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The existence of three idempotent IMOLS
Discrete Mathematics 262 (2003) 1 – 16
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The existence of three idempotent IMOLS R. Julian R. Abela; ∗ , Beiliang Dub a School
b Department
of Mathematics, University of New South Wales, Sydney 2052, Australia of Mathematics, Suzhou University, Suzhou 215006, People’s Republic of China
Received 26 June 2000; received in revised form 6 August 2001; accepted 5 November 2001
Abstract In this paper it is shown that an idempotent TD(5; m) − TD(5; n) exists whenever the known necessary condition m ¿ 4n+1 is satis3ed, except when (m; n)=(6; 1) and possible when (m; n)= (10; 1). For m ¡ 60 and n 6 10, we also indicate where several idempotent TD(k; m)−TD(k; n)’s for k = 6; 7 can be found. c 2002 Elsevier Science B.V. All rights reserved. Keywords: Transversal design; Incomplete transversal design; Idempotent incomplete transversal design; Quasi-di
1. Introduction Let v and hi ; i = 1; : : : ; n, be positive integers. An incomplete transversal design (ITD), TD(k; v)− 16i6r TD(k; hi ), is a quadruple (X; G; B; H) satisfying the following conditions: 1. X is a set of kv elements called points and G = {G1 ; G2 ; : : : ; Gk } is a partition of X into k disjoint v-element subsets called groups. 2. B is a collection of k-element subsets of X called blocks, each containing one element from each Gs . 3. H = {H1 ; H2 ; : : : ; Hn } is a collection of disjoint subsets of X (called holes) and Hj contains hj elements from each group Gs . 4. Any two points in di
Corresponding author. Fax: +61-294492814. E-mail addresses: [email protected] (R.J.R. Abel), [email protected] (B. Du).
c 2002 Elsevier Science B.V. All rights reserved. 0012-365X/03/$ - see front matter PII: S 0 0 1 2 - 3 6 5 X ( 0 2 ) 0 0 5 1 7 - 4
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A holey parallel class in such an ITD is a set of v − 16i6r hi blocks containing every point outside the holes once and no points from the holes. If an ITD contains a holey parallel class, it is called idempotent. TD(k; v) − TD(k; h)’s with several disjoint holey parallel classes can be useful in recursive constructions for obtaining larger idempotent ITDs; we use the notation TD s (k; v) − TD s (k; h) to denote such an ITD with s disjoint holey parallel classes. When s = 1 (equivalent to an idempotent ITD) the notation TD∗ (k; v) − TD∗ (k; h) is more commonly used. Simple counting arguments imply (1) if a TD(k; v)−TD(k; h) exists, then v¿(k −1)h and (2) if a TD(k; v) − TD(k; h1 ) − TD(k; h2 ) exists then v¿(k − 1)h1 + h2 . From the second of these, a necessary condition for an idempotent TD(k; v) − TD(k; h) to exist is v¿(k − 1)h + 1. In [17], Du showed that for k = 5, the required condition v¿4h + 1 is suKcient for the existence of an idempotent TD(5; v) − TD(5; h) when h¿53. In this paper, we examine the case h653 and establish that the condition v¿4h+1 is suKcient here in all but two cases, namely when (v; h) = (6; 1) (nonexistent, for example, see [20,18]) or (10; 1). 2. Direct constructions All the direct constructions in this section come from quasi-di
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QDM: (a; b; c; d; e)T and (b; a; d; c; e)T . Combining these columns with the 3rst column ( for (v; h) = (50; 3)) and the 3rst two (in other cases) gives a (v; 5; 1; 1; h) QDM over Zv−h . In all cases the number of columns without a blank entry, that is, (v − h)− (k − 2)h = v − 4h is ¿1; hence these ITDs are all idempotent. (v; h) = (32; 2): 0 0 0 0 —
0 15 16 1 —
0 1 26 21 0
29 5 23 12 0
— 11 27 0 0
— 22 7 20 22 13 10 — 0 0
2 28 7 — 0
12 26 4 6 0
19 27 1 15 0
14 24 16 17 0
9 21 18 8 0
18 23 3 29 0
10 13 11 24 0
8 15 5 14 0
25 4 9 2 0
6 17 19 25 0
3 16 28 20 0
0 0 0 0 —
0 0 21 6 31 14 32 29 27 13 17 1 5 12 23 4 10 15 24 8 18 8 30 18 25 1 24 14 9 26 1 21 15 16 13 10 27 22 19 33 — 0 0 0 0 0 0 0 0 0
2 9 17 6 0
16 25 4 5 0
22 11 20 0 0
17 30 28 32 0
7 26 3 31 0
27 2 13 15 0
7 17 23 19 0
9 24 32 27 0
(v; h) = (36; 2):
— 19 7 12 0
— 20 28 18 11 23 29 — 0 0
3 33 2 — 0
26 28 24 21 0
13 31 18 12 0
(v; h) = (38; 4): 0 0 0 0 —
0 17 18 1 —
0 1 14 5 —
20 23 17 6 0
— — — — 25 5 18 19 8 3 14 3 16 33 26 4 25 30 0 — 0 0 0 0 0
12 16 31 10 0
29 0 7 22 0
14 30 22 1 10 33 5 2 9 — — — 0 0 0
15 21 11 29 0
4 11 28 20 0
32 6 8 1 0
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(v; h) = (44; 10): 0 0 0 0 0 0 17 3 5 7 0 18 21 11 5 0 1 28 22 10 — — — — — — 14 11 23 0
0 11 14 15 —
— — 33 30 29 15 7 12 11 31 27 33 5 31 8 13 15 — — — 0 0 0 0 0
13 0 — — — — — — — 22 1 26 32 4 9 25 20 21 29 24 1 0 28 17 4 19 26 32 16 14 25 30 21 10 9 7 0 0 0 0 0 0 0 0 0 18 10 6 5 28 3 23 24 2 16 17 8 19 27 20 6 3 12 22 2 18 — — — — — — — 0 0 0 0 0 0 0
(v; h) = (49; 3): 0 0 0 0 0
0 0 23 17 24 20 1 23 — —
39 21 33 34 23 40 27 32 0 0
37 1 8 9 0
41 6 20 3 0
23 16 43 28 24 38 3 30 25 29 9 36 42 5 4 34 18 24 35 28 31 6 14 39 38 16 43 44 0 0 0 0 0 0 0
7 18 11 35 4 19 31 14 21 10 15 26 11 5 33 37 0 0 0 0
— 2 45 36 0
— 15 42 30 0
— 45 10 20 26 27 40 12 19 12 22 25 13 — — — 0 0 0 0
22 17 8 44 32 13 17 2 7 1 41 29 0 0 0
(v; h) = (50; 3): 0 0 0 0 —
0 11 25 8 —
0 10 17 18 0
24 41 44 0 0
19 26 46 1 0
40 2 11 37 0
20 5 41 16 0
13 38 43 23 0
4 12 33 2 0
30 18 24 34 0
32 36 13 27 0
25 23 39 15 0
42 16 28 20 0
45 14 29 38 0
33 9 26 21 0
44 39 31 35 0
46 43 9 45 0
27 28 22 7 0
1 35 10 3 0
31 11 30 12 0
34 6 32 19 0
37 8 42 — 0
15 21 4 — 0
17 — 3 7 36 8 — 14 0 0
0 12 19 20 0
6 40 35 41 26 13 28 23 10 32 37 19 43 18 27 31 17 38 7 30 33 2 5 33 29 12 6 36 4 15 1 27 26 16 9 3 43 42 31 17 22 35 8 34 0 0 0 0 0 0 0 0 0 0 0
— — 29 22 40 6 5 25 0 0
(v; h) = (50; 4): 0 0 0 23 0 24 0 1 — —
0 31 44 41 —
R.J.R. Abel, B. Du / Discrete Mathematics 262 (2003) 1 – 16
34 4 20 24 42 — 36 25 11 5 14 45 18 13 40 28 44 37 32 7 24 11 2 0 0 0 0 0 0 0
— 21 14 39 0
— 16 21 23 0
— 29 38 10 0
8 22 15 44 41 25 — — 0 0
9 3 45 — 0
5
1 39 30 — 0
(v; h) = (52; 6): 0 0 0 0 0 0 23 37 17 4 0 24 18 40 7 0 1 33 45 40 — — — — 0 25 3 11 — 27 28 24 22 31 26 21 32 1 12 43 33 0 0 0 0
13 40 12 30 39 1 16 9 36 35 5 15 44 38 41 0 30 5 37 35 24 4 19 14 0 0 0 0 0 0
37 44 31 19 21 18 32 26 29 3 2 28 10 15 27 34 0 0 0 0
— — — — — 6 10 20 7 8 42 43 23 41 29 14 33 45 38 17 2 34 18 11 25 22 6 39 13 45 8 20 23 36 9 17 42 16 — — — — — — 0 0 0 0 0 0 0 0 0 0 0
Lemma 2.2. There exists a TD∗ (5; v) − TD∗ (5; h) for any (v; h) ∈ {(24; 5); (32; 7); (51; 4)}. Proof. For (v; h) = (24; 5), replace each column (a; b; c; d; e)T in the array below by the following four: (a; b; c; d; e)T , (b; a; d; c; e)T , (c; d; a; b; e)T , and (d; c; b; a; e)T . Then add one extra column (0; 0; 0; 0; −)T . This gives a (19; 5; 1; 1; 5)-QDM over Z19 : 0 6 4 9 —
0 1 7 11 0
— — 9 15 6 10 4 13 0 0
— — — 3 17 12 2 5 8 14 16 18 0 0 0
For (v; h) = (51; 4), a (47; 5; 1; 1; 4)-QDM is obtainable as follows: each column (a; b; c; d; e)T in the array below (except the last one) generates 10 columns, by applying the automorphism group of order 10 generated by T1 ; T2 where T1 (a; b; c; d; e)T = (e; d; c; b; a)T and T2 (a; b; c; d; e)T = (b; c; d; e; a)T . The last given column generates 3ve columns by applying the group of order 5 generated by T2 . The resulting 55 columns form the required QDM: — 0 17 25 38
— 0 0 0 0 0 16 32 36 7 12 19 13 12 3 10 45 46 6 3 32 27 37 5 7
For (v; h) = (32; 7), a (25; 5; 1; 1; 7)-QDM is obtained by applying the group of order 5 generated by T1 (where T1 (a; b; c; d; e)T = (b; c; d; e; a)T ) to the following seven
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columns, and then adding four additional columns of the form (0; u; 2u; 3u; 4u) for u = 0; 5; 15; 20: — — 0 0 4 6 18 8 1 9
— 0 7 1 14
— 0 9 6 23
— 0 10 13 4
— 0 12 23 21
— 0 18 17 13
Lemma 2.3. There exists a TD∗ (6; v) − TD∗ (6; h) for (v; h) ∈ {(28; 2); (30; 4); (38; 7); (40; 2); (42; 4); (43; 5); (44; 7); (45; 7); (46; 8)}. Proof. All of these designs are obtained from QDMs over Zv−h . To each column (a; b; c; d; e; f)T in the arrays below, we apply the automorphism group of order 9 generated by T1 ; T2 where T1 (a; b; c; d; e; f)T = (b; c; a; f; d; e)T , T2 (a; b; c; d; e; f)T = (wc; wa; wb; wf; wd; we)T and w is a given cube root of unity in Zv−h . Here, we take w = 3; 5; 10; 7, respectively, for v − h = 26; 31; 37; 38. In all cases, each column of A1 generates just three distinct columns under this automorphism group, each column of A2 generates nine distinct columns, and each column of A3 generates no columns other than itself. (v; h) = (28; 2): 0 1 0 11 13 9 0 21 — 3 — 7 A1 = ; 1 0 24 0 9 13 8 0 3 — 20 —
0 0 6 19 23 15 A2 = : 6 4 20 6 1 7
(v; h) = (30; 4): 0 13 — A1 = 1 9 3
7 11 21 ; 0 13 —
0 0 0 2 14 24 23 — 3 A2 = ; 6 4 20 8 7 11 — 23 15
0 0 0 A3 = : 0 0 0
(v; h) = (38; 7): — 0 0 A1 = 17 22 23
17 22 23 — 0 0
11 24 27 ; 11 24 27
— 0 11 16 21 23 A2 = 0 — 16 11 23 21
— 5 11 0 29 20
0 29 20 : — 5 11
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(v; h) = (40; 2): 0 19 — A1 = 1 11 7
33 0 5 1 21 0 17 7 3 — 35 11 ; 0 37 0 35 19 27 0 17 — 31 — 5
0 10 37 A2 = 12 9 0
0 0 2 25 17 3 : 32 22 3 5 25 29
(v; h) = (42; 4): 0 29 5 19 15 35 — 13 17 A1 = ; 1 0 26 11 19 30 7 — 20
0 0 0 0 22 28 32 12 13 — 37 5 A2 = ; 16 30 36 10 18 7 6 30 — 31 13 27
0 0 0 A3 = : 0 0 0
(v; h) = (43; 5): 0 19 — A1 = 14 2 22
11 0 9 7 0 23 1 — 25 ; 0 12 0 19 18 0 — 8 —
0 16 3 A2 = 10 20 —
0 0 0 18 26 7 — 25 35 : 22 4 32 13 7 21 11 25 23
(v; h) = (44; 7): — 0 0 A1 = 19 13 5
19 4 13 3 5 30 — 27 0 11 0 36
27 31 11 14 36 29 ; 4 31 3 14 30 29
— 15 — 0 11 0 22 6 13 A2 = 15 — 25 11 0 33 6 22 36
25 33 36 : — 0 13
(v; h) = (45; 7): 0 32 19 10 — 34 A1 = ; 2 0 22 19 14 —
A2 =
0 0 0 0 22 6 18 20 1 — 31 — 8 36 22 30 6 23 34 35 — 37 — 1
0 32 15 ; 26 27 23
0 0 0 A3 = : 0 0 0
7
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(v; h) = (46; 8): 0 33 0 5 1 15 19 21 0 17 7 29 — 3 — 35 11 13 A1 = ; 1 0 14 0 22 1 11 19 2 0 2 7 7 — 22 — 14 11
0 0 0 0 15 21 9 5 31 — — 1 A2 = : 8 28 36 10 23 6 0 19 — 27 5 —
The 3nal construction in this section was found by Colbourn [11]. Lemma 2.4. There exists a TD∗ (6; 19) − TD∗ (6; 2). Proof. Multiply all columns in the following array by 1 and −1. The result is a (17; 6; 1; 0; 2)-QDM. — 0 0 0 0 0 0 0 — 1 2 3 4 5 1 1 — 4 7 11 8 8 15 14 — 13 16 7 10 3 6 16 — 10 4 13 4 10 8 15 — 6
0 6 14 12 8 3
0 7 12 8 15 5
0 8 2 11 12 1:
3. Recursive constructions We need some known constructions for ITDs, which are mainly the working corollaries of Theorems 1.1 and 1.2 in [10] and Lemma 2.2 in [15]. So, we state the following lemmas without proof. Lemma 3.1. Suppose that v = mt + 16j6t mj and that the following designs exist: a TD(6; t), a TD∗ (5; m + m1 ) − TD∗ (5; s) − TD∗ (5; m1 ) and TD(5; m + mj ) − TD(5; s) − TD(5; mj ) ( for 26j6t). Then: 1. There exists a TD∗ (5; v) − TD∗ (5; st) if a TD∗ (5; 16j6t mj ) exists. 2. There exists a TD∗ (v; 16j6t mj )) if s = 0. 3. There exists a TD∗ (5; v)−TD∗ (5; m) if a TD∗ (5; 16j6t mj )) exists and m1 = s = 0. 4. There exists a TD∗ (5; v) − TD∗ (5; m + m1 ) if a TD∗ (5; 16j6t mj ) − TD∗ (5; m1 ) exists and s = 0. Lemma 3.2. Let v = mt + m1 + m2 where for i = 1; 2, mi = 16j6t mij . Suppose, there exist a TD(7; t) and TD(5; m + m1k + m2j ) − TD(5; m1k ) − TD(5; m2j ) (for 26k6t, 16j6t). Suppose also, there exist TD∗ (5; m2 ) and TD∗ (5; m + m11 + m2j ) − TD∗ (5; m11 ) − TD∗ (5; m2j ) ( for 16j6t). Then: 1. There exists a TD∗ (5; v) − TD∗ (5; m1 ). ∗ 2. If m21 = 0, and a TD (5; 16k6t m1k ) − TD∗ (5; m11 ) exists, then there exists a TD∗ (5; v) − TD∗ (5; m + m11 ).
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Lemma 3.3. Let v = mt + 16i6d wi . If there exist a TD(5 + d; t), a TD(5; m), a ∗ ∗ TD (5; m + w1 ) − TD (5; w1 ) and TD(5; m + wi ) − TD(5; wi ) ( for 26i6d), then: 1. There exists a TD∗ (5; v)−TD∗ (5; t) if a TD∗ (5; m+ 16i6d wi ) exists and m¿3wi (26i6d). 2. There exists a TD∗ (5; v) − TD∗ (5; m + 16i6d wi ). 3. There exists a TD∗ (5; v) − TD∗ (5; m + w1 ) if a TD∗ (5; m + 16i6d wi ) − TD∗ (5; w1 ) exists. Lemma 3.4. Let v = mt + 16i6d wi + 26j6t mj . If there exist a TD(6 + d; t), a TD(5; m) and TD(5; m+mj )−TD(5; mj ) and TD(5; m+wi +mj )−TD(5; wi )−TD(5; mj ) ( for 26i 6 d and 26j 6 t) and TD∗ (5m + w1 + mj ) − TD∗ (5; w1 ) − TD∗ (5; mj ) ( for 16j 6 t), then: 1. A TD∗ (5; v) − TD∗ (5; 26j6t mj ) exists if a TD∗ (5; m + 16i6d wi ) exists. 2. A TD∗ (5; v) − TD∗ (5; m + 16i6d wi ) exists if a TD∗ (5; 26j6t mj ) exists. 3. A TD∗ (5; v + m1 ) − TD∗ (5; m + m1 + 16i6d wi ) exists if a TD∗ (5; 16j6t mj ) − TD∗ (5; m1 ) exists. Lemma 3.5. (1) Suppose a TD(10; t) exists, n is even, 26n62t; 06k62t; k = 3 or 4, and k = 21, if t = 11. Then for v = 7t + k + n, there exists a TD∗ (5; v) − TD∗ (5; n). (2) Suppose there exist a TD(10; t) and a TD∗ (5; t + 1), n is odd, 36n6 2t + 1; 06k62t and v = 7t + k + n. Then there exists a TD∗ (5; v) − TD∗ (5; n) if k = 1; 2; 3; 5; 9. Lemma 3.6. (1) Suppose the following designs exist: a TD(8; t) a TD∗ (5; s1 ), a TD∗ (5; s2 ), and a TD∗ (5; s3 ), where 06s2 ; s3 6t and 16s1 6t. Then there exists a TD∗ (5; 8t + s1 + s2 + s3 ) − TD∗ (5; 2t). (2) Suppose the following designs exist: a TD(8; t), a TD∗ (5; s1 ), a TD∗ (5; s2 ), and a TD∗ (5; n1 ), where 06s1 ; s2 ; 6t. Then, there exists a TD∗ (5; 7t + s1 + s2 + n1 ) − TD∗ (5; t + n1 ) if either (a) n1 ¡t or (b) n1 = t and s1 + s2 ¿0. We also need the following constructions for incomplete transversal designs, which are mainly the working corollaries of Theorem 3.1 in [16]. So, we state the following lemmas without proof. Lemma 3.7. Suppose the following designs exist: a TDs (5; t) − TDs (5; u), a TD(5; m), a TD∗ (5; m + m1 ) − TD∗ (5; m1 ) and TD(5; m + mj ) − TD(5; mj ) ( for 26j6s). Then, a ∗ TD (5; mt + 16j6s mj ) − TD∗ (5; t) exists if a TD∗ (5; mu + 16j6s mj ) − TD∗ (5; u) exists and m¿3mj ( for 26j6s). Lemma 3.8. Suppose there exist a TD(6; t) − TD(6; u), a TD(5; m), a TD∗ (5; m + m1 ) − TD∗ (5; m1 ) and TD(5; m + mj ) − TD(5; mj ) (for 26j6u). Then, a TD∗ (5; mt + ∗ ∗ ∗ 16j6u mj ) − TD (5; t) exists if a TD (5; mu + 16j6u mj ) − TD (5; u) exists and m¿3mj ( for 26j6u).
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Lemma 3.9. Suppose there exist a TD(6; t) − TD(6; u), a TD∗ (5; m + m1 ) − TD∗ (5; m1 ) and TD(5; m + mj ) − TD(5; m j ) ( for 26j6t). Suppose also, there exists a TD∗ (5; 16j6t mj ) − TD∗ (5; 16j6u mj ). Then there exists a TD∗ (5; mt+ ∗ ∗ ∗ 16j6t mj ) − TD (5; t) if a TD (5; mu + 16j6u mj ) − TD (5; u) exists and m¿3mj ( for 26j 6 t). Finally, we state a few useful lemmas which are fairly obvious: Lemma 3.10. If there exists TD∗ (5; v) − TD∗ (5; u) and TD∗ (5; u) − TD∗ (5; n), then there exists a TD∗ (5; v) − TD∗ (5; n). Lemma 3.11. Suppose v = mn and TD∗ (k; m); TD∗ (k; n) both exist. Then there exists a TD∗ (k; v) − TD∗ (k; n). Lemma 3.12. If there exists a TD(6; v) − TD(6; n), then there exists a TDn (5; v) − TDn (5; n) and hence also a TD∗ (5; v) − TD∗ (5; n).
4. Main result In this section, we make use of the constructions in the previous section to obtain TD∗ (5; v) − TD∗ (5; n)’s with v¿4n + 1. We obtain the following result: Theorem 4.1. For any integer n¿1, there exists a TD∗ (5; v) − TD∗ (5; n) if and only if v¿4n + 1. In some cases, the smallest designs are used to deal with larger ones, so it will be convenient to deal with the smallest values of n 3rst. Using this approach, we only need to consider values of v satisfying 4n + 16v616n + 4. For if v¿16n + 5, and both TD∗ (5; v) − TD∗ (5; 4n + 1) and TD∗ (5; 4n + 1) − TD∗ (5; n) exist, then a TD∗ (5; v) − TD∗ (5; n) can be obtained by Lemma 3.10. First of all, we mention the following theorem which was obtained by the second author in [17]. Theorem 4.2. For any integer n¿53, there exists a TD∗ (5; v) − TD∗ (5; n) if and only if v¿4n + 1. Therefore, we only need to consider the cases 26n653. Remark 4.3. Colbourn and Dinitz [14, p. 164] provide a list of values (v; n) for which TD(6; v)−TD(6; n) is unknown when v¿5n and n650. As mentioned in Lemma 3.12, deleting a group from TD(6; v)−TD(6; n) gives a TDn (5; v)−TDn (5; n) and hence also a TD∗ (5; v) − TD∗ (5; n). Therefore, for n¿5, we often only check the existence of
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TD∗ (5; v) − TD∗ (5; n) when either (1) 4n¡v¡5n or (2) v¿5n and TD(6; v) − TD(6; n) is unknown. In some cases however, in particular for v¡50, we point out where a known TD∗ (k; v) − TD∗ (k; n) can be found. Lemma 4.4. (a) There exist a TD∗ (7; v) − TD∗ (7; 2) for v ∈ {41; 47; 55} and a TD∗ (6; v) − TD∗ (6; 2) for v ∈ {15; 19; 28; 29; 33; 35; 40}. (b) There exists a TD∗ (5; v) − TD∗ (5; 2) for 96v636. Proof. For the values in (a), QDM solutions are given in [2] for v = 41; 47; 55, in [4] for v = 29, in Section 2, for v = 19; 28; 40, in [6] for n = 15; 33; 35 and in [6] for other values. For v = 22; 30; 34, we can use Lemma 3.1(2) since these values equal 4t + 2 for t = 5; 7; 8. For all the remaining values in (b), QDM solutions exist. For v = 9; 24; 29; 34 see [8]; for v = 10; 11; 12 see [9,19,16], respectively. For v = 32; 36, see Section 2; for v = 18; 23 see [13], and for all other values see [6]. Lemma 4.5. (a) There exist a TD∗ (7; v) − TD∗ (7; 3) for v ∈ {19; 40; 46}, and a TD∗ (6; v) − TD∗ (6; 3) for v ∈ {16; 22; 24; 25; 28; 34; 42; 45}. (b) There exists a TD∗ (5; v) − TD∗ (5; 3) for 136v652. Proof. For the values in (a), QDM solutions are given in [2] for v = 19; 40; 46, in [6] for v = 25; 36; 41; 45 and in [4] for other values. Now consider the values in (b). For v = 13; 18; 23; 33; 38; 43; 48, constructions appear in Lemmas 2.3 and 2.13 of [8]; if v ∈ {31; 35; 39; 47; 51} then v is of the form 4t + 3 where a TD(6; t) exists; hence Lemma 3.1(2) works for these values, and also for 52 = 7:7 + 3. The remaining values are all obtainable from QDMs: for v = 15; 17; 27, see [13], for v = 49; 50, see Section 2, and for other values, see [6]. Lemma 4.6. (a) There exists a TD∗ (7; v) − TD∗ (7; 4) for v ∈ {29; 39; 45}, a TD(6; 20) − TD(6; 4) and a TD∗ (6; v) − TD∗ (6; 4) for v ∈ {21; 23; 25; 27; 28; 30; 31; 33; 34; 35; 37; 39; 40; 41; 42; 43; 45; 46; 47}. (b) There exists a TD∗ (5; v) − TD∗ (5; 4) for 176v668. Proof. First we deal with the values in (a); all these come from QDMs. For v = 29 see [6], and for v = 39; 45 see [2]. For v = 20, see [7]. For v = 21; 25; 27; 28; 31; 33; 34; 40; 43; 46, see [4]. For v = 23; 35; 37; 41; 47, see [6]. For v = 30; 42, see Section 2. Now we consider the values in (b). For v = 17, see [6], for v = 19, see [13], for v = 18; 22; 26; 30, see [21] (or [14, p. 450]) and for v = 38; 50; 51, see Section 2. If v ∈ {24; 32; 36; 48; 52; 53; 56; 59; 60; 67; 68}, then v can be written as either 4t + 4; 7t + 4 or 11t + 4 where a TD(6; t) exists; hence Lemma 3.1(2) works for these values. If v ∈ {44; 49; 55; 57; 63; 64; 65}, then v can be written as 4t +u where TD(6; t); TD∗ (5; u) both exist and 06u¡t; use Lemma 3.1(3) for these values. Finally, v = 54; 58; 61; 62; 66 are obtainable using Lemma 3.2 with m = 7; t = 7; m1j ∈ {0; 1} and m2j ∈ {0; 1; 2} and 16j6t m1j = 4.
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Lemma 4.7. (a) There exists a TD∗ (7; v)−TD∗ (7; 5) for v ∈ {44; 58}, and a TD∗ (6; v)− TD∗ (6; 5) for v ∈ {26; 27; 30; 31; 32; 35; 36; 38; 40; 41; 43; 44; 45; 46; 50; 51; 56}. (b) There exists a TD∗ (5; v) − TD∗ (5; 5) for 216v684. Proof. The designs in (a) come from QDMs; they can be found in [2] for v = 44; 58, in [6] for v = 27, in [14, p. 118] for v = 30, in [4] for v = 26; 32; 38; 44, in Section 2 for v = 43, and in [1] for v = 31; 36; 41; 46; 50; 51; 56. For v = 35; 40; 45, use Lemma 3.11. By Remark 4.3, it is only necessary to consider v ∈ {21–24; 28; 29; 31–34; 37; 39; 41; 42; 46–53; 56; 57; 78; 79; 81} for the values in (b). For v = 22; 23; 28 and 34, see [6]; for v = 24, see Section 2. For v = 33; 54; 57; 77; 82, use Lemma 3.1(2) with m = 4 or 7. For v = 21; 29; 37; 39; 49; 52; 53; 57; 59; 79; 81, apply Lemma 3.1(4) with m = 4. For v = 47; 48; 56, use Lemma 3.3 with m = 4. Finally, for v = 42 = 7:5+(7 = 3:2+1:1), use Lemma 3.1(1) with m = 7; t = 5; s = 1, and for 78 = 7:9 + (5 = 5:1) + (9 = 4:2 + 1:1) + (1 = 1:1), apply Lemma 3.5(2). Lemma 4.8. (a) There exist a TD∗ (7; v) − TD∗ (7; 6) for v ∈ {41; 43; 47; 51; 57; 79}, a TD(6; 30) − TD(6; 6) and a TD∗ (6; v) − TD∗ (6; 6) for v ∈ {31; 33; 35; 36; 37; 39; 40; 41; 45; 46; 50}. (b) There exists a TD∗ (5; v) − TD∗ (5; 6) for 256v6100. Proof. A TD∗ (7; v) − TD∗ (7; 6) can be found in [2] for v = 41; 51; 57; 79, in [12] for v = 43 and in [6] for v = 47. TD∗ (6; v) − TD∗ (6; 6)s are given for v = 31; 33; 37; 45; 49 in [4] and for v = 39; 48 in [6]. A TD(6; 30) − TD(6; 6) and TD∗ (6; v) − TD∗ (6; 6)s for v = 35; 36; 40; 41; 46; 50 can be found in [1] or [3]. By Remark 4.3, it is only necessary to consider v ∈ {25–30; 32–36; 38; 40–42; 44; 46; 50–54; 57; 58; 79–82} for the values in (b). For v = 25, see [14, p. 295]; for v = 26–28; 32; 39; 44; 53, see [6]; for v = 29, see [13]; and for v = 52, see Section 2. For v = 34; 38; 42; 50; 54; 58, apply Lemma 3.1(2) with m = 4; for v = 79–82, apply Lemma 3.5 with t = 9. Lemma 4.9. (a) There exists a TD∗ (6; v) − TD∗ (6; 7) for 366v646; v ∈= {39; 41}. (b) There exists a TD∗ (5; v) − TD∗ (5; 7) for 296v6116. Proof. A TD∗ (6; v) − TD∗ (6; 7) is given in [4] for v = 36; 37; 42; 43; 46, in [6] for v = 40, and in Section 2 for v = 38; 44; 45. By Remark 4.3, it is only necessary to consider v ∈ {29–34; 37; 39; 41; 43; 46–48; 52; 53}. For v = 32, see Section 2; for v = 41, see [5]. For v = 29; 33 apply Lemma 3.1(1) with m = 4; t = 7; for v = 31; 39; 47, apply Lemma 3.3(1) with m = 4; t = 7; 8 or 9 and for v = 52; 53, apply Lemma 3.7 with m = t = 7; u = 1; s = 4 and mi = 0 or 1. For v = 30; 34; 45; 48, see [6]. Lemma 4.10. (a) There exists a TD∗ (6; v) − TD∗ (6; 8) for v ∈ {40; 41; 46; 47}. (b) There exists a TD∗ (5; v) − TD∗ (5; 8) for 336v6132. Proof. (a) See [6] for v = 41; 47, Section 2 for v = 46, while for v = 40, we can use Lemma 3.11. By Remark 4.3, it is only necessary to consider v ∈ {33–39; 42–45; 52; 53}
R.J.R. Abel, B. Du / Discrete Mathematics 262 (2003) 1 – 16
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for the values in (b). For v = 34; 38; 42, see [21]; for v = 43 = 7:5+(8 = 3:2+2:1), apply Lemma 3.1(2), and for 45 = 8:5 + (5 = 2:2 + 1:1) apply Lemma 3.1(3). For v = 35; 36, apply Lemma 3.3(1) with m = 4; t = 8. For v = 33; 37; 39; 44; 52, apply Lemma 3.1(1) with m = 4; t = 8; s = 1 or Lemma 3.1(2) with m = 4; t = 9 or 11. For v = 36; 53, apply Lemma 3.1(4) with m = 4 or 7, t = 7. Lemma 4.11. There exists a TD∗ (5; v) − TD∗ (5; 9) for 376v6148. Proof. By Remark 4.3, it is only necessary to consider v ∈{37–44; 47; 50; 53; 54; 56; 60}. For v = 38; 42, see [6]. For v = 37; 41; 43; 44, apply Lemma 3.1(1) with m = 4; t = 9; for v = 47, apply Lemma 3.1(4) with m = 8; t = 5; mi ∈ {0; 1; 2}. For v = 39; 40, apply Lemma 3.3(1) with m = 4; t = 9. For v = 50; 53; 54, apply Lemma 3.1(3) with m = 9; t = 5; mi ∈ {0; 1; 2}. For v = 56; 60, apply Lemma 3.4(2) with m = t = 7; d = 2; w1 = w2 = 1 and mi ∈ {0; 1; 2}. Lemma 4.12. There exists a TD∗ (5; v) − TD∗ (5; 10) for 416v6164. Proof. By Remark 4.3, it is only necessary to consider v ∈ {41–50; 52–57; 60; 62; 68; 69; 76; 78; 79}. (In [14], TD(6; v) − TD(6; 10) was incorrectly stated to be known for v = 68; 69; 76; 79.). For v = 41, see [14, p. 925]; for v = 43; 53 see [6] and for v = 44, see Section 2. For v = 45; 50; 54; 55; 62 apply Lemma 3.1(2) with t = 5; m ∈ {7; 8; 9}, or m = 4; t ∈ {11; 13}. For 57 = 6:7 + (10 = 3:2 + 4:1) + (5 = 5:1) apply Lemma 3.2(1) with m = 6; m1; j ∈ {1; 2} and m2; j ∈ {0; 1}. For 56 = 7:7 + (2 = 2:1) + (5 = 5:1), and 60 = 7:7 + (2 = 2:1) + (9 = 4:2 + 1:1), use Lemma 3.4(3). Finally, 52 = 9:5 + (7 = 3:2 + 1:1); 49 = 8:5 + (9 = 4:2 + 1:1); 46 = 9:5 + 1 and 68 = 8:7 + (12 = 6:2) can be handled by Lemma 3.1(4) with m = 9 or 8; t = 5 or 7; m1 = 1 or 2, and mi ∈ {0; 1; 2} for other i. For v = 42; 47; 48 use Lemmas 3.8 and 3.9 with (t; u; m) = (10; 2; 4) and mj = 0 or 1) for all j. The required TD(6; 10) − TD(6; 2) comes from [9]. The remaining values are all obtainable from Lemma 3.6(2) with n1 = 7 or 9. Lemma 4.13. There exists a TD∗ (5; v) − TD∗ (5; 11) for 456v6180. Proof. By Remark 4.3, it is only necessary to consider v ∈ {45–58; 60–66; 69; 70; 72; 76}. For 456v655; v = 46; 47; 48; 50; 54 apply Lemma 3.1(3) with m = 4; t = 11. For v = 48, apply Lemma 3.3(1) with m = 4; t = 11; for v = 54 = 6:7 + 11 + 1, apply Lemma 3.6, and for v = 50, see [6]. For v = 46; 47, we can apply Lemma 3.7 with m = 4; t = 11; u = 2; a TD3 (5; 11) − TD3 (5; 2) is given in [19]. For all exceptions in the range [53,76], except 56, 57, 62, 70, we can apply Lemma 3.6(2); here, we take t = 7; n = 4 when v663 and t = 8; n = 3 for v¿64. For 57 = 7:7 + 2 + 1 + 5; 62 = 7:7 + 1 + 1 + 11 and 70 = 7:8 + 1 + 1 + 12 apply Lemma 3.4(3) with (m; t; w1 ; w2 ; m1 ) = (7; 7; 2; 1; 1); (7; 7; 1; 1; 2) or (7; 8; 1; 1; 2); here, for j¿2; mj ∈ {0; 1} for v = 57 and mj ∈ {1; 2} for v = 62; 70. Finally, for v = 56, a TD∗ (6; v) − TD∗ (6; 11) can be found in [1]. From Remark 4.3 and Lemmas 4.4 – 4.13 we have:
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Table 1
t
Range for n using Lemma 3.6
Range for v using Lemma 3.6
7 8 9 11 13 16 17 19 23 25 27
76n614 86n616 96n618 116n622 136n626 166n632 176n634 196n638 236n646
[47 + n; 54 + n] [53 + n; 64 + n] [59 + n; 72 + n] [71 + n; 86 + n] [83 + n; 104 + n] [101 + n; 128 + n] [107 + n; 136 + n] [119 + n; 152 + n] [143 + n; 184 + n]
Range for n using Lemma 3.5
Range for v using Lemma 3.5
n622 n626 n632 n634 n638 n646 n650 n654
[87 + n; 97 + n] [101 + n; 117 + n] [122 + n; 144 + n] [129 + n; 153 + n] [143 + n; 171 + n] [171 + n; 207 + n] [185 + n; 225 + n] [199 + n; 243 + n]
Theorem 4.14. There exists a TD∗ (5; v) − TD∗ (5; n) for v¿4n and 26n611. We now consider 126n653. First of all, we treat larger values of v: Lemma 4.15. There exists a TD∗ (5; v) − TD∗ (5; n) for v¿5n and 126n653. Proof. For nearly all these values, we can use either Lemma 3.5 or Lemma 3.6(2) with t a prime power 6n; for Lemma 3.6(2), we also require n62t. Table 1 indicates ranges covered for v when 47 + n6v6243 + n. Note that Lemma 3.5 works for 7t + 10 + n6v69t + n and 116t except for t = 11; v = 98 + n. Also Lemma 3.6(2) with t¿7; 56s1 +s2 62t; n = t+n1 works for v in the range [7t+5+(n−t); 9t+(n−t)] or [6t +6+n; 8t +n] except for t = 7; v ∈ {51+n; 53+n; 55+n} and t ∈ {8; 9}; v = 6t +11. For n¡46 and v close to 5n, it is frequently necessary to use Lemma 3.6(2); for larger v, Lemma 3.5 works. Finally, whenever n653 and 243 + n¡v¡16n + 5, Lemma 3.5 with t a prime power, 316t689 works. The values with n¿12 that are not covered by the table are for (1) (v; n) = (51 + 12; 12), (2) v = 65 + n and 126n616, (3) n = 12 and 98 + n6v6100 + n and (4) 126n615 and 118 + n6v6121 + n. For (v; n) = (51 + 12; 12), we can use Lemma 3.1(4) with m = 11; t = 5 since 63 = 11:5 + (2:1 + 3:2). For (v; n) ∈ {(65 + n; n) : n = 12 − 16}, we can use Lemma 3.1(2) with m = 13; t = 5, since 65 + n = 13:5 + n; here n = 3:4; 2:2 + 3:3; 2:3 + 2:4; 3:4 + 1:3; 4:4, respectively. For 98 + 126v6100 + 12 and n = 12, we have v = 7:13 + 12 + m2 for 76m2 69; apply Lemma 3.2 with (m; t) = (7; 13); m1 = 12 and mi; j ∈ {0; 1} for all i; j. For the remaining values 118 + n6v6121 + n and 126n615, we have v = 7:16 + d + n for 66d69; apply Lemma 3.4(1) with (m; t) = (7; 16); wi = 1 for i6d and mj ∈ {0; 1} for 26j 6 16. Lemma 4.16. A TD∗ (5; v) − TD∗ (5; n) exists for v = 4n + k; n¿5; n ∈= {6; 10; 14; 18; 22}, and 16k6n; k = 2; 3; 6; 10.
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Table 2
n 12 13 14 15 16 17 18
u 2 2 2 2 2 2 2
n 19 20 21 22 23 24 25
u 2 2 2 3 2 2 3
n 26 27 28 29 30 31 32
u 2 3 2 3 3 3 3
n 33 34 35 36 37 38 39
u 3 5 4 5 3 5 6
n 40 41 42 43 44 45 46
u 3 3 3 2 3 4 3
n 47 48 49 50 51 52 53
u 4 6 3 4 4 6 6
Proof. Apply Lemma 3.1(2) with (t; s; m; mj ) = (n; 0; 4; 0 or 1). Lemma 4.17. There exists a TD∗ (5; v) − TD∗ (5; n) for v = 4n + k, whenever 16k6n and n ∈ {14; 16; 18; 22}. Proof. Apply Lemma 3.6(1) with t = n=2. All values are covered except k = 4, which can be taken care of by Lemma 3.7 with (t; u; m; mj ) = (n; 2; 4; 0 or 1). The required TD4 (5; n) − TD4 (5; 2) and TD4 (5; 22) − TD4 (5; 3) come from QDMs mentioned in the next lemma. We now prove: Lemma 4.18. There exists a TD∗ (5; v)−TD∗ (5; n) for v = 4n+k, whenever k ∈ {2; 3; 4; 6; 10} and 126n653. Proof. First, we consider cases (1) n¿18; k610, and (2) 146n617; k66. For the values of (n; u) in Table 2, a (n − u; 5; 1; 1; u)-QDM is known; most of these appear in [6], but a few are in Section 2, in [13] or in [16]. In all cases, this QDM gives a TD∗ (5; n) − TD∗ (5; u) with n − 4u disjoint partial parallel classes; further, we always have n − 4u¿k for the cases being considered. Hence, we can apply Lemma 3.7 with ∗ m = 4; t = n and mj ∈ {0; 1}; m = k. The required TD (5; mu + 16j6t j 16i6d mj ) − TD∗ (5; u)’s exist by Lemmas 4.4–4.8. For n = 14 or 16 and k = 10, see Lemma 4.17. For n = 13; 15; 17; k = 10; 4n + k equals 7t + 7 + (n − t); for t = 7; 8; 9; apply Lemma 3.6. For (n; k) = (12; 10), we have v = 58 andcan apply Lemma 3.4 with m = t = 7; w1 = w2 = 2; m1 = 1; mj ∈ {0; 1} for j¿2 and 16i6d mj = 5. Finally for (n; k) = (12; 6) or (13; 6), we have v = 54 = 6:7 + (12 = 5:2 + 2:1) or v = 58 = 9:5 + (13 = 3:3 + 2:2); here Lemma 3.2(2) with m = 6 or 9 works. From Lemmas 4.15 – 4.18 we have Theorem 4.19. There exists a TD∗ (5; v) − TD∗ (5; n) for v¿4n and 126n653. Combining Theorems 4.2, 4.14 and 4.19, we have now proved Theorem 4.1, restated here for convenience.
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Lemma 4.20. For any integer n¿2, there exists a TD∗ (5; v) − TD∗ (5; n) if and only if v¿4n + 1. References [1] R.J.R. Abel, A.M. Assaf, Modi3ed group divisible designs with block size 5 and = 1, Discrete Math. 256 (2002) 1–22. [2] R.J.R. Abel, F.E. Bennett, The existence of perfect Mendelsohn designs with block size 7, Discrete Math. 190 (1998) 1–14. [3] R.J.R. Abel, F.E. Bennett, G. Ge, Almost resolvable perfect Mendelsohn designs with block size 3ve, Discrete Appl. Math. 116 (2002) 1–15. [4] R.J.R. Abel, F.E. Bennett, H. Zhang, Perfect Mendelsohn designs with block size 6, J. Statist. Plann. Inference 86(2) (2000) 287–319. [5] R.J.R. Abel, F.E. Bennett, H. Zhang, L. Zhu, Existence of HSOLSSOMs with types hn and 1n u1 , Ars. Combin. 55 (2000) 97–115. [6] R.J.R. Abel, C.J. Colbourn, J. Yin, H. Zhang, Existence of incomplete transversal designs with block size 5 and any index lambda, Designs, Codes Cryptogr. 10 (1997) 275 – 307. [7] R.J.R. Abel, M. Greig, Some New RBIBDs with block size 5 and PBDs with block sizes = 1 mod 5, Austral. J. Combin. 15 (1997) 177–202. [8] F.E. Bennett, J. Yin, H. Zhang, R.J.R. Abel, Perfect Mendelsohn packing designs with block size 3ve, Designs, Codes Cryptogr. 16 (1998) 5 – 22. [9] A.E. Brouwer, Four MOLS of order 10 with a hole of size 2, J. Statist. Plann. Inference 10 (1984) 203–205. [10] A.E. Brouwer, G.H.J. van Rees, More mutually orthogonal latin squares, Discrete Math. 39 (1982) 263–281. [11] C.J. Colbourn, Personal communication. [12] C.J. Colbourn, Construction techniques for mutually orthogonal latin squares, in: C.J. Colbourn, E.S. Mahmoodian (Eds.), Combinatorics Advances, Kluwer Academic Press, Dordrecht, 1995, pp. 27–48. [13] C.J. Colbourn, Some direct constructions for incomplete transversal designs, J. Statist. Plann Inference 56 (1996) 93–104. [14] C.J. Colbourn, J.H. Dinitz (Eds.), C.R.C. Handbook of Combinatorial Designs, CRC Press, Boca Raton FL, 1996. [15] J.H. Dinitz, D.R. Stinson, MOLS with holes, Discrete Math. 44 (1983) 145–154. [16] B. Du, On the existence of incomplete transversal designs with block size 5, Discrete Math. 135 (1994) 81–92. [17] B. Du, On the existence of three incomplete idempotent MOLS, Austral. J. Combin. 12 (1995) 193–199. [18] D.R. Stinson, A short proof of the non-existence of a pair of orthogonal latin squares of order 6, J. Combin. Theory 36(A) (1984) 373–376. [19] D.R. Stinson, L. Zhu, On sets of three MOLS with holes, Discrete Math. 54 (1985) 321–328. [20] G. Tarry, Le probleme de 36 oKciers, C. R. Assoc. Franc. Avanc. Sci. Nat. 1 (1900) 122–123. [21] S.P. Wang, On self-orthogonal latin squares and patrial transversals, of latin squares, Ph.D. Thesis, Ohio State University.
www.elsevier.com/locate/disc
The existence of three idempotent IMOLS R. Julian R. Abela; ∗ , Beiliang Dub a School
b Department
of Mathematics, University of New South Wales, Sydney 2052, Australia of Mathematics, Suzhou University, Suzhou 215006, People’s Republic of China
Received 26 June 2000; received in revised form 6 August 2001; accepted 5 November 2001
Abstract In this paper it is shown that an idempotent TD(5; m) − TD(5; n) exists whenever the known necessary condition m ¿ 4n+1 is satis3ed, except when (m; n)=(6; 1) and possible when (m; n)= (10; 1). For m ¡ 60 and n 6 10, we also indicate where several idempotent TD(k; m)−TD(k; n)’s for k = 6; 7 can be found. c 2002 Elsevier Science B.V. All rights reserved. Keywords: Transversal design; Incomplete transversal design; Idempotent incomplete transversal design; Quasi-di
1. Introduction Let v and hi ; i = 1; : : : ; n, be positive integers. An incomplete transversal design (ITD), TD(k; v)− 16i6r TD(k; hi ), is a quadruple (X; G; B; H) satisfying the following conditions: 1. X is a set of kv elements called points and G = {G1 ; G2 ; : : : ; Gk } is a partition of X into k disjoint v-element subsets called groups. 2. B is a collection of k-element subsets of X called blocks, each containing one element from each Gs . 3. H = {H1 ; H2 ; : : : ; Hn } is a collection of disjoint subsets of X (called holes) and Hj contains hj elements from each group Gs . 4. Any two points in di
Corresponding author. Fax: +61-294492814. E-mail addresses: [email protected] (R.J.R. Abel), [email protected] (B. Du).
c 2002 Elsevier Science B.V. All rights reserved. 0012-365X/03/$ - see front matter PII: S 0 0 1 2 - 3 6 5 X ( 0 2 ) 0 0 5 1 7 - 4
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A holey parallel class in such an ITD is a set of v − 16i6r hi blocks containing every point outside the holes once and no points from the holes. If an ITD contains a holey parallel class, it is called idempotent. TD(k; v) − TD(k; h)’s with several disjoint holey parallel classes can be useful in recursive constructions for obtaining larger idempotent ITDs; we use the notation TD s (k; v) − TD s (k; h) to denote such an ITD with s disjoint holey parallel classes. When s = 1 (equivalent to an idempotent ITD) the notation TD∗ (k; v) − TD∗ (k; h) is more commonly used. Simple counting arguments imply (1) if a TD(k; v)−TD(k; h) exists, then v¿(k −1)h and (2) if a TD(k; v) − TD(k; h1 ) − TD(k; h2 ) exists then v¿(k − 1)h1 + h2 . From the second of these, a necessary condition for an idempotent TD(k; v) − TD(k; h) to exist is v¿(k − 1)h + 1. In [17], Du showed that for k = 5, the required condition v¿4h + 1 is suKcient for the existence of an idempotent TD(5; v) − TD(5; h) when h¿53. In this paper, we examine the case h653 and establish that the condition v¿4h+1 is suKcient here in all but two cases, namely when (v; h) = (6; 1) (nonexistent, for example, see [20,18]) or (10; 1). 2. Direct constructions All the direct constructions in this section come from quasi-di
R.J.R. Abel, B. Du / Discrete Mathematics 262 (2003) 1 – 16
3
QDM: (a; b; c; d; e)T and (b; a; d; c; e)T . Combining these columns with the 3rst column ( for (v; h) = (50; 3)) and the 3rst two (in other cases) gives a (v; 5; 1; 1; h) QDM over Zv−h . In all cases the number of columns without a blank entry, that is, (v − h)− (k − 2)h = v − 4h is ¿1; hence these ITDs are all idempotent. (v; h) = (32; 2): 0 0 0 0 —
0 15 16 1 —
0 1 26 21 0
29 5 23 12 0
— 11 27 0 0
— 22 7 20 22 13 10 — 0 0
2 28 7 — 0
12 26 4 6 0
19 27 1 15 0
14 24 16 17 0
9 21 18 8 0
18 23 3 29 0
10 13 11 24 0
8 15 5 14 0
25 4 9 2 0
6 17 19 25 0
3 16 28 20 0
0 0 0 0 —
0 0 21 6 31 14 32 29 27 13 17 1 5 12 23 4 10 15 24 8 18 8 30 18 25 1 24 14 9 26 1 21 15 16 13 10 27 22 19 33 — 0 0 0 0 0 0 0 0 0
2 9 17 6 0
16 25 4 5 0
22 11 20 0 0
17 30 28 32 0
7 26 3 31 0
27 2 13 15 0
7 17 23 19 0
9 24 32 27 0
(v; h) = (36; 2):
— 19 7 12 0
— 20 28 18 11 23 29 — 0 0
3 33 2 — 0
26 28 24 21 0
13 31 18 12 0
(v; h) = (38; 4): 0 0 0 0 —
0 17 18 1 —
0 1 14 5 —
20 23 17 6 0
— — — — 25 5 18 19 8 3 14 3 16 33 26 4 25 30 0 — 0 0 0 0 0
12 16 31 10 0
29 0 7 22 0
14 30 22 1 10 33 5 2 9 — — — 0 0 0
15 21 11 29 0
4 11 28 20 0
32 6 8 1 0
4
R.J.R. Abel, B. Du / Discrete Mathematics 262 (2003) 1 – 16
(v; h) = (44; 10): 0 0 0 0 0 0 17 3 5 7 0 18 21 11 5 0 1 28 22 10 — — — — — — 14 11 23 0
0 11 14 15 —
— — 33 30 29 15 7 12 11 31 27 33 5 31 8 13 15 — — — 0 0 0 0 0
13 0 — — — — — — — 22 1 26 32 4 9 25 20 21 29 24 1 0 28 17 4 19 26 32 16 14 25 30 21 10 9 7 0 0 0 0 0 0 0 0 0 18 10 6 5 28 3 23 24 2 16 17 8 19 27 20 6 3 12 22 2 18 — — — — — — — 0 0 0 0 0 0 0
(v; h) = (49; 3): 0 0 0 0 0
0 0 23 17 24 20 1 23 — —
39 21 33 34 23 40 27 32 0 0
37 1 8 9 0
41 6 20 3 0
23 16 43 28 24 38 3 30 25 29 9 36 42 5 4 34 18 24 35 28 31 6 14 39 38 16 43 44 0 0 0 0 0 0 0
7 18 11 35 4 19 31 14 21 10 15 26 11 5 33 37 0 0 0 0
— 2 45 36 0
— 15 42 30 0
— 45 10 20 26 27 40 12 19 12 22 25 13 — — — 0 0 0 0
22 17 8 44 32 13 17 2 7 1 41 29 0 0 0
(v; h) = (50; 3): 0 0 0 0 —
0 11 25 8 —
0 10 17 18 0
24 41 44 0 0
19 26 46 1 0
40 2 11 37 0
20 5 41 16 0
13 38 43 23 0
4 12 33 2 0
30 18 24 34 0
32 36 13 27 0
25 23 39 15 0
42 16 28 20 0
45 14 29 38 0
33 9 26 21 0
44 39 31 35 0
46 43 9 45 0
27 28 22 7 0
1 35 10 3 0
31 11 30 12 0
34 6 32 19 0
37 8 42 — 0
15 21 4 — 0
17 — 3 7 36 8 — 14 0 0
0 12 19 20 0
6 40 35 41 26 13 28 23 10 32 37 19 43 18 27 31 17 38 7 30 33 2 5 33 29 12 6 36 4 15 1 27 26 16 9 3 43 42 31 17 22 35 8 34 0 0 0 0 0 0 0 0 0 0 0
— — 29 22 40 6 5 25 0 0
(v; h) = (50; 4): 0 0 0 23 0 24 0 1 — —
0 31 44 41 —
R.J.R. Abel, B. Du / Discrete Mathematics 262 (2003) 1 – 16
34 4 20 24 42 — 36 25 11 5 14 45 18 13 40 28 44 37 32 7 24 11 2 0 0 0 0 0 0 0
— 21 14 39 0
— 16 21 23 0
— 29 38 10 0
8 22 15 44 41 25 — — 0 0
9 3 45 — 0
5
1 39 30 — 0
(v; h) = (52; 6): 0 0 0 0 0 0 23 37 17 4 0 24 18 40 7 0 1 33 45 40 — — — — 0 25 3 11 — 27 28 24 22 31 26 21 32 1 12 43 33 0 0 0 0
13 40 12 30 39 1 16 9 36 35 5 15 44 38 41 0 30 5 37 35 24 4 19 14 0 0 0 0 0 0
37 44 31 19 21 18 32 26 29 3 2 28 10 15 27 34 0 0 0 0
— — — — — 6 10 20 7 8 42 43 23 41 29 14 33 45 38 17 2 34 18 11 25 22 6 39 13 45 8 20 23 36 9 17 42 16 — — — — — — 0 0 0 0 0 0 0 0 0 0 0
Lemma 2.2. There exists a TD∗ (5; v) − TD∗ (5; h) for any (v; h) ∈ {(24; 5); (32; 7); (51; 4)}. Proof. For (v; h) = (24; 5), replace each column (a; b; c; d; e)T in the array below by the following four: (a; b; c; d; e)T , (b; a; d; c; e)T , (c; d; a; b; e)T , and (d; c; b; a; e)T . Then add one extra column (0; 0; 0; 0; −)T . This gives a (19; 5; 1; 1; 5)-QDM over Z19 : 0 6 4 9 —
0 1 7 11 0
— — 9 15 6 10 4 13 0 0
— — — 3 17 12 2 5 8 14 16 18 0 0 0
For (v; h) = (51; 4), a (47; 5; 1; 1; 4)-QDM is obtainable as follows: each column (a; b; c; d; e)T in the array below (except the last one) generates 10 columns, by applying the automorphism group of order 10 generated by T1 ; T2 where T1 (a; b; c; d; e)T = (e; d; c; b; a)T and T2 (a; b; c; d; e)T = (b; c; d; e; a)T . The last given column generates 3ve columns by applying the group of order 5 generated by T2 . The resulting 55 columns form the required QDM: — 0 17 25 38
— 0 0 0 0 0 16 32 36 7 12 19 13 12 3 10 45 46 6 3 32 27 37 5 7
For (v; h) = (32; 7), a (25; 5; 1; 1; 7)-QDM is obtained by applying the group of order 5 generated by T1 (where T1 (a; b; c; d; e)T = (b; c; d; e; a)T ) to the following seven
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columns, and then adding four additional columns of the form (0; u; 2u; 3u; 4u) for u = 0; 5; 15; 20: — — 0 0 4 6 18 8 1 9
— 0 7 1 14
— 0 9 6 23
— 0 10 13 4
— 0 12 23 21
— 0 18 17 13
Lemma 2.3. There exists a TD∗ (6; v) − TD∗ (6; h) for (v; h) ∈ {(28; 2); (30; 4); (38; 7); (40; 2); (42; 4); (43; 5); (44; 7); (45; 7); (46; 8)}. Proof. All of these designs are obtained from QDMs over Zv−h . To each column (a; b; c; d; e; f)T in the arrays below, we apply the automorphism group of order 9 generated by T1 ; T2 where T1 (a; b; c; d; e; f)T = (b; c; a; f; d; e)T , T2 (a; b; c; d; e; f)T = (wc; wa; wb; wf; wd; we)T and w is a given cube root of unity in Zv−h . Here, we take w = 3; 5; 10; 7, respectively, for v − h = 26; 31; 37; 38. In all cases, each column of A1 generates just three distinct columns under this automorphism group, each column of A2 generates nine distinct columns, and each column of A3 generates no columns other than itself. (v; h) = (28; 2): 0 1 0 11 13 9 0 21 — 3 — 7 A1 = ; 1 0 24 0 9 13 8 0 3 — 20 —
0 0 6 19 23 15 A2 = : 6 4 20 6 1 7
(v; h) = (30; 4): 0 13 — A1 = 1 9 3
7 11 21 ; 0 13 —
0 0 0 2 14 24 23 — 3 A2 = ; 6 4 20 8 7 11 — 23 15
0 0 0 A3 = : 0 0 0
(v; h) = (38; 7): — 0 0 A1 = 17 22 23
17 22 23 — 0 0
11 24 27 ; 11 24 27
— 0 11 16 21 23 A2 = 0 — 16 11 23 21
— 5 11 0 29 20
0 29 20 : — 5 11
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(v; h) = (40; 2): 0 19 — A1 = 1 11 7
33 0 5 1 21 0 17 7 3 — 35 11 ; 0 37 0 35 19 27 0 17 — 31 — 5
0 10 37 A2 = 12 9 0
0 0 2 25 17 3 : 32 22 3 5 25 29
(v; h) = (42; 4): 0 29 5 19 15 35 — 13 17 A1 = ; 1 0 26 11 19 30 7 — 20
0 0 0 0 22 28 32 12 13 — 37 5 A2 = ; 16 30 36 10 18 7 6 30 — 31 13 27
0 0 0 A3 = : 0 0 0
(v; h) = (43; 5): 0 19 — A1 = 14 2 22
11 0 9 7 0 23 1 — 25 ; 0 12 0 19 18 0 — 8 —
0 16 3 A2 = 10 20 —
0 0 0 18 26 7 — 25 35 : 22 4 32 13 7 21 11 25 23
(v; h) = (44; 7): — 0 0 A1 = 19 13 5
19 4 13 3 5 30 — 27 0 11 0 36
27 31 11 14 36 29 ; 4 31 3 14 30 29
— 15 — 0 11 0 22 6 13 A2 = 15 — 25 11 0 33 6 22 36
25 33 36 : — 0 13
(v; h) = (45; 7): 0 32 19 10 — 34 A1 = ; 2 0 22 19 14 —
A2 =
0 0 0 0 22 6 18 20 1 — 31 — 8 36 22 30 6 23 34 35 — 37 — 1
0 32 15 ; 26 27 23
0 0 0 A3 = : 0 0 0
7
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(v; h) = (46; 8): 0 33 0 5 1 15 19 21 0 17 7 29 — 3 — 35 11 13 A1 = ; 1 0 14 0 22 1 11 19 2 0 2 7 7 — 22 — 14 11
0 0 0 0 15 21 9 5 31 — — 1 A2 = : 8 28 36 10 23 6 0 19 — 27 5 —
The 3nal construction in this section was found by Colbourn [11]. Lemma 2.4. There exists a TD∗ (6; 19) − TD∗ (6; 2). Proof. Multiply all columns in the following array by 1 and −1. The result is a (17; 6; 1; 0; 2)-QDM. — 0 0 0 0 0 0 0 — 1 2 3 4 5 1 1 — 4 7 11 8 8 15 14 — 13 16 7 10 3 6 16 — 10 4 13 4 10 8 15 — 6
0 6 14 12 8 3
0 7 12 8 15 5
0 8 2 11 12 1:
3. Recursive constructions We need some known constructions for ITDs, which are mainly the working corollaries of Theorems 1.1 and 1.2 in [10] and Lemma 2.2 in [15]. So, we state the following lemmas without proof. Lemma 3.1. Suppose that v = mt + 16j6t mj and that the following designs exist: a TD(6; t), a TD∗ (5; m + m1 ) − TD∗ (5; s) − TD∗ (5; m1 ) and TD(5; m + mj ) − TD(5; s) − TD(5; mj ) ( for 26j6t). Then: 1. There exists a TD∗ (5; v) − TD∗ (5; st) if a TD∗ (5; 16j6t mj ) exists. 2. There exists a TD∗ (v; 16j6t mj )) if s = 0. 3. There exists a TD∗ (5; v)−TD∗ (5; m) if a TD∗ (5; 16j6t mj )) exists and m1 = s = 0. 4. There exists a TD∗ (5; v) − TD∗ (5; m + m1 ) if a TD∗ (5; 16j6t mj ) − TD∗ (5; m1 ) exists and s = 0. Lemma 3.2. Let v = mt + m1 + m2 where for i = 1; 2, mi = 16j6t mij . Suppose, there exist a TD(7; t) and TD(5; m + m1k + m2j ) − TD(5; m1k ) − TD(5; m2j ) (for 26k6t, 16j6t). Suppose also, there exist TD∗ (5; m2 ) and TD∗ (5; m + m11 + m2j ) − TD∗ (5; m11 ) − TD∗ (5; m2j ) ( for 16j6t). Then: 1. There exists a TD∗ (5; v) − TD∗ (5; m1 ). ∗ 2. If m21 = 0, and a TD (5; 16k6t m1k ) − TD∗ (5; m11 ) exists, then there exists a TD∗ (5; v) − TD∗ (5; m + m11 ).
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Lemma 3.3. Let v = mt + 16i6d wi . If there exist a TD(5 + d; t), a TD(5; m), a ∗ ∗ TD (5; m + w1 ) − TD (5; w1 ) and TD(5; m + wi ) − TD(5; wi ) ( for 26i6d), then: 1. There exists a TD∗ (5; v)−TD∗ (5; t) if a TD∗ (5; m+ 16i6d wi ) exists and m¿3wi (26i6d). 2. There exists a TD∗ (5; v) − TD∗ (5; m + 16i6d wi ). 3. There exists a TD∗ (5; v) − TD∗ (5; m + w1 ) if a TD∗ (5; m + 16i6d wi ) − TD∗ (5; w1 ) exists. Lemma 3.4. Let v = mt + 16i6d wi + 26j6t mj . If there exist a TD(6 + d; t), a TD(5; m) and TD(5; m+mj )−TD(5; mj ) and TD(5; m+wi +mj )−TD(5; wi )−TD(5; mj ) ( for 26i 6 d and 26j 6 t) and TD∗ (5m + w1 + mj ) − TD∗ (5; w1 ) − TD∗ (5; mj ) ( for 16j 6 t), then: 1. A TD∗ (5; v) − TD∗ (5; 26j6t mj ) exists if a TD∗ (5; m + 16i6d wi ) exists. 2. A TD∗ (5; v) − TD∗ (5; m + 16i6d wi ) exists if a TD∗ (5; 26j6t mj ) exists. 3. A TD∗ (5; v + m1 ) − TD∗ (5; m + m1 + 16i6d wi ) exists if a TD∗ (5; 16j6t mj ) − TD∗ (5; m1 ) exists. Lemma 3.5. (1) Suppose a TD(10; t) exists, n is even, 26n62t; 06k62t; k = 3 or 4, and k = 21, if t = 11. Then for v = 7t + k + n, there exists a TD∗ (5; v) − TD∗ (5; n). (2) Suppose there exist a TD(10; t) and a TD∗ (5; t + 1), n is odd, 36n6 2t + 1; 06k62t and v = 7t + k + n. Then there exists a TD∗ (5; v) − TD∗ (5; n) if k = 1; 2; 3; 5; 9. Lemma 3.6. (1) Suppose the following designs exist: a TD(8; t) a TD∗ (5; s1 ), a TD∗ (5; s2 ), and a TD∗ (5; s3 ), where 06s2 ; s3 6t and 16s1 6t. Then there exists a TD∗ (5; 8t + s1 + s2 + s3 ) − TD∗ (5; 2t). (2) Suppose the following designs exist: a TD(8; t), a TD∗ (5; s1 ), a TD∗ (5; s2 ), and a TD∗ (5; n1 ), where 06s1 ; s2 ; 6t. Then, there exists a TD∗ (5; 7t + s1 + s2 + n1 ) − TD∗ (5; t + n1 ) if either (a) n1 ¡t or (b) n1 = t and s1 + s2 ¿0. We also need the following constructions for incomplete transversal designs, which are mainly the working corollaries of Theorem 3.1 in [16]. So, we state the following lemmas without proof. Lemma 3.7. Suppose the following designs exist: a TDs (5; t) − TDs (5; u), a TD(5; m), a TD∗ (5; m + m1 ) − TD∗ (5; m1 ) and TD(5; m + mj ) − TD(5; mj ) ( for 26j6s). Then, a ∗ TD (5; mt + 16j6s mj ) − TD∗ (5; t) exists if a TD∗ (5; mu + 16j6s mj ) − TD∗ (5; u) exists and m¿3mj ( for 26j6s). Lemma 3.8. Suppose there exist a TD(6; t) − TD(6; u), a TD(5; m), a TD∗ (5; m + m1 ) − TD∗ (5; m1 ) and TD(5; m + mj ) − TD(5; mj ) (for 26j6u). Then, a TD∗ (5; mt + ∗ ∗ ∗ 16j6u mj ) − TD (5; t) exists if a TD (5; mu + 16j6u mj ) − TD (5; u) exists and m¿3mj ( for 26j6u).
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Lemma 3.9. Suppose there exist a TD(6; t) − TD(6; u), a TD∗ (5; m + m1 ) − TD∗ (5; m1 ) and TD(5; m + mj ) − TD(5; m j ) ( for 26j6t). Suppose also, there exists a TD∗ (5; 16j6t mj ) − TD∗ (5; 16j6u mj ). Then there exists a TD∗ (5; mt+ ∗ ∗ ∗ 16j6t mj ) − TD (5; t) if a TD (5; mu + 16j6u mj ) − TD (5; u) exists and m¿3mj ( for 26j 6 t). Finally, we state a few useful lemmas which are fairly obvious: Lemma 3.10. If there exists TD∗ (5; v) − TD∗ (5; u) and TD∗ (5; u) − TD∗ (5; n), then there exists a TD∗ (5; v) − TD∗ (5; n). Lemma 3.11. Suppose v = mn and TD∗ (k; m); TD∗ (k; n) both exist. Then there exists a TD∗ (k; v) − TD∗ (k; n). Lemma 3.12. If there exists a TD(6; v) − TD(6; n), then there exists a TDn (5; v) − TDn (5; n) and hence also a TD∗ (5; v) − TD∗ (5; n).
4. Main result In this section, we make use of the constructions in the previous section to obtain TD∗ (5; v) − TD∗ (5; n)’s with v¿4n + 1. We obtain the following result: Theorem 4.1. For any integer n¿1, there exists a TD∗ (5; v) − TD∗ (5; n) if and only if v¿4n + 1. In some cases, the smallest designs are used to deal with larger ones, so it will be convenient to deal with the smallest values of n 3rst. Using this approach, we only need to consider values of v satisfying 4n + 16v616n + 4. For if v¿16n + 5, and both TD∗ (5; v) − TD∗ (5; 4n + 1) and TD∗ (5; 4n + 1) − TD∗ (5; n) exist, then a TD∗ (5; v) − TD∗ (5; n) can be obtained by Lemma 3.10. First of all, we mention the following theorem which was obtained by the second author in [17]. Theorem 4.2. For any integer n¿53, there exists a TD∗ (5; v) − TD∗ (5; n) if and only if v¿4n + 1. Therefore, we only need to consider the cases 26n653. Remark 4.3. Colbourn and Dinitz [14, p. 164] provide a list of values (v; n) for which TD(6; v)−TD(6; n) is unknown when v¿5n and n650. As mentioned in Lemma 3.12, deleting a group from TD(6; v)−TD(6; n) gives a TDn (5; v)−TDn (5; n) and hence also a TD∗ (5; v) − TD∗ (5; n). Therefore, for n¿5, we often only check the existence of
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TD∗ (5; v) − TD∗ (5; n) when either (1) 4n¡v¡5n or (2) v¿5n and TD(6; v) − TD(6; n) is unknown. In some cases however, in particular for v¡50, we point out where a known TD∗ (k; v) − TD∗ (k; n) can be found. Lemma 4.4. (a) There exist a TD∗ (7; v) − TD∗ (7; 2) for v ∈ {41; 47; 55} and a TD∗ (6; v) − TD∗ (6; 2) for v ∈ {15; 19; 28; 29; 33; 35; 40}. (b) There exists a TD∗ (5; v) − TD∗ (5; 2) for 96v636. Proof. For the values in (a), QDM solutions are given in [2] for v = 41; 47; 55, in [4] for v = 29, in Section 2, for v = 19; 28; 40, in [6] for n = 15; 33; 35 and in [6] for other values. For v = 22; 30; 34, we can use Lemma 3.1(2) since these values equal 4t + 2 for t = 5; 7; 8. For all the remaining values in (b), QDM solutions exist. For v = 9; 24; 29; 34 see [8]; for v = 10; 11; 12 see [9,19,16], respectively. For v = 32; 36, see Section 2; for v = 18; 23 see [13], and for all other values see [6]. Lemma 4.5. (a) There exist a TD∗ (7; v) − TD∗ (7; 3) for v ∈ {19; 40; 46}, and a TD∗ (6; v) − TD∗ (6; 3) for v ∈ {16; 22; 24; 25; 28; 34; 42; 45}. (b) There exists a TD∗ (5; v) − TD∗ (5; 3) for 136v652. Proof. For the values in (a), QDM solutions are given in [2] for v = 19; 40; 46, in [6] for v = 25; 36; 41; 45 and in [4] for other values. Now consider the values in (b). For v = 13; 18; 23; 33; 38; 43; 48, constructions appear in Lemmas 2.3 and 2.13 of [8]; if v ∈ {31; 35; 39; 47; 51} then v is of the form 4t + 3 where a TD(6; t) exists; hence Lemma 3.1(2) works for these values, and also for 52 = 7:7 + 3. The remaining values are all obtainable from QDMs: for v = 15; 17; 27, see [13], for v = 49; 50, see Section 2, and for other values, see [6]. Lemma 4.6. (a) There exists a TD∗ (7; v) − TD∗ (7; 4) for v ∈ {29; 39; 45}, a TD(6; 20) − TD(6; 4) and a TD∗ (6; v) − TD∗ (6; 4) for v ∈ {21; 23; 25; 27; 28; 30; 31; 33; 34; 35; 37; 39; 40; 41; 42; 43; 45; 46; 47}. (b) There exists a TD∗ (5; v) − TD∗ (5; 4) for 176v668. Proof. First we deal with the values in (a); all these come from QDMs. For v = 29 see [6], and for v = 39; 45 see [2]. For v = 20, see [7]. For v = 21; 25; 27; 28; 31; 33; 34; 40; 43; 46, see [4]. For v = 23; 35; 37; 41; 47, see [6]. For v = 30; 42, see Section 2. Now we consider the values in (b). For v = 17, see [6], for v = 19, see [13], for v = 18; 22; 26; 30, see [21] (or [14, p. 450]) and for v = 38; 50; 51, see Section 2. If v ∈ {24; 32; 36; 48; 52; 53; 56; 59; 60; 67; 68}, then v can be written as either 4t + 4; 7t + 4 or 11t + 4 where a TD(6; t) exists; hence Lemma 3.1(2) works for these values. If v ∈ {44; 49; 55; 57; 63; 64; 65}, then v can be written as 4t +u where TD(6; t); TD∗ (5; u) both exist and 06u¡t; use Lemma 3.1(3) for these values. Finally, v = 54; 58; 61; 62; 66 are obtainable using Lemma 3.2 with m = 7; t = 7; m1j ∈ {0; 1} and m2j ∈ {0; 1; 2} and 16j6t m1j = 4.
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Lemma 4.7. (a) There exists a TD∗ (7; v)−TD∗ (7; 5) for v ∈ {44; 58}, and a TD∗ (6; v)− TD∗ (6; 5) for v ∈ {26; 27; 30; 31; 32; 35; 36; 38; 40; 41; 43; 44; 45; 46; 50; 51; 56}. (b) There exists a TD∗ (5; v) − TD∗ (5; 5) for 216v684. Proof. The designs in (a) come from QDMs; they can be found in [2] for v = 44; 58, in [6] for v = 27, in [14, p. 118] for v = 30, in [4] for v = 26; 32; 38; 44, in Section 2 for v = 43, and in [1] for v = 31; 36; 41; 46; 50; 51; 56. For v = 35; 40; 45, use Lemma 3.11. By Remark 4.3, it is only necessary to consider v ∈ {21–24; 28; 29; 31–34; 37; 39; 41; 42; 46–53; 56; 57; 78; 79; 81} for the values in (b). For v = 22; 23; 28 and 34, see [6]; for v = 24, see Section 2. For v = 33; 54; 57; 77; 82, use Lemma 3.1(2) with m = 4 or 7. For v = 21; 29; 37; 39; 49; 52; 53; 57; 59; 79; 81, apply Lemma 3.1(4) with m = 4. For v = 47; 48; 56, use Lemma 3.3 with m = 4. Finally, for v = 42 = 7:5+(7 = 3:2+1:1), use Lemma 3.1(1) with m = 7; t = 5; s = 1, and for 78 = 7:9 + (5 = 5:1) + (9 = 4:2 + 1:1) + (1 = 1:1), apply Lemma 3.5(2). Lemma 4.8. (a) There exist a TD∗ (7; v) − TD∗ (7; 6) for v ∈ {41; 43; 47; 51; 57; 79}, a TD(6; 30) − TD(6; 6) and a TD∗ (6; v) − TD∗ (6; 6) for v ∈ {31; 33; 35; 36; 37; 39; 40; 41; 45; 46; 50}. (b) There exists a TD∗ (5; v) − TD∗ (5; 6) for 256v6100. Proof. A TD∗ (7; v) − TD∗ (7; 6) can be found in [2] for v = 41; 51; 57; 79, in [12] for v = 43 and in [6] for v = 47. TD∗ (6; v) − TD∗ (6; 6)s are given for v = 31; 33; 37; 45; 49 in [4] and for v = 39; 48 in [6]. A TD(6; 30) − TD(6; 6) and TD∗ (6; v) − TD∗ (6; 6)s for v = 35; 36; 40; 41; 46; 50 can be found in [1] or [3]. By Remark 4.3, it is only necessary to consider v ∈ {25–30; 32–36; 38; 40–42; 44; 46; 50–54; 57; 58; 79–82} for the values in (b). For v = 25, see [14, p. 295]; for v = 26–28; 32; 39; 44; 53, see [6]; for v = 29, see [13]; and for v = 52, see Section 2. For v = 34; 38; 42; 50; 54; 58, apply Lemma 3.1(2) with m = 4; for v = 79–82, apply Lemma 3.5 with t = 9. Lemma 4.9. (a) There exists a TD∗ (6; v) − TD∗ (6; 7) for 366v646; v ∈= {39; 41}. (b) There exists a TD∗ (5; v) − TD∗ (5; 7) for 296v6116. Proof. A TD∗ (6; v) − TD∗ (6; 7) is given in [4] for v = 36; 37; 42; 43; 46, in [6] for v = 40, and in Section 2 for v = 38; 44; 45. By Remark 4.3, it is only necessary to consider v ∈ {29–34; 37; 39; 41; 43; 46–48; 52; 53}. For v = 32, see Section 2; for v = 41, see [5]. For v = 29; 33 apply Lemma 3.1(1) with m = 4; t = 7; for v = 31; 39; 47, apply Lemma 3.3(1) with m = 4; t = 7; 8 or 9 and for v = 52; 53, apply Lemma 3.7 with m = t = 7; u = 1; s = 4 and mi = 0 or 1. For v = 30; 34; 45; 48, see [6]. Lemma 4.10. (a) There exists a TD∗ (6; v) − TD∗ (6; 8) for v ∈ {40; 41; 46; 47}. (b) There exists a TD∗ (5; v) − TD∗ (5; 8) for 336v6132. Proof. (a) See [6] for v = 41; 47, Section 2 for v = 46, while for v = 40, we can use Lemma 3.11. By Remark 4.3, it is only necessary to consider v ∈ {33–39; 42–45; 52; 53}
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for the values in (b). For v = 34; 38; 42, see [21]; for v = 43 = 7:5+(8 = 3:2+2:1), apply Lemma 3.1(2), and for 45 = 8:5 + (5 = 2:2 + 1:1) apply Lemma 3.1(3). For v = 35; 36, apply Lemma 3.3(1) with m = 4; t = 8. For v = 33; 37; 39; 44; 52, apply Lemma 3.1(1) with m = 4; t = 8; s = 1 or Lemma 3.1(2) with m = 4; t = 9 or 11. For v = 36; 53, apply Lemma 3.1(4) with m = 4 or 7, t = 7. Lemma 4.11. There exists a TD∗ (5; v) − TD∗ (5; 9) for 376v6148. Proof. By Remark 4.3, it is only necessary to consider v ∈{37–44; 47; 50; 53; 54; 56; 60}. For v = 38; 42, see [6]. For v = 37; 41; 43; 44, apply Lemma 3.1(1) with m = 4; t = 9; for v = 47, apply Lemma 3.1(4) with m = 8; t = 5; mi ∈ {0; 1; 2}. For v = 39; 40, apply Lemma 3.3(1) with m = 4; t = 9. For v = 50; 53; 54, apply Lemma 3.1(3) with m = 9; t = 5; mi ∈ {0; 1; 2}. For v = 56; 60, apply Lemma 3.4(2) with m = t = 7; d = 2; w1 = w2 = 1 and mi ∈ {0; 1; 2}. Lemma 4.12. There exists a TD∗ (5; v) − TD∗ (5; 10) for 416v6164. Proof. By Remark 4.3, it is only necessary to consider v ∈ {41–50; 52–57; 60; 62; 68; 69; 76; 78; 79}. (In [14], TD(6; v) − TD(6; 10) was incorrectly stated to be known for v = 68; 69; 76; 79.). For v = 41, see [14, p. 925]; for v = 43; 53 see [6] and for v = 44, see Section 2. For v = 45; 50; 54; 55; 62 apply Lemma 3.1(2) with t = 5; m ∈ {7; 8; 9}, or m = 4; t ∈ {11; 13}. For 57 = 6:7 + (10 = 3:2 + 4:1) + (5 = 5:1) apply Lemma 3.2(1) with m = 6; m1; j ∈ {1; 2} and m2; j ∈ {0; 1}. For 56 = 7:7 + (2 = 2:1) + (5 = 5:1), and 60 = 7:7 + (2 = 2:1) + (9 = 4:2 + 1:1), use Lemma 3.4(3). Finally, 52 = 9:5 + (7 = 3:2 + 1:1); 49 = 8:5 + (9 = 4:2 + 1:1); 46 = 9:5 + 1 and 68 = 8:7 + (12 = 6:2) can be handled by Lemma 3.1(4) with m = 9 or 8; t = 5 or 7; m1 = 1 or 2, and mi ∈ {0; 1; 2} for other i. For v = 42; 47; 48 use Lemmas 3.8 and 3.9 with (t; u; m) = (10; 2; 4) and mj = 0 or 1) for all j. The required TD(6; 10) − TD(6; 2) comes from [9]. The remaining values are all obtainable from Lemma 3.6(2) with n1 = 7 or 9. Lemma 4.13. There exists a TD∗ (5; v) − TD∗ (5; 11) for 456v6180. Proof. By Remark 4.3, it is only necessary to consider v ∈ {45–58; 60–66; 69; 70; 72; 76}. For 456v655; v = 46; 47; 48; 50; 54 apply Lemma 3.1(3) with m = 4; t = 11. For v = 48, apply Lemma 3.3(1) with m = 4; t = 11; for v = 54 = 6:7 + 11 + 1, apply Lemma 3.6, and for v = 50, see [6]. For v = 46; 47, we can apply Lemma 3.7 with m = 4; t = 11; u = 2; a TD3 (5; 11) − TD3 (5; 2) is given in [19]. For all exceptions in the range [53,76], except 56, 57, 62, 70, we can apply Lemma 3.6(2); here, we take t = 7; n = 4 when v663 and t = 8; n = 3 for v¿64. For 57 = 7:7 + 2 + 1 + 5; 62 = 7:7 + 1 + 1 + 11 and 70 = 7:8 + 1 + 1 + 12 apply Lemma 3.4(3) with (m; t; w1 ; w2 ; m1 ) = (7; 7; 2; 1; 1); (7; 7; 1; 1; 2) or (7; 8; 1; 1; 2); here, for j¿2; mj ∈ {0; 1} for v = 57 and mj ∈ {1; 2} for v = 62; 70. Finally, for v = 56, a TD∗ (6; v) − TD∗ (6; 11) can be found in [1]. From Remark 4.3 and Lemmas 4.4 – 4.13 we have:
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Table 1
t
Range for n using Lemma 3.6
Range for v using Lemma 3.6
7 8 9 11 13 16 17 19 23 25 27
76n614 86n616 96n618 116n622 136n626 166n632 176n634 196n638 236n646
[47 + n; 54 + n] [53 + n; 64 + n] [59 + n; 72 + n] [71 + n; 86 + n] [83 + n; 104 + n] [101 + n; 128 + n] [107 + n; 136 + n] [119 + n; 152 + n] [143 + n; 184 + n]
Range for n using Lemma 3.5
Range for v using Lemma 3.5
n622 n626 n632 n634 n638 n646 n650 n654
[87 + n; 97 + n] [101 + n; 117 + n] [122 + n; 144 + n] [129 + n; 153 + n] [143 + n; 171 + n] [171 + n; 207 + n] [185 + n; 225 + n] [199 + n; 243 + n]
Theorem 4.14. There exists a TD∗ (5; v) − TD∗ (5; n) for v¿4n and 26n611. We now consider 126n653. First of all, we treat larger values of v: Lemma 4.15. There exists a TD∗ (5; v) − TD∗ (5; n) for v¿5n and 126n653. Proof. For nearly all these values, we can use either Lemma 3.5 or Lemma 3.6(2) with t a prime power 6n; for Lemma 3.6(2), we also require n62t. Table 1 indicates ranges covered for v when 47 + n6v6243 + n. Note that Lemma 3.5 works for 7t + 10 + n6v69t + n and 116t except for t = 11; v = 98 + n. Also Lemma 3.6(2) with t¿7; 56s1 +s2 62t; n = t+n1 works for v in the range [7t+5+(n−t); 9t+(n−t)] or [6t +6+n; 8t +n] except for t = 7; v ∈ {51+n; 53+n; 55+n} and t ∈ {8; 9}; v = 6t +11. For n¡46 and v close to 5n, it is frequently necessary to use Lemma 3.6(2); for larger v, Lemma 3.5 works. Finally, whenever n653 and 243 + n¡v¡16n + 5, Lemma 3.5 with t a prime power, 316t689 works. The values with n¿12 that are not covered by the table are for (1) (v; n) = (51 + 12; 12), (2) v = 65 + n and 126n616, (3) n = 12 and 98 + n6v6100 + n and (4) 126n615 and 118 + n6v6121 + n. For (v; n) = (51 + 12; 12), we can use Lemma 3.1(4) with m = 11; t = 5 since 63 = 11:5 + (2:1 + 3:2). For (v; n) ∈ {(65 + n; n) : n = 12 − 16}, we can use Lemma 3.1(2) with m = 13; t = 5, since 65 + n = 13:5 + n; here n = 3:4; 2:2 + 3:3; 2:3 + 2:4; 3:4 + 1:3; 4:4, respectively. For 98 + 126v6100 + 12 and n = 12, we have v = 7:13 + 12 + m2 for 76m2 69; apply Lemma 3.2 with (m; t) = (7; 13); m1 = 12 and mi; j ∈ {0; 1} for all i; j. For the remaining values 118 + n6v6121 + n and 126n615, we have v = 7:16 + d + n for 66d69; apply Lemma 3.4(1) with (m; t) = (7; 16); wi = 1 for i6d and mj ∈ {0; 1} for 26j 6 16. Lemma 4.16. A TD∗ (5; v) − TD∗ (5; n) exists for v = 4n + k; n¿5; n ∈= {6; 10; 14; 18; 22}, and 16k6n; k = 2; 3; 6; 10.
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Table 2
n 12 13 14 15 16 17 18
u 2 2 2 2 2 2 2
n 19 20 21 22 23 24 25
u 2 2 2 3 2 2 3
n 26 27 28 29 30 31 32
u 2 3 2 3 3 3 3
n 33 34 35 36 37 38 39
u 3 5 4 5 3 5 6
n 40 41 42 43 44 45 46
u 3 3 3 2 3 4 3
n 47 48 49 50 51 52 53
u 4 6 3 4 4 6 6
Proof. Apply Lemma 3.1(2) with (t; s; m; mj ) = (n; 0; 4; 0 or 1). Lemma 4.17. There exists a TD∗ (5; v) − TD∗ (5; n) for v = 4n + k, whenever 16k6n and n ∈ {14; 16; 18; 22}. Proof. Apply Lemma 3.6(1) with t = n=2. All values are covered except k = 4, which can be taken care of by Lemma 3.7 with (t; u; m; mj ) = (n; 2; 4; 0 or 1). The required TD4 (5; n) − TD4 (5; 2) and TD4 (5; 22) − TD4 (5; 3) come from QDMs mentioned in the next lemma. We now prove: Lemma 4.18. There exists a TD∗ (5; v)−TD∗ (5; n) for v = 4n+k, whenever k ∈ {2; 3; 4; 6; 10} and 126n653. Proof. First, we consider cases (1) n¿18; k610, and (2) 146n617; k66. For the values of (n; u) in Table 2, a (n − u; 5; 1; 1; u)-QDM is known; most of these appear in [6], but a few are in Section 2, in [13] or in [16]. In all cases, this QDM gives a TD∗ (5; n) − TD∗ (5; u) with n − 4u disjoint partial parallel classes; further, we always have n − 4u¿k for the cases being considered. Hence, we can apply Lemma 3.7 with ∗ m = 4; t = n and mj ∈ {0; 1}; m = k. The required TD (5; mu + 16j6t j 16i6d mj ) − TD∗ (5; u)’s exist by Lemmas 4.4–4.8. For n = 14 or 16 and k = 10, see Lemma 4.17. For n = 13; 15; 17; k = 10; 4n + k equals 7t + 7 + (n − t); for t = 7; 8; 9; apply Lemma 3.6. For (n; k) = (12; 10), we have v = 58 andcan apply Lemma 3.4 with m = t = 7; w1 = w2 = 2; m1 = 1; mj ∈ {0; 1} for j¿2 and 16i6d mj = 5. Finally for (n; k) = (12; 6) or (13; 6), we have v = 54 = 6:7 + (12 = 5:2 + 2:1) or v = 58 = 9:5 + (13 = 3:3 + 2:2); here Lemma 3.2(2) with m = 6 or 9 works. From Lemmas 4.15 – 4.18 we have Theorem 4.19. There exists a TD∗ (5; v) − TD∗ (5; n) for v¿4n and 126n653. Combining Theorems 4.2, 4.14 and 4.19, we have now proved Theorem 4.1, restated here for convenience.
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Lemma 4.20. For any integer n¿2, there exists a TD∗ (5; v) − TD∗ (5; n) if and only if v¿4n + 1. References [1] R.J.R. Abel, A.M. Assaf, Modi3ed group divisible designs with block size 5 and = 1, Discrete Math. 256 (2002) 1–22. [2] R.J.R. Abel, F.E. Bennett, The existence of perfect Mendelsohn designs with block size 7, Discrete Math. 190 (1998) 1–14. [3] R.J.R. Abel, F.E. Bennett, G. Ge, Almost resolvable perfect Mendelsohn designs with block size 3ve, Discrete Appl. Math. 116 (2002) 1–15. [4] R.J.R. Abel, F.E. Bennett, H. Zhang, Perfect Mendelsohn designs with block size 6, J. Statist. Plann. Inference 86(2) (2000) 287–319. [5] R.J.R. Abel, F.E. Bennett, H. Zhang, L. Zhu, Existence of HSOLSSOMs with types hn and 1n u1 , Ars. Combin. 55 (2000) 97–115. [6] R.J.R. Abel, C.J. Colbourn, J. Yin, H. Zhang, Existence of incomplete transversal designs with block size 5 and any index lambda, Designs, Codes Cryptogr. 10 (1997) 275 – 307. [7] R.J.R. Abel, M. Greig, Some New RBIBDs with block size 5 and PBDs with block sizes = 1 mod 5, Austral. J. Combin. 15 (1997) 177–202. [8] F.E. Bennett, J. Yin, H. Zhang, R.J.R. Abel, Perfect Mendelsohn packing designs with block size 3ve, Designs, Codes Cryptogr. 16 (1998) 5 – 22. [9] A.E. Brouwer, Four MOLS of order 10 with a hole of size 2, J. Statist. Plann. Inference 10 (1984) 203–205. [10] A.E. Brouwer, G.H.J. van Rees, More mutually orthogonal latin squares, Discrete Math. 39 (1982) 263–281. [11] C.J. Colbourn, Personal communication. [12] C.J. Colbourn, Construction techniques for mutually orthogonal latin squares, in: C.J. Colbourn, E.S. Mahmoodian (Eds.), Combinatorics Advances, Kluwer Academic Press, Dordrecht, 1995, pp. 27–48. [13] C.J. Colbourn, Some direct constructions for incomplete transversal designs, J. Statist. Plann Inference 56 (1996) 93–104. [14] C.J. Colbourn, J.H. Dinitz (Eds.), C.R.C. Handbook of Combinatorial Designs, CRC Press, Boca Raton FL, 1996. [15] J.H. Dinitz, D.R. Stinson, MOLS with holes, Discrete Math. 44 (1983) 145–154. [16] B. Du, On the existence of incomplete transversal designs with block size 5, Discrete Math. 135 (1994) 81–92. [17] B. Du, On the existence of three incomplete idempotent MOLS, Austral. J. Combin. 12 (1995) 193–199. [18] D.R. Stinson, A short proof of the non-existence of a pair of orthogonal latin squares of order 6, J. Combin. Theory 36(A) (1984) 373–376. [19] D.R. Stinson, L. Zhu, On sets of three MOLS with holes, Discrete Math. 54 (1985) 321–328. [20] G. Tarry, Le probleme de 36 oKciers, C. R. Assoc. Franc. Avanc. Sci. Nat. 1 (1900) 122–123. [21] S.P. Wang, On self-orthogonal latin squares and patrial transversals, of latin squares, Ph.D. Thesis, Ohio State University.