 , (10, 11, 4, 9, 13)3 } and B6 = {(1, 8, 7, 14, 10)3 , (3, 7, 11, 14, 15)3 , (2, 9, 14, 8, 12)3 , (1, 9, 15, 7, 12)3 , (5, 13, 14, 4, 12)3 , (6, 9, 12, 5, 8)3 , (2, 10, 13, 12, 1)3 , (4, 8, 15, 12, 10)3 , (3, 13, 15, 2, 8)3 , (11, 15, 5, 10, 8)3 , (3, 12, 14, 1, 10)3 , (3, 9, 10, 7, 2)3 , (1, 11, 13, 8, 3)3 , (10, 15, 6, 7, 5)3 , (2, 12, 11, 9, 5)3 , (4, 7, 13, 6, 14)3 , (6, 11, 8, 9, 7)3 , (10, 11, 4, 9, 13)3 }. Then |T (B1 ) ∩ T (B2 )| = 0, |T (B1 ) ∩ T (B3 )| = 4, |T (B1 ) ∩ T (B4 )| = 8, |T (B1 ) ∩ T (B5 )| = 12, |T (B1 ) ∩ T (B6 )| = 14 and |T (B1 ) ∩ T (B1 )| = 18. From those results, we have {0, 4, 8, 12, 14, 18} ⊆ I3 (K15 \ K6 ). The graph K15 can be regarded as a union of graphs K15 \ K6 and K6 . Therefore, we have I3 (15) ⊇ I3 (K15 \ K6 ) + I3 (K6 ) ⊇ {0, 4, 8, 12, 14, 18} + J3 (K6 ) = {0, 4, 8, 12, 14, 18} + {0, 1, 2, 3} = {0, 1, 2, . . . , 21} = J3 (15). ⋄. Lemma 2.33 {0, 21} ⊆ I3 (K16 \ K6 ) and I3 (16) = J3 (16).. Proof: I3 (K16 \ K6 ) ⊇ I3 (K5,5,5 ) + 2 · I3 (K6 ) ⊇ {0, 15} + 2 · J3 (K6 ) = 41.

(46) {0, 15} + 2 · {0, 1, 2, 3} = {0, 15} + {0, 1, 2, . . . , 6} ⊇ {0, 21}. I3 (16) ⊇ I3 (K5,5,5 ) + 3 · I3 (K6 ) ⊇ {0, 5, 15} + 3 · J3 (K6 ) = {0, 5, 15} + 3 · {0, 1, 2, 3} = {0, 5, 15} + {0, 1, 2, . . . , 9}) = {0, 1, 2, . . . , 24} = J3 (16). ⋄. Lemma 2.34 I3 (20) = J3 (20) and I3 (21) = J3 (21).. Proof: I3 (20) ⊇ I3 (K5,5,5 ) + 2 · I3 (K10 \ K5 ) + I3 (K10 ) ⊇ {0, 7, 15} + 2 · I3 (K10 \ K5 ) + J3 (K10 ) = {0, 7, 15} + 2 · {0, 7} + {0, 1, 2, . . . , 9} = {0, 7, 15} + {0, 7, 14} + {0, 1, 2, . . . , 9} = {0, 1, 2, . . . , 38} = J3 (20). I3 (21) ⊇ I3 (K5,5,5 ) + 2 · I3 (K11 \ K6 ) + I3 (K11 ) = I3 (K5,5,5 ) + 2 · I3 (K11 \ K6 ) + J3 (K11 ) ⊇ {0, 7, 15} + 2 · {0, 4, 8} + {0, 1, 2, . . . , 11} = {0, 7, 15} + {0, 4, 8, 12, 16} + {0, 1, 2, . . . , 11} = {0, 1, 2, . . . , 42} = J3 (21). ⋄. Lemma 2.35 I3 (25) = J3 (25) and I3 (26) = J3 (26).. Proof: I3 (25) ⊇ I3 (K5,5,5,5 ) + 3 · I3 (K10 \ K5 ) + I3 (K10 ) ⊇ I3 (K5,5,5,5 ) + 3 · I3 (K10 \K5 )+J3 (K10 ) = {0, 30}+3·{0, 7}+{0, 1, 2} ⊇ {0, 30}+{0, 7, 14, 21}+ {0, 1, 2, . . . , 9} = {0, 1, 2, . . . , 60} = J3 (25). I3 (26) ⊇ I3 (K5,5,5,5 ) + 3 · I3 (K11 \ K6 ) + I3 (K11 ) = I3 (K5,5,5,5 ) + 3 · I3 (K11 \ 42.

(47) K6 ) + J3 (K11 ) ⊇ {0, 30} + 3 · {0, 4, 8} + {0, 1, 2, . . . , 11} = {0, 30} + {0, 4, 8, 12, 16, 20, 24} + {0, 1, 2, . . . , 11} = {0, 1, 2, . . . 65} = J3 (26). ⋄. 3. Main Theorem. For proving the Main Theorem, we use an extremely useful construction: Construction of GDDs [8].. 43.

(48) .. Figure 11: Construction of GDD. Theorem 3.1 Ii (n) = Ji (n) for n ≡ 0, 1, 5, 6 (mod 30).. Proof. The cases when n = 5 and 6 follow from Lemmas 2.2, 2.3, 2.12 and 2.24, respectively. Assume that n ≥ 30. Let n = 30k + a with k ≥ 1 and a ∈ {0, 1, 5, 6}. Start from a GDD(6k, 2, 3) by Theorem 1.21. Give each 44.

(49) point of the GDD weight 5. By Lemma 2.6, 2.17 and 2.29, there is a pair of Ti -designs of K5,5,5 with α common triangles, α ∈ {0, 15}. Then, apply Figure 11 to obtain a pair of Ti -designs of K3k·10 with. ∑x i=1. α common blocks,. where x is the number of blocks of the GDD(6k, 2, 3) and. x=. 1 6k(6k − 1) ·( − 3k) = k(6k − 1) − k = 2k(3k − 1) 3 2. .. Case 1. a = 0 and n = 30k = 10(3k). By Lemmas 2.4, 2.15 and 2.27, take a pair of Ti -designs of K10 with γ common triangles, γ ∈ Ji (K10 ). There are a pair of Ti -designs of order 30k with. ∑x i=1. α+. ∑3k i=1. γ common tri-. angles. Thus, we have Ii (30k) ⊇ x·Ii (K5,5,5 )+3k·Ii (K10 ) ⊇ x·{0, 15}+ 3k · Ji (K10 ) = x · {0, 15} + 3k · {0, 1, 2, . . . , 9} = {0, 15, 30, . . . , 15x} + {0, 1, 2, . . . , 27k} = {0, 1, 2, . . . , 15x + 27k} = {0, 1, 2, . . . , 3k(30k − 1)} = Ji (30k).. 45.

(50) K10 K10 3k. .. K10 Figure 12: Decomposition of Case 1. Case 2. a = 1 and n = 30k + 1 = 10(3k) + 1. By Lemmas 2.5, 2.16 and 2.28, take 3k pair of Ti -designs of K11 with γ common triangles, γ ∈ Ji (K11 ). There are a pair of Ti -designs of order 30k + 1 with. ∑x i=1. α+. ∑3k i=1. γ. common triangles. Thus, we have Ii (30k + 1) ⊇ x · Ii (K5,5,5 ) + 3k · Ii (K11 ) ⊇ x · {0, 15} + 3k · Ji (K11 ) = x · {0, 15} + 3k · {0, 1, 2, . . . , 11} = {0, 15, 30, . . . , 15x} + {0, 1, 2, . . . , 33k} = {0, 1, 2, . . . , 15x + 33k} = {0, 1, 2, . . . , 3k(30k + 1)} = Ji (30k + 1).. 46.

(51) K11 K11 3k K11 . Figure 13: Decomposition of Case 2. Case 3. a = 5 and n = 30k + 5 = 10(3k) + 5. By Lemmas 2.8, 2.20 and 2.32, take a pair of Ti -designs of K15 with β common triangles, β ∈ Ji (15), take 3k − 1 pair of Ti -designs of K15 \ K5 with γ common triangles, γ ∈ {0, 19}. There are a pair of Ti -design of order 30k+5 with ∑x i=1. α+β+. ∑3k−1 i=1. γ common triangles. Thus, we have I(30k + 5) ⊇. x·Ji (K5,5,5 )+Ii (15)+(3k−1)·Ii (K15 \K5 ) ⊇ x·{0, 15}+Ji (15)+(3k−1)· {0, 19} = {0, 15, 30, . . . , 15x} + {0, 1, 2, . . . , 21} + {0, 19, 38, . . . , 19(3k − 1)} = {0, 1, 2, . . . , 15x + 21 + 19(3k − 1)} = {0, 1, 2, . . . , (15k + 2) · (6k + 1)} = Ji (30k + 5).. 47.

(52) K15 K15 \ K5 3k. K15 \ K5. .. Figure 14: Decomposition of Case 3. Case 4. a = 6 and n = 30k + 6 = 10(3k) + 6. By Lemmas 2.9, 2.21 and 2.33, take a pair of Ti -designs of K16 with β common triangles, β ∈ Ji (16), take 3k − 1 pair of Ti -designs of K16 \ K6 with γ common triangles, γ ∈ {0, 21}. There are a pair of Ti -design of order 30k+6 with ∑x i=1. α+β+. ∑3k−1 i=1. γ common triangles. Thus, we have Ii (30k + 6) ⊇. x·Ii (K5,5,5 )+Ii (16)+(3k−1)·Ii (K16 \K6 ) ⊇ x·{0, 15}+Ji (16)+(3k−1)· {0, 21} = {0, 15, 30, . . . , 15x} + {0, 1, 2, . . . , 24} + {0, 21, 42, . . . , 21(3k − 1)} = {0, 1, 2, . . . , 15x + 24 + 21(3k − 1)} = {0, 1, 2, . . . , (15k + 3) · (6k + 1)} = Ji (30k + 6).. 48.

(53) K16 K16 \ K6 3k. K16 \ K6. .. Figure 15: Decomposition of Case 4. This completes the proof. ⋄. Theorem 3.2 Ii (n) = Ji (n) for n ≡ 10, 11, 15, 16 (mod 30).. Proof. The cases when n = 10, 11, 15 and 16 follow from Lemmas 2.4, 2.5, 2.8, 2.9, 2.15, 2.16, 2.20, 2.21, 2.27, 2.28, 2.32 and 2.33, respectively. Assume that n ≥ 30. Let n = 30k + a with k ≥ 1 and a ∈ {10, 11, 15, 16}. Start from a GDD(6k + 2, 2, 3) by Theorem 1.21. Give each point of the GDD weight 5. By Lemma 2.6, 2.17 and 2.29, there is a pair of Ti -designs of K5,5,5 with α common triangles, α ∈ Ji (K5,5,5 ). Then, apply Figure 11 to obtain a pair of Ti -designs of K(3k+1)·10 with. ∑x i=1. α common triangles, where x is the number. of blocks of the GDD(6k + 2, 2, 3) and 49.

(54) x=. 1 (6k + 2)(6k + 1) ·( − (3k + 1)) = 2k(3k + 1) 3 2. ... Case 1. a = 10 and n = 30k + 10 = 10(3k + 1). By Lemmas 2.4, 2.15 and 2.27, take 3k +1 pair of Ti -designs of K10 with γ common triangles, γ ∈ Ji (K10 ). There are a pair of Ti -designs of order 30k + 10 with ∑3k+1 i=1. ∑x i=1. α+. γ common triangles. Thus, we have I(30k + 10) ⊇ x · Ii (K5,5,5 ) +. (3k +1)·Ii (K10 ) ⊇ x·{0, 15}+(3k +1)·Ji (K10 ) = {0, 15, 30, . . . , 15x}+ (3k+1)·{0, 1, 2, . . . , 9} = {0, 15, 30, . . . , 15x}+{0, 1, 2, . . . , 9(3k+1)} = {0, 1, 2, . . . , 15x + 9(3k + 1)} = {0, 1, 2, . . . , (3k + 1) · (30k + 9)} = Ji (30k + 10). K10 K10 3k + 1. .. K10 Figure 16: Decomposition of Case 1. 50.

(55) Case 2. a = 11 and n = 30k + 11 = 10(3k + 1) + 1. By Lemmas 2.5, 2.16 and 2.28, take 3k +1 pair of Ti -designs of K11 with γ common triangles, γ ∈ Ji (K11 ). There are a pair of Ti -designs of order 30k + 11 with ∑3k+1 i=1. ∑x i=1. α+. γ common triangles. Thus, we have I(30k + 11) ⊇ x · Ii (K5,5,5 ) +. (3k +1)·Ii (K11 ) ⊇ x·{0, 15}+(3k +1)·Ji (K11 ) = {0, 15, 30, . . . , 15x}+ (3k + 1) · {0, 1, 2, . . . , 11} = {0, 15, 30, . . . , 15x} + {0, 1, 2, . . . , 11(3k + 1)} = {0, 1, 2, . . . , 15x+11(3k+1)} = {0, 1, 2, . . . , (30k+11)·(3k+1)} = J(30k + 11). K11 K11 3k + 1 K11 . Figure 17: Decomposition of Case 2. Case 3. a = 15 and n = 30k + 15 = 10(3k + 1) + 5. By Lemmas 2.8, 2.20 and 2.32, take a pair of Ti -designs of K15 with β common triangles, β ∈ Ji (15), take 3k pair of Ti -designs of K15 \ K5 with γ common triangles, γ ∈ Ji (K15 \ K5 ). There are a pair of Ti -designs of order 51.

(56) n = 30k + 15 with. ∑x i=1. α+β+. ∑3k i=1. γ common triangles. Thus,. we have I(30k + 15) ⊇ x · Ii (K5,5,5 ) + Ii (15) + (3k) · Ii (K15 \ K5 ) ⊇ x · {0, 15}+Ji (15)+(3k)·{0, 19} = {0, 15, 30, . . . , 15x}+{0, 1, 2, . . . , 21}+ {0, 19, 38, . . . , 19(3k)} = {0, 1, 2, . . . , 15x + 21 + 19(3k)} = {0, 1, 2, . . . , (6k + 3) · (15k + 7)} = Ji (30k + 15). K15 K15 \ K5 3k + 1. K15 \ K5. .. Figure 18: Decomposition of Case 3. Case 4. a = 16 and n = 30k + 16 = 10(3k + 1) + 6. By Lemmas 2.9, 2.21 and 2.33, take a pair of Ti -designs of K16 with β common triangles, β ∈ Ji (16), take 3k pair of Ti -designs of K16 \ K6 with γ common triangles, γ ∈ Ji (K16 \ K6 ). There are a pair of Ti -designs of order 30k + 16 with. ∑x i=1. α+β+. ∑3k i=1. γ common triangles. Thus, we. have Ii (30k + 16) ⊇ x · Ii (K5,5,5 ) + Ii (16) + (3k) · Ii (K16 \ K6 ) ⊇ x · {0, 15}+Ji (16)+(3k)·{0, 21} = {0, 15, 30, . . . , 15x}+{0, 1, 2, . . . , 24}+ 52.

(57) {0, 21, 42, . . . , 21(3k)} = {0, 1, 2, . . . , 15x + 24 + 21(3k)} = {0, 1, 2, . . . , (15k + 8) · (6k + 3)} = Ji (30k + 16). K16 K16 \ K6 3k + 1. K16 \ K6. .. Figure 19: Decomposition of Case 4. This completes the proof. ⋄. Theorem 3.3 Ii (n) = Ji (n) for n ≡ 20, 21, 25, 26 (mod 30).. Proof. The cases when n = 20, 21, 25 and 26 follow from Lemmas 2.10, 2.11, 2.22, 2.23, 2.33 and 2.34. Assume that n ≥ 50. Let n = 30k + a with k ≥ 1 and a ∈ {20, 21, 25, 26}. Start from a GDD(6k + 4, {2, 4∗ }, 3) from Theorem 1.23. Give each point of the GDD weight 5. By Lemma 2.6, 2.17 and 2.29, there is a pair of Ti -designs of K5,5,5 with α common triangles, α ∈ J(K5,5,5 ). Then, apply Figure 11 to obtain a pair of Ti -design of K(3k+2)·10,20 with 53. ∑x i=1. α.

(58) common triangles, where x is the number of blocks of the GDD(6k + 4, 2, 3) and. x=. 1 (6k + 4)(6k + 3) ·( − (3k + 6)) = 6k(k + 1) 3 2. .. Case 1. a = 20 and n = 30k + 20 = 10(3k + 2). By Lemmas 2.10, 2.22 and 2.33, take a pair of Ti -designs of K20 with β common triangles, β ∈ Ji (20), take 3k pair of Ti -designs of K10 with γ common triangles, γ ∈ Ji (K10 ). There are a pair of Ti -designs of order 30k + 20 with ∑x i=1. α+β+. ∑3k i=1. γ common triangles. Thus, we have Ii (30k + 20) ⊇. x · Ii (K5,5,5 ) + Ii (20) + (3k) · Ii (K10 ) ⊇ x · {0, 15} + Ji (20) + (3k) · Ji (K10 ) = {0, 15, 30, . . . , 15x}+{0, 1, 2, . . . , 38}+(3k)·{0, 1, 2, . . . , 9} = {0, 15, 30, . . . , 15x} + {0, 1, 2, . . . , 38} + {0, 1, 2, . . . , 9(3k)} = {0, 1, 2, . . . , 15x+38+9(3k)} = {0, 1, 2, . . . , (3k+2)·(30k+19)} = Ji (30k+20).. 54.

(59) K10 K10 3k. K10 .. K20 Figure 20: Decomposition of Case 1. Case 2. a = 21 and n = 30k + 21 = 10(3k + 2) + 1. By Lemmas 2.10, 2.22 and 2.33, take a pair of Ti -designs of K21 with β common triangles, β ∈ Ji (21), take 3k pair of Ti -designs of K11 with γ common triangles, γ ∈ Ji (K11 ). There are a pair of Ti -designs of order 30k + 21 with. ∑x i=1. α+β +. ∑3k i=1. γ common triangles.. Thus,. we have Ii (30k + 21) ⊇ x · Ii (K5,5,5 ) + Ii (21) + (3k) · Ii (K11 ) ⊇ x · {0, 15}+Ji (21)+(3k)·Ji (K11 ) = {0, 15, 30, . . . , 15x}+{0, 1, 2, . . . , 42}+ (3k) · {0, 1, 2, . . . , 11} = {0, 15, 30, . . . , 15x} + {0, 1, 2, . . . , 42} + {0, 1, 2, . . . , 11(3k)} = {0, 1, 2, . . . , 15x + 42 + 11(3k)} = {0, 1, 2, . . . , (3k + 2) · (30k + 21)} = J(30k + 21).. 55.

(60) K11 K11 3k. K11 K21 . Figure 21: Decomposition of Case 2. Case 3. a = 25 and n = 30k + 25 = 10(3k + 2) + 5. By Lemmas 2.11, 2.23 and 2.34, take a pair of Ti -designs of K25 with β common triangles, β ∈ Ji (25), take 3k pair of Ti -designs of K15 \ K5 with γ common triangles, γ ∈ Ji (K15 \ K5 ). There are a pair of Ti -designs of order 30k + 25 with. ∑x i=1. α+β+. ∑3k i=1. γ common triangles. Thus, we have. Ii (30k + 25) ⊇ x · Ii (K5,5,5 ) + Ii (25) + (3k) · Ii (K15 \ K5 ) ⊇ x · {0, 15} + Ji (25) + (3k) · {0, 19} = {0, 15, 30, . . . , 15x} + {0, 1, 2, . . . , 60} + {0, 19, 38, . . . , 19(3k)} = {0, 1, 2, v . . . , 15x + 60 + 19(3k)} = {0, 1, 2, . . . , (6k + 5) · (10k + 12)} = Ji (30k + 25).. 56.

(61) K15 \ K5 K15 \ K5 K15 \ K5. 3k. K25. .. Figure 22: Decomposition of Case 3. Case 4. a = 26 and n = 30k + 26 = 10(3k + 2) + 6. By Lemmas 2.11, 2.23 and 2.34, take a pair of Ti -designs of K26 with β common triangles, β ∈ Ji (26), take 3k pair of Ti -designs of K16 \ K6 with γ common triangles, γ ∈ Ji (K16 \ K6 ). There are a pair of Ti -designs of order 30k + 26 with ∑x i=1. α+β +. ∑3k i=1. γ common triangles. Thus, we have Ii (30k +26) ⊇ x·. Ii (K5,5,5 )+Ii (26)+(3k)·Ii (K16 \K6 ) ⊇ x·{0, 15}+Ji (26)+(3k)·{0, 21} = {0, 15, 30, . . . , 15x}+{0, 1, 2, . . . , 65}+{0, 21, 42, . . . , 21(3k)} = {0, 1, 2, . . . , 15x+65+21(3k)} = {0, 1, 2, . . . , (6k+5)·(15k+13)} = Ji (30k+26).. 57.

(62) K16 \ K6 K16 \ K6 K16 \ K6. 3k. K26. .. Figure 23: Decomposition of Case 4. This completes the proof. ⋄. Combining the Theorems 3.1, 3.2 ans 3.3, we obtain the Main Theorem.. Theorem 3.4 Ii (n) = Ji (n) for n ≡ 0, 1 (mod 5), n ≥ 6 and I1 (5) = J1 (5).. References [1] J. C. Bermond and J. Schonheim, G-decomposition of Kn , where G has four vertices or less, Discrete Math., 19(1977), 113-120. [2] E. J. Billington and D. L. Kreher, The intersection problem for small G-designs, Australasian Journal of Combinatorics, 12 (1995), 239-258.. 58.

(63) [3] E. J. Billington, Emine Sul. Yazici, C. C. Lindner, The triangle intersection problem for K4 − e designs, Utilitas Mathematica, 73 (2007), 3-21. [4] Yeow Meng Chee, Steiner Triple Systems Intersecting in Pairwise Disjoint Blocks, The Electronic J. of Combin., 11(2004), ♯R27. [5] C. J. Colbourn, D. G. Hoffman, R. Rees, A new class of group divisible designs with block size three, J. Combin. Theory Ser. A, 59(1992), 73-89. [6] S. El-Zanati and C. A. Rodger, Blocking sets in G-design, Ars Combin., 35 (1993), 237-251. [7] C. C. Lindner and C. A. Rodger, Design theory, Boca Raton: CRC Press series on discrete mathematics and its applications, 1997. [8] R. M. Wilson, Constructions and used of pairwise balanced designs, Math. Centre Tracts, 55(1974), 18-41.. 59.

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