The existence of perfect T(K1,2p) -triple systems

Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

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The existence of perfect T(K1,2p )-triple systems夡 Yuanyuan Liu, Qingde Kang∗ Institute of Mathematics, Hebei Normal University, Shijiazhuang 050016, PR China

A R T I C L E

I N F O

Article history: Received 25 January 2007 Received in revised form 26 November 2007 Accepted 10 March 2008 Available online 19 March 2008 Keywords: T(G)-Triple Perfect T(G)-triple system Star graph

A B S T R A C T

Let G be a subgraph of Kn . The graph obtained from G by replacing each edge with a 3-cycle whose third vertex is distinct from other vertices in the configuration is called a T(G)-triple. An edge-disjoint decomposition of 3Kn into copies of T(G) is called a T(G)-triple system of order n. If, in each copy of T(G) in a T(G)-triple system, one edge is taken from each 3-cycle (chosen so that these edges form a copy of G) in such a way that the resulting copies of G form an edge-disjoint decomposition of Kn , then the T(G)-triple system is said to be perfect. The set of positive integers n for which a perfect T(G)-triple system exists is called its spectrum. Earlier papers by authors including Billington et al. determined the spectra for cases where G is any subgraph of K4 . In this paper, we will focus on star graphs K1,2p and discuss the existence for perfect T(K1,2p )-triple systems. In particular, for odd prime power p, the spectrum is completely determined. © 2008 Elsevier B.V. All rights reserved.

1. Introduction Denote an edge in Kn on vertices x and y by xy or yx, and denote a 3-cycle on vertices x, y, z by (x, y, z) or (x, z, y) (or any cyclic shift of these). Let G be a subgraph of Kn . Let T(G) be a collection of triples obtained by replacing each edge ab ∈ E(G) with a triple (a, b, c), where c ∈/ V(G) and c does not occur in any other triple of T(G). The graph formed in this way, by taking a triangle or triple on each edge of G, will be called a T(G)-triple. In a T(G)-triple, the vertices and edges in G and T(G) − G are called interior and exterior, respectively. A T(G)-triple system of order n is denoted by T(G, n) = (X, B), where X is the vertex set of Kn and B is an edge-disjoint collection of T(G)-triples which partitions the edge set of 3Kn . If the interior edges of the T(G)-triples, which form the copies of G, partition the edge set of Kn on X, then (X, B) is said to be a perfect T(G)-triple system. The spectrum for perfect T(G)-triple systems is the set of all positive integers n for which there exists a perfect T(G)-triple system of order n. ¨ ukçifçi ¨ The concepts of T(G)-triple, T(G)-triple system and perfect T(G)-triple system were first introduced in Kuç and Lindner (2004) and Lindner and Rosa (2008). A -fold triple system of order n is a pair (X, T), where X is the vertex set of Kn and T is a collection of edge disjoint triangles (or triples) which partitions the edge set of Kn . It is easy to see that all T(G)-triples are formed ¨ ukçifçi ¨ by piecing the triples of 3-fold triple system and the ”interior” of all these T(G)-triples form a G-decomposition of Kn . In Kuç ¨ ukçifçi ¨ and Lindner (2004), a hexagon triple system, i.e. T(K3 , v), is researched. Kuç and Linder have shown that the spectrum of perfect T(K3 , v) is the same as that of Steiner triple systems. Subsequently, Lucia Gionfriddo proved that every Steiner triple system is the ”interior” of certain perfect T(K3 , v). In Lindner and Rosa (2008), Linder proved the spectrum of perfect T(K4 , v) is precisely the set of v ≡ 1 mod 12. When v ≡ 1 mod 12, we could piece three triples of a 3-fold triple system of order v at a time to form a copy of K4 so that the collection of these copies is a partition of the edge set of Kv .

夡 ∗

Research supported by NSFC Grant 10671055 and NSFHB A2007000230. Corresponding author. E-mail address: [email protected] (Q. Kang).

0378-3758/$ - see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.jspi.2008.03.031

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Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

Billington et al. (2005) and Billington and Lindner (2006) gave the spectrum of perfect T(G, v) for any subgraph G of K4 . Then, for any subgraph G of K5 with six edges or less, the spectrum of perfect T(G, v) also has been determined in Kang et al. (to appear). These researches can be regarded as a kind of extension for T(K3 , v) and T(K4 , v). On the other hand, as we know, various star decompositions are interesting and have wide application in filing theory (see Tazawa, 1979). So we further considered the existence of perfect T(K1,k , v) in Liu et al. (to appear), where k was a prime power. In this paper, we continue the research for the perfect T(K1,k , v), where k ≡ 2 mod 4. A holey T(G)-triple system with m h-holes, denoted by T(G, hm ) briefly, is a pair ({S1 , . . . , Sm }, A), where each Si is a h-set (or hole), these Si are pairwise disjoint, and A is an edge-disjoint collection of T(G)-triples, each edge joining the vertices in distinct holes appears once as an interior edge and twice as an exterior edge in these T(G)-triples. An incomplete T(H)-triple system on the set X − Y, denoted by T(H, v : h), is a triple (X, Y, C), where Y ⊂ X, |X| = v, |Y| = h and C is an edge-disjoint collection of T(H)-triples, each edge on X excluding that on Y appears once as an interior edge and twice as an exterior edge in these T(G)-triples. The following necessary condition is trivial. Lemma 1.1. Let H be a simple graph with 2e edges. If there exists a T(H, v), then 4e|v(v − 1) and v is odd. In particular, the orders v ≡ 1 mod 4e and the orders v ≡ e mod 4e for e ≡ 1 mod 4, or the orders v ≡ 3e mod 4e for e ≡ 3 mod 4 satisfy these necessary conditions. Remark. Let H be a simple graph with 2e edges and e is the prime power, there exists a T(H, v) if and only if v ≡ 1 mod 4e and v ≡ e mod 4e for e ≡ 1 mod 4, or the orders v ≡ 3e mod 4e for e ≡ 3 mod 4. Below, in order to express the block in a T(K1,k , v) briefly, we use one of these two forms listed right.

Sometimes, the vertex sequences a1 , a2 , . . . , ak and b1 , b2 , . . . , bk can be replaced by some integer intervals. Let c and d be integers and c  d. The so-called integer interval [c, d] represents the ordered set (c, c + 1, . . . , d − 1, d). If c  d and c ≡ d mod t, the generalized integer interval [c, d]t represents the ordered set (c, c + t, . . . , d − t, d). Also, denote −[c, d] = (−c, −(c + 1), . . . , −(d − 1), −d) and −[c, d]t = (−c, −(c + t), . . . , −(d − t), −d). Furthermore, in the following constructions the base bloke B mod (m, n) means {B + (i, j) : i ∈ Zm , j ∈ Zn }, B mod (m, −) means {B + (i, 0) : i ∈ Zm }, and B mod (−, n) means {B + (0, j) : j ∈ Zn }. 2. Main constructions Theorem 2.1 (Liu et al., to appear). There exists a T(K1,k , 2km + 1) for any integers k, m  1. Theorem 2.2. Let H be a simple graph with 2p edges. If there exist T(H, 5p), T(H, 9p), T(H, (2p)3 ) and T(H, 5p : p), then there exists a T(H, 4mp + p) for all m  1. Construction. Take the vertex set (Z2m × Z2p ) ∪ ({∞} × Zp ). There are (4m + 1) · (4m + 1)p − 1/4 blocks in a T(H, 4mp + p). From Colbourn and Dinitz (2006), for m  3, there exist 3-GDD(2m ) = (Z2m , {Gj : 0  j  m − 1}, B)

for 3(m − 2),

where Gj = {2j + 1, 2j + 2}, 0  j  m − 1 3-GDD(2m−2 41 ) = (Z2m , {Gj : 0  j  m − 2}, B)

for 3|(m − 2),

where G0 = {1, 2, 3, 4} and Gj = {2j + 3, 2j + 4}, 1  j  m − 2.

Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

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For the group G0 , let ((G0 ×Z2p )∪({∞}×Zp ), A0 ) be a T(H, 5p) if 3(m−2) or a T(H, 9p) if 3|(m−2). For each group Gj , j = 0, there exists a T(H, 5p : p)=((Gj ×Z2p )∪({∞}×Zp ), {∞}×Zp ), Aj ). For each triple B ∈ B, there exists a T(H, (2p)3 )=({{a}×Z2p : a ∈ B}, CB ). Then, ⎛ ⎞ ⎛ ⎞ s   ⎝ ⎠ ⎝ = CB ∪ Aj ⎠ j=0 B∈B forms a T(H, 4mp + p), where s = m − 1 if 3(m − 2) or s = m − 2 if 3|(m − 2). Proof. First, we have the following enumeration:  ⎧ 4p 5(5p − 1) ⎪ + 4p2 ⎨ if 3(m − 2), 2 4 = 6p − 1 for j = 0, |Aj | = | A0 | = ⎪ 2p ⎩ 9(9p − 1) if 3|(m − 2), 4 ⎧ m  ⎪ 22 ⎪ ⎪ 3 ⎪ 2m(m − 1) 2 ⎪ (2p)2 ⎨ = if 3(m − 2), 2 3 3  = 6p, |CB | = |B| = m−2 ⎪ 2p ⎪ ⎪ 22 + 8(m − 2) ⎪ ⎪ 2(m − 2)(m + 1) 2 ⎩ = if 3|(m − 2), 3 3 ⎧ 5(5p − 1) 2m(m − 1) ⎪ ⎨ · 6p + + (6p − 1)(m − 1) 4pm + p − 1 3 4 . = (4m + 1) · || = 2(m − 2)(m + 1) 9(9p − 1) ⎪ 4 ⎩ · 6p + + (6p − 1)(m − 2) 3 4 The number || is just the number of blocks in a T(H, 4mp + p). Furthermore, ∀i = i ∈ Zp , {(∞, i), (∞, i )} appears in three blocks of A0 , where exactly one edge is interior. ∀x ∈ Z2m , ∃Gj containing x ⇒ ∀i ∈ Z2p , i ∈ Zp , {(x, i), (∞, i )} appears in three blocks of Aj , where exactly one edge is interior. ∀(x, i) = (x , i ) ∈ Z2m × Z2p , x ∈ Gj and x ∈ Gj , if j = j , then ∃B∈B such that x, x ∈B ⇒ {(x, i), (x , i )} appears in three blocks of CB , where exactly one edge is interior. if j = j , then x, x ∈ Gj ⇒ {(x, i), (x , i )} appears in three blocks of Aj , where exactly one edge is interior.  Theorem 2.3. Let H be a simple graph with 2p edges. If there exist T(H, 7p), T(H, 11p), T(H, (2p)3 ) and T(H, 7p : 3p), then there exists a T(H, 4mp + 3p) for all m  1. Proof. Its construction and proof are similar to Theorem 2.2.



Below, the notation Di,j represents the set of the differences (pure- for i = j, or else mixed-) from Zp × {i} to Zp × {j}, where i, j ∈ Z4 . Then d ∈ Di,j means that all pairs {(xi , (x + d)j ) : x ∈ Zp } occur in the design. In order to simplify, the elements (t, 0), (t, 1), (t, 2), (t, 3) may be denoted by t0 , t1 , t2 , t3 or t, t, t, t, respectively. And the intervals [t0 , s0 ], [t1 , s1 ], [t2 , s2 ], [t3 , s3 ] may be denoted by [t, s], [t, s], [t, s], [t, s], respectively. Theorem 2.4. There exists a T(K1,2p , (2p)3 ) for any integer p  1. Proof. Take the vertex set Z2p × Z3 . The block set is {Ai mod (−, 3) : 0  i  2p − 1}, where Ai = (i0 : 01 , 11 , . . . , (2p − 1)1 ; i2 , (1 + i)2 , (2 + i)2 , . . . , (2p − 1 + i)2 ). Then, for the interior, the pairs {i0 , j1 }, 0  i, j  2p − 1 appear in all Ai , which give all pairs in D0,1 . Therefore, Ai mod (−, 3) will give all pairs in D1,2 and D2,0 . As for the exterior, the pairs {j1 , (i + j)2 } and {(i + j)2 , i0 }, 0  i, j  2p − 1 appear in all Ai , which give all pairs in D1,2 and D2,0 . Therefore, Ai mod (−, 3) will give all pairs in D2,0 , D0,1 and D0,1 , D1,2 , respectively.  Theorem 2.5. There exists a T(K1,2p , 5p : p) for any odd integer p. Construction. Take the vertex set (Zp × Z4 ) ∪ {∞0 , · · · , ∞p−1 } with a p-hole {∞0 , . . . , ∞p−1 }. The total 6p − 1 blocks are {Ai , A i : 1  i  p − 1} ∪ {Bj mod (p, −) : 1  j  4} ∪ {C}. These blocks Ai , A i , Bj and C are listed below.  [0, p − 1] [0, p − 1] , Ai = ∞i [i, i + p − 1] [i, i + p − 1]

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[0, p − 1] [0, p − 1] , [i, i + p − 1] [i, i + p − 1]   ⎛ p−1 [1, p − 1] 0 0 1, ⎜ 2 0 B1 = ⎜   ⎝ p−1 −[1, p − 1] ∞0 0 − 1, 2   ⎛ p−1 0 0 1, [1, p − 1] ⎜ 2 ⎜   B2 = ⎜0 ⎝ p−1 0 ∞0 − 1, −[1, p − 1] 2   ⎛ p−1 [1, p − 1] 0 1, ∞ 0 ⎜ 2 B3 = ⎜   ⎝0 p−1 0 0 − 1, −[1, p − 1] 2   ⎛ p−1 1, ∞0 0 [1, p − 1] ⎜ 2 ⎜   B4 = ⎜0 ⎝ p−1 0 0 − 1, −[1, p − 1] 2  [0, p − 1] [0, p − 1] C = ∞0 . [0, p − 1] [0, p − 1]  A i = ∞i

 1,

p−1 2



(∞2 , ∞4 , . . . , ∞p−1 )

⎞ ⎟ ⎟, ⎠

  p−1 1, 2

⎞

(∞1 , ∞3 , . . . , ∞p−2 )  1,

p−1 2



(∞2 , ∞4 , . . . , ∞p−1 )  1,

p−1 2

⎟ ⎟ ⎟, ⎠

⎞ ⎟ ⎟, ⎠



(∞1 , ∞3 , . . . , ∞p−2 )

⎞ ⎟ ⎟ ⎟, ⎠

Proof. Firstly, each pair containing ∞r (0  r  p − 1) occurs once as an interior edge and twice as an exterior edge, as required. Next, each difference in Zp × Z4 appears once as an interior edges, as listed below:     p−1 p−1 B1 --- D0,0 : 1, , D0,2 : [1, p − 1], D0,1 : 1, ∪ {0}, D0,3 : 0, 2 2     p−1 p+1 , D1,3 : [1, p − 1] ∪ {0}, D0,1 : , p − 1 , D1,2 : 0, B2 --- D1,1 : 1, 2 2     p−1 p−1 , D1,2 : [1, p − 1], D2,3 : 1, , D0,2 : 0, B3 --- D2,2 : 1, 2 2     p−1 p+1 , D0,3 : [1, p − 1], D2,3 : , p − 1 ∪ {0}. B4 --- D3,3 : 1, 2 2 Finally, note that {|i − (−i)| : 1  i  p − 1/2} = [1, p − 1/2] and {i − (−i) : 1  i  p − 1} = [1, p − 1], and each difference in Zp × Z4 appears twice as an exterior edge, as follows:  Ai --- D0,3 : [1, p − 1], D1,2 : [1, p − 1], i



A i --- D0,2 : [1, p − 1],

D1,3 : [1, p − 1],

i

   p−1 ∪ 1, B1 --- D0,0 : 1, 2    p−1 ∪ 1, B2 --- D1,1 : 1, 2    p−1 B3 --- D2,2 : 1, ∪ 1, 2    p−1 ∪ 1, B4 --- D3,3 : 1, 2

 p−1 , 2  p−1 , 2  p−1 , 2  p−1 , 2

C--- D0,2 : 0,



D1,3 : 0.

D0,3 : [1, p − 1],

D2,3 : [1, p − 1] ∪ {0},

D1,2 : [1, p − 1],

D2,3 : [1, p − 1],

D0,2 : [1, p − 1],

D0,1 : [1, p − 1] ∪ {0},

D1,3 : [1, p − 1] ∪ {0},

Theorem 2.6. There exists a T(K1,2p , 7p : 3p) for any odd integer p.

D0,2 : 0,

D0,1 : 0,

D0,1 : [1, p − 1],

D0,3 : 0,

D2,3 : 0,

D1,2 : 0,

D0,3 : 0,

D1,2 : 0,

Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

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Construction. Take the vertex set (Zp × Z4 ) ∪ {∞0 , . . . , ∞3p−1 } with a 3p-hole {∞0 , . . . , ∞3p−1 }. The total 10p − 1 blocks are {Ai : 0  i  p − 1, A i : 1  i  p − 1} ∪ {Bj , B j : 0  j  2p − 1} ∪ {Ck mod (p, −) : 1  k  4}. These blocks Ai , A i , Bj , B j and Ck are listed below.  Ai = ∞i

 [0, p − 1] [0, p − 1] [0, p − 1] [0, p − 1] , A i = ∞i , [i, i + p − 1] [i, i + p − 1] [i, i + p − 1] [i, i + p − 1]   [0, p − 1] [0, p − 1] [0, p − 1] [0, p − 1] , B j = ∞j , Bj = ∞j [j, j + p − 1] [j, j + p − 1] [j, j + p − 1] [j, j + p − 1]     ⎞ ⎛ p−1 p−1 [0, p − 1] 0 1, 1, ⎟ ⎜ 2 2 ⎟, C1 = ⎜   ⎠ ⎝0 p−1 (∞p , ∞p+1 , . . . , ∞2p−1 ) (∞1 , ∞2 , . . . , ∞p−1/2 ) ∞0 − 1, 2     ⎞ ⎛ p−1 p−1 1, 1, ∞ [0, p − 1] 0 ⎟ ⎜ 2 2 ⎟ ⎜   C2 = ⎜ 0 ⎟, ⎠ ⎝ p−1 0 − 1, (∞p , ∞p+1 , . . . , ∞2p−1 ) (∞p+1/2 , ∞p+3/2 , . . . , ∞p−1 ) 2   ⎞ ⎛   p−1 p−1 0 1, 1, [0, p − 1] ⎟ ⎜ 2 2 ⎟, C3 = ⎜ ⎠ ⎝0 p−1 ] (∞2p , ∞2p+1 , . . . , ∞3p−1 ) (∞1 , ∞2 , . . . , ∞p−1/2 ) ∞0 −[1, 2     ⎛ ⎞ p−1 p−1 1, ∞ [0, p − 1] 1, 0 ⎜ ⎟ 2 2 ⎜ ⎟   C4 = ⎜ 0 ⎟. ⎝ ⎠ p−1 0 − 1, (∞2p , ∞2p+1 , . . . , ∞3p−1 ) (∞p+1/2 , ∞p+3/2 , . . . , ∞p−1 ) 2

Proof. Firstly, each pair containing ∞r (0  r  3p − 1) occurs once as an interior edge and twice as an exterior edge, as required. Next, each difference in Zp × Z4 appears once as an interior edge, as listed below:     p−1 p−1 , D0,2 : [0, p − 1], D0,1 : 1, ∪ {0}, C1 --- D0,0 : 1, 2 2     p−1 p+1 , D1,3 : [0, p − 1], D0,1 : ,p − 1 , C2 --- D1,1 : 1, 2 2     p−1 p−1 , D1,2 : [0, p − 1], D2,3 : 1, ∪ {0}, C3 --- D2,2 : 1, 2 2     p−1 p+1 , D0,3 : [0, p − 1], D2,3 : ,p − 1 . C4 --- D3,3 : 1, 2 2 Finally, each difference in Zp × Z4 appears twice as an exterior edges, as follows:         p−1 p−1 p−1 p−1 ∪ 1, , C2 --- D1,1 : 1, ∪ 1, , D0,1 : 0, C1 --- D0,0 : 1, 2 2 2 2         p−1 p−1 p−1 p−1 ∪ 1, , C4 --- D3,3 : 1, ∪ 1, , D2,3 : 0, C3 --- D2,2 : 1, 2 2 2 2  Ai --- D0,1 : [0, p − 1], D2,3 : [0, p − 1], i



A i --- D0,1 : [1, p − 1],

D2,3 : [1, p − 1],

i



Bj --- D0,3 : [0, p − 1] ∪ [0, p − 1],

D1,2 : [0, p − 1] ∪ [0, p − 1],

B j --- D0,2 : [0, p − 1] ∪ [0, p − 1],

D1,3 : [0, p − 1] ∪ [0, p − 1].

j





j

Theorem 2.7. There exists a T(K1,2p , 7p) for any positive integer p ≡ 3 mod 8.

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Construction. Let p = 8t + 3. For t = 0, the following blocks form a T(K1,6 , 21) on Z7 × Z3 . (00 : 01 , 11 , 21 , 31 , 41 , 30 ; 62 , 52 , 42 , 32 , 22 , 40 ),

(01 : 02 , 12 , 22 , 32 , 42 , 52 ; 60 , 50 , 40 , 30 , 20 , 10 ),

(02 : 12 , 22 , 32 , 00 , 10 , 20 ; 62 , 52 , 42 , 61 , 51 , 41 ),

(00 : 10 , 20 , 12 , 22 , 32 , 42 ; 60 , 50 , 11 , 31 , 51 , 01 ),

(01 : 10 , 20 , 11 , 21 , 31 , 62 ; 12 , 32 , 61 , 51 , 41 , 00 ) mod (7, −). Below, for t  1, take the vertex set (Z14t+5 × Z4 ) ∪ {∞}. The block set consists of the following seven base blocks module 14t + 5.  0 [1, 7t + 2] [0, 3t] [0, 3t] [1, 3t] ∞ B1 = 0 , 7t + 3 14t + 4 −[1, 7t + 2] −[1, 3t + 1] −[7t + 2, 10t + 2] −[7t + 4, 10t + 3]  [0, 7t + 2] [7t + 4, 14t + 4] 7t + 3 [1, 2t + 1] , B2 = 0 ∞ −[1, 2t + 1] [0, 14t + 4]2 [3, 14t + 3]2  ∞ [1, 7t + 3] [7t + 4, 11t + 4] [2t + 2, 7t + 2] , B3 = 0 0 −[0, 7t + 2] [2, 8t + 2]2 −[2t + 2, 7t + 2]  ∞ [1, 7t + 2] [7t + 3, 11t + 4] [2t + 2, 7t + 2] , B4 = 0 7t + 3 −[1, 7t + 2] [−7t − 1, t + 1]2 −[2t + 2, 7t + 2]  [7t + 2, 14t + 4] [1, 2t + 1] [0, 7t + 1] , B5 = 0 −[1, 2t + 1] −[1, 7t + 2] −[7t + 3, 14t + 5]  [0, 7t + 2] [7t + 3, 14t + 4] 7t + 2 [1, 2t] , B6 = 0 ∞ −[1, 2t] −[0, 7t + 2] −[7t + 4, 14t + 5]  [1, 7t + 1] [7t + 3, 11t + 4] ∞ [2t + 1, 7t + 2] B7 = 0 . 14t + 4 −[2t + 1, 7t + 2] −[2, 7t + 2] [−7t − 2, t]2 Proof. Firstly, each pair containing ∞ occurs once as an interior edge and twice as an exterior edge, as required. Next, each difference in Z14t+5 × Z4 appears once as an interior edge, as listed below: B1 --- D0,0 : [1, 7t + 2],

D1,0 : [11t + 5, 14t + 5],

D2,0 : [11t + 5, 14t + 5],

D3,0 : [11t + 5, 14t + 4] ∪ {0}, B2 --- D1,2 : [0, 14t + 4],

D1,1 : [1, 2t + 1],

B3 --- D1,0 : [1, 11t + 4],

D1,1 : [2t + 2, 7t + 2],

B4 --- D2,0 : [1, 11t + 4],

D2,2 : [2t + 2, 7t + 2],

B5 --- D2,3 : [0, 14t + 4],

D2,2 : [1, 2t + 1],

B6 --- D3,0 : 7t + 2,

D3,3 : [1, 2t],

B7 --- D3,3 : [2t + 1, 7t + 2],

D1,3 : [0, 14t + 4],

D3,0 : [1, 7t + 1] ∪ [7t + 3, 11t + 4].

Finally, each difference in Z14t+5 × Z4 appears twice as an exterior edge, as follows: B1 --- D0,0 : [1, 7t + 2] ∪ [1, 7t + 2], D3,0 : [7t + 2, 10t + 2],

D2,0 : [1, 3t + 1] ∪ [7t + 3, 10t + 3],

D1,2 : [8t + 4, 14t + 4]2 ,

B2 --- D1,1 : [1, 2t + 1] ∪ [10t + 3, 14t + 3]2 , D2,0 : [0, 7t + 2], B3 --- D1,0 : 0,

D1,3 : 1,

D1,0 : [0, 14t + 4]2 ,

D1,3 : [3, 14t + 3]2 ,

D2,3 : [7t + 4, 14t + 4],

D1,3 : [7t + 3, 14t + 5],

D1,2 : [2, 8t + 2]2 ,

D1,1 : [2t + 2, 7t + 2] ∪ [1, 10t + 1]2 , B4 --- D2,3 : [−7t − 1, t + 1]2 ∪ {7t + 3},

B5 --- D2,2 : [1, 2t + 1] ∪ [10t + 3, 14t + 3]2 ,

D2,0 : [3t + 2, 7t + 2],

D3,0 : [1, 14t + 5]2 ,

D1,2 : [1, 7t + 2],

D2,2 : [2t + 2, 7t + 2] ∪ [1, 10t + 1]2 ,

D3,0 : [2, 14t + 4]2 ,

D1,0 : 1,

D2,3 : [t + 3, 7t + 3]2 ∪ [−7t, −t − 2]2 ,

D1,0 : [2, 14t + 4]2 ,

D3,0 : [10t + 3, 14t + 4], D2,0 : [7t + 3, 14t + 4],

D1,3 : [0, 14t + 4]2 ,

D1,2 : [7t + 3, 14t + 5],

Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

B6 --- D3,3 : [1, 2t] ∪ [10t + 5, 14t + 3]2 , D1,2 : [1, 14t + 5]2 ,

D2,3 : [0, 7t + 2],

D3,0 : [0, 7t + 1],

D1,0 : [1, 14t + 3]2 ,

B7 --- D3,3 : [2t + 1, 7t + 2] ∪ [1, 10t + 3]2 , D1,0 : [3, 14t + 3]2 ,

4117

D1,3 : [1, 7t + 2],

D2,0 : [10t + 4, 14t + 5].

D2,3 : [−t, 7t + 2]2 ,



Theorem 2.8. There exists a T(K1,2p , 7p) for any positive integer p ≡ 7 mod 8. Construction. Let p = 8t + 7. Take the vertex set (Z14t+11 × Z4 ) ∪ {∞0 , ∞1 , ∞2 , ∞3 , ∞4 }. The total (14t + 11) × 7 + 7 blocks are {Ai : 1  i  7} ∪ {Bj mod (14t + 11, −) : 1  j  7}. These blocks Ai , Bj are listed below.  A1 = ∞0  A2 = ∞1  A3 = ∞2

∞1 ∞4

∞2 ∞3

[0, 7t + 4] [7t + 5, 14t + 10] [0, 2t] , [7t + 6, 14t + 10] [0, 7t + 5] [0, 2t]

∞2 ∞0

∞3 ∞4

[0, 7t + 5] [7t + 5, 14t + 10]

∞3 ∞1

∞4 ∞0

[0, 14t + 10] [1, 14t + 11]

[7t + 6, 14t + 10] [0, 7t + 4] [4t + 2, 6t + 2] , [4t + 1, 6t + 1]

[2t + 1, 4t + 1] , [2t + 4, 4t + 4]

[7t + 5, 14t + 10] [6t + 3, 8t + 3] , [0, 7t + 5] [6t + 1, 8t + 1]  ∞0 ∞1 [0, 7t + 5] [7t + 6, 14t + 10] [8t + 4, 10t + 4] , A5 = ∞4 [0, 7t + 4] [8t + 4, 10t + 4] ∞3 ∞2 [7t + 5, 14t + 10]  [0, 7t + 6] [7t + 7, 14t + 10] [10t + 5, 12t + 7] A6 = ∞0 , [7t + 4, 14t + 10] [0, 7t + 3] [10t + 5, 12t + 7]  14t + 8 14t + 9 14t + 10 [0, 14t + 10] [12t + 8, 14t + 7] , A7 = ∞1 0 1 2 [1, 14t + 11] [12t + 11, 14t + 10]  [1, 7t + 6] [7t + 7, 11t + 7] U P M R [1, 5t + 3] , B1 = 0 V Q N S −[1, 5t + 3] −[0, 7t + 5] [2, 8t + 2]2  1 7t + 6 7t + 7 0 [5t + 4, 7t + 5] [1, 7t + 5] [7t + 8, 14t + 10] , B2 = 0 ∞1 ∞2 −[5t + 4, 7t + 5] [2, 14t + 10]2 [5, 14t + 9]2 1 ∞0  7t + 7 ∞3 ∞4 [1, 7t + 5] [1, 3t + 1] [1, 3t + 1] ∞0 B3 = 0 14t + 10 ∞1 0 0 −[1, 7t + 5] −[7t + 7, 10t + 7] −[7t + 6, 10t + 6]  [1, 7t + 4] [7t + 5, 14t + 10] ∞4 [5t + 3, 7t + 5] , B4 = 0 0 −[5t + 3, 7t + 5] −[2, 7t + 5] −[7t + 6, 14t + 11]  0 0 [1, 5t + 2] [1, 7t + 5] [7t + 6, 11t + 9] ∞2 , B5 = 0 14t + 10 ∞4 ∞3 −[1, 5t + 2] −[1, 7t + 5] [−7t − 4, t + 2]2  0 0 [5t + 4, 7t + 5] [1, 7t + 5] [7t + 6, 14t + 9] ∞1 B6 = 0 , 14t + 10 ∞2 ∞3 −[5t + 4, 7t + 5] −[1, 7t + 5] −[7t + 7, 14t + 10]  ∞2 0 7t + 5 [1, 5t + 3] [1, 7t + 3] [7t + 6, 11t + 9] ∞3 . B7 = 0 7t + 6 14t + 10 ∞4 ∞0 −[1, 5t + 3] −[2, 7t + 4] [−7t − 5, t + 1]2  A4 = ∞3

∞4 ∞2

∞0 ∞1

[0, 7t + 4] [7t + 6, 14t + 10]

[1, 3t + 3] , −[2, 3t + 4]

The undetermined variable in B1 are listed as follows, where k ∈ Z14t+11 . ⎧ ∞ ⎧∞ 1 2 ⎪ ⎪ for 0  k  2t, for 0  k  4t + 1, ⎪ ⎪ ⎪ ⎪ 3 14t + 10 ⎪ ⎪  ⎪  ⎪ ⎨ ⎨ P ∞0 ∞1 U = for 2t + 1  k  10t + 4, for 4t + 2  k  12t + 7, = ⎪ ⎪ Q 0 3 V ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∞0 ⎩ ∞4 for 10t + 5  k  14t + 10, for 12t + 8  k  14t + 10, 0 0 ⎧∞ ⎧ ∞ 3 4  ⎪  ⎪ for 0  k  6t + 2, for 0  k  8t + 3, ⎨ ⎨ R M 14t + 9 0 =  =  ∞ ∞ ⎪ ⎪ S N 2 3 ⎩ ⎩ for 6t + 3  k  14t + 10, for 8t + 4  k  14t + 10. 14t + 10 14t + 9

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Proof. Firstly, in the following the interior and edges occurring in the structures in the top row are scattered in the  table  exterior b a rest of the table. Note that a is equivalent to b . c c

Further, each pair containing ∞r (0  r  4) occurs once as an interior edge and twice as an exterior edge, as required. Next, each difference in Z14t+11 × Z4 appears once as an interior edge, as listed below: B1 --- D1,1 : [1, 5t + 3],

D1,0 : [1, 11t + 7],

B2 --- D1,1 : [5t + 4, 7t + 5], B3 --- D0,0 : [1, 7t + 5],

D1,2 : [1, 14t + 10],

B4 --- D2,2 : [5t + 3, 7t + 5],

D1,0 : [11t + 8, 14t + 10],

D2,3 : [1, 14t + 10],

D2,0 : [0, 11t + 9],

B6 --- D3,3 : [5t + 4, 7t + 5], B7 --- D3,3 : [1, 5t + 3],

D1,3 : 1,

D3,0 : [11t + 10, 14t + 10] ∪ {7t + 4},

D2,0 : [11t + 10, 14t + 10],

B5 --- D2,2 : [1, 5t + 2],

D1,0 : 0,

D1,2 : 0,

D1,3 : [2, 14t + 10],

D3,0 : 0,

D2,3 : 0,

D3,0 : [1, 7t + 3] ∪ [7t + 5, 11t + 9],

D1,3 : 0.

        P M R Finally, each difference in Z14t+11 × Z4 appears twice as an exterior edge. In fact, the exterior edges in U V , Q , N , S and all Ai (1  i  7) are just all pairs in D2,3 : 7t + 6, 7t + 6, 7t + 7, D1,0 : 0, 14t + 9,

D2,0 : 1, 7t + 6,

D1,2 : 0, 14t + 10,

D3,0 : 7t + 5, 14t + 10,

D1,3 : 3.

In other base blocks, we have B1 --- D1,1 : [1, 5t + 3] ∪ [4t + 5, 14t + 9]2 , D3,0 : [1, 14t + 11]2 ,

B2 --- D1,1 : [5t + 4, 7t + 5] ∪ [1, 4t + 3]2 , D2,0 : [1, 7t + 5],

D1,3 : [7t + 6, 14t + 11],

D1,2 : [2, 8t + 2]2 ,

D2,0 : [3t + 5, 7t + 5], D1,0 : [2, 14t + 10]2 ∪ {1},

D2,3 : [7t + 8, 14t + 10],

B3 --- D0,0 : [1, 7t + 5] ∪ [1, 7t + 5], D3,0 : [7t + 6, 10t + 6],

D1,3 : [5, 14t + 9]2 ,

D3,0 : 0,

D2,0 : [7t + 7, 10t + 7] ∪ [2, 3t + 4] ∪ {0},

D1,0 : 0, 1,

D2,3 : [t + 4, 7t + 4]2 ∪ [−7t − 3, −t − 3]2 ,

D1,2 : [8t + 4, 14t + 8]2 , B4 --- D2,2 : [5t + 3, 7t + 5] ∪ [1, 4t + 5]2 , D3,0 : [2, 14t + 8]2 ,

D2,0 : [7t + 6, 14t + 9],

D1,3 : [0, 14t + 10]2 ,

D2,3 : 0,

D1,2 : [7t + 6, 14t + 11],

Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

B5 --- D2,2 : [1, 5t + 2] ∪ [4t + 7, 14t + 9]2 , D1,0 : [2, 14t + 10]2 ,

D3,0 : [10t + 7, 14t + 10],

B6 --- D3,3 : [5t + 4, 7t + 5] ∪ [1, 4t + 3]2 , D1,2 : [1, 14t + 9]2 , B7 --- D1,3 : [1, 7t + 5],

D1,2 : [1, 7t + 5],

4119

D2,3 : [−7t − 4, t + 2]2 ,

D2,0 : 14t + 10,

D2,3 : [1, 7t + 5],

D3,0 : [1, 7t + 4],

D1,3 : 1,

D1,0 : [3, 14t + 9]2 , D2,3 : [−t − 1, 7t + 5]2 ,

D3,3 : [1, 5t + 3] ∪ [4t + 5, 14t + 9]2 .

D1,0 : [3, 14t + 7]2 ,

D2,0 : [10t + 8, 14t + 11],



Theorem 2.9. There exists a T(K1,2p , 11p) for any positive integer p ≡ 7 mod 8. Construction. Let p = 8t + 7. Take the vertex set (Z22t+19 × Z4 ) ∪ {∞}. The block set consists of the following 11 base blocks module 22t + 19.  ∞ [1, 11t + 9] [2t + 3, 7t + 6] , B1 = 0 0 −[1, 11t + 9] −[13t + 12, 18t + 15]  [0, 5t + 3] ∞ [1, 11t + 9] , B2 = 0 1 −[1, 11t + 9] [0, 10t + 6]2  ∞ [1, 11t + 9] [0, 5t + 3] B3 = 0 , 11t + 10 −[1, 11t + 9] −[1, 5t + 4]  [t + 1, 6t + 4] ∞ [1, 11t + 9] B4 = 0 , 11t + 9 −[1, 11t + 9] −[t + 1, 6t + 4]  [1, 7t + 6] [1, 2t + 2] [0, 7t + 5] B5 = 0 , −[1, 7t + 6] −[11t + 11, 18t + 16] −[11t + 10, 13t + 11]  11t + 10 [5t + 4, 10t + 8] [11t + 11, 22t + 18] , B6 = 0 [3, 22t + 17]2 ∞ [10t + 8, 20t + 16]2  [1, 11t + 10] [11t + 11, 15t + 13] [10t + 9, 11t + 9] B7 = 0 , [2, 8t + 6]2 [20t + 18, 22t + 18]2 −[0, 11t + 9]  [11t + 9, 21t + 17] [5t + 4, 11t + 8] B8 = 0 , −[5t + 5, 11t + 9] −[11t + 10, 21t + 18]  [1, 11t + 9] [11t + 10, 15t + 12] 0 [21t + 18, 22t + 18] , B9 = 0 11t + 10 −[21t + 19, 22t + 19] −[1, 11t + 9] [−11t − 8, −3t − 4]2  [11t + 10, 22t + 18] [6t + 5, 11t + 9] B10 = 0 , −[6t + 5, 11t + 9] −[11t + 11, 22t + 19]  [0, 11t + 8] [11t + 10, 15t + 12] 11t + 9 [0, t] . B11 = 0 ∞ −[0, t] −[1, 11t + 9] [−11t − 9, −3t − 5]2 Proof. Firstly, each pair containing ∞ occurs once as an interior edge and twice as an exterior edge, as required. Next, each difference in Z22t+19 × Z4 appears once as an interior edge, as listed below: B1 --- D0,0 : [1, 11t + 9],

D0,2 : [2t + 3, 7t + 6],

B2 --- D1,1 : [1, 11t + 9],

D1,2 : [0, 5t + 3],

B3 --- D2,2 : [1, 11t + 9],

D2,3 : [0, 5t + 3],

B4 --- D3,3 : [1, 11t + 9],

D1,3 : [16t + 15, 21t + 18],

B5 --- D0,1 : [0, 7t + 5],

D0,3 : [1, 7t + 6],

D0,2 : [1, 2t + 2],

B6 --- D1,2 : [5t + 4, 10t + 8] ∪ [11t + 10, 22t + 18], B7 --- D0,1 : [7t + 6, 22t + 18], B8 --- D2,3 : [5t + 4, 21t + 17],

D1,2 : [10t + 9, 11t + 9],

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B9 --- D0,2 : [7t + 7, 22t + 19],

D2,3 : [21t + 18, 22t + 18],

B10 --- D1,3 : [1, 16t + 14], B11 --- D0,3 : [7t + 7, 22t + 19],

D1,3 : [21t + 19, 22t + 19].

Finally, each difference in Z22t+19 × Z4 appears twice as an exterior edge, as follows: B1 --- D0,0 : [1, 11t + 9] ∪ [1, 11t + 9],

D1,0 : 0,

D3,0 : [13t + 12, 18t + 15],

B2 --- D1,1 : [1, 11t + 9] ∪ [1, 11t + 9],

D1,3 : 1,

D1,0 : [0, 10t + 6]2 ,

B3 --- D2,2 : [1, 11t + 9] ∪ [1, 11t + 9],

D2,0 : [17t + 15, 22t + 18] ∪ {11t + 10},

D2,3 : [−(3t + 2), 7t + 4]2 , D2,0 : [0, 5t + 3],

D3,0 : [12t + 12, 22t + 18]2 , B4 --- D3,3 : [1, 11t + 9] ∪ [1, 11t + 9],

D2,3 : [t + 1, 6t + 4] ∪ {11t + 10},

D1,2 : [10t + 11, 20t + 17]2 , B5 --- D2,0 : [1, 7t + 6] ∪ [11t + 11, 18t + 16],

D3,0 : [11t + 10, 13t + 11],

D1,2 : [8t + 8, 22t + 18]2 ,

D2,3 : [−(11t + 7), 3t + 3]2 ∪ [7t + 6, 11t + 8]2 , B6 --- D1,0 : [10t + 8, 20t + 16]2 ,

D1,3 : [3, 22t + 17]2 ,

D2,0 : [5t + 4, 10t + 8],

D2,3 : [11t + 11, 22t + 18], B7 --- D1,3 : [11t + 10, 22t + 19], D1,0 : [20t + 18, 22t + 18]2 , B8 --- D2,0 : [11t + 10, 17t + 14],

D1,2 : [2, 8t + 6]2 ,

D3,0 : [1, 22t + 19]2 ,

D2,0 : [7t + 7, 11t + 9] ∪ [10t + 9, 11t + 9], D1,2 : [11t + 10, 21t + 18],

D3,0 : [2, 12t + 10]2 ,

D1,3 : [0, 20t + 16]2 , B9 --- D1,2 : [1, 11t + 9] ∪ [21t + 19, 22t + 19], D3,0 : [18t + 16, 22t + 18] ∪ {11t + 9}, B10 --- D2,3 : [6t + 5, 11t + 9],

D3,0 : [0, 11t + 8],

B11 --- D2,3 : [0, t] ∪ [3t + 5, 11t + 9]2 , D1,0 : [1, 22t + 17]2 ,

D2,3 : [−11t − 8, −3t − 4]2 ∪ {11t + 10}, D1,0 : [2, 22t + 18]2 ,

D1,3 : [20t + 18, 22t + 18]2 ,

D1,2 : [1, 10t + 9]2 ,

D2,0 : [18t + 17, 22t + 19],

D1,2 : [20t + 19, 22t + 19]2 .

D1,0 : [1, 22t + 17]2 ,

D1,3 : [1, 11t + 9],



Theorem 2.10. There exists a T(K1,2p , 11p) for any p ≡ 3 mod 8. Construction. Let p = 8t + 3. For t = 0, the following 88 blocks form a T(K1,6 , 33) on Z11 × Z3 . (00 : 10 , 20 , 30 , 40 , 50 , 01 ; 100 , 90 , 80 , 70 , 60 , 102 ), (00 : 11 , 21 , 31 , 41 , 51 , 61 ; 92 , 82 , 72 , 62 , 52 , 42 ) mod (11, 3), (01 : 72 , 82 , 10 , 20 , 30 , 40 ; 33 , 23 , 12 , 32 , 52 , 72 ), (02 : 70 , 80 , 90 , 100 , 11 , 21 ; 31 , 21 , 11 , 01 , 13 , 33 ) mod (11, −). Below, for t  1, take the vertex set (Z22t+8 × Z4 ) ∪ {∞}. The block set consists of the following 11 base blocks module 22t + 8, where the entry a(b) in base block B mean that B + i contains the entry a + i, when 0  i  11t + 3, and the entry b + i, when 11t + 4  i  22t + 7.     ⎛ ⎞ 11t + 3 11t + 5 1, , 11t + 3 ∞ [2t + 1, 7t + 2] 2 2 ⎜ ⎟       B1 = ⎝0 ⎠, 11t + 7 33t + 12 11t + 3 − , 11t + 4 −[13t + 5, 18t + 6] − 1, 2 2 2     ⎛ ⎞ 11t + 3 11t + 5 , 11t + 3 1, ∞ [0, 5t + 1] ⎟ ⎜ 2 2 ⎜ ⎟       B2 = ⎜ 0 ⎟, ⎝ ⎠ 11t + 3 11t + 7 33t + 12 , 11t + 4 − 1, − [0, 10t + 2]2 2 2 2

Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

⎛ ⎜ B3 = ⎝0 ⎛ ⎜ ⎜ B4 = ⎜0 ⎝  B5 = 0  B6 = 0  B7 = 0  B8 = 0  B9 = 0  B10 = 0  B11 = 0

4121



   ⎞ 11t + 3 11t + 5 1, , 11t + 3 ∞ [1, 5t + 2] 2 ⎟   2     ⎠, 11t + 7 33t + 12 11t + 3 − , 11t + 4 −[2, 5t + 3] − 1, 2 2 2     ⎞ 11t + 3 11t + 5 , 11t + 3 1, ∞ [t + 1, 6t + 2] ⎟ 2 2 ⎟      ⎟,  ⎠ 33t + 12 11t + 3 11t + 7 , 11t + 4 −[t + 2, 6t + 3] − 1, − 2 2 2 0(11t + 4) [1, 2t] [1, 5t + 1] [1, 7t + 3] 0(∞) −[11t + 5, 13t + 4] −[1, 5t + 1] −[11t + 6, 18t + 8] 11t + 4(0) [5t + 2, 11t + 3] [11t + 4, 21t + 6] , ∞(0) [10t + 4, 22t + 6]2 [1, 20t + 5]2 [21t + 7, 22t + 7] [2t + 2, 11t + 3] [11t + 4, 17t + 6] , [0, 12t + 4]2 [20t + 7, 22t + 7]2 −[2t + 2, 11t + 3] [5t + 3, 11t + 3] [11t + 4, 21t + 7] 11t + 4(0) , ∞(22t + 7) −[5t + 4, 11t + 4] −[11t + 4, 21t + 7]

[20t + 7, 22t + 7] [18t + 6, 22t + 6]2

,

[1, 11t + 4] [11t + 5, 15t + 5] , −[0, 11t + 3] [−11t − 2, −3t − 2]2 [6t + 3, 11t + 3] [1, 11t + 4] 0(11t + 4) , 22t + 7(∞) −[6t + 4, 11t + 4] −[1, 11t + 4] [0, t] [11t + 5, 15t + 4] 0 [11t + 4, 22t + 7] . 11t + 3 −[11t + 5, 22t + 8] −[1, t + 1] [−11t − 3, −3t − 5]2

0 [21t + 8, 22t + 7] 11t + 4 −[21t + 8, 22t + 7]

Proof. Firstly, each pair containing ∞ occurs once as an interior edge and twice as an exterior edge, as required. Next, each difference in Z22t+8 ×Z4 appears once as an interior edge, as listed below, where {d}∗ means that all pairs {(x, x+d) : 0  x  11t+3} occur in the design, {d} means that all pairs {(x, x + d) : 11t + 4  x  22t + 7} occur in the design. B1 --- D0,0 : [1, 11t + 3],

D2,0 : [15t + 6, 20t + 7],

B2 --- D1,1 : [1, 11t + 3],

D1,2 : [0, 5t + 1],

B3 --- D2,2 : [1, 11t + 3],

D2,3 : [1, 5t + 2],

B4 --- D3,3 : [1, 11t + 3],

D1,3 : [16t + 6, 21t + 7],

B5 --- D0,0 : {11t + 4},

D1,0 : [1, 2t + 1] ∪ [17t + 7, 22t + 7] ∪ {0}∗ ,

D2,0 : [20t + 8, 22t + 7],

D3,0 : [15t + 5, 22t + 7], B6 --- D1,1 : {11t + 4},

D1,0 : {0} ,

B7 --- D1,2 : [21t + 7, 22t + 7], B8 --- D2,2 : {11t + 4}, B9 --- D2,0 : [0, 15t + 5], B10 --- D2,3 : {0}∗ ,

D1,2 : [5t + 2, 21t + 6],

D1,0 : [2t + 2, 17t + 6],

D2,3 : [5t + 3, 21t + 7] ∪ {0} , D2,3 : [21t + 8, 22t + 7],

D3,3 : {11t + 4},

B11 --- D3,0 : 0, [11t + 5, 15t + 4],

D1,3 : [11t + 5, 16t + 5],

D3,0 : [1, 11t + 4],

D1,3 : [1, 11t + 4] ∪ [21t + 8, 22t + 8].

Finally, each difference in Z22t+8 × Z4 appears twice as an exterior edge, as follows: B1 --- D0,0 : [1, 11t + 4] ∪ [1, 11t + 3],

D3,0 : [13t + 5, 18t + 6],

B2 --- D1,1 : [1, 11t + 4] ∪ [1, 11t + 3],

D1,0 : [0, 10t + 2]2 ,

D2,3 : [−3t, 7t + 2]2 ,

D2,0 : [0, 5t + 1],

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B3 --- D2,2 : [1, 11t + 4] ∪ [1, 11t + 3],

D2,0 : [17t + 5, 22t + 6],

B4 --- D3,3 : [1, 11t + 4] ∪ [1, 11t + 3],

D2,3 : [t + 2, 6t + 3],

B5 --- D3,0 : [2, 4t + 2]2 ∪ [11t + 5, 13t + 4] ∪ {0}∗ ,

D1,2 : [12t + 6, 22t + 6]2 ,

D1,3 : [20t + 7, 22t + 7] ∪ {0}∗ ,

D2,0 : [5t + 2, 11t + 3],

D1,0 : [10t + 4, 22t + 6]2 ,

D3,0 : {0} ,

D2,3 : [11t + 5, 21t + 7],

B7 --- D1,3 : [20t + 7, 22t + 7]2 ∪ [11t + 5, 20t + 6], D3,0 : [4t + 4, 22t + 6]2 ,

D1,2 : [0, 12t + 4]2 ,

D2,3 : [21t + 8, 22t + 8],

D2,0 : [5t + 2, 11t + 4],

B8 --- D2,0 : [11t + 4, 17t + 4] ∪ {22t + 7} , D : [1, 12t + 1] ∪ {22t + 7} , D 3,0

D1,2 : [10t + 3, 20t + 5]2 ,

D2,0 : [1, 5t + 1] ∪ [11t + 6, 18t + 8],

D2,3 : [7t + 4, 11t + 2]2 ∪ [−(11t + 1), 3t + 3]2 , B6 --- D1,3 : [1, 20t + 5]2 ∪ {0} ,

D3,0 : [12t + 3, 22t + 5]2 ,

D1,2 : [11t + 4, 21t + 7],

1,3 : [0, 20t + 6]2 ,

2

B9 --- D2,3 : [−11t − 2, −3t − 2]2 ∪ {11t + 4}, D3,0 : [18t + 7, 22t + 7] ∪ {11t + 4}, B10 --- D2,3 : [6t + 4, 11t + 4], D1,2 : [1, 10t + 1]2 ,

D1,2 : [0, 11t + 3] ∪ [21t + 8, 22t + 7],

D1,3 : [20t + 8, 22t + 6]2 ,

D1,3 : [1, 11t + 4],

D3,0 : {22t + 7}∗ ,

D1,0 : [1, 22t + 7]2 , D2,0 : {22t + 7}∗ ,

D1,0 : [0, 22t + 6]2 ,

B11 --- D2,3 : [1, t + 1] ∪ [3t + 5, 11t + 5]2 , D1,2 : [20t + 7, 22t + 7]2 ,

D3,0 : [0, 11t + 3],

D1,0 : [1, 22t + 7]2 .

D2,0 : [18t + 9, 22t + 8] ∪ {11t + 5},



Theorem 2.11. There exists a T(K1,2p , 5p) for any positive integer p ≡ 5 mod 8. Construction. Let p = 8t + 5. For t = 0, the following base blocks mod 5 form a T(K1,10 , 25) on Z5 × Z5 : (00 : 13 , 23 , 42 , 21 , 31 , 10 , 02 , 11 , 32 , 03 ; 44 , 20 , 22 , 43 , 14 , 34 , 04 , 12 , 33 , 01 ), (01 : 12 , 22 , 23 , 11 , 44 , 00 , 14 , 21 , 13 , 02 ; 34 , 03 , 31 , 42 , 43 , 40 , 41 , 33 , 04 , 20 ), (00 : 14 , 22 , 43 , 41 , 44 , 30 , 34 , 24 , 33 , 12 ; 02 , 11 , 23 , 21 , 04 , 01 , 13 , 31 , 42 , 10 ), (03 : 11 , 21 , 22 , 23 , 24 , 44 , 14 , 13 , 12 , 01 ; 43 , 02 , 31 , 40 , 42 , 41 , 00 , 32 , 04 , 10 ), (04 : 31 , 21 , 22 , 23 , 24 , 00 , 14 , 03 , 12 , 01 ; 44 , 42 , 33 , 34 , 43 , 20 , 41 , 13 , 02 , 10 ), (02 : 23 , 13 , 12 , 21 , 24 , 04 , 14 , 11 , 22 , 03 ; 43 , 42 , 20 , 32 , 31 , 34 , 40 , 01 , 44 , 30 )

mod (5, −).

Below, for t  1, take the vertex set (Z10t+5 × Z4 ) ∪ {∞1 , ∞2 , ∞3 , ∞4 , ∞5 }. The total (10t + 5) × 5 + 5 blocks are {Ai : 1  i  5} ∪ {Bj mod (10t + 5, −) : 1  j  5}. These blocks Ai , Bj are listed below.  A1 = ∞0  A2 = ∞1

[0, 6t + 2] , [1, 6t + 3]

∞1 ∞4

∞2 ∞3

[0, 10t + 4] [0, 10t + 4]

∞2 ∞0

∞3 ∞4

[0, 10t + 4] [6t + 3, 10t + 4] [0, 2t] , [a, 10t + 4 + a] [6t + 3, 10t + 4] [0, 2t]

[2t + 1, 5t + 3] [5t + 4, 8t + 3] , [7t + 2, 10t + 4] [0, 3t − 1]  ∞4 ∞0 [0, 10t + 4] [8t + 4, 10t + 5] [1, 4t + 1] A4 = ∞3 , ∞2 ∞1 [b, 10t + 4 + b] [8t + 3, 10t + 4] [0, 4t]  ∞0 ∞1 [0, 10t + 4] [4t + 2, 5t + 1] [5t + 2, 10t + 4] A5 = ∞4 , ∞3 ∞2 [1, 10t + 5] [9t + 5, 10t + 4] [0, 5t + 2]  [0, 4t + 1] [5t + 4, 10t + 4] 5t + 2 U P [1, 3t + 2] [t + 1, 5t + 2] B1 = 0 , −[1, 3t + 2] −[t + 2, 5t + 3] −[0, 4t + 1] −[5t + 4, 10t + 4] 5t + 3 V Q  A3 = ∞2

∞3 ∞1

∞4 ∞0

0 10t + 3

1 [2, 10t + 4] 10t + 4 [0, 10t + 2]

Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

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 0

[3t + 3, 5t + 2] [5t + 3, 10t + 3] [5t + 5, 10t + 4] [0, 4t + 3] −[1, 4t + 4] −[3t + 3, 5t + 2] −[5t + 3, 10t + 3] −[5t + 6, 10t + 5] 10t + 4 5t + 3 5t + 3 5t + 4 0 , ∞1 ∞0 ∞4 ∞3 ∞2  [1, 5t + 2] [1, 10t + 1]2 [2, 10t + 4]2 [9t + 5, 10t + 4] ∞0 B3 = 0 −[1, 5t + 2] [0, 5t] [1, 5t + 2] [8t + 4, 10t + 2]2 0 0 ∞3 ∞4 ∞1 5t + 2 ∞2 10t + 3 5t + 3  [4, 10t + 4]2 [5t + 4, 6t + 3] [1, 5t + 2] [1, 8t − 1]2 [8t + 3, 10t + 3]2 B4 = 0 [3, 2t + 1]2 −[1, 5t + 2] [0, 4t − 1] [4t + 1, 5t + 1] [5t + 5, 10t + 5] ∞0 0 ∞2 2 ∞4 2 , 4 ∞3 4t 5t + 3 5t + 5 ∞1  X Y [1, 10t + 3]2 [5t + 3, 6t + 1] 0 ∞1 ∞2 ∞3 2t + 4 [1, 5t + 2] B5 = 0 . −[1, 5t + 2] W Z [0, 5t + 1] [0, 2t − 4]2 ∞0 1 5t + 2 6t + 4 ∞4 B2 =

The undetermined variable in A2 , A4 , B1 , B5 are listed as follows, where k ∈ Z10t+5 . 5t + 4 5t + 5 (t odd) or (t even), b = 1 (t odd) or 5t + 4(t even). 2 2  ⎧ ∞ ⎧∞ 2 ⎪ 4 for 0  k  2t, ⎪ ⎪ for 0  k  4t + 1, ⎪ ⎪ 5t + 1 ⎪ ⎪ ⎪ ⎪ 5t + 3 ⎪ ⎪  ⎪  ⎪ ⎨ ⎨ ∞1 U P ∞3 = for 2t + 1  k  6t + 2, = for 4t + 2  k  8t + 3, ⎪ ⎪ V 0 Q ⎪ ⎪ 10t + 4 ⎪ ⎪  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∞2 ⎩ ∞0 for 8t + 4  k  10t + 4. for 6t + 3  k  10t + 4. 5t + 1 1 ⎞ ⎧ ⎛ [2, 5t + 3] ⎧ ⎛ [5t + 5, 10t + 4] ⎞ 2 2    ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝ 15t + 5 ⎠ (t odd), ⎪ ⎝ 15t + 7 , 10t + 4 ⎠ (t odd), ⎪ 5t + 3, ⎪ ⎪  ⎪  ⎪ ⎨ ⎨ 2 2 Y X = ⎛ . = ⎛ ⎞ ⎞ ⎪ ⎪ Z W ⎪ ⎪ [5t + 4, 10t + 4] [2, 5t + 2] ⎪ ⎪ 2 2     ⎪⎝ ⎪⎝ ⎪ ⎪ ⎠ (t even). ⎪ ⎪ 15t + 8 15t + 6 ⎠ (t even), ⎩ ⎩ 5t + 3, , 10t + 4 2 2 a=

Proof. Firstly, in the following table the interior and exterior edges occurring in the structures in the top row are scattered in the rest of the table.

Further, each pair containing ∞r (0  r  4) occurs once as an interior edge and twice as an exterior edge, as required. Next, each difference in Z10t+5 × Z4 appears once as an interior edge, as listed below: B1 --- D0,0 : [1, 3t + 2],

D1,0 : [5t + 3, 9t + 4],

B2 --- D3,0 : [6t + 2, 10t + 4] ∪ [0, 5t + 2], B3 --- D1,1 : [1, 5t + 2],

D2,0 : [1, 5t + 1] ∪ [6t + 4, 10t + 5] ∪ {5t + 3},

D0,0 : [3t + 3, 5t + 2],

D1,2 : [0, 10t + 2] ∪ {10t + 4},

D1,0 : [0, 5t + 2],

D1,0 : [9t + 5, 10t + 4],

D2,0 : 5t + 2,

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B4 --- D2,2 : [1, 5t + 2],

D2,3 : [0, 8t] ∪ [8t + 2, 10t + 4],

B5 --- D3,3 : [1, 5t + 2],

D1,3 : [0, 10t + 4],

D2,0 : [5t + 4, 6t + 3],

D3,0 : [5t + 3, 6t + 1],

D1,2 : 10t + 3,

D2,3 : 8t + 1.

    P Finally, each difference in Z10t+5 × Z4 appears twice as an exterior edge. In fact, the exterior edges in U V , Q and Ai (1  i  5) are just all pairs in D2,3 : 0;

D1,2 : −a, −b;

D1,3 : 10t + 3, 10t + 4;

D1,0 : 10t + 4;

D3,0 : 0, 1;

D2,0 : 5t + 4, 5t + 2.

In other base blocks, we have B1 --- D0,0 : [1, 3t + 2] ∪ [4t + 1, 10t + 3]2 , D1,3 : [0, 8t + 2]2 ,

D1,2 : [0, 8t + 2]2 ∪ {10t + 4},

B2 --- D2,0 : [1, 4t + 4] ∪ [5t + 6, 10t + 5], D2,3 : [1, 8t + 7]2 ,

B4 --- D2,2 : [1, 5t + 2] ∪ [1, 5t + 2], D1,3 : [5t + 4, 10t + 4],

D0,0 : [3t + 3, 5t + 2] ∪ [1, 4t − 1]2 ,

D1,2 : [1, 10t − 1]2 ,

D1,0 : [0, 5t] ∪ {5t + 2},

D2,0 : [5t + 4, 10t + 4],

D1,0 : [0, 4t + 1] ∪ {5t + 2},

D2,3 : [2, 10t + 2]2 ,

D1,0 : [5t + 3, 10t + 3],

D1,3 : [1, 10t + 1]2 ,

B3 --- D1,1 : [1, 5t + 2] ∪ [1, 5t + 2], D1,2 : 10t + 3,

D3,0 : [t + 2, 10t + 4],

D1,3 : [0, 5t + 3] ∪ [8t + 4, 10t + 2]2 ,

D2,3 : [5t + 3, 10t + 4],

D3,0 : [2, t + 1],

D1,2 : [0, 5t] ∪ [8t + 4, 10t + 2]2 ∪ {10t + 1},

D2,0 : [0, 5t + 1] ∪ {5t + 3, 5t + 5},

D3,0 : [5t + 3, 6t + 3] ∪ [6t + 5, 10t + 4], B5 --- D3,3 : [1, 5t + 2] ∪ [1, 5t + 2],

D1,0 : [4t + 2, 5t + 1] ∪ {5t + 1},

D2,3 : [1, 5t + 2] ∪ [8t + 9, 10t + 5]2 ∪ {10t + 4},

D3,0 : [0, 5t + 2] ∪ {6t + 4}, D1,0 : [5t + 3, 10t + 4], D2,0 : [4t + 5, 5t + 3],    ⎧ 15t + 4 15t + 8 ⎪ 5t + 2, ∪ , 10t + 4 (t even), ⎪ ⎨ 2 2 D1,2 =      ⎪ ⎪ ⎩ 5t + 1, 15t + 3 ∪ 15t + 7 , 10t + 3 (t odd). 2 2 Theorem 2.12. There exists a T(K1,2p , 5p) for any positive integer p ≡ 1 mod 8. Construction. Let p = 8t + 1. For t = 0, a T(K1,2 , 5) can be taken as (0 : 1, 2; 4, 3) mod 5. For t = 1, a T(K1,18 , 45) can be constructed on (Z9 × Z4 ) ∪ {∞1 , . . . , ∞9 }. Its 55 blocks are listed as follows, where 0  i  3, 0  j  2. (0 : ∞6 , ∞9 , 3, 0, 2, 4, 6, 1, ∞2 , 5, ∞5 , 6, 4, 7, 1, 2, 3, 4; 8, 7, 1, 0, 1, 2, 3, ∞3 , 0, ∞4 , 8, 7, 6, 5, 8, 7, 6, 5) + i, (5 : ∞7 , ∞8 , 8, 5, 7, 0, 2, 6, ∞2 , 1, ∞5 , 2, 0, 3, 6, 7, 8, 0; 3, 4, 6, 5, 6, 7, 8, ∞3 , 5, ∞4 , 4, 3, 2, 1, 4, 3, 2, 1) + j, (0 : ∞2 , ∞4 , 3, 4, 7, 8, 0, 1, 2, ∞7 , 3, 1, 2, ∞9 , 2, 3, 4, 3; 8, 0, 5, 4, 2, 1, 0, 8, 7, 7, 6, ∞8 , ∞6 , 6, 7, 6, 5, 5) + i, (5 : ∞2 , ∞4 , 8, 0, 3, 4, 5, 6, 7, ∞7 , 8, 6, 7, ∞9 , 7, 8, 0, 8; 4, 5, 1, 0, 7, 6, 5, 4, 3, 3, 2, ∞8 , ∞6 , 2, 3, 2, 1, 1) + j, (0 : 7, 0, 4, 2, ∞3 , ∞5 , ∞4 , ∞7 , ∞9 , 1, 3, 2, 4, 1, 1, 2, 3, 4; 3, 4, 7, ∞2 , 8, 8, 3, 0, 8, ∞8 , ∞6 , 5, 6, 5, 8, 7, 6, 5) mod 9, (0 : 6, 1, 0, 5, ∞2 , ∞3 , 8, ∞4 , 1, 3, ∞6 , ∞8 , 7, 4, 1, 2, 3, 4; 5, 6, ∞1 , 0, 8, 2, ∞5 , 8, ∞7 , ∞9 , 3, 5, 3, 6, 8, 7, 6, 5) mod 9, (0 : 0, 1, 5, 6, 7, 8, 7, 8, 5, 0, 2, 6, 4, 5, 6, ∞6 , ∞8 , 1; 8, 7, 4, 3, 2, 1, 1, 0, ∞1 , ∞2 , ∞3 , ∞4 , ∞5 , ∞7 , ∞9 , 2, 3, 8) mod 9,

Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

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(4 : ∞2 , ∞4 , 7, 8, 2, 3, 4, 5, 6, ∞7 , 7, 5, 6, ∞9 , 6, 7, 8, 7; 3, 4, 0, 8, 6, 5, 4, 3, 2, 2, 1, ∞8 , ∞6 , 1, 2, 1, 0, 0), (8 : ∞2 , ∞4 , 2, 3, 6, 7, 8, 0, 1, ∞7 , 2, 0, 1, ∞9 , 1, 2, 3, 2; 7, 8, 4, 3, 1, 0, 8, 7, 6, 6, 5, ∞8 , ∞6 , 5, 6, 5, 4, 4), (4 : ∞1 , ∞8 , 7, 4, 6, 8, 1, 5, ∞2 , 0, ∞5 , 1, 8, 2, 5, 6, 7, 8; 8, 3, 5, 4, 5, 6, 7, ∞3 , 4, ∞4 , 3, 2, 1, 0, 3, 2, 1, 0), (8 : ∞1 , ∞7 , 2, 8, 1, 3, 5, 0, ∞2 , 4, ∞5 , 5, 3, 6, 0, 1, 2, 3; 3, 6, 0, 8, 0, 1, 2, ∞3 , 8, ∞4 , 7, 6, 5, 4, 7, 6, 5, 4), (∞1 : ∞4 , ∞5 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 5, 6, 7; ∞7 , ∞6 , 1, 2, 3, 4, 5, 6, 7, 8, 0, 0, 1, 2, 3, 5, 6, 7), (∞1 : ∞2 , ∞3 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 5, 6, 7; ∞9 , ∞8 , 3, 4, 5, 6, 7, 8, 0, 1, 2, 4, 5, 6, 7, 0, 1, 2), (∞2 : ∞3 , ∞4 , ∞5 , ∞6 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 4, 5, 6, 7, 8; ∞1 , ∞9 , ∞8 , ∞7 , 5, 6, 7, 8, 0, 1, 2, 3, 4, 3, 4, 5, 6, 7), (∞3 : ∞4 , ∞5 , ∞6 , ∞7 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4; ∞2 , ∞1 , ∞9 , ∞8 , 5, 6, 7, 8, 0, 1, 2, 3, 4, 6, 7, 8, 0, 1), (∞4 : ∞5 , ∞6 , ∞7 , ∞8 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 4, 5, 6, 7, 8; ∞3 , ∞2 , ∞1 , ∞9 , 3, 4, 5, 6, 7, 8, 0, 1, 2, 4, 5, 6, 7, 8), (∞5 : ∞6 , ∞7 , ∞8 , ∞9 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 8, 0, 1, 2, 3; ∞4 , ∞3 , ∞2 , ∞1 , 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8), (∞6 : ∞7 , ∞8 , ∞9 , ∞1 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 4, 5, 6, 7, 8; ∞5 , ∞4 , ∞3 , ∞2 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 3, 4, 5, 6, 7), (∞7 : ∞8 , ∞9 , ∞1 , ∞2 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4; ∞6 , ∞5 , ∞4 , ∞3 , 6, 7, 8, 0, 1, 2, 3, 4, 5, 7, 8, 0, 1, 2), (∞8 : ∞9 , ∞1 , ∞2 , ∞3 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 8, 0, 1, 2, 3; ∞7 , ∞6 , ∞5 , ∞4 , 1, 2, 3, 4, 5, 6, 7, 8, 0, 7, 8, 0, 1, 2), (∞9 : ∞1 , ∞2 , ∞3 , ∞4 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 4, 5, 6, 7, 8; ∞8 , ∞7 , ∞6 , ∞5 , 7, 8, 0, 1, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6). Below, for t  2, take the vertex set (Z10t−1 × Z4 ) ∪ {∞0 , ∞1 , . . . , ∞8 }. The total (10t − 1) × 5 + 10 blocks are {Ai : 1  i  10} ∪ {Bj mod (10t − 1, −) : 1  j  5}. These blocks Ai , Bj are listed below.  A1 = ∞1  A2 = ∞2  A3 = ∞3  A4 = ∞4  A5 = ∞5  A6 = ∞6

[6t + 1, 10t − 1] [1, 2t] , [6t, 10t − 2] [0, 2t − 1]

∞2 ∞0

∞3 ∞8

∞4 ∞7

∞5 ∞6

[0, 10t − 2] [a, 10t − 2 + a]

∞3 ∞1

∞4 ∞0

∞5 ∞8

∞6 ∞7

0 [1, 10t − 2] [2t + 1, 5t + 1] [5t + 2, 8t − 1] , 10t − 2 [0, 10t − 3] [7t − 2, 10t − 2] [0, 3t − 3]

∞4 ∞2

∞5 ∞1

∞6 ∞0

∞7 ∞8

[0, 5t] [5t − 2, 10t − 2]

∞5 ∞3

∞6 ∞2

∞7 ∞1

∞8 ∞0

0 [1, 10t − 2] [4t, 5t − 2] 10t − 2 [0, 10t − 3] [9t, 10t − 2]

∞6 ∞4

∞7 ∞3

∞8 ∞2

∞0 ∞1

[0, 10t − 2] [0, 10t − 2]

∞7 ∞5

∞8 ∞4

∞0 ∞3

∞1 ∞2

0 10t − 2

[5t + 1, 10t − 2] [0, 5t − 3]

[8t, 10t − 2] [0, 4t − 1] , [3t − 1, 5t − 3] [5t − 2, 9t − 3] [5t − 1, 10t − 2] , [0, 5t − 1]

[6t + 1, 10t − 2] [0, 2t] , [t + 2, 5t − 1] [5t, 7t]

[1, 10t − 2] [0, 10t − 3]

[2t + 1, 8t − 1] , [2t − 1, 8t − 3]

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 A7 = ∞7  A8 = ∞8  A9 = ∞0  A10 = ∞0

∞8 ∞6

∞0 ∞5

∞1 ∞4

∞2 ∞3

[0, 10t − 2] [1, 10t − 1]

∞0 ∞7

∞1 ∞6

∞2 ∞5

∞3 ∞4

0 1 10t − 3 10t − 2

∞1 ∞8

∞2 ∞7

[0, 10t − 4] [3, 10t − 1]

∞3 ∞6

∞4 ∞5

[8t, 10t − 2] [8t − 2, 10t − 4]

[2, 10t − 2] [4t, 10t − 2] , [0, 10t − 4] [4t − 4, 10t − 6]

10t − 3 10t − 2 [0, 5t − 1] 1 2 [5t − 1, 10t − 2]

[0, 10t − 2] [b, 10t − 2 + b]

[1, 5t − 1] [3, 10t − 7]2 [0, 10t − 2]2 −[1, 5t − 1] [1, 5t − 4] [0, 5t − 1] ∞4 U P 1 ∞1 10t − 2 , 0 ∞3 10t − 2 V Q ∞2 0

[8t, 9t − 3] [6t, 8t − 6]2

[10t − 8, 10t − 4]2 [10t − 5, 10t − 3]

 B2 =

[4, 10t − 6]2 [5t + 2, 6t] [1, 5t − 1] [5, 10t − 1]2 −[1, 5t − 1] [2, 5t − 1] [5t + 2, 10t − 3] [5, 2t + 1]2 ∞3 ∞6 ∞8 1 3 2 4 ∞4 , 10t − 2 1 0 10t − 2 ∞7 ∞5 5t 5t + 1 0

 B3 = 0  B4 =

Y Z

[9, 10t − 3]2 [4, 5t − 2]

[5t, 6t] [0, 2t]2

0 ∞0

∞1 2t + 2

[2t, 5t − 1] [5t + 2, 10t − 2] [0, 4t − 2] [3, t + 1] −[4, t + 2] −[2t + 1, 5t] −[5t + 2, 10t − 2] −[0, 4t − 2] 1 2 ∞8 5t − 2 M R , ∞7 ∞5 10t − 4 5t + 1 N S 0

 B5 =

X [1, 5t − 1] −[1, 5t − 1] W

2 ∞1

∞2 10t − 2

∞2 1

c ∞4

∞3 0

[2t, 5t − 1] −[2t, 5t − 1]

[5t, 10t − 2] [5t + 2, 10t − 2] [1, 2t − 1] [0, 4t − 3] −[1, 4t − 2] −[5t, 10t − 2] −[5t + 3, 10t − 1] −[1, 2t − 1] 4t − 2 5t + 1 5t − 1 5t 5t + 1 ∞5 ∞7 . ∞2 ∞3 ∞4 ∞6 ∞8 6t 5t − 2 0

[5t, 6t] , [0, t]

[0, 6t] , [0, 6t]

 B1 =

0 1 [2, 4t − 1] , 10t − 3 10t − 2 [0, 4t − 3]

5t ∞0

1 ∞6

3 ∞8

∞5 3

∞6 10t − 3

0 ∞1

The undetermined variable in A1 , A10 , B1 , B3 , B4 are listed as follows, where k ∈ Z10t−1 . a = 5t (t odd) or 1 (t even), b = (5t + 1)/2 (t odd) or (5t + 2)/2 (t even), c = 5t − 1 (t odd) or 5t (t even). ⎧∞ ⎧∞ 8 6 ⎪ ⎪ for 0  k  2t, for 0  k  4t − 1, ⎪ ⎪ ⎪ ⎪ 10t −3 10t − 5 ⎪ ⎪  ⎪  ⎪ ⎨ ⎨ U P ∞5 ∞7 = = for 2t + 1  k  6t, for 4t  k  8t − 1, ⎪  5t ⎪ V Q ⎪ ⎪  10t − 3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∞ ∞ 0 6 ⎩ ⎩ for 6t + 1  k  10t − 2, for 8t  k  10t − 2, 5t − 1 10t − 3 ⎧∞ ⎧∞ 2 4 ⎪ ⎪ for 0  k  2t, for 0  k  4t − 1, ⎪ ⎪ ⎪ 5t ⎪ ⎪ ⎪  5t − 3  ⎪  ⎪ ⎨ ⎨ R M ∞1 ∞3 = = for 2t + 1  k  6t, for 4t  k  8t − 1, ⎪ ⎪ S N −2 ⎪ ⎪  10t  5t − 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∞ ∞ 0 2 ⎩ ⎩ for 6t + 1  k  10t − 2, for 8t  k  10t − 2, 0 5t − 3 ⎞ ⎧ ⎛ [2, 5t − 3] ⎧ ⎛ [5t + 1, 10t − 2] ⎞ 2    2  ⎪ ⎪ ⎪ ⎪ ⎝ ⎝ ⎠ ⎠ (t odd), ⎪ ⎪ 15t − 3 15t − 1 (t odd), ⎪ ⎪ ⎪ 5t, , 10t − 3 ⎪  ⎪  ⎪ ⎨ ⎨ 2 2 Y X = ⎛ = ⎛ ⎞ ⎞ ⎪ ⎪ Z W ⎪ ⎪ [5t + 2, 10t − 2] [2, 5t − 2]2 ⎪ ⎪ 2     ⎪ ⎪ ⎪ ⎪ ⎝ ⎠ ⎠ (t even). ⎝ ⎪ ⎪ 15t − 4 15t − 2 (t even), ⎩ ⎩ , 10t − 3 5t, 2 2

∞7 5

,

Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

4127

Proof. Firstly, in the following table the interior and exterior edges occurring in the structures in the left-hand column are scattered in the rest of the table.

Further, each pair containing ∞r (0  r  8) occurs once as an interior edge and twice as an exterior edge, as required. Next, each difference in Z10t−1 × Z4 appears once as an interior edge, as listed below: B1 --- D1,1 : [1, 5t − 1],

D1,2 : [0, 10t − 6] ∪ {10t − 4, 10t − 2},

D1,0 : [8t, 9t − 3],

D1,3 : [10t − 8, 10t − 2]2 , B2 --- D2,2 : [1, 5t − 1],

D2,3 : [1, 10t − 5] ∪ {10t − 3, 0},

B3 --- D3,3 : [1, 5t − 1],

D1,3 : [1, 10t − 1]2 ∪ [2, 10t − 10]2 ,

B4 --- D1,0 : [5t, 8t − 1] ∪ [9t − 2, 10t − 2],

D2,0 : [5t + 2, 6t],

D1,2 : 10t − 5, 10t − 3,

D2,3 : 10t − 4, 10t − 2,

D3,0 : [5t, 6t],

D2,0 : [1, 5t − 3] ∪ [6t + 1, 10t − 1] ∪ {5t + 1},

D0,0 : [2t, 5t − 1], B5 --- D3,0 : [1, 5t − 1] ∪ [6t + 1, 10t − 1],

D1,0 : [0, 5t − 1],

D0,0 : [1, 2t − 1],

D2,0 : 5t − 1, 5t − 2, 5t.         P M R Finally, each difference in Z10t−1 × Z4 appears twice as an exterior edge. In fact, the exterior edges in U V , Q , N , S and Ai (1  i  10) are just all pairs in

D1,2 : −a, 5t − 2, −b, 10t − 5, 10t − 3, D2,0 : 0, 1, 5t + 2,

D1,3 : 10t − 2, 10t − 3, 5t,

D3,0 : 10t − 2, 10t − 3, 0, 1, 5t + 1,

D2,3 : 1, 10t − 4,

D1,0 : 5t − 1, 5t − 1.

In other base blocks, we have B1 --- D1,1 : [1, 5t − 1] ∪ [1, 5t − 1], D1,3 : [0, 5t − 1] ∪ [6t, 8t − 6]2 , B2 --- D2,2 : [1, 5t − 1] ∪ [1, 5t − 1],

D1,0 : [0, 5t − 4] ∪ [10t − 5, 10t − 2], D2,0 : [5t + 2, 10t − 3],

D3,0 : [t + 3, 2t] ∪ {1, 2, 3},

D2,3 : [5t, 10t − 1],

D2,0 : [2, 5t + 1] ∪ {10t − 2},

D2,3 : 10t − 2,

D1,3 : [5t + 1, 10t − 4],

D1,2 : [0, 5t − 3] ∪ [8t − 2, 10t − 6]2 ∪ {10t − 2},

D3,0 : [5t − 1, 10t − 4],

D1,0 : [4t − 1, 5t − 2] ∪ {5t − 3},

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B3 --- D3,3 : [1, 5t − 1] ∪ [1, 5t − 1],

D2,3 : [2, 5t − 1] ∪ [8t − 3, 10t − 1]2 ,

D3,0 : [4, 5t − 2] ∪ {0},

D1,0 : [5t, 10t − 6], D2,0 : [4t, 5t], D1,3 : 10t − 6, 10t − 4, 10t − 2,    ⎧ 15t − 6 15t − 2 ⎪ 5t − 1, , , 10t − 3 (t even), ⎪ ⎨ 2 2 D1,2 =     ⎪ ⎪ ⎩ 5t, 15t − 5 , 15t − 1 , 10t − 2 (t odd). 2 2 B4 --- D3,0 : [5t + 2, 10t − 2] ∪ [2t + 1, 5t] ∪ [2, t + 2], D1,3 : [8t − 4, 10t − 8]2 ∪ [0, 6t − 2]2 ,

D1,0 : [0, 4t − 2] ∪ {5t − 2},

D2,3 : [2, 10t − 6]2 ,

D1,2 : [0, 8t − 4]2 ∪ {10t − 4},

D0,0 : [2t, 5t − 1] ∪ [1, 6t − 1]2 , B5 --- D2,0 : [1, 4t − 1] ∪ [5t + 3, 10t − 1] ∪ {5t + 1}, D0,0 : [1, 2t − 1] ∪ [6t + 1, 10t − 3]2 ,

D1,0 : [5t, 10t − 2],

D1,3 : [1, 10t − 3]2 ,

D2,3 : [1, 8t − 5]2 ,

D1,2 : [1, 10t − 7]2 .



Theorem 2.13. There exists a T(K1,2p , 9p) for any positive integer p ≡ 1 mod 8. Construction. Let p = 8t + 1. For t = 0, a T(K1,2 , 9) can be constructed on Z9 , as follows. (0 : 1, 2; 8, 7), (0 : 3, 4; 6, 5) mod 9. Below, for t  1, take the vertex set (Z18t+2 × Z4 ) ∪ {∞}. The block set consists of the following 9 base blocks modulo 18t + 2, where the entry a(b) in base block B mean that B + i contains the entry a + i, when 0  i  9t, and the entry b + i, when 9t + 1  i  18t + 1.     ⎛ ⎞ 9t 9t + 2 , 2t + 1, 9t [0, 9t] ⎟ ∞ ⎜ 2 2 ⎜ ⎟     B1 = ⎜ 0  ⎟,  ⎝ ⎠ 27t + 3 9t 9t + 4 − 2t + 1, , 9t + 1 [0, 18t]2 − 2 2 2     ⎞ ⎛ 9t 9t + 2 2t + 1, [1, 9t + 1] ⎟ ∞ , 9t ⎜ 2 2 ⎟ ⎜     B2 = ⎜0  ⎟,  ⎠ ⎝ 27t + 3 9t 9t + 4 − 2t + 1, , 9t + 1 [−18t, 0]2 − 2 2 2     ⎞ ⎛ 9t + 2 9t , 9t [1, 7t] 1, ∞ 9t + 1 2 ⎟ ⎜     2  B3 = ⎝ 0  ⎠, 9t + 4 27t + 3 9t − , 9t + 1 −[2, 7t + 1] 9t + 2 − 1, 2 2 2     ⎛ ⎞ 9t 9t + 2 ∞ 1, , 9t [9t + 2, 13t + 1] [9t + 1, 12t + 1] 2 2 ⎜ ⎟      B4 = ⎝ 0  ⎠, 9t 9t + 4 27t + 3 − 1, − [0, 6t]2 , 9t + 1 [2, 8t]2 2 2 2  [2t + 1, 9t] [9t + 1, 18t + 1] 0(9t + 1) B5 = 0 , 18t + 1(∞) −[2t + 2, 9t + 1] −[9t + 2, 18t + 2]  [1, 9t] [14t + 1, 18t + 1] (9t + 1)(0) [12t + 2, 15t + 1] , B6 = 0 ∞(0) [6t + 2, 12t]2 −[9t + 2, 18t + 1] [10t + 1, 18t + 1]2  [1, 5t] [9t + 1, 18t + 1] 0(9t + 1) [1, 2t] , B7 = 0 0(∞) −[1, 2t] −[1, 5t] [0, 18t]2  [7t + 1, 9t] [4t + 2, 9t] 0 9t + 1(0) [1, 9t + 1] B8 = 0 , 9t + 1 ∞(18t + 1) −[0, 9t] −[7t + 2, 9t + 1] −[4t + 1, 9t − 1]  [1, 3t] [1, 2t] [9t + 2, 18t + 2] [0, 2t] . B9 = 0 −[1, 2t + 1] −[1, 3t] −[1, 2t] [−9t, 9t]2 Proof. Firstly, each pair containing ∞ occurs once as an interior edge and twice as an exterior edge, as required. Next, each difference in Z18t+2 × Z4 appears once as an interior edge, as listed below, where {d}∗ means that all pairs {(x, x + d) : 0  x  9t} occur in the design, {d} means that all pairs {(x, x + d) : 9t + 1  x  18t + 1} occur in the design. B1 --- D1,1 : [2t + 1, 9t],

D1,2 : [0, 9t],

B2 --- D3,3 : [2t + 1, 9t],

D2,3 : [9t + 1, 18t + 1],

Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

D2,0 : 9t + 1,

B3 --- D2,2 : [1, 9t],

D2,3 : [1, 7t],

B4 --- D0,0 : [1, 9t],

D1,0 : [5t + 1, 9t],

D3,0 : [6t + 1, 9t + 1],

B5 --- D3,3 : {9t + 1},

D2,3 : {0}∗ ,

D1,3 : [1, 16t + 1],

B6 --- D0,0 : {9t + 1},

D1,0 : {0} ,

D3,0 : [3t + 1, 6t],

D2,0 : [1, 4t + 1] ∪ [9t + 2, 18t + 1],

D1,0 : [1, 5t] ∪ [9t + 1, 18t + 1] ∪ {0}∗ ,

B7 --- D1,1 : [1, 2t] ∪ {9t + 1}, B8 --- D2,2 : {9t + 1},

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D2,3 : [7t + 1, 9t] ∪ {0} ,

B9 --- D1,3 : [16t + 2, 18t + 2],

D1,2 : [9t + 1, 18t + 1],

D3,0 : [1, 3t] ∪ [9t + 2, 18t + 2],

D2,0 : [4t + 2, 9t] ∪ {0},

D3,3 : [1, 2t].

Finally, each difference in Z18t+2 × Z4 appears twice as an exterior edge, as follows: B1 --- D1,1 : [1, 4t + 1]2 ∪ [4t + 2, 9t + 1] ∪ [2t + 1, 9t],

D1,0 : [0, 18t]2 ,

D2,0 : [0, 9t],

B2 --- D3,3 : [1, 4t + 1]2 ∪ [4t + 2, 9t + 1] ∪ [2t + 1, 9t],

D1,3 : [0, 18t]2 ,

D1,2 : [9t + 1, 18t + 1],

B3 --- D2,2 : [1, 9t + 1] ∪ [1, 9t],

D1,2 : 9t,

D2,0 : [11t + 1, 18t],

D3,0 : [4t + 1, 18t − 1]2 ,

D1,0 : 18t + 1, B4 --- D0,0 : [1, 9t + 1] ∪ [1, 9t],

D3,0 : [10t + 2, 18t]2 ,

D1,0 : [12t + 2, 18t + 2]2 ,

D1,3 : [6t + 1, 13t + 1], D3,0 : [0, 9t] ∪ {18t + 1}∗ ,

B5 --- D2,3 : [2t + 2, 9t + 1], D1,0 : [1, 18t + 1]2 ,

D2,0 : {18t + 1}∗ ,

D1,2 : [1, 14t − 1]2 ,

B6 --- D1,0 : [6t + 2, 12t]2 ∪ [1, 8t + 1]2 , D3,0 : [9t + 2, 18t + 1] ∪ {0} , D : [3t + 1, 6t] ∪ {0} , D : [−(9t − 1), 9t − 1] , D : [0, 4t], 1,3

B7 --- D1,1 : [1, 2t] ∪ [2, 4t]2 , D3,0 : [2, 10t]2 ∪ {0}∗ ,

2,3

2

D1,3 : [13t + 2, 18t + 1] ∪ {0}∗ ,

1,2

D1,2 : [0, 18t]2 ,

D2,0 : [1, 9t + 1],

D2,0 : [9t + 1, 11t] ∪ {18t + 1} , D1,2 : [4t + 1, 9t − 1], D3,0 : [1, 4t − 1]2 ∪ {9t + 1} ∪ {18t + 1} , D1,3 : [1, 18t + 1]2 , D1,0 : [8t + 3, 18t − 1]2 ,

B8 --- D2,3 : [9t + 1, 18t + 2],

B9 --- D2,3 : [1, 2t + 1] ∪ [−9t, 9t]2 , D1,2 : [14t + 1, 18t + 1]2 ,

D3,3 : [1, 2t] ∪ [2, 4t]2 ,

D1,0 : [2, 6t]2 ,

D1,3 : [1, 3t],

D2,0 : [9t + 2, 18t + 2].



Theorem 2.14. There exists a T(K1,2p , 9p) for any positive integer p ≡ 5 mod 8. Construction. Let p = 8t + 5. Take the vertex set (Z18t+11 × Z4 ) ∪ {∞}. The block set consists of the following 9 base blocks module 18t + 11.  [0, t − 1] ∞ [1, 9t + 5] [1, 6t + 4] , B1 = 0 9t + 5 −[1, 9t + 5] −[9t + 7, 15t + 10] −[1, t]  [0, 4t + 1] [14t + 8, 18t + 10] [t, 9t + 4] B2 = 0 , −[t + 1, 9t + 5] −[9t + 5, 13t + 6] [10t + 6, 18t + 10]2  [4t + 4, 9t + 6] [9t + 6, 11t + 6] ∞ [1, 9t + 5] B3 = 0 , [1, 4t + 1]2 0 −[1, 9t + 5] −[4t + 3, 9t + 5]  18t + 10 [0, 9t + 5] [11t + 7, 18t + 9] B4 = 0 , ∞ [0, 18t + 10]2 [4t + 3, 18t + 7]2  [1, 9t + 5] [9t + 6, 14t + 9] [1, 2t + 1] B5 = 0 , −[1, 9t + 5] [−9t − 4, t + 2]2 −[1, 2t + 1]  [2t + 2, 9t + 5] ∞ [0, 9t + 4] B6 = 0 , 18t + 10 −[1, 9t + 5] −[2t + 2, 9t + 5]

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 B7 = 0  B8 = 0  B9 = 0

∞ 9t + 6 [0, 7t + 2] 2 9t + 6 −[0, 7t + 2]

[0, 9t + 4] , −[1, 9t + 5]

[1, 5t + 2] [9t + 6, 18t + 10] [7t + 3, 9t + 5] , −[7t + 3, 9t + 5] −[1, 5t + 2] −[9t + 7, 18t + 11] 9t + 5 ∞

[5t + 3, 9t + 5] −[5t + 3, 9t + 5]

[1, 9t + 6] [1, 18t + 11]2

[9t + 7, 12t + 6] [−9t − 3, −3t − 5]2

.

Proof. Firstly, each pair containing ∞ occurs once as an interior edge and twice as an exterior edge, as required. Next, each difference in Z18t+11 × Z4 appears once as an interior edge, as listed below: B1 --- D0,0 : [1, 9t + 5],

D3,0 : [12t + 7, 18t + 10],

B2 --- D1,0 : [9t + 7, 17t + 11] ∪ [1, 4t + 3], B3 --- D1,1 : [1, 9t + 5],

D1,0 : [17t + 12, 18t + 11],

D2,0 : [14t + 10, 18t + 11],

D1,0 : [4t + 4, 9t + 6],

D1,2 : [9t + 6, 11t + 6],

B4 --- D1,2 : [0, 9t + 5] ∪ [11t + 7, 18t + 10], B5 --- D2,2 : [1, 9t + 5],

D2,0 : [1, 2t + 1] ∪ [9t + 6, 14t + 9],

B6 --- D2,3 : [0, 9t + 4],

D2,0 : [2t + 2, 9t + 5],

B7 --- D1,3 : [11t + 9, 18t + 11], B8 --- D1,3 : [1, 11t + 8],

D3,0 : [0, 9t + 4] ∪ {9t + 6},

D3,3 : [1, 5t + 2],

B9 --- D3,3 : [5t + 3, 9t + 5],

D2,3 : [9t + 5, 18t + 10],

D3,0 : [9t + 7, 12t + 6] ∪ {9t + 5}.

Finally, each difference in Z18t+11 × Z4 appears twice as an exterior edge, as follows: B1 --- D0,0 : [1, 9t + 5] ∪ [1, 9t + 5], D2,3 : [−9t − 3, 3t + 3]2 , B2 --- D2,0 : [t + 1, 9t + 5],

D3,0 : [1, 8t + 5]2 ∪ [9t + 5, 13t + 6],

D2,3 : [t + 4, 9t + 6]2 ,

D1,3 : [1, 4t + 1]2 ∪ [9t + 6, 14t + 8],

D3,0 : [8t + 7, 18t + 11]2 ,

B4 --- D1,0 : [0, 18t + 10]2 ,

D2,3 : [9t + 6, 11t + 6],

D1,3 : [4t + 3, 18t + 7]2 ,

B5 --- D2,2 : [1, 9t + 5] ∪ [1, 9t + 5], D3,0 : [13t + 7, 18t + 10], B6 --- D2,3 : 18t + 10,

D2,3 : [−9t − 4, t + 2]2 ,

D1,2 : [4t + 7, 18t + 11]2 , B8 --- D2,3 : [7t + 3, 9t + 5],

B9 --- D1,3 : [0, 18t + 10]2 ,

D2,3 : [11t + 7, 18t + 9],

D1,2 : [1, 2t + 1],

D1,2 : [2t + 2, 9t + 5],

D1,0 : [4t + 4, 18t + 10]2 ,

B7 --- D2,3 : [0, 7t + 2] ∪ {9t + 5},

D1,2 : [1, 4t + 5]2 ,

D2,0 : [0, 9t + 5],

D1,0 : [2, 4t + 2]2 ,

D2,0 : [9t + 6, 18t + 10],

D3,0 : [2, 18t + 10]2 ,

D1,2 : [2, 16t + 10]2 ,

D1,3 : [14t + 9, 18t + 11],

B3 --- D1,1 : [1, 9t + 5] ∪ [1, 9t + 5], D1,0 : 0,

D2,0 : [1, t] ∪ [9t + 6, 15t + 10],

D1,2 : [16t + 12, 18t + 10]2 ,

D1,3 : [1, 9t + 5] ∪ {18t + 9}, D1,0 : [1, 18t + 9]2 ,

D2,0 : 0,

D3,3 : [1, 5t + 2] ∪ [8t + 7, 18t + 9]2 ,

D3,0 : [0, 9t + 4],

D1,0 : [1, 18t + 9]2 , D3,3 : [5t + 3, 9t + 5] ∪ [1, 8t + 5]2 ,

D2,0 : [15t + 11, 18t + 10],

D1,2 : [9t + 6, 18t + 11].



D2,3 : [3t + 5, 9t + 3]2 ,

Y. Liu, Q. Kang / Journal of Statistical Planning and Inference 138 (2008) 4111 -- 4131

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3. Conclusion Theorem 3.1. There exists a T(K1,2p , v) for v ≡ 1 mod 4p. Proof. By Theorem 2.1.  Theorem 3.2. For p ≡ 1 mod 4, a T(K1,2p , v) exists for v ≡ p mod 4p. Proof. By Theorems 2.4, 2.5, 2.11--2.14, and the recursive construction in Theorem 2.2.  Theorem 3.3. For p ≡ 3 mod 4, a T(K1,2p , v) exists for v ≡ 3p mod 4p. Proof. By Theorems 2.4, 2.6--2.10 and the recursive construction in Theorem 2.3.  Theorem 3.4. (1). For any prime power p ≡ 1 mod 4, the spectrum for perfect T(K1,2p , v) is v ≡ 1, p mod 4p, (2). For any prime power p ≡ 3 mod 4, the spectrum for perfect T(K1,2p , v) is v ≡ 1, 3p mod 4p. Proof. By Theorems 3.1--3.3 and Lemma 1.1.  Acknowledgments The authors are thankful to the referees for their valuable suggestions. References Billington, E.J., Lindner, C.C., 2006. Perfect triple configurations from subgraphs of K4 : the remaining cases. Bull. Inst. Combin. Appl. 47, 77--90. Billington, E.J., Lindner, C.C., Rosa, A., 2005. Lambda-fold complete graph decompositions into perfect four-triple configurations. Australas. J. Combin. 32, 323--330. Colbourn, C.J., Dinitz, J.H., 2006. The CRC Handbook of Combinatorial Designs. CRC Press Inc., New York. Kang, Q., Zhang, Y., Hou, Y., to appear. Perfect T(H)-triple systems for subgraphs H of K5 with six edges or less. Bull. Inst. Combin. Appl. ¨ ukçifçi, ¨ Kuç S., Lindner, C.C., 2004. Perfect hexagon triple systems. Discrete Math. 279, 325--335. Lindner, C.C., Rosa, A., 2008. Perfect dexagon triple systems. Discrete Math. 308, 214--219. Liu, Y., Kang, Q., Li, M., to appear. The existence for perfect T(K1,k )-triple systems. Ars Combin. Tazawa, S., 1979. Claw-decomposition and evenly-partite-claw-decomposition of complete multipartite graphs. Hiroshima Math. J. 9, 503--531.