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A survey on resolvable group divisible designs with block size four
Discrete Mathematics 279 (2004) 225 – 245
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A survey on resolvable group divisible designs with block size four Gennian Gea;1 , Alan C.H. Lingb;2 a Department
of Mathematics, Zhejiang University, Hangzhou 310027, Zhejiang, People’s Republic of China b Department of Computer Science, University of Vermont, Burlington, VT 05405, USA Received 30 October 2002; received in revised form 24 April 2003; accepted 9 June 2003
Abstract In this paper, a survey for the existence of resolvable group divisible designs (RGDDs) with block size four, group-type hn and index unity is presented. It is proved that there exists a 4-RGDD of type hn if and only if n ¿ 4, hn ≡ 0 (mod 4) and h(n − 1) ≡ 0 (mod 3) with a handful of possible exceptions. c 2003 Elsevier B.V. All rights reserved. MSC: Primary 05B05 Keywords: Group divisible designs; Resolvable group divisible designs; Incomplete resolvable group divisible designs; 4-frames
1. Introduction A group divisible design (GDD) is a triple (X; G; B) which satis8es the following properties: 1. G is a partition of a set X (of points) into subsets called groups, 2. B is a set of subsets of X (called blocks) such that a group and a block contain at most one common point, 3. every pair of points from distinct groups occurs in a unique block.
1 2
Researcher supported in part by YNSFC Grant 10001026. Researcher supported by an ARO grant 19-01-1-0406 and a DOE grant. E-mail address: [email protected] (G. Ge).
c 2003 Elsevier B.V. All rights reserved. 0012-365X/$ - see front matter doi:10.1016/S0012-365X(03)00274-7
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The group-type (or type) of the GDD is the multiset {|G|: G ∈ G}. We usually use an “exponential” notation to describe group-type: a type 1i 2j 3k : : : denotes i occurrences of 1; j occurrences of 2, etc. A GDD (X; G; B) will be referred to as a K-GDD if |B| ∈ K for every block B in B. When K = {k}, we simply write k for K. An -parallel class of blocks in a GDD (X; G; B) is a subset B ⊆ B such that each point x ∈ X is contained in exactly blocks in B . When = 1, we will employ the usual term parallel class. If the block set B can be partitioned into -parallel classes, then the GDD is called -resolvable (or just resolvable if = 1). A GDD (X; G; B) is called A-resolvable if its block set B admits a partition into subsets B1 ; B2 ; : : : ; Br where for each i = 1; 2; : : : ; r there is an i ∈ A such that each point x ∈ X is contained in exactly i blocks in Bi . It is not diIcult to see that an A-resolvable GDD with K = {k} must be uniform. Resolvable group divisible designs have been instrumental in the construction of other types of designs. Many researchers have been involved in investigating the existence of resolvable group divisible designs. Simple counting arguments show that if there is a k-RGDD of type hn , then n ¿ k; hn ≡ 0 (mod k)
and
h(n − 1) ≡ 0 (mod k − 1): The necessary conditions for the existence of a k-RGDD of type hn have been proved to be suIcient for k = 3 (see [1,16,19]), with the de8nite exception of 3-RGDDs of types 23 , 26 and 63 . However, the case for k = 4 has remained open for a long time and we have the following known results (see [8,10,12–14,17,20,22–24]). Theorem 1.1. The necessary conditions for the existence of a 4-RGDD(hn ), namely, n ¿ 4, hn ≡ 0 (mod 4) and h(n − 1) ≡ 0 (mod 3), are also su
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Theorem 1.2. The necessary conditions for the existence of a 4-RGDD(hn ), namely, n ¿ 4, hn ≡ 0 (mod 4) and h(n − 1) ≡ 0 (mod 3), are also su
h ∈ {6; 30} ∪ {n: 66 6 n 6 438} and u ∈ {7; 23; 27; 35; 39; 47}; 450 6 h 6 2190 and u ∈ {7; 23; 27; 39; 47}; h ∈ {42; 54} ∪ {n: 2202 6 n 6 11238} and u ∈ {23; 27}; h = 18 and u ∈ {15; 17; 23; 27}.
For recursion, we will also need the following 4-frames. Lemma 2.2 (Ge et al. [9,12]). There exists a 4-frame of type 24u 121 with u = 5; 6 or 7. A transversal design (TD) TD(k; n) is a GDD of group type nk and block size k. A resolvable TD(k; n) (denoted by RTD(k; n)) is equivalent to a TD(k + 1; n). It is well
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known that the existence of a TD(k; n) is equivalent to the existence of k − 2 mutually orthogonal Latin squares (MOLS) of order n. In this paper, we mainly employ the following known results on TDs. Lemma 2.3 (Colbourn and Dinitz [3]). 1. An RTD(4; n) exists for all n ¿ 4 except for n = 6 and possibly excepting n = 10. 2. A TD(6; n) exists for all n ¿ 5 except possibly for n ∈{6,10,14,18,22}. 3. A TD(7; n) exists for all n ¿ 7 except possibly for n ∈{10,14,15,18,20,22,26,30,34, 38,46,60,62}. 4. A TD(8; n) exists for all n ¿ 7 except possibly for n ∈{10,12,14,15,18,20,21,22,26, 28,30,33,34,35,38,39,42,44,46,51,52,54,58,60,62,66,68,74}. 5. A TD(q + 1; q) exists, where q is a prime power. An incomplete group divisible design (IGDD) with block size k and index unity is a quadruple (X; G; H; B) which satis8es the following properties: 1. G = {G1 ; G2 ; : : : ; Gn } is a partition of a set X (of points) into subsets called groups, 2. H is a subset of X called a hole, 3. B is a collection of subsets of X , called blocks, such that a group and a block contain at most one common point, 4. every pair of points from distinct groups is either in H or occurs in a unique block but not both. We denote this design by k-IGDD(T ) where T is the type and de8ned by the multiset {(|Gi |; |Gi ∩ H |): 1 6 i 6 n}. As with GDDs, we shall use an “exponential” notation to describe the type. When H = ∅, an IGDD of type {(|Gi |; 0): 1 6 i 6 n} is just a GDD of type {|Gi |: 1 6 i 6 n}. A k-IGDD is said to be resolvable and denoted by k-IRGDD if its blocks can be partitioned into parallel classes and partial parallel classes, the latter partitioning X \ H . In this paper, we shall only use IRGDDs of such types as (h; 0)m−n (h; h)n where h ¿ 0 and m ¿ n ¿ 0. So, we shall simply use h(m; n) to denote the type. An IRGDD of type h(m; 1) is just an RGDD of type hm . Lemma 2.4. There exists a 4-IRGDD of type 6(12; 2) . Proof. Let the point set be (Z30 × {0; 1}) ∪ {w0 ; w1 ; w2 ; x0 ; x1 ; x2 ; y0 ; y1 ; y2 ; z0 ; z1 ; z2 }, let the groups be generated by {(0; i); (5; i); : : : ; (25; i)}, i ∈ {0; 1} and let the hole set be {w0 ; w1 ; w2 ; x0 ; x1 ; x2 ; y0 ; y1 ; y2 ; z0 ; z1 ; z2 }. The two partial parallel classes missing the hole {w0 ; w1 ; w2 ; x0 ; x1 ; x2 ; y0 ; y1 ; y2 ; z0 ; z1 ; z2 } are generated by {(0; 0); (3; 0); (0; 1); (13; 1)}. The following base blocks in each column form a parallel class by +10 mod 30. Note that wi + j = wi+j mod 3 . That happens similarly to xi ; yi ; zi for i ∈ {0; 1; 2}: {(1; 0); (10; 0); (13; 1); (19; 1)}, {(0; 0); (14; 1); (17; 1); (26; 1)}, {(6; 0); (18; 0); (10; 1); (12; 1)}, {(5; 0); (27; 0); (29; 0); (20; 1)}, {(3; 0); (7; 0); (14; 0); w0 }, {(2; 1); (3; 1); (25; 1); w0 },
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{(2; 0); (1; 1); (27; 1); x0 }, {(3; 0); (4; 0); (11; 1); x0 }, {(9; 0); (14; 1); (28; 1); y0 }, {(2; 0); (16; 0); (18; 1); y0 }, {(5; 0); (6; 1); (25; 1); z0 }, {(1; 0); (18; 0); (29; 1); z0 }. To obtain our main results, we shall use the following basic constructions, whose proofs can be found in [6]. Construction 2.5 (Breaking up groups). If there exists a k-RGDD of type (hm)u and a k-RGDD of type hm , then there exists a k-RGDD of type hmu and a k-IRGDD of type h(mu; m) . Construction 2.6 (Weighting). Let (X; G; B) be a GDD, and let w : X → Z + ∪ {0} be a weight function on X . Suppose that for each block B ∈ B, there exists a k-frame of type {w(x): x ∈ B}. Then there is a k-frame of type { x∈Gi w(x): Gi ∈ G}. Construction 2.7 (InNating RGDDs by RTDs). If there exists a k-RGDD of type hu and an RTD(k; m), then there exists a k-RGDD of type (mh)u . Construction 2.8 (Frame constructions). Suppose there is a k-frame with type T = {ti : i = 1; 2; : : : ; n}. Suppose also that t|ti and that there exists a k-RGDDof type n t 1+ti =t for i =1; 2; : : : ; n. Then there exists a k-RGDD of type t u where u=1+ i=1 ti =t. Construction 2.9 (Generalized frame constructions). Suppose there is a k-frame with type T = {ti : i = 1; 2; : : : ; n}. Let t|ti and b ¿ 0. If there exists a k-IRGDD of type b) t (ti =t+b; of type t (u+b; tn =t+b) where n for i = 1; 2; : : : ; n − 1, then there exists a k-IRGDD tn =t+b u = i=1 ti =t. Furthermore, if a k-RGDD of type t exists, then a k-RGDD of type t u+b exists. The following construction is established in [20]. Construction 2.10 (Constructions using color classes). Suppose there exists a k-RGDD of type gu , a k-frame of type (mg)v where u ¿ m + 1, and that there exists an RTD(k; mv). Then there exists a k-RGDD of type (mg)uv . 3. Direct constructions Lemma 3.1. There exists a 4-RGDD of type 810 . Proof. Let the point set be Z20 × Z4 , and let the group set be {{j; j + 10} × Z4 : j = 1; 2; : : : ; 10}. Below are the required base blocks: {(2; 1); (3; 4); (5; 4); (16; 4)}, {(2; 1); (4; 1); (5; 1); (16; 2)}, {(3; 1); (5; 3); (6; 4); (17; 2)}, {(3; 1); (4; 3); (6; 1); (17; 4)}, {(13; 1); (14; 4); (18; 3); (20; 2)}, {(14; 1); (15; 3); (19; 4); (1; 3)}, {(7; 1); (9; 2); (10; 2); (15; 4)}, {(8; 1); (10; 2); (11; 3); (16; 2)}, {(3; 1); (6; 3); (11; 3); (19; 3)}, {(4; 1); (7; 2); (12; 3); (20; 3)}, {(8; 1); (12; 2); (17; 3); (1; 3)}, {(9; 1); (13; 4); (18; 4); (2; 4)}.
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Here, all the above base blocks are developed by (+2 mod 20; +1 mod 4). Note that these blocks are organized into 3 columns. Each of the 2 blocks in the 8rst column (left-handed) gives a parallel class when developed by (+4 mod 20; +1 mod 4). The blocks in the second column form a parallel class when developed by ( –, +1 mod 4), and so do the blocks in the last column. Lemma 3.2. There exists a 4-RGDD of type 1010 . Proof. Let the point set be (Z30 × Z3 ) ∪ {∞0 ; ∞1 ; : : : ; ∞8 ; ∞} and let the group set be {{(i; j); (3 + i; j); : : : ; (27 + i; j)}: i; j ∈ Z3 } ∪ {{∞0 ; ∞1 ; : : : ; ∞8 ; ∞}}. Below are the required base blocks {(0; 0); (1; 0); (5; 0); (2; 1)}, {(3; 0); (10; 0); (6; 1); (26; 1)}, {(4; 0); (18; 0); (12; 1); (25; 1)}, {(7; 0); (15; 0); (19; 1); (21; 1)}, {(8; 0); (27; 0); (17; 1); (16; 2)}, {(9; 0); (22; 1); (20; 2); ∞0 }, {(11; 0); (29; 1); (24; 2); ∞3 }, {(13; 0); (23; 1); (28; 2); ∞6 }, {(14; 0); (14; 1); (14; 2); ∞}.
Developing the 8rst 8 blocks ( –, mod 3) plus the last block gives an initial parallel class. Then, we develop this parallel class (+1 mod 30, – ) to get the RGDD as required. Here, ∞i +(j; k)=∞i+(k mod 3) and ∞+(j; k)=∞ for i ∈ {0; 3; 6}, j ∈ Z30 and k ∈ Z3 . Lemma 3.3. For each even n, 6 6 n 6 14, there exists a 4-RGDD of type 18n . Proof. Let the point set be (Z2(n−1) ∪ {∞1 ; ∞2 }) × Z9 , and let the group set be {{j; j + (n − 1)} × Z9 : j = 1; : : : ; n − 1} ∪ {{∞1 ; ∞2 } × Z9 }. Below are the required base blocks: n = 6: {(3; 1); (6; 1); (7; 1); (9; 4)}, {(4; 1); (8; 5); (10; 5); (1; 2)}, {(7; 1); (8; 2); (9; 7); (10; 6)}, {(∞1 ; 1); (4; 1); (8; 2); (10; 3)}, {(∞1 ; 1); (5; 6); (7; 4); (9; 9)}, {(∞1 ; 1); (2; 8); (4; 7); (1; 5)}, {(∞2 ; 1); (2; 1); (5; 2); (1; 3)}, {(∞2 ; 1); (2; 4); (3; 8); (6; 6)}, {(∞2 ; 1); (3; 5); (5; 7); (6; 9)}: n = 8: {(2; 1); (5; 1); (14; 1); (1; 2)}, {(2; 1); (4; 7); (13; 6); (14; 2)}, {(3; 1); (8; 5); (12; 5); (14; 8)}, {(3; 1); (4; 1); (6; 2); (12; 2)}, {(3; 1); (5; 5); (8; 8); (9; 3)}, {(5; 1); (10; 7); (11; 5); (13; 3)}, {(∞1 ; 1); (9; 1); (10; 7); (13; 4)}, {(∞1 ; 1); (6; 2); (10; 6); (12; 8)}, {(∞1 ; 1); (4; 9); (6; 5); (9; 3)}, {(∞2 ; 1); (7; 1); (8; 3); (11; 2)}, {(∞2 ; 1); (7; 6); (11; 8); (1; 5)}, {(∞2 ; 1); (2; 7); (7; 9); (1; 4)}: n = 10: {(6; 1); (7; 1); (11; 1); (18; 2)}, {(6; 1); (12; 4); (14; 2); (16; 8)}, {(4; 1); (8; 6); (9; 5); (16; 1)}, {(14; 1); (16; 1); (17; 2); (1; 3)}, {(4; 1); (11; 5); (17; 3); (18; 7)}, {(7; 1); (13; 5); (15; 8); (17; 4)}, {(2; 1); (3; 4); (10; 1); (13; 1)}, {(2; 1); (5; 5); (7; 9); (10; 6)}, {(2; 1); (5; 6); (6; 3); (10; 9)}, {(∞1 ; 1); (9; 1); (12; 4); (15; 2)}, {(∞1 ; 1); (3; 6); (8; 3); (9; 8)}, {(∞1 ; 1); (3; 7); (14; 9); (1; 5)}, {(∞2 ; 1); (4; 1); (5; 3); (8; 2)}, {(∞2 ; 1); (13; 8); (15; 7); (1; 5)}, {(∞2 ; 1); (11; 6); (12; 4); (18; 9)}: n = 12: {(3; 1); (15; 1); (18; 1); (22; 2)}, {(2; 1); (11; 4); (18; 2); (1; 7)}, {(8; 1); (11; 2); (14; 4); (16; 1)}, {(5; 1); (17; 3); (19; 5); (21; 8)}, {(2; 1); (6; 1); (19; 1); (20; 3)}, {(6; 1); (7; 1); (10; 5); (14; 4)}, {(7; 1); (10; 4); (12; 3); (17; 6)}, {(15; 1); (16; 5); (20; 8); (22; 6)}, {(∞1 ; 1); (4; 1); (9; 2); (1; 3)}, {(∞1 ; 1); (3; 8); (8; 5); (9; 6)}, {(∞2 ; 1); (5; 1); (13; 2); (21; 4)}, {(∞2 ; 1); (4; 5); (12; 9); (13; 7)},
{(3; 1); (6; 6); (18; 2); (19; 1)}, {(8; 1); (14; 5); (21; 2); (1; 6)}, {(5; 1); (11; 6); (13; 7); (15; 3)}, {(9; 1); (10; 7); (12; 7); (22; 6)}, {(∞1 ; 1); (2; 9); (7; 4); (20; 7)}, {(∞2 ; 1); (4; 6); (16; 3); (17; 8)}:
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n = 14: {(8; 1); (11; 1); (18; 1); (1; 2)}, {(2; 1); (3; 4); (18; 8); (1; 4)}, {(9; 1); (13; 3); (17; 8); (25; 4)}, {(2; 1); (14; 1); (19; 1); (22; 2)}, {(9; 1); (10; 5); (12; 7); (20; 1)}, {(4; 1); (16; 6); (26; 7); (1; 5)}, {(5; 1); (6; 1); (7; 2); (23; 4)}, {(6; 1); (7; 9); (17; 4); (26; 3)}, {(6; 1); (12; 2); (18; 8); (21; 6)}, {(3; 1); (17; 2); (21; 1); (24; 3)}, {(4; 1); (19; 8); (23; 5); (25; 4)}, {(2; 1); (10; 5); (22; 7); (24; 1)}, {(4; 1); (10; 1); (13; 4); (15; 2)}, {(8; 1); (13; 9); (15; 5); (24; 7)}, {(11; 1); (14; 9); (19; 3); (23; 7)}, {(∞1 ; 1); (9; 1); (12; 5); (16; 3)}, {(∞1 ; 1); (5; 9); (11; 2); (16; 6)}, {(∞1 ; 1); (3; 7); (5; 4); (7; 8)}, {(∞2 ; 1); (20; 1); (25; 3); (26; 5)}, {(∞2 ; 1); (14; 8); (21; 2); (22; 7)}, {(∞2 ; 1); (8; 6); (15; 4); (20; 9)}:
Here, we 8rst develop the blocks in each column ( –, mod 9) to get three parallel classes. Then, we develop these parallel classes (mod 2(n − 1), – ) to get the RGDD as required. Lemma 3.4. For each n ∈{17,18,23}, there exists a 4-RGDD of type 12n . Proof. Let the point set be (Z4(n−1) ∪{∞1 ; ∞2 ; ∞3 ; ∞4 })×Z3 , and let the group set be {{j; j+(n−1); j+2(n−1); j+3(n−1)}×Z3 : j=1; : : : ; n−1}∪{{∞1 ; ∞2 ; ∞3 ; ∞4 }×Z3 }. Below are the required base blocks. n = 17:
{(42; 1); (47; 2); (55; 3); (61; 2)}, {(17; 1); (24; 1); (39; 1); (51; 1)}, {(6; 1); (30; 2); (34; 2); (45; 1)}, {(29; 1); (32; 2); (53; 3); (58; 2)}, {(3; 1); (11; 3); (12; 2); (36; 2)}, {(14; 1); (16; 1); (25; 1); (33; 1)}, {(19; 1); (31; 3); (54; 1); (60; 2)}, {(26; 1); (28; 3); (46; 1); (59; 1)}, {(2; 1); (8; 1); (15; 2); (62; 3)}, {(4; 1); (18; 3); (41; 3); (48; 2)}, {(21; 1); (22; 1); (40; 3); (43; 3)}, {(35; 1); (52; 2); (56; 1); (57; 2)}, {(9; 1); (27; 1); (37; 1); (63; 2)}, {(23; 1); (38; 2); (49; 3); (∞1 ; 1)}, {(20; 1); (50; 2); (64; 3); (∞2 ; 1)}, {(5; 1); (7; 2); (44; 3); (∞3 ; 1)}, {(10; 1); (13; 3); (1; 2); (∞4 ; 1)}:
n = 18: {(27; 1); (33; 2); (34; 3); (43; 1)}, {(29; 1); (55; 1); (60; 1); (61; 3)}, {(16; 1); (20; 3); (44; 3); (52; 3)}, {(13; 1); (50; 2); (58; 1); (68; 2)}, {(19; 1); (24; 3); (54; 2); (65; 3)}, {(3; 1); (6; 1); (15; 1); (59; 3)}, {(18; 1); (47; 1); (63; 2); (67; 2)}, {(21; 1); (23; 1); (28; 2); (48; 2)}, {(5; 1); (14; 3); (26; 2); (40; 1)}, {(31; 1); (37; 1); (53; 3); (66; 3)}, {(2; 1); (25; 2); (45; 1); (56; 3)}, {(35; 1); (39; 2); (42; 1); (57; 1)}, {(11; 1); (30; 1); (41; 1); (51; 1)}, {(7; 1); (8; 1); (32; 3); (46; 3)}, {(10; 1); (12; 3); (38; 2); (∞1 ; 1)}, {(9; 1); (17; 2); (36; 3); (∞2 ; 1)}, {(49; 1); (62; 2); (64; 3); (∞3 ; 1)}, {(4; 1); (22; 2); (1; 3); (∞4 ; 1)}: n = 23: {(23; 1); (26; 1); (34; 1); (74; 2)}, {(29; 1); (54; 1); (64; 2); (82; 1)}, {(8; 1); (20; 3); (38; 3); (57; 1)}, {(18; 1); (46; 2); (75; 1); (77; 3)}, {(3; 1); (58; 2); (84; 3); (85; 2)}, {(2; 1); (37; 3); (83; 1); (87; 3)}, {(9; 1); (14; 1); (30; 1); (55; 3)}, {(17; 1); (62; 2); (71; 1); (72; 1)}, {(39; 1); (42; 3); (56; 3); (88; 3)},
{(11; 1); (16; 3); (31; 1); (41; 1)}, {(12; 1); (25; 1); (36; 3); (59; 3)}, {(22; 1); (35; 3); (80; 3); (86; 3)}, {(6; 1); (53; 1); (67; 3); (68; 1)}, {(21; 1); (47; 3); (49; 3); (70; 2)}, {(28; 1); (40; 2); (44; 2); (76; 1)}, {(10; 1); (48; 2); (60; 2); (81; 3)}, {(7; 1); (52; 3); (61; 3); (66; 1)}, {(19; 1); (27; 2); (43; 1); (50; 3)},
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{(15; 1); (24; 2); (32; 1); (51; 1)}, {(4; 1); (73; 2); (79; 3); (∞1 ; 1)}, {(45; 1); (63; 2); (65; 3); (∞2 ; 1)}, {(5; 1); (78; 2); (1; 3); (∞3 ; 1)}, {(13; 1); (33; 2); (69; 3); (∞4 ; 1)}: Here, we 8rst develop these blocks ( –, mod 3) to get a parallel class. Then, we develop this parallel class (mod 4(n − 1), – ) to get the RGDD as required. Lemma 3.5. There exists a 4-RGDD of type 246 . Proof. Let the point set be (Z10 ∪ {∞1 ; ∞2 }) × Z12 , and let the group set be {{j; j + 5} × Z12 : j = 1; : : : ; 5} ∪ {{∞1 ; ∞2 } × Z12 }. Below are the required base blocks, which are divided into 4 parts. {(2; 1); (3; 1); (4; 2); (5; 1)}, {(2; 1); (3; 6); (4; 12); (5; 9)}, {(∞1 ; 1); (6; 1); (7; 3); (8; 6)}, {(∞1 ; 1); (6; 2); (8; 8); (10; 4)}, {(∞2 ; 1); (9; 1); (10; 5); (1; 3)}, {(∞2 ; 1); (7; 2); (9; 9); (1; 6)}, {(2; 1); (3; 9); (6; 6); (8; 10)}, {(2; 1); (3; 8); (6; 1); (9; 11)}, {(∞1 ; 1); (4; 7); (7; 10); (10; 9)}, {(∞1 ; 1); (4; 11); (7; 5); (10; 12)}, {(∞2 ; 1); (5; 4); (9; 10); (1; 8)}, {(∞2 ; 1); (5; 7); (8; 11); (1; 12)}: Here, we 8rst develop the 3 blocks in each part ( –, mod 12) to get four parallel classes. Then, we develop these parallel classes (mod 10, – ) to get the RGDD as required. Lemma 3.6. For each n ∈{7,11}, there exists a 4-RGDD of type 24n . Proof. Let the point set be (Z4(n−1) ∪{∞1 ; ∞2 ; ∞3 ; ∞4 })×Z6 , and let the group set be {{j; j+(n−1); j+2(n−1); j+3(n−1)}×Z6 : j=1; : : : ; n−1}∪{{∞1 ; ∞2 ; ∞3 ; ∞4 }×Z6 }. Below are the required base blocks: n = 7:
{(3; 1); (10; 1); (14; 1); (1; 3)}, {(15; 1); (16; 3); (19; 3); (24; 3)}, {(5; 1); (8; 6); (12; 3); (22; 4)}, {(∞1 ; 1); (7; 1); (18; 2); (20; 3)}, {(∞2 ; 1); (2; 1); (4; 3); (23; 4)}, {(∞3 ; 1); (13; 1); (17; 6); (21; 4)}, {(∞4 ; 1); (6; 1); (9; 2); (11; 5)},
{(2; 1); (9; 5); (10; 2); (23; 5)}, {(3; 1); (16; 2); (17; 2); (1; 1)}, {(5; 1); (6; 2); (7; 6); (22; 6)}, {(∞1 ; 1); (8; 6); (11; 4); (15; 5)}, {(∞2 ; 1); (4; 5); (13; 6); (18; 2)}, {(∞3 ; 1); (12; 5); (20; 3); (21; 2)}, {(∞4 ; 1); (14; 6); (19; 3); (24; 4)}:
n = 11: {(4; 1); (15; 1); (16; 1); (17; 3)}, {(12; 1); (15; 1); (37; 1); (38; 6)}, {(21; 1); (28; 1); (35; 3); (40; 5)}, {(6; 1); (19; 6); (23; 1); (25; 6)}, {(6; 1); (9; 3); (14; 1); (25; 4)}, {(10; 1); (14; 4); (18; 4); (1; 1)}, {(11; 1); (27; 3); (29; 3); (32; 6)}, {(8; 1); (13; 1); (16; 6); (39; 4)}, {(2; 1); (3; 4); (18; 5); (24; 4)}, {(2; 1); (4; 4); (11; 3); (28; 1)}, {(7; 1); (10; 5); (22; 6); (38; 6)}, {(5; 1); (32; 1); (34; 5); (40; 2)}, {(26; 1); (30; 3); (37; 6); (1; 4)}, {(3; 1); (7; 6); (24; 5); (35; 3)}, {(∞1 ; 1); (12; 1); (19; 5); (34; 3)}, {(∞1 ; 1); (21; 2); (22; 6); (36; 4)}, {(∞2 ; 1); (8; 1); (13; 2); (20; 3)}, {(∞2 ; 1); (17; 6); (30; 4); (31; 5)},
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{(∞3 ; 1); (31; 1); (33; 2); (39; 3)}, {(∞3 ; 1); (20; 6); (26; 4); (29; 5)}, {(∞4 ; 1); (5; 1); (23; 2); (36; 3)}, {(∞4 ; 1); (9; 5); (27; 4); (33; 6)}: Here, we 8rst develop the blocks in each column ( –, mod 6) to get two parallel classes. Then, we develop these parallel classes (mod 4(n − 1), – ) to get the RGDD as required. Lemma 3.7. There exists a 4-RGDD of type 366 . Proof. Let the point set be (Z10 ∪ {∞1 ; ∞2 }) × Z18 , and let the group set be {{j; j + 5} × Z18 : j = 1; : : : ; 5} ∪ {{∞1 ; ∞2 } × Z18 }. Below are the required base blocks, which are divided into 6 parts. {(2; 1); (3; 1); (8; 1); (9; 2)}, {(8; 1); (9; 4); (10; 1); (1; 5)}, {(4; 1); (6; 5); (10; 14); (1; 8)}, {(∞1 ; 1); (4; 1); (7; 2); (1; 3)}, {(∞1 ; 1); (5; 4); (6; 9); (7; 7)}, {(∞1 ; 1); (2; 5); (3; 12); (5; 8)}, {(∞2 ; 1); (5; 1); (6; 3); (10; 5)}, {(∞2 ; 1); (2; 2); (3; 8); (4; 4)}, {(∞2 ; 1); (7; 6); (8; 14); (9; 13)}, {(4; 1); (7; 1); (10; 13); (1; 5)}, {(4; 1); (6; 13); (8; 8); (1; 14)}, {(2; 1); (4; 6); (8; 16); (10; 13)}, {(∞1 ; 1); (2; 6); (3; 15); (9; 17)}, {(∞1 ; 1); (2; 10); (5; 18); (9; 13)}, {(∞1 ; 1); (3; 14); (6; 16); (7; 11)}, {(∞2 ; 1); (5; 18); (6; 11); (8; 10)}, {(∞2 ; 1); (3; 16); (7; 12); (10; 7)}, {(∞2 ; 1); (5; 9); (9; 17); (1; 15)}:
Here, we 8rst develop the 3 blocks in each part ( –, mod 18) to get six parallel classes. Then, we develop these parallel classes (mod 10, – ) to get the RGDD as required.
4. Rees constructions In this section, we shall employ the elegant and powerful constructions established by Rees in [16], which are dramatically used in [17,18]. Construction 4.1 (Rees [16], Construciton 1). Let (X; G; B) be an A-resolvable k-GDD of type gu in which for each i ∈ A there are ri i -parallel classes of blocks. Suppose that there is a TD(u; h) admitting H as a group of automorphisms acting sharply transitively on the points of each group. Let Hj be a collection of subsets of H , there being ri such subsets of size i for each i ∈ A, and suppose that the collection {Hj ∗ : ∈ H; j = 1; 2; : : : ; i ri } is resolvable on H . Then there is a resolvable GDD of type (hg)u . Construction 4.2 (Rees [17], Corollary 2.2). If there is an -resolvable k-GDD of type gu and a TD(u; ), then there is a resolvable k-GDD of type ( g)u . Lemma 4.3. There exists a [1; 1; 2; 2; 4]-resolvable 4-GDD of type 66 . Hence, there exists a 5-resolvable and a {1; 9}-resolvable 4-GDD of the same type. Proof. Let the point set be Z36 and let the group set be {{j; j + 6; : : : ; j + 30}: j = 0; 1; : : : ; 5}. Below are the required base blocks. {1; 4; 11; 18}, {1; 12; 28; 33}, {1; 2; 6; 29}, {4; 7; 30; 32}, {1; 20; 21; 35}:
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Here, all the base blocks are developed by +2 mod 36. The 8rst block generates one parallel class when developed by +4 mod 36 and gives 2 such classes in total. Each of the second and the third block generates one 2-resolvable parallel class when developed by +2 mod 36. The remaining 2 base blocks generate a 4-resolvable parallel class. Lemma 4.4. There exist 4-RGDDs of types 306 and 546 . Proof. Apply Construction 4.2 with a 5-resolvable 4-GDD of type 66 coming from Lemma 4.3 and a TD(6; 5), we obtain a 4-RGDD of type 306 . For 4-RGDD of type 546 , apply Construction 4.1 with a {1; 9}-resolvable 4-GDD of type 66 coming from Lemma 4.3 and a TD(6; 9) by taking H = GF(9), H1 = {0} and H2 = H . Lemma 4.5. There exists a 4-RGDD of type (6h)6 for any h ¿ 7, (h; 5) = 1 and h ≡ 1; 5 (mod 6). Proof. Apply Construction 4.1 with a [1; 1; 2; 2; 4]-resolvable 4-GDD of type 66 coming from Lemma 4.3 and a TD(6; h). Since h ¿ 7, (h; 5) = 1 and h ≡ 1; 5 (mod 6), we may take H = Zh . Let H1 = H2 = {0}, H3 = H4 = {0; 1} and H5 = {0; 1; 2; 3}. It remains to be shown that the collection {Hj + : ∈ Zh ; j = 1; 2; 3; 4; 5} is resolvable. We need the following notations. De8ne Ma = {{2i + a + 1; 2i + a + 2}}. Then all the translates of H3 (and i=0;1;:::;(h−3)=2
H4 ) can be partitioned into 3 classes, where the 8rst class consists of (h − 1)=2 blocks of size 2 missing the point a, and the second class consists of (h − 1)=2 blocks of size 2 missing the point a + 1 and the third class has one block of the form {a; a + 1}. Let K be a subset of H . De8ne S(K) = H \ K. We consider the partition in the following three cases. When h = 12t + 5: {H5 + 4i + j: i ∈ {0; 1; : : : ; 3t}, {12t + 4 + j}} for j = 0; 1; 2; 3, M3 ∪ {{3}}, M4 ∪ {{4}}, M5 ∪ {{5}}, M6 ∪ {{6}}, {{12t + 4; 0; 1; 2}; {3; 4}; {5; 6}} ∪ S({12t + 4; 0; 1; 2; 3; 4; 5; 6}), and H . When h = 12t + 11: {H5 + 4i + j − 3 : i ∈ {0; 1; : : : ; 3t + 1}; {12t + 5 + j}, {12t + 6 + j; 12t + 7 + j}} for j = 0; 1; 2; 3, M12t+3 ∪ {{12t + 6; 12t + 7; 12t + 8; 12t + 9}} ∪ {{12t + 3}} \ {{12t + 6; 12t + 7}; {12t + 8; 12t + 9}}, M12t+4 ∪{{12t +7; 12t +8; 12t +9; 12t +10}}∪{{12t +4}}\{{12t +7; 12t +8}; {12t + 9; 12t + 10}}, {{12t + 5; 12t + 6; 12t + 7; 12t + 8}; {12t + 3; 12t + 4}} ∪ S({12t + 3; 12t + 4; 12t + 5; 12t + 6; 12t + 7; 12t + 8}), M0 ∪ {{0}}, M1 ∪ {{1}}, and
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{{0; 1}} ∪ S({0; 1}). When h = 6t + 1 and t ¿ 2: i=0; 1; :::; t−1 {{0; 1; 2; 3} + 6i + j} i=0; 1; :::; t−1 {{0; 1} + 6i + j + 4} ∪ {{6t + j}} for j = 0; 1; 2; 3; 4; 5, {{6t; 0; 1; 2}} ∪ M5 ∪ {{5}} \ {{6t; 0}; {1; 2}}, M6 ∪ {{6}}, {{6t; 0}; {1; 2}; {3; 4}; {5; 6}} ∪ S({6t; 0; 1; 2; 3; 4; 5; 6}), and H. When h = 7: {{0; 1; 2; 3} + j} ∪ {{4; 5} + j} ∪ {{6} + j} for j = 0; 1; : : : ; 6, {{0; 1}; {3; 4}; {5; 6}; {2}}, {{1; 2}; {4; 5}; {6; 0}; {3}}, and {2; 3} ∪ S({2; 3}). Theorem 4.6. If h ≡ 6 (mod 12) and h ¿ 18, then there exists a 4-RGDD of type h6 . Proof. For h ≡ 0 (mod 18), apply Construction 2.7 with a 4-RGDD of type 186 coming from Lemma 3.3 or a 4-RGDD of type 546 coming from Lemma 4.4. For h ≡ 0 (mod 18), apply Construction 2.7 with a 4-RGDD of type 306 coming from Lemma 4.4 or a 4-RGDD coming from Lemma 4.5. Theorem 4.7. If h ≡ 0 (mod 24), then there exists a 4-RGDD of type h9 . Proof. From Theorem 2.1, we have a 4-frame of type (h=8)9 , which is also a 8resolvable 4-GDD of the same type by Lemma 1.1 of [17]. Now, apply Construction 4.2 with = 8, k = 4 and u = 9. 5. 4-RGDD(hn ) with h ≡ 2; 10 (mod 12) In this section, we discuss the existence of 4-RGDD(hn ) with h ≡ 2; 10 (mod 12). First, we deal with the case when h = 10. Lemma 5.1. If n ∈{46,82,100,118,334}, then there exists a 4-RGDD of type 10n . Proof. Start from a 4-frame of type (90)t where t ∈ {5; 9; 11; 13; 37} coming from Theorem 2.1 and apply Construction 2.8. Adjoining 10 in8nite points and using 4-RGDD of type 1010 coming from Lemma 3.2 to 8ll in the holes gives the required 4-RGDDs of type 109t+1 . Lemma 5.2. If n ∈{70,130}, then there exists a 4-RGDD of type 10n . Proof. From Theorem 1.1, we have a 4-RGDD of type 100u for u ∈ {7; 13}. Start from this 4-RGDD and apply Construction 2.5 with a 4-RGDD of type 1010 coming from Lemma 3.2, we obtain the 4-RGDDs of type 1010u as desired.
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Lemma 5.3. There exist 4-RGDDs of types 10142 , 10184 and 10214 . Proof. First, from [12, Lemma 2.5] we have a 4-IRGDD of type 2(28; 4) . InNating it by an RTD(4; 5) gives a 4-IRGDD of type 10(28; 4) . Similarly, we can get 4-RGDDs of types 1016 and 1022 from 4-RGDDs of types 216 and 222 , respectively. For 4-RGDD(10142 ), take a TD(6; 20) and truncate one group to size 15. This gives a {5; 6}-GDD of type 205 151 . Apply Construction 2.6 with weight 12, adjoin 40 in8nite points and apply Construction 2.9 with a 4-RGDD(1022 ) and a 4-IRGDD of type 10(28; 4) , we then obtain the design as desired. For 4-RGDD(10184 ), take a TD(8; 11) and delete a block for a {7; 8}-GDD of type 8 10 . Now, give points in one group weight 12 and rest 24 for a 4-frame of type 2407 1201 . Add 40 in8nite points and 8ll in holes to obtain the desired design. Here, we need 4-frames of types 246 121 and 247 121 coming from Lemma 2.2, as well as a 4-frame of type 247 coming from Theorem 2.1. For 4-RGDD(10214 ), start from an RTD(9; 23) and truncate in 1 group and 3 blocks to obtain a {5; 6; 7; 8; 9}-GDD of type 208 151 . InNate by 12 to obtain a 4-frame of type 2408 1801 . Add 40 in8nite points and 8ll in holes to obtain the desired design. Lemma 5.4. If n ∈{178,202,238,250,346}, then there exists a 4-RGDD of type 10n . Proof. The proof is similar to that of Lemma 5.3. Here, we 8rst start from a 4-RGDD of type 1004 and apply Construction 2.5 with a 4-RGDD of type 1010 to get a 4-IRGDD of type 10(30+10; 10) . Then, take a TD(x + 1; 25) with x ∈ {5; 6; 7; 8; 11}, truncate one group to size y with y ∈ {0; 5; 10; 15}, apply Construction 2.6 with weight 12, adjoin 100 in8nite points and apply Construction 2.9, we obtain the design as desired. We list the suitable parameters such that n = 30x + 6y=5 + 10 below: n = 178: x = 5, n = 238: x = 7, n = 346: x = 11,
y = 15; n = 202: x = 6, y = 10; y = 15; n = 250: x = 8, y = 0; y = 5.
Lemma 5.5. Let h ≡ 2; 10 (mod 12) and h ¿ 14. If n ≡ 4 (mod 6) and n ∈ {10; 70; 82; 118; 142}, then there exists a 4-RGDD of type hn . Proof. For n = 3 · 7 + 1; 3 · 35 + 1, apply Construction 2.7 with a 4-RGDD of type 222 or 2106 coming from Theorem 1.1. For other values of n = 3s + 1 with odd s ¿ 5 and s ∈ {7; 23; 27; 35; 39; 47}, we have a 4-frame of type (3h)s from Theorem 2.1. Apply Construction 2.8 with a 4-RGDD(h4 ) coming from Lemma 2.3, we obtain the desired 4-RGDD of type h3s+1 . Lemma 5.6. Let h ≡ 2; 10 (mod 12) and h ¿ 14. If n ∈{118,142}, then there exists a 4-RGDD of type hn . Proof. For n = 142, starting from a 4-RGDD of type (4h)7 coming from Theorem 1.1 and applying Construction 2.5 with a 4-RGDD of type h4 coming from Lemma 2.3, we obtain a 4-IRGDD of type h(24+4; 4) . Then, take a TD(6; 2h) and truncate one group to
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size 3h=2. This gives a {5; 6}-GDD of type (2h)5 (3h=2)1 . Apply Construction 2.6 with weight 12, adjoin 4h in8nite points and apply Construction 2.9 with a 4-RGDD(h22 ) coming from Lemma 5.5 and a 4-IRGDD of type h(24+4; 4) , we then obtain the desired design. For n = 118, we 8rst construct a 5-GDD of type 88 121 on Z64 ∪ {∞1 ; ∞2 ; : : : ; ∞12 }. The required base blocks are listed below: {1; 36; 45; 48; 63}, {1; 2; 27; 32}, {1; 14; 20; 42}, {7; 11; 21; 64}: Here, all the base blocks are developed by +1 mod 64. The second block generates one parallel class when developed by +4 mod 64 and gives 4 such classes in total. The last two blocks together generate one parallel class when developed by +8 mod 64 and give 8 such classes in total. Hence, we have 12 parallel classes of blocks of size 4. This gives the desired 5-GDD of type 88 121 . Then, starting from a 4-RGDD of type (4h)4 and applying Construction 2.5 with a 4-RGDD of type h4 , we obtain a 4-IRGDD of type h(12+4; 4) . Take the above 5-GDD and apply Construction 2.6 with weight 3h=2, adjoin 4h in8nite points and apply Construction 2.9 with a 4-RGDD(h22 ) and a 4-IRGDD of type h(12+4; 4) , we obtain the desired design. Theorem 5.7. If h ≡ 2; 10 (mod 12), then the necessary conditions for the existence of a 4-RGDD(hn ), namely, n ¿ 4 and n ≡ 4 (mod 6) are also su
6. 4-RGDD(hn ) with h ≡ 6 (mod 12) In this section, we discuss the existence of 4-RGDD(hn ) with h ≡ 6 (mod 12). First, we improve the known results on the existence of 4-RGDD(6n ). Lemma 6.1. If n ∈{52,58,62,74,102,114,124}, then there exists a 4-RGDD of type 6n .
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Proof. For n ∈{52,62,102}, start from a 4-frame of type 60s with s ∈ {5; 6; 10} coming from Theorem 2.1, add 12 in8nite points and apply Construction 2.9 with a 4-RGDD(612 ) coming from Theorem 1.1 and a 4-IRGDD of type 6(10+2; 2) coming from Lemma 2.4 to 8ll in holes, we obtain the desired designs. For n = 58, start from a TD(6; 5) and truncate one group to 3 points to obtain a {5; 6}-GDD of type 55 31 . Applying Construction 2.6 with weight 12, adjoining 12 points and 8lling in holes with a 4-RGDD(68 ) and a 4-IRGDD of type 6(10+2; 2) , we obtain the desired design. For n = 74, start from a TD(6; 7), remove a block and use one of the deleted point to rede8ne the groups, we obtain a {5; 6}-GDD of type 56 61 . Applying Construction 2.6 with weight 12, adjoining 12 points and 8lling in holes with a 4-RGDD(614 ) and a 4-IRGDD of type 6(10+2; 2) , we obtain the desired design. For n = 114, start from an RTD(6; 11), remove 10 points from the 8rst group and take one truncated parallel class to rede8ne the groups, we obtain a {5; 6; 11}-GDD of type 510 61 . Applying Construction 2.6 with weight 12, adjoining 12 points and 8lling in holes with a 4-RGDD(614 ) and a 4-IRGDD of type 6(10+2; 2) , we obtain the desired design. For n = 124, start from a TD(6; 11), truncate one group to size 6 and use the deleted point to rede8ne the groups, we obtain a {5; 6; 11}-GDD of type 511 61 . Applying Construction 2.6 with weight 12, adjoining 12 points and 8lling in holes with a 4-RGDD(614 ) and a 4-IRGDD of type 6(10+2; 2) , we obtain the desired design. In the remainder of this section, we shall focus mainly on the existence of 4-RGDDs of type 18n . First, we establish a preliminary bound and then reduce it. Denote N18 = {n: a 4-RGDD of type 18n exists}. We need the following working lemma. Lemma 6.2. Let x and s be nonnegative integers. Suppose a TD(6 + x; m) exists. Suppose also that there exists a 4-RGDD(182b=3+s ) and 4-IRGDDs of types 18(2m=3+s; s) and 18(2ai =3+s; s) when x ¿ 0, where 0 6 b 6 m and 0 6 ai 6 m for i ∈ {1; 2; : : : ; x}. If n = (10m + 2(a1 + a2 + · · · + ax ))=3 + (2b=3 + s), then there exists a 4-RGDD of type 18n . Proof. Truncate any x + 1 groups in the TD(6 + x; m) to sizes a1 ; a2 ; : : : ; ax and b. This gives a {5; 6; : : : ; 6 + x}-GDD with groups of sizes m; a1 ; a2 ; : : : ; ax and b. Apply Construction 2.6 with weight 12, adjoin 18s in8nite points and apply Construction 2.9 with a 4-RGDD(182b=3+s ) and 4-IRGDDs of types 18(2m=3+s; s) and 18(2ai =3+s; s) , we then obtain the design as required. Here we need 4-frames of types 12u for u ∈ {5; 6; : : : ; 6 + x} as input designs, which all come from Theorem 2.1. The proof is complete. Lemma 6.3. Let T = {t: t ¿ 2; t = 11; 13}, x be an integer and x ¿ 0. Suppose a TD(6 + x; 9t) exists. Suppose also that there exists a 4-RGDD(182c+4 ), where 0 6 c 6 3t. If bi = 0 or bi 6 t and bi ∈ T for i ∈ {1; 2; : : : ; x}, then 30t + 6(b1 + b2 + · · · + bx ) + (2c + 4) ∈ N18 .
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Proof. First, for each t ∈ T apply Construction 2.9 with a 4-frame of type 542t+1 coming from Theorem 2.1 and a 4-RGDD of type 184 coming from Lemma 2.3, which is also a 4-IRGDD of type 18(4; 1) , we get a 4-IRGDD of type 18(6t+4; 4) . Then, apply Lemma 6.2 with m = 9t, s = 4, b = 3c and ai = 9bi for i ∈ {1; 2; : : : ; x}, we obtain the 4-RGDDs as required. Lemma 6.4. If n ¿ 256, then n ∈ N18 . Proof. From Lemma 2.3, we have a TD(8; 9t) for any t ¿ 8. Apply Lemma 6.3 with x = 2, t ¿ 8, c ∈ {0; 1; 2}, 0 6 bi 6 t and bi ∈ {1; 11; 13} for i ∈ {1; 2}, we then have [30t + 6 × 2 + 4; 30t + 6 × (2t − 4) + 4] = [30t + 16; 42t − 20] ⊂ N18 noting that [2; 2t − 4] ⊂ {b1 + b2 : 0 6 bi 6 t; bi = 1; 11; 13; i = 1; 2}. Here, we need a 4-RGDD of type 182c+4 for c ∈ {0; 1; 2} whose existence can be found in Lemma 2.3 or Lemma 3.3. It is not diIcult to check that the above intervals [30t + 16; 42t − 20] overlap when t runs over [8; ∞) \ {11; 13; 19}. This completes the proof. Lemma 6.5. If 118 6 n 6 254, then n ∈ N18 . Proof. The proof is similar to that of Lemma 6.4. Here, starting from a TD(15; 27) and applying Lemma 6.3 with x =9, t =3, c ∈ {0; 1; 2}, and bi ∈ {0; 2; 3} for i ∈ {1; 2; : : : ; 9}, we then have [18 · 5 + 6 × 4 + 4; 18 · 5 + 6 × 27 + 4] = [118; 256] ⊂ N18 noting that [4; 27] ⊂ {b1 +b2 +· · ·+b9 : bi ∈ {0; 2; 3}; i =1; 2; : : : ; 9}. This completes the proof. Now, we treat the small orders of n. Lemma 6.6. If n ∈{4,6,8,10,12,14,16,24,32,40,48}, then n ∈ N18 . Proof. For n=4, the design comes from Lemma 2.3. For n ∈{6,8,10,12,14}, the designs come from Lemma 3.3. For other values of n, from Lemma 2.3, we have a 4-RGDD of type (18m)4 for each m ∈ {4; 6; 8; 10; 12}. Start from these 4-RGDDs and apply Construction 2.5 with a 4-RGDD of type 18m , we obtain the 4-RGDDs of type 184m as desired. Lemma 6.7. If n ∈{22,26,34,46,50,52,56,58}, then n ∈ N18 . Proof. Start from a 4-frame of type (18s)t where s ∈ {3; 5; 7; 9} and t ∈ {5; 7; 11; 17; 19} coming from Theorem 2.1 and apply Construction 2.8. Adjoining 18 in8nite points and using 4-RGDDs of type 18u for u ∈ {4; 6; 8; 10} coming from Lemma 6.6 to 8ll in the holes, we obtain the required 4-RGDDs. The suitable parameters such that n = st + 1 are listed below: n = 22: s = 3, t = 7; n = 50: s = 7, n = 26: s = 5, t = 5; n = 52: s = 3, n = 34: s = 3, t = 11; n = 56: s = 5, n = 46: s = 9, t = 5; n = 58: s = 3,
t = 7; t = 17; t = 11; t = 19.
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Lemma 6.8. n ∈{20,28,30,36,42,44,54,60}, then n ∈ N18 . Proof. Apply Construction 2.10 with g = 18, u ∈ {4; 6; 12}, m = 1 and v ∈ {5; 7; 9; 11}. Here, we need 4-RGDDs of type 18u , 4-frames of type 18v and an RTD(4; v), which come from Lemma 6.6, Theorem 2.1 and Lemma 2.3, respectively. It is easy to see that u ¿ 4 ¿ 2 = m + 1, so we can get the designs as desired. The suitable parameters such that n = uv are listed below: n = 20: u = 4, v = 5; n = 42: u = 6, v = 7; n = 28: u = 4, v = 7; n = 44: u = 4, v = 11; n = 30: u = 6, v = 5; n = 54: u = 6, v = 9; n = 36: u = 4, v = 9; n = 60: u = 12, t = 5. Lemma 6.9. [64; 88] ⊂ N18 . Proof. First, starting from a 4-RGDD of type 724 coming from Lemma 2.3 and applying Construction 2.5 with a 4-RGDD of type 184 coming from Lemma 2.3, we obtain a 4-IRGDD of type 18(12+4; 4) . Then, take a TD(7; 9) and apply Construction 2.6 with weight 24 to the points of the 8rst 5 groups, weight 0 or 24 to all the points of the sixth group and weight 0; 12 or 24 to the points in the last group. We obtain a 4-frame of type 2165 (18u)1 or 2166 (18u)1 with u ∈ {0; 2; 4; : : : ; 12}. Adjoin 72 in8nite points and apply Construction 2.9 with a 4-RGDD of type 18u+4 and a 4-IRGDD of type 18(12+4; 4) , we obtain a 4-RGDD of type 1860+u+4 or 1872+u+4 . Here, we need the input 4-frames of types 245 , 245 121 , 246 , 246 121 and 247 , which come from Theorem 2.1 or Lemma 2.2. Lemma 6.10. [90; 116] ⊂ N18 . Proof. First, starting from a 4-RGDD of type 725 coming from Theorem 1.1 and applying Construction 2.5 with a 4-RGDD of type 184 coming from Lemma 2.3, we obtain a 4-IRGDD of type 18(16+4; 4) . Similarly, we can get a 4-IRGDD of type 18(12+4; 4) from 4-RGDDs of types 724 and 184 . Then, take a TD(8; 24) and apply Construction 2.6 with weight 12 to the points of the 8rst 5 groups, weight 12 to 18 or 24 points of both the sixth and the seventh group and weight 0 to the remaining points in these two groups, and weight 0 or 12 to the points in the last group. We obtain a 4-frame of type 2885 (18u)1 , 2885 2161 (18u)1 , 2885 2162 (18u)1 , 2886 2161 (18u)1 , or 2167 (18u)1 with u ∈ {0; 2; 4; : : : ; 12}. Adjoin 72 in8nite points and apply Construction 2.9 with a 4-RGDD of type 18u+4 and 4-IRGDDs of types 18(12+4; 4) and 18(16+4; 4) , we obtain the designs as desired. Theorem 6.11. If h ≡ 6 (mod 12), then the necessary conditions for the existence of a 4-RGDD(hn ), namely, n ¿ 4 and n ≡ 0 (mod 2) are also su
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Construction 2.7 with a 4-RGDD of type 6n and an RTD(4; h=6) coming from Lemma 2.3. For h ¿ 30 and n ∈ {4; 6}, the designs come from Lemma 2.3 and Theorem 4.6. This leaves h ¿ 30 and n ∈ {54; 68} to be considered. For h ¿ 30 and n = 54, starting from a 4-RGDD of type 129 coming from Theorem 1.1 and applying Construction 2.7 with an RTD(4; h=2), we obtain a 4-RGDD of type (6h)9 . Applying Construction 2.5 with a 4-RGDD of type h6 gives the desired designs. For h ¿ 30 and n = 68, starting from a 4-RGDD of type (4h)4 and applying Construction 2.5 with a 4-RGDD of type h4 , we obtain a 4-IRGDD of type h(12+4; 4) . Then, take a TD(6; h) and truncate one group into size h=3. Applying Construction 2.6 with weight 12 gives a 4-frame of type (12h)5 (4h)1 . Adjoining 4h in8nite points and applying Construction 2.9 with a 4-RGDD of type h8 and a 4-IRGDD of type h(12+4; 4) , we obtain the designs as desired. 7. 4-RGDD(hn ) with h ≡ 0 (mod 12) In this section, we discuss the existence of 4-RGDD(hn ) with h ≡ 0 (mod 12). From Construction 2.7, we need mainly work on the cases when h = 24; 36 or 72. Lemma 7.1. Let h ∈ {24; 36; 72}. If n ∈{4,5,6,7,8,10}, then there exists a 4-RGDD of type hn . Proof. For n ∈ {4; 5}, the required designs come from Theorem 1.1. For n = 6 and h ∈ {24; 36}, the required designs come from Lemma 3.5 or Lemma 3.7. For n = 6 and h = 72, apply Construction 2.7 with a 4-RGDD of type 186 coming from Lemma 3.3 and an RTD(4; 4). For n=7 and h=24, the required design comes from Lemma 3.6. For n = 7 and h ∈ {36; 72}, apply Construction 2.7 with a 4-RGDD of type 47 coming from Theorem 1.1. For n = 8 and h ∈ {24; 36; 72}, apply Construction 2.7 with a 4-RGDD of type 38 coming from Theorem 1.1. For n = 10 and h = 24, apply Construction 2.7 with a 4-RGDD of type 610 coming from Theorem 1.1. For n = 10 and h ∈ {36; 72}, apply Construction 2.7 with a 4-RGDD of type 410 coming from Theorem 1.1. Lemma 7.2. Let h ∈ {24; 36; 72}. Suppose a TD(7; m) exists. Suppose also there exist 4-frames of type ht for t = m, a2 and a 4-RGDD(ha1 +1 ), where 3 6 a1 ¡ m and 5 6 a2 6 m. If n = 5m + a1 + a2 + 1, then there exists a 4-RGDD of type hn . Proof. Truncate 2 groups in the TD(7; m) to sizes a1 and a2 . Take a deleted point from the group of size a1 to rede8ne groups. This gives a {5; 6; 7; m; a2 }-GDD with groups of sizes 5; 6 and a1 . Apply Construction 2.6 with weight h, add h in8nite points and apply Construction 2.8 with 4-RGDDs of types h6 , h7 coming from Lemma 7.1, and type ha1 +1 , we then get the design as desired. Here we also need 4-frames of types hu for u ∈ {5; 6; 7} as input designs, which all come from Theorem 2.1. The proof is complete. Lemma 7.3. Let h ∈ {24; 36; 72}. If n ¿ 84, then there exists a 4-RGDD of type hn .
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Proof. From Lemma 2.3, we have a TD(7; m) or a TD(7; m + 1) for any m ¿ 15. Apply Lemma 7.2 with m ¿ 15, a1 = 3 and a2 ∈ [5; 11] ∪ [13; m] or a1 = 4 and a2 = 11 noting that we do not have a 4-frame of type 3612 , we then obtain a 4-RGDD of type hn for each n ∈ [5m + 9; 6m + 4]. Here, we need a 4-RGDD of type h4 or h5 to 8ll in holes, which comes from Lemma 7.1. It is not diIcult to check that the intervals [5m + 9; 6m + 4] overlap when m runs over [15; ∞). Lemma 7.4. Let h ∈ {24; 36; 72}. If 49 6 n 6 83, then there exists a 4-RGDD of type hn . Proof. Start with a TD(10; 9) and truncate 5 groups to size ai , where ai ∈ {0; 3; 4; 5; 6; 7; 9} and 1 6 i 6 5. This gives a {5; 6; 7; 8; 9; 10}-GDD of type 95 a11 a12 · · · a15 . Give weight h to each point of this GDD to obtain a 4-frame of type (9h)5 (ha1 )1 (ha2 )1 · · · (ha5 )1 . Adding h in8nite points and using 4-RGDDs of types hu for u ∈ {4; 5; 6; 7; 8; 10} coming from Lemma 7.1 to 8ll in the holes gives the desired 4-RGDDs. Lemma 7.5. Let h ∈ {24; 36; 72}. If 39 6 n 6 48, then there exists a 4-RGDD of type hn . Proof. The proof is similar to that of Lemma 7.4. Here, we start with a TD(7; 7) and truncate 2 groups to size ai , where ai ∈ {0; 3; 4; : : : ; 7} and 1 6 i 6 2. Lemma 7.6. There exists a 4-RGDD of type h38 for h ∈ {24; 36; 72}. Proof. Start with a TD(6; 7) and delete 5 points from a block so as to form a {5; 6}-GDD of type 65 71 . Give weight h to the points of this GDD to obtain a 4-frame of type (6h)5 (7h)1 . Adding h in8nite points and using 4-RGDDs of types hu for u=7; 8 coming from Lemma 7.1 to 8ll in the holes gives the desired 4-RGDD of type h38 . Lemma 7.7. Let h ∈ {24; 36; 72}. If n ∈{19,21,22,25,26,29,31,33,34,37}, then there exists a 4-RGDD of type hn . Proof. Start from a 4-frame of type (hs)t where s ∈ {3; 4; 5} and t ∈ {5; 6; 7; 8; 9; 10; 11} coming from Theorem 2.1 and apply Construction 2.8. Adding h in8nite points and using 4-RGDDs of types hu for u = 4; 5; 6 coming from Lemma 7.1 to 8ll in the holes gives the desired 4-RGDDs. We list the suitable parameters such that n = st + 1 below: n = 19: n = 21: n = 22: n = 25: n = 26:
s = 3, s = 4, s = 3, s = 4, s = 5,
t = 6; t = 5; t = 7; t = 6; t = 5;
n = 29: n = 31: n = 33: n = 34: n = 37:
s = 4, s = 3, s = 4, s = 3, s = 4,
t = 7; t = 10; t = 8; t = 11; t = 9.
Lemma 7.8. Let h ∈ {24; 36; 72}. If n ∈{12,16,20,24,28,32,36}, then there exists a 4-RGDD of type hn .
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Proof. For each given n, a 4-RGDD of type 3n exists from Theorem 1.1. Apply Construction 2.7 to these 4-RGDDs with an RTD(4; h=3), we get the results as desired. Lemma 7.9. Let h ∈ {24; 36; 72}. If n ∈ {30; 35}, then there exists a 4-RGDD of type hn . Proof. Starting from a 4-RGDD of type (6h)5 or (7h)5 coming from Theorem 1.1 and apply Construction 2.5 with a 4-RGDD of type h6 or h7 coming from Lemma 7.1, we obtain the desired designs. Combining Lemma 7.1 and Lemmas 7.3–7.9 and applying Construction 2.7, we have the following result. Lemma 7.10. Let h ∈ {24; 36; 72; 120}. If n ¿ 4 and n ∈ {9; 11; 13; 14; 15; 17; 18; 23; 27}, then there exists a 4-RGDD of type hn . Lemma 7.11. If h ≡ 0 (mod 12) and h ∈ {12; 60}, then there exists a 4-RGDD of type h27 . Proof. For h = 180, take a TD(7; h=3) and truncate 2 groups to size h=4. This gives a {5; 6; 7}-GDD of type (h=3)5 (h=4)2 . Give weight 12 to each point of this GDD to obtain a 4-frame of type (4h)5 (3h)2 . Adding h in8nite points and using 4-RGDDs of types hu for u ∈ {4; 5} coming from Theorem 1.1 to 8ll in the holes gives the desired 4-RGDDs. For a 4-RGDD of type 18027 , apply Construction 2.7 with a 4-RGDD of type 3627 . Lemma 7.12. If h ≡ 24; 72 (mod 96), then there exists a 4-RGDD of type h15 . Proof. From [21], we have an 8-RGDD of type 815 , which can be extended to a 9-GDD of type 815 161 . Similarly, we can obtain a 5-GDD of type (h=8)8 (7h=24)1 from a 4-RGDD of type (h=8)8 coming from Theorem 1.1. Now, in the 9-GDD of type 815 161 we give weight h=8 to the points of groups of size 8 and weight 7h=24 to the points of the group of size 16, to obtain a 5-GDD of type h15 (14h=3)1 . Removing the distinguished group of this 5-GDD gives the 4-RGDD of type h15 as desired. Lemma 7.13. Let h ∈ {72; 120}. If n ∈{11,13,17,23}, then there exists a 4-RGDD of type hn . Proof. Start with a TD(h=12 + 1; n) and adjoin an in8nite point ∞ to the groups, then delete a 8nite point so as to form a {h=12 + 1; n + 1}-GDD of type (h=12)n n1 . Note that each block of size n + 1 intersects the group of size n in the in8nite point ∞ and each block of size h=12 + 1 intersects the group of size n, but certainly not in ∞. Now, in the group of size n, we give ∞ weight 4(n − 1) and give the remaining points weight h=3 − 4 for a total weight of h(n − 1)=3. Give all other points in the GDD weight 12. The result is a 5-GDD of type hn (h(n − 1)=3)1 , where we make use of input designs
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from 5-GDDs of types 12h=12 (h=3 − 4)1 (from a 4-RGGD(12h=12 )) and 12n (4(n − 1))1 (from a 4-RGDD(12n )). Removing the distinguished group of this 5-GDD gives the 4-RGDD of type hn as desired. Lemma 7.14. There exist 4-RGDDs of types 2414 , 2418 , 3613 , 7214 , 7218 , 12014 and 12018 . Proof. Starting from 4-RGDDs of types 614 , 618 , 413 and applying Construction 2.7, we obtain the designs as desired. Combining Theorem 1.1, Lemmas 3.4 and 3.6, Theorem 4.7 and Lemmas 7.10–7.14 and applying Construction 2.7, we have the following result. Theorem 7.15. If h ≡ 0 (mod 12), then the necessary conditions for the existence of a 4-RGDD(hn ), namely, n ¿ 4 are also su
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[14] D.L. Kreher, A.C.H. Ling, R.S. Rees, C.W.H. Lam, A note on {4}-GDDs of type 210 , Discrete Math. 261 (2003) 373–376. [15] E.R. Lamken, W.H. Mills, R.S. Rees, Resolvable minimum coverings with quadruples, J. Combin. Designs 6 (1998) 431–450. [16] R.S. Rees, Two new direct product type constructions for resolvable group divisible designs, J. Combin. Designs 1 (1993) 15–26. [17] R.S. Rees, Group divisible designs with block size k having k + 1 groups, for k = 4 and 5, J. Combin. Designs 8 (2000) 363–386. [18] R.S. Rees, Truncated transversal designs: a new lower bound on the number of idempotent MOLS of side n, J. Combin. Theory Ser. A 90 (2000) 257–266. [19] R.S. Rees, D.R. Stinson, On resolvable group-divisible designs with block size 3, Ars Combin. 23 (1987) 107–120. [20] R.S. Rees, D.R. Stinson, Frames with block size four, Canad. J. Math. 44 (1992) 1030–1049. [21] E. Seiden, A method of construction of resolvable BIBD, SankhyRa Ser. A 25 (1963) 393–394. [22] H. Shen, Resolvable group divisible designs with block size 4, J. Combin. Math. Combin. Comput. 1 (1987) 125–130. [23] H. Shen, On the existence of nearly Kirkman systems, Ann. Discrete Math. 52 (1992) 511–518. [24] H. Shen, J. Shen, Existence of resolvable group divisible designs with block size four I, Discrete Math. 254 (2002) 513–525. [25] R.M. Wilson, Constructions and uses of pairwise balanced designs, Math. Centre Tracts 55 (1974) 18–41.
www.elsevier.com/locate/disc
A survey on resolvable group divisible designs with block size four Gennian Gea;1 , Alan C.H. Lingb;2 a Department
of Mathematics, Zhejiang University, Hangzhou 310027, Zhejiang, People’s Republic of China b Department of Computer Science, University of Vermont, Burlington, VT 05405, USA Received 30 October 2002; received in revised form 24 April 2003; accepted 9 June 2003
Abstract In this paper, a survey for the existence of resolvable group divisible designs (RGDDs) with block size four, group-type hn and index unity is presented. It is proved that there exists a 4-RGDD of type hn if and only if n ¿ 4, hn ≡ 0 (mod 4) and h(n − 1) ≡ 0 (mod 3) with a handful of possible exceptions. c 2003 Elsevier B.V. All rights reserved. MSC: Primary 05B05 Keywords: Group divisible designs; Resolvable group divisible designs; Incomplete resolvable group divisible designs; 4-frames
1. Introduction A group divisible design (GDD) is a triple (X; G; B) which satis8es the following properties: 1. G is a partition of a set X (of points) into subsets called groups, 2. B is a set of subsets of X (called blocks) such that a group and a block contain at most one common point, 3. every pair of points from distinct groups occurs in a unique block.
1 2
Researcher supported in part by YNSFC Grant 10001026. Researcher supported by an ARO grant 19-01-1-0406 and a DOE grant. E-mail address: [email protected] (G. Ge).
c 2003 Elsevier B.V. All rights reserved. 0012-365X/$ - see front matter doi:10.1016/S0012-365X(03)00274-7
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The group-type (or type) of the GDD is the multiset {|G|: G ∈ G}. We usually use an “exponential” notation to describe group-type: a type 1i 2j 3k : : : denotes i occurrences of 1; j occurrences of 2, etc. A GDD (X; G; B) will be referred to as a K-GDD if |B| ∈ K for every block B in B. When K = {k}, we simply write k for K. An -parallel class of blocks in a GDD (X; G; B) is a subset B ⊆ B such that each point x ∈ X is contained in exactly blocks in B . When = 1, we will employ the usual term parallel class. If the block set B can be partitioned into -parallel classes, then the GDD is called -resolvable (or just resolvable if = 1). A GDD (X; G; B) is called A-resolvable if its block set B admits a partition into subsets B1 ; B2 ; : : : ; Br where for each i = 1; 2; : : : ; r there is an i ∈ A such that each point x ∈ X is contained in exactly i blocks in Bi . It is not diIcult to see that an A-resolvable GDD with K = {k} must be uniform. Resolvable group divisible designs have been instrumental in the construction of other types of designs. Many researchers have been involved in investigating the existence of resolvable group divisible designs. Simple counting arguments show that if there is a k-RGDD of type hn , then n ¿ k; hn ≡ 0 (mod k)
and
h(n − 1) ≡ 0 (mod k − 1): The necessary conditions for the existence of a k-RGDD of type hn have been proved to be suIcient for k = 3 (see [1,16,19]), with the de8nite exception of 3-RGDDs of types 23 , 26 and 63 . However, the case for k = 4 has remained open for a long time and we have the following known results (see [8,10,12–14,17,20,22–24]). Theorem 1.1. The necessary conditions for the existence of a 4-RGDD(hn ), namely, n ¿ 4, hn ≡ 0 (mod 4) and h(n − 1) ≡ 0 (mod 3), are also su
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Theorem 1.2. The necessary conditions for the existence of a 4-RGDD(hn ), namely, n ¿ 4, hn ≡ 0 (mod 4) and h(n − 1) ≡ 0 (mod 3), are also su
h ∈ {6; 30} ∪ {n: 66 6 n 6 438} and u ∈ {7; 23; 27; 35; 39; 47}; 450 6 h 6 2190 and u ∈ {7; 23; 27; 39; 47}; h ∈ {42; 54} ∪ {n: 2202 6 n 6 11238} and u ∈ {23; 27}; h = 18 and u ∈ {15; 17; 23; 27}.
For recursion, we will also need the following 4-frames. Lemma 2.2 (Ge et al. [9,12]). There exists a 4-frame of type 24u 121 with u = 5; 6 or 7. A transversal design (TD) TD(k; n) is a GDD of group type nk and block size k. A resolvable TD(k; n) (denoted by RTD(k; n)) is equivalent to a TD(k + 1; n). It is well
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known that the existence of a TD(k; n) is equivalent to the existence of k − 2 mutually orthogonal Latin squares (MOLS) of order n. In this paper, we mainly employ the following known results on TDs. Lemma 2.3 (Colbourn and Dinitz [3]). 1. An RTD(4; n) exists for all n ¿ 4 except for n = 6 and possibly excepting n = 10. 2. A TD(6; n) exists for all n ¿ 5 except possibly for n ∈{6,10,14,18,22}. 3. A TD(7; n) exists for all n ¿ 7 except possibly for n ∈{10,14,15,18,20,22,26,30,34, 38,46,60,62}. 4. A TD(8; n) exists for all n ¿ 7 except possibly for n ∈{10,12,14,15,18,20,21,22,26, 28,30,33,34,35,38,39,42,44,46,51,52,54,58,60,62,66,68,74}. 5. A TD(q + 1; q) exists, where q is a prime power. An incomplete group divisible design (IGDD) with block size k and index unity is a quadruple (X; G; H; B) which satis8es the following properties: 1. G = {G1 ; G2 ; : : : ; Gn } is a partition of a set X (of points) into subsets called groups, 2. H is a subset of X called a hole, 3. B is a collection of subsets of X , called blocks, such that a group and a block contain at most one common point, 4. every pair of points from distinct groups is either in H or occurs in a unique block but not both. We denote this design by k-IGDD(T ) where T is the type and de8ned by the multiset {(|Gi |; |Gi ∩ H |): 1 6 i 6 n}. As with GDDs, we shall use an “exponential” notation to describe the type. When H = ∅, an IGDD of type {(|Gi |; 0): 1 6 i 6 n} is just a GDD of type {|Gi |: 1 6 i 6 n}. A k-IGDD is said to be resolvable and denoted by k-IRGDD if its blocks can be partitioned into parallel classes and partial parallel classes, the latter partitioning X \ H . In this paper, we shall only use IRGDDs of such types as (h; 0)m−n (h; h)n where h ¿ 0 and m ¿ n ¿ 0. So, we shall simply use h(m; n) to denote the type. An IRGDD of type h(m; 1) is just an RGDD of type hm . Lemma 2.4. There exists a 4-IRGDD of type 6(12; 2) . Proof. Let the point set be (Z30 × {0; 1}) ∪ {w0 ; w1 ; w2 ; x0 ; x1 ; x2 ; y0 ; y1 ; y2 ; z0 ; z1 ; z2 }, let the groups be generated by {(0; i); (5; i); : : : ; (25; i)}, i ∈ {0; 1} and let the hole set be {w0 ; w1 ; w2 ; x0 ; x1 ; x2 ; y0 ; y1 ; y2 ; z0 ; z1 ; z2 }. The two partial parallel classes missing the hole {w0 ; w1 ; w2 ; x0 ; x1 ; x2 ; y0 ; y1 ; y2 ; z0 ; z1 ; z2 } are generated by {(0; 0); (3; 0); (0; 1); (13; 1)}. The following base blocks in each column form a parallel class by +10 mod 30. Note that wi + j = wi+j mod 3 . That happens similarly to xi ; yi ; zi for i ∈ {0; 1; 2}: {(1; 0); (10; 0); (13; 1); (19; 1)}, {(0; 0); (14; 1); (17; 1); (26; 1)}, {(6; 0); (18; 0); (10; 1); (12; 1)}, {(5; 0); (27; 0); (29; 0); (20; 1)}, {(3; 0); (7; 0); (14; 0); w0 }, {(2; 1); (3; 1); (25; 1); w0 },
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{(2; 0); (1; 1); (27; 1); x0 }, {(3; 0); (4; 0); (11; 1); x0 }, {(9; 0); (14; 1); (28; 1); y0 }, {(2; 0); (16; 0); (18; 1); y0 }, {(5; 0); (6; 1); (25; 1); z0 }, {(1; 0); (18; 0); (29; 1); z0 }. To obtain our main results, we shall use the following basic constructions, whose proofs can be found in [6]. Construction 2.5 (Breaking up groups). If there exists a k-RGDD of type (hm)u and a k-RGDD of type hm , then there exists a k-RGDD of type hmu and a k-IRGDD of type h(mu; m) . Construction 2.6 (Weighting). Let (X; G; B) be a GDD, and let w : X → Z + ∪ {0} be a weight function on X . Suppose that for each block B ∈ B, there exists a k-frame of type {w(x): x ∈ B}. Then there is a k-frame of type { x∈Gi w(x): Gi ∈ G}. Construction 2.7 (InNating RGDDs by RTDs). If there exists a k-RGDD of type hu and an RTD(k; m), then there exists a k-RGDD of type (mh)u . Construction 2.8 (Frame constructions). Suppose there is a k-frame with type T = {ti : i = 1; 2; : : : ; n}. Suppose also that t|ti and that there exists a k-RGDDof type n t 1+ti =t for i =1; 2; : : : ; n. Then there exists a k-RGDD of type t u where u=1+ i=1 ti =t. Construction 2.9 (Generalized frame constructions). Suppose there is a k-frame with type T = {ti : i = 1; 2; : : : ; n}. Let t|ti and b ¿ 0. If there exists a k-IRGDD of type b) t (ti =t+b; of type t (u+b; tn =t+b) where n for i = 1; 2; : : : ; n − 1, then there exists a k-IRGDD tn =t+b u = i=1 ti =t. Furthermore, if a k-RGDD of type t exists, then a k-RGDD of type t u+b exists. The following construction is established in [20]. Construction 2.10 (Constructions using color classes). Suppose there exists a k-RGDD of type gu , a k-frame of type (mg)v where u ¿ m + 1, and that there exists an RTD(k; mv). Then there exists a k-RGDD of type (mg)uv . 3. Direct constructions Lemma 3.1. There exists a 4-RGDD of type 810 . Proof. Let the point set be Z20 × Z4 , and let the group set be {{j; j + 10} × Z4 : j = 1; 2; : : : ; 10}. Below are the required base blocks: {(2; 1); (3; 4); (5; 4); (16; 4)}, {(2; 1); (4; 1); (5; 1); (16; 2)}, {(3; 1); (5; 3); (6; 4); (17; 2)}, {(3; 1); (4; 3); (6; 1); (17; 4)}, {(13; 1); (14; 4); (18; 3); (20; 2)}, {(14; 1); (15; 3); (19; 4); (1; 3)}, {(7; 1); (9; 2); (10; 2); (15; 4)}, {(8; 1); (10; 2); (11; 3); (16; 2)}, {(3; 1); (6; 3); (11; 3); (19; 3)}, {(4; 1); (7; 2); (12; 3); (20; 3)}, {(8; 1); (12; 2); (17; 3); (1; 3)}, {(9; 1); (13; 4); (18; 4); (2; 4)}.
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Here, all the above base blocks are developed by (+2 mod 20; +1 mod 4). Note that these blocks are organized into 3 columns. Each of the 2 blocks in the 8rst column (left-handed) gives a parallel class when developed by (+4 mod 20; +1 mod 4). The blocks in the second column form a parallel class when developed by ( –, +1 mod 4), and so do the blocks in the last column. Lemma 3.2. There exists a 4-RGDD of type 1010 . Proof. Let the point set be (Z30 × Z3 ) ∪ {∞0 ; ∞1 ; : : : ; ∞8 ; ∞} and let the group set be {{(i; j); (3 + i; j); : : : ; (27 + i; j)}: i; j ∈ Z3 } ∪ {{∞0 ; ∞1 ; : : : ; ∞8 ; ∞}}. Below are the required base blocks {(0; 0); (1; 0); (5; 0); (2; 1)}, {(3; 0); (10; 0); (6; 1); (26; 1)}, {(4; 0); (18; 0); (12; 1); (25; 1)}, {(7; 0); (15; 0); (19; 1); (21; 1)}, {(8; 0); (27; 0); (17; 1); (16; 2)}, {(9; 0); (22; 1); (20; 2); ∞0 }, {(11; 0); (29; 1); (24; 2); ∞3 }, {(13; 0); (23; 1); (28; 2); ∞6 }, {(14; 0); (14; 1); (14; 2); ∞}.
Developing the 8rst 8 blocks ( –, mod 3) plus the last block gives an initial parallel class. Then, we develop this parallel class (+1 mod 30, – ) to get the RGDD as required. Here, ∞i +(j; k)=∞i+(k mod 3) and ∞+(j; k)=∞ for i ∈ {0; 3; 6}, j ∈ Z30 and k ∈ Z3 . Lemma 3.3. For each even n, 6 6 n 6 14, there exists a 4-RGDD of type 18n . Proof. Let the point set be (Z2(n−1) ∪ {∞1 ; ∞2 }) × Z9 , and let the group set be {{j; j + (n − 1)} × Z9 : j = 1; : : : ; n − 1} ∪ {{∞1 ; ∞2 } × Z9 }. Below are the required base blocks: n = 6: {(3; 1); (6; 1); (7; 1); (9; 4)}, {(4; 1); (8; 5); (10; 5); (1; 2)}, {(7; 1); (8; 2); (9; 7); (10; 6)}, {(∞1 ; 1); (4; 1); (8; 2); (10; 3)}, {(∞1 ; 1); (5; 6); (7; 4); (9; 9)}, {(∞1 ; 1); (2; 8); (4; 7); (1; 5)}, {(∞2 ; 1); (2; 1); (5; 2); (1; 3)}, {(∞2 ; 1); (2; 4); (3; 8); (6; 6)}, {(∞2 ; 1); (3; 5); (5; 7); (6; 9)}: n = 8: {(2; 1); (5; 1); (14; 1); (1; 2)}, {(2; 1); (4; 7); (13; 6); (14; 2)}, {(3; 1); (8; 5); (12; 5); (14; 8)}, {(3; 1); (4; 1); (6; 2); (12; 2)}, {(3; 1); (5; 5); (8; 8); (9; 3)}, {(5; 1); (10; 7); (11; 5); (13; 3)}, {(∞1 ; 1); (9; 1); (10; 7); (13; 4)}, {(∞1 ; 1); (6; 2); (10; 6); (12; 8)}, {(∞1 ; 1); (4; 9); (6; 5); (9; 3)}, {(∞2 ; 1); (7; 1); (8; 3); (11; 2)}, {(∞2 ; 1); (7; 6); (11; 8); (1; 5)}, {(∞2 ; 1); (2; 7); (7; 9); (1; 4)}: n = 10: {(6; 1); (7; 1); (11; 1); (18; 2)}, {(6; 1); (12; 4); (14; 2); (16; 8)}, {(4; 1); (8; 6); (9; 5); (16; 1)}, {(14; 1); (16; 1); (17; 2); (1; 3)}, {(4; 1); (11; 5); (17; 3); (18; 7)}, {(7; 1); (13; 5); (15; 8); (17; 4)}, {(2; 1); (3; 4); (10; 1); (13; 1)}, {(2; 1); (5; 5); (7; 9); (10; 6)}, {(2; 1); (5; 6); (6; 3); (10; 9)}, {(∞1 ; 1); (9; 1); (12; 4); (15; 2)}, {(∞1 ; 1); (3; 6); (8; 3); (9; 8)}, {(∞1 ; 1); (3; 7); (14; 9); (1; 5)}, {(∞2 ; 1); (4; 1); (5; 3); (8; 2)}, {(∞2 ; 1); (13; 8); (15; 7); (1; 5)}, {(∞2 ; 1); (11; 6); (12; 4); (18; 9)}: n = 12: {(3; 1); (15; 1); (18; 1); (22; 2)}, {(2; 1); (11; 4); (18; 2); (1; 7)}, {(8; 1); (11; 2); (14; 4); (16; 1)}, {(5; 1); (17; 3); (19; 5); (21; 8)}, {(2; 1); (6; 1); (19; 1); (20; 3)}, {(6; 1); (7; 1); (10; 5); (14; 4)}, {(7; 1); (10; 4); (12; 3); (17; 6)}, {(15; 1); (16; 5); (20; 8); (22; 6)}, {(∞1 ; 1); (4; 1); (9; 2); (1; 3)}, {(∞1 ; 1); (3; 8); (8; 5); (9; 6)}, {(∞2 ; 1); (5; 1); (13; 2); (21; 4)}, {(∞2 ; 1); (4; 5); (12; 9); (13; 7)},
{(3; 1); (6; 6); (18; 2); (19; 1)}, {(8; 1); (14; 5); (21; 2); (1; 6)}, {(5; 1); (11; 6); (13; 7); (15; 3)}, {(9; 1); (10; 7); (12; 7); (22; 6)}, {(∞1 ; 1); (2; 9); (7; 4); (20; 7)}, {(∞2 ; 1); (4; 6); (16; 3); (17; 8)}:
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n = 14: {(8; 1); (11; 1); (18; 1); (1; 2)}, {(2; 1); (3; 4); (18; 8); (1; 4)}, {(9; 1); (13; 3); (17; 8); (25; 4)}, {(2; 1); (14; 1); (19; 1); (22; 2)}, {(9; 1); (10; 5); (12; 7); (20; 1)}, {(4; 1); (16; 6); (26; 7); (1; 5)}, {(5; 1); (6; 1); (7; 2); (23; 4)}, {(6; 1); (7; 9); (17; 4); (26; 3)}, {(6; 1); (12; 2); (18; 8); (21; 6)}, {(3; 1); (17; 2); (21; 1); (24; 3)}, {(4; 1); (19; 8); (23; 5); (25; 4)}, {(2; 1); (10; 5); (22; 7); (24; 1)}, {(4; 1); (10; 1); (13; 4); (15; 2)}, {(8; 1); (13; 9); (15; 5); (24; 7)}, {(11; 1); (14; 9); (19; 3); (23; 7)}, {(∞1 ; 1); (9; 1); (12; 5); (16; 3)}, {(∞1 ; 1); (5; 9); (11; 2); (16; 6)}, {(∞1 ; 1); (3; 7); (5; 4); (7; 8)}, {(∞2 ; 1); (20; 1); (25; 3); (26; 5)}, {(∞2 ; 1); (14; 8); (21; 2); (22; 7)}, {(∞2 ; 1); (8; 6); (15; 4); (20; 9)}:
Here, we 8rst develop the blocks in each column ( –, mod 9) to get three parallel classes. Then, we develop these parallel classes (mod 2(n − 1), – ) to get the RGDD as required. Lemma 3.4. For each n ∈{17,18,23}, there exists a 4-RGDD of type 12n . Proof. Let the point set be (Z4(n−1) ∪{∞1 ; ∞2 ; ∞3 ; ∞4 })×Z3 , and let the group set be {{j; j+(n−1); j+2(n−1); j+3(n−1)}×Z3 : j=1; : : : ; n−1}∪{{∞1 ; ∞2 ; ∞3 ; ∞4 }×Z3 }. Below are the required base blocks. n = 17:
{(42; 1); (47; 2); (55; 3); (61; 2)}, {(17; 1); (24; 1); (39; 1); (51; 1)}, {(6; 1); (30; 2); (34; 2); (45; 1)}, {(29; 1); (32; 2); (53; 3); (58; 2)}, {(3; 1); (11; 3); (12; 2); (36; 2)}, {(14; 1); (16; 1); (25; 1); (33; 1)}, {(19; 1); (31; 3); (54; 1); (60; 2)}, {(26; 1); (28; 3); (46; 1); (59; 1)}, {(2; 1); (8; 1); (15; 2); (62; 3)}, {(4; 1); (18; 3); (41; 3); (48; 2)}, {(21; 1); (22; 1); (40; 3); (43; 3)}, {(35; 1); (52; 2); (56; 1); (57; 2)}, {(9; 1); (27; 1); (37; 1); (63; 2)}, {(23; 1); (38; 2); (49; 3); (∞1 ; 1)}, {(20; 1); (50; 2); (64; 3); (∞2 ; 1)}, {(5; 1); (7; 2); (44; 3); (∞3 ; 1)}, {(10; 1); (13; 3); (1; 2); (∞4 ; 1)}:
n = 18: {(27; 1); (33; 2); (34; 3); (43; 1)}, {(29; 1); (55; 1); (60; 1); (61; 3)}, {(16; 1); (20; 3); (44; 3); (52; 3)}, {(13; 1); (50; 2); (58; 1); (68; 2)}, {(19; 1); (24; 3); (54; 2); (65; 3)}, {(3; 1); (6; 1); (15; 1); (59; 3)}, {(18; 1); (47; 1); (63; 2); (67; 2)}, {(21; 1); (23; 1); (28; 2); (48; 2)}, {(5; 1); (14; 3); (26; 2); (40; 1)}, {(31; 1); (37; 1); (53; 3); (66; 3)}, {(2; 1); (25; 2); (45; 1); (56; 3)}, {(35; 1); (39; 2); (42; 1); (57; 1)}, {(11; 1); (30; 1); (41; 1); (51; 1)}, {(7; 1); (8; 1); (32; 3); (46; 3)}, {(10; 1); (12; 3); (38; 2); (∞1 ; 1)}, {(9; 1); (17; 2); (36; 3); (∞2 ; 1)}, {(49; 1); (62; 2); (64; 3); (∞3 ; 1)}, {(4; 1); (22; 2); (1; 3); (∞4 ; 1)}: n = 23: {(23; 1); (26; 1); (34; 1); (74; 2)}, {(29; 1); (54; 1); (64; 2); (82; 1)}, {(8; 1); (20; 3); (38; 3); (57; 1)}, {(18; 1); (46; 2); (75; 1); (77; 3)}, {(3; 1); (58; 2); (84; 3); (85; 2)}, {(2; 1); (37; 3); (83; 1); (87; 3)}, {(9; 1); (14; 1); (30; 1); (55; 3)}, {(17; 1); (62; 2); (71; 1); (72; 1)}, {(39; 1); (42; 3); (56; 3); (88; 3)},
{(11; 1); (16; 3); (31; 1); (41; 1)}, {(12; 1); (25; 1); (36; 3); (59; 3)}, {(22; 1); (35; 3); (80; 3); (86; 3)}, {(6; 1); (53; 1); (67; 3); (68; 1)}, {(21; 1); (47; 3); (49; 3); (70; 2)}, {(28; 1); (40; 2); (44; 2); (76; 1)}, {(10; 1); (48; 2); (60; 2); (81; 3)}, {(7; 1); (52; 3); (61; 3); (66; 1)}, {(19; 1); (27; 2); (43; 1); (50; 3)},
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{(15; 1); (24; 2); (32; 1); (51; 1)}, {(4; 1); (73; 2); (79; 3); (∞1 ; 1)}, {(45; 1); (63; 2); (65; 3); (∞2 ; 1)}, {(5; 1); (78; 2); (1; 3); (∞3 ; 1)}, {(13; 1); (33; 2); (69; 3); (∞4 ; 1)}: Here, we 8rst develop these blocks ( –, mod 3) to get a parallel class. Then, we develop this parallel class (mod 4(n − 1), – ) to get the RGDD as required. Lemma 3.5. There exists a 4-RGDD of type 246 . Proof. Let the point set be (Z10 ∪ {∞1 ; ∞2 }) × Z12 , and let the group set be {{j; j + 5} × Z12 : j = 1; : : : ; 5} ∪ {{∞1 ; ∞2 } × Z12 }. Below are the required base blocks, which are divided into 4 parts. {(2; 1); (3; 1); (4; 2); (5; 1)}, {(2; 1); (3; 6); (4; 12); (5; 9)}, {(∞1 ; 1); (6; 1); (7; 3); (8; 6)}, {(∞1 ; 1); (6; 2); (8; 8); (10; 4)}, {(∞2 ; 1); (9; 1); (10; 5); (1; 3)}, {(∞2 ; 1); (7; 2); (9; 9); (1; 6)}, {(2; 1); (3; 9); (6; 6); (8; 10)}, {(2; 1); (3; 8); (6; 1); (9; 11)}, {(∞1 ; 1); (4; 7); (7; 10); (10; 9)}, {(∞1 ; 1); (4; 11); (7; 5); (10; 12)}, {(∞2 ; 1); (5; 4); (9; 10); (1; 8)}, {(∞2 ; 1); (5; 7); (8; 11); (1; 12)}: Here, we 8rst develop the 3 blocks in each part ( –, mod 12) to get four parallel classes. Then, we develop these parallel classes (mod 10, – ) to get the RGDD as required. Lemma 3.6. For each n ∈{7,11}, there exists a 4-RGDD of type 24n . Proof. Let the point set be (Z4(n−1) ∪{∞1 ; ∞2 ; ∞3 ; ∞4 })×Z6 , and let the group set be {{j; j+(n−1); j+2(n−1); j+3(n−1)}×Z6 : j=1; : : : ; n−1}∪{{∞1 ; ∞2 ; ∞3 ; ∞4 }×Z6 }. Below are the required base blocks: n = 7:
{(3; 1); (10; 1); (14; 1); (1; 3)}, {(15; 1); (16; 3); (19; 3); (24; 3)}, {(5; 1); (8; 6); (12; 3); (22; 4)}, {(∞1 ; 1); (7; 1); (18; 2); (20; 3)}, {(∞2 ; 1); (2; 1); (4; 3); (23; 4)}, {(∞3 ; 1); (13; 1); (17; 6); (21; 4)}, {(∞4 ; 1); (6; 1); (9; 2); (11; 5)},
{(2; 1); (9; 5); (10; 2); (23; 5)}, {(3; 1); (16; 2); (17; 2); (1; 1)}, {(5; 1); (6; 2); (7; 6); (22; 6)}, {(∞1 ; 1); (8; 6); (11; 4); (15; 5)}, {(∞2 ; 1); (4; 5); (13; 6); (18; 2)}, {(∞3 ; 1); (12; 5); (20; 3); (21; 2)}, {(∞4 ; 1); (14; 6); (19; 3); (24; 4)}:
n = 11: {(4; 1); (15; 1); (16; 1); (17; 3)}, {(12; 1); (15; 1); (37; 1); (38; 6)}, {(21; 1); (28; 1); (35; 3); (40; 5)}, {(6; 1); (19; 6); (23; 1); (25; 6)}, {(6; 1); (9; 3); (14; 1); (25; 4)}, {(10; 1); (14; 4); (18; 4); (1; 1)}, {(11; 1); (27; 3); (29; 3); (32; 6)}, {(8; 1); (13; 1); (16; 6); (39; 4)}, {(2; 1); (3; 4); (18; 5); (24; 4)}, {(2; 1); (4; 4); (11; 3); (28; 1)}, {(7; 1); (10; 5); (22; 6); (38; 6)}, {(5; 1); (32; 1); (34; 5); (40; 2)}, {(26; 1); (30; 3); (37; 6); (1; 4)}, {(3; 1); (7; 6); (24; 5); (35; 3)}, {(∞1 ; 1); (12; 1); (19; 5); (34; 3)}, {(∞1 ; 1); (21; 2); (22; 6); (36; 4)}, {(∞2 ; 1); (8; 1); (13; 2); (20; 3)}, {(∞2 ; 1); (17; 6); (30; 4); (31; 5)},
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{(∞3 ; 1); (31; 1); (33; 2); (39; 3)}, {(∞3 ; 1); (20; 6); (26; 4); (29; 5)}, {(∞4 ; 1); (5; 1); (23; 2); (36; 3)}, {(∞4 ; 1); (9; 5); (27; 4); (33; 6)}: Here, we 8rst develop the blocks in each column ( –, mod 6) to get two parallel classes. Then, we develop these parallel classes (mod 4(n − 1), – ) to get the RGDD as required. Lemma 3.7. There exists a 4-RGDD of type 366 . Proof. Let the point set be (Z10 ∪ {∞1 ; ∞2 }) × Z18 , and let the group set be {{j; j + 5} × Z18 : j = 1; : : : ; 5} ∪ {{∞1 ; ∞2 } × Z18 }. Below are the required base blocks, which are divided into 6 parts. {(2; 1); (3; 1); (8; 1); (9; 2)}, {(8; 1); (9; 4); (10; 1); (1; 5)}, {(4; 1); (6; 5); (10; 14); (1; 8)}, {(∞1 ; 1); (4; 1); (7; 2); (1; 3)}, {(∞1 ; 1); (5; 4); (6; 9); (7; 7)}, {(∞1 ; 1); (2; 5); (3; 12); (5; 8)}, {(∞2 ; 1); (5; 1); (6; 3); (10; 5)}, {(∞2 ; 1); (2; 2); (3; 8); (4; 4)}, {(∞2 ; 1); (7; 6); (8; 14); (9; 13)}, {(4; 1); (7; 1); (10; 13); (1; 5)}, {(4; 1); (6; 13); (8; 8); (1; 14)}, {(2; 1); (4; 6); (8; 16); (10; 13)}, {(∞1 ; 1); (2; 6); (3; 15); (9; 17)}, {(∞1 ; 1); (2; 10); (5; 18); (9; 13)}, {(∞1 ; 1); (3; 14); (6; 16); (7; 11)}, {(∞2 ; 1); (5; 18); (6; 11); (8; 10)}, {(∞2 ; 1); (3; 16); (7; 12); (10; 7)}, {(∞2 ; 1); (5; 9); (9; 17); (1; 15)}:
Here, we 8rst develop the 3 blocks in each part ( –, mod 18) to get six parallel classes. Then, we develop these parallel classes (mod 10, – ) to get the RGDD as required.
4. Rees constructions In this section, we shall employ the elegant and powerful constructions established by Rees in [16], which are dramatically used in [17,18]. Construction 4.1 (Rees [16], Construciton 1). Let (X; G; B) be an A-resolvable k-GDD of type gu in which for each i ∈ A there are ri i -parallel classes of blocks. Suppose that there is a TD(u; h) admitting H as a group of automorphisms acting sharply transitively on the points of each group. Let Hj be a collection of subsets of H , there being ri such subsets of size i for each i ∈ A, and suppose that the collection {Hj ∗ : ∈ H; j = 1; 2; : : : ; i ri } is resolvable on H . Then there is a resolvable GDD of type (hg)u . Construction 4.2 (Rees [17], Corollary 2.2). If there is an -resolvable k-GDD of type gu and a TD(u; ), then there is a resolvable k-GDD of type ( g)u . Lemma 4.3. There exists a [1; 1; 2; 2; 4]-resolvable 4-GDD of type 66 . Hence, there exists a 5-resolvable and a {1; 9}-resolvable 4-GDD of the same type. Proof. Let the point set be Z36 and let the group set be {{j; j + 6; : : : ; j + 30}: j = 0; 1; : : : ; 5}. Below are the required base blocks. {1; 4; 11; 18}, {1; 12; 28; 33}, {1; 2; 6; 29}, {4; 7; 30; 32}, {1; 20; 21; 35}:
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Here, all the base blocks are developed by +2 mod 36. The 8rst block generates one parallel class when developed by +4 mod 36 and gives 2 such classes in total. Each of the second and the third block generates one 2-resolvable parallel class when developed by +2 mod 36. The remaining 2 base blocks generate a 4-resolvable parallel class. Lemma 4.4. There exist 4-RGDDs of types 306 and 546 . Proof. Apply Construction 4.2 with a 5-resolvable 4-GDD of type 66 coming from Lemma 4.3 and a TD(6; 5), we obtain a 4-RGDD of type 306 . For 4-RGDD of type 546 , apply Construction 4.1 with a {1; 9}-resolvable 4-GDD of type 66 coming from Lemma 4.3 and a TD(6; 9) by taking H = GF(9), H1 = {0} and H2 = H . Lemma 4.5. There exists a 4-RGDD of type (6h)6 for any h ¿ 7, (h; 5) = 1 and h ≡ 1; 5 (mod 6). Proof. Apply Construction 4.1 with a [1; 1; 2; 2; 4]-resolvable 4-GDD of type 66 coming from Lemma 4.3 and a TD(6; h). Since h ¿ 7, (h; 5) = 1 and h ≡ 1; 5 (mod 6), we may take H = Zh . Let H1 = H2 = {0}, H3 = H4 = {0; 1} and H5 = {0; 1; 2; 3}. It remains to be shown that the collection {Hj + : ∈ Zh ; j = 1; 2; 3; 4; 5} is resolvable. We need the following notations. De8ne Ma = {{2i + a + 1; 2i + a + 2}}. Then all the translates of H3 (and i=0;1;:::;(h−3)=2
H4 ) can be partitioned into 3 classes, where the 8rst class consists of (h − 1)=2 blocks of size 2 missing the point a, and the second class consists of (h − 1)=2 blocks of size 2 missing the point a + 1 and the third class has one block of the form {a; a + 1}. Let K be a subset of H . De8ne S(K) = H \ K. We consider the partition in the following three cases. When h = 12t + 5: {H5 + 4i + j: i ∈ {0; 1; : : : ; 3t}, {12t + 4 + j}} for j = 0; 1; 2; 3, M3 ∪ {{3}}, M4 ∪ {{4}}, M5 ∪ {{5}}, M6 ∪ {{6}}, {{12t + 4; 0; 1; 2}; {3; 4}; {5; 6}} ∪ S({12t + 4; 0; 1; 2; 3; 4; 5; 6}), and H . When h = 12t + 11: {H5 + 4i + j − 3 : i ∈ {0; 1; : : : ; 3t + 1}; {12t + 5 + j}, {12t + 6 + j; 12t + 7 + j}} for j = 0; 1; 2; 3, M12t+3 ∪ {{12t + 6; 12t + 7; 12t + 8; 12t + 9}} ∪ {{12t + 3}} \ {{12t + 6; 12t + 7}; {12t + 8; 12t + 9}}, M12t+4 ∪{{12t +7; 12t +8; 12t +9; 12t +10}}∪{{12t +4}}\{{12t +7; 12t +8}; {12t + 9; 12t + 10}}, {{12t + 5; 12t + 6; 12t + 7; 12t + 8}; {12t + 3; 12t + 4}} ∪ S({12t + 3; 12t + 4; 12t + 5; 12t + 6; 12t + 7; 12t + 8}), M0 ∪ {{0}}, M1 ∪ {{1}}, and
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{{0; 1}} ∪ S({0; 1}). When h = 6t + 1 and t ¿ 2: i=0; 1; :::; t−1 {{0; 1; 2; 3} + 6i + j} i=0; 1; :::; t−1 {{0; 1} + 6i + j + 4} ∪ {{6t + j}} for j = 0; 1; 2; 3; 4; 5, {{6t; 0; 1; 2}} ∪ M5 ∪ {{5}} \ {{6t; 0}; {1; 2}}, M6 ∪ {{6}}, {{6t; 0}; {1; 2}; {3; 4}; {5; 6}} ∪ S({6t; 0; 1; 2; 3; 4; 5; 6}), and H. When h = 7: {{0; 1; 2; 3} + j} ∪ {{4; 5} + j} ∪ {{6} + j} for j = 0; 1; : : : ; 6, {{0; 1}; {3; 4}; {5; 6}; {2}}, {{1; 2}; {4; 5}; {6; 0}; {3}}, and {2; 3} ∪ S({2; 3}). Theorem 4.6. If h ≡ 6 (mod 12) and h ¿ 18, then there exists a 4-RGDD of type h6 . Proof. For h ≡ 0 (mod 18), apply Construction 2.7 with a 4-RGDD of type 186 coming from Lemma 3.3 or a 4-RGDD of type 546 coming from Lemma 4.4. For h ≡ 0 (mod 18), apply Construction 2.7 with a 4-RGDD of type 306 coming from Lemma 4.4 or a 4-RGDD coming from Lemma 4.5. Theorem 4.7. If h ≡ 0 (mod 24), then there exists a 4-RGDD of type h9 . Proof. From Theorem 2.1, we have a 4-frame of type (h=8)9 , which is also a 8resolvable 4-GDD of the same type by Lemma 1.1 of [17]. Now, apply Construction 4.2 with = 8, k = 4 and u = 9. 5. 4-RGDD(hn ) with h ≡ 2; 10 (mod 12) In this section, we discuss the existence of 4-RGDD(hn ) with h ≡ 2; 10 (mod 12). First, we deal with the case when h = 10. Lemma 5.1. If n ∈{46,82,100,118,334}, then there exists a 4-RGDD of type 10n . Proof. Start from a 4-frame of type (90)t where t ∈ {5; 9; 11; 13; 37} coming from Theorem 2.1 and apply Construction 2.8. Adjoining 10 in8nite points and using 4-RGDD of type 1010 coming from Lemma 3.2 to 8ll in the holes gives the required 4-RGDDs of type 109t+1 . Lemma 5.2. If n ∈{70,130}, then there exists a 4-RGDD of type 10n . Proof. From Theorem 1.1, we have a 4-RGDD of type 100u for u ∈ {7; 13}. Start from this 4-RGDD and apply Construction 2.5 with a 4-RGDD of type 1010 coming from Lemma 3.2, we obtain the 4-RGDDs of type 1010u as desired.
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Lemma 5.3. There exist 4-RGDDs of types 10142 , 10184 and 10214 . Proof. First, from [12, Lemma 2.5] we have a 4-IRGDD of type 2(28; 4) . InNating it by an RTD(4; 5) gives a 4-IRGDD of type 10(28; 4) . Similarly, we can get 4-RGDDs of types 1016 and 1022 from 4-RGDDs of types 216 and 222 , respectively. For 4-RGDD(10142 ), take a TD(6; 20) and truncate one group to size 15. This gives a {5; 6}-GDD of type 205 151 . Apply Construction 2.6 with weight 12, adjoin 40 in8nite points and apply Construction 2.9 with a 4-RGDD(1022 ) and a 4-IRGDD of type 10(28; 4) , we then obtain the design as desired. For 4-RGDD(10184 ), take a TD(8; 11) and delete a block for a {7; 8}-GDD of type 8 10 . Now, give points in one group weight 12 and rest 24 for a 4-frame of type 2407 1201 . Add 40 in8nite points and 8ll in holes to obtain the desired design. Here, we need 4-frames of types 246 121 and 247 121 coming from Lemma 2.2, as well as a 4-frame of type 247 coming from Theorem 2.1. For 4-RGDD(10214 ), start from an RTD(9; 23) and truncate in 1 group and 3 blocks to obtain a {5; 6; 7; 8; 9}-GDD of type 208 151 . InNate by 12 to obtain a 4-frame of type 2408 1801 . Add 40 in8nite points and 8ll in holes to obtain the desired design. Lemma 5.4. If n ∈{178,202,238,250,346}, then there exists a 4-RGDD of type 10n . Proof. The proof is similar to that of Lemma 5.3. Here, we 8rst start from a 4-RGDD of type 1004 and apply Construction 2.5 with a 4-RGDD of type 1010 to get a 4-IRGDD of type 10(30+10; 10) . Then, take a TD(x + 1; 25) with x ∈ {5; 6; 7; 8; 11}, truncate one group to size y with y ∈ {0; 5; 10; 15}, apply Construction 2.6 with weight 12, adjoin 100 in8nite points and apply Construction 2.9, we obtain the design as desired. We list the suitable parameters such that n = 30x + 6y=5 + 10 below: n = 178: x = 5, n = 238: x = 7, n = 346: x = 11,
y = 15; n = 202: x = 6, y = 10; y = 15; n = 250: x = 8, y = 0; y = 5.
Lemma 5.5. Let h ≡ 2; 10 (mod 12) and h ¿ 14. If n ≡ 4 (mod 6) and n ∈ {10; 70; 82; 118; 142}, then there exists a 4-RGDD of type hn . Proof. For n = 3 · 7 + 1; 3 · 35 + 1, apply Construction 2.7 with a 4-RGDD of type 222 or 2106 coming from Theorem 1.1. For other values of n = 3s + 1 with odd s ¿ 5 and s ∈ {7; 23; 27; 35; 39; 47}, we have a 4-frame of type (3h)s from Theorem 2.1. Apply Construction 2.8 with a 4-RGDD(h4 ) coming from Lemma 2.3, we obtain the desired 4-RGDD of type h3s+1 . Lemma 5.6. Let h ≡ 2; 10 (mod 12) and h ¿ 14. If n ∈{118,142}, then there exists a 4-RGDD of type hn . Proof. For n = 142, starting from a 4-RGDD of type (4h)7 coming from Theorem 1.1 and applying Construction 2.5 with a 4-RGDD of type h4 coming from Lemma 2.3, we obtain a 4-IRGDD of type h(24+4; 4) . Then, take a TD(6; 2h) and truncate one group to
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size 3h=2. This gives a {5; 6}-GDD of type (2h)5 (3h=2)1 . Apply Construction 2.6 with weight 12, adjoin 4h in8nite points and apply Construction 2.9 with a 4-RGDD(h22 ) coming from Lemma 5.5 and a 4-IRGDD of type h(24+4; 4) , we then obtain the desired design. For n = 118, we 8rst construct a 5-GDD of type 88 121 on Z64 ∪ {∞1 ; ∞2 ; : : : ; ∞12 }. The required base blocks are listed below: {1; 36; 45; 48; 63}, {1; 2; 27; 32}, {1; 14; 20; 42}, {7; 11; 21; 64}: Here, all the base blocks are developed by +1 mod 64. The second block generates one parallel class when developed by +4 mod 64 and gives 4 such classes in total. The last two blocks together generate one parallel class when developed by +8 mod 64 and give 8 such classes in total. Hence, we have 12 parallel classes of blocks of size 4. This gives the desired 5-GDD of type 88 121 . Then, starting from a 4-RGDD of type (4h)4 and applying Construction 2.5 with a 4-RGDD of type h4 , we obtain a 4-IRGDD of type h(12+4; 4) . Take the above 5-GDD and apply Construction 2.6 with weight 3h=2, adjoin 4h in8nite points and apply Construction 2.9 with a 4-RGDD(h22 ) and a 4-IRGDD of type h(12+4; 4) , we obtain the desired design. Theorem 5.7. If h ≡ 2; 10 (mod 12), then the necessary conditions for the existence of a 4-RGDD(hn ), namely, n ¿ 4 and n ≡ 4 (mod 6) are also su
6. 4-RGDD(hn ) with h ≡ 6 (mod 12) In this section, we discuss the existence of 4-RGDD(hn ) with h ≡ 6 (mod 12). First, we improve the known results on the existence of 4-RGDD(6n ). Lemma 6.1. If n ∈{52,58,62,74,102,114,124}, then there exists a 4-RGDD of type 6n .
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Proof. For n ∈{52,62,102}, start from a 4-frame of type 60s with s ∈ {5; 6; 10} coming from Theorem 2.1, add 12 in8nite points and apply Construction 2.9 with a 4-RGDD(612 ) coming from Theorem 1.1 and a 4-IRGDD of type 6(10+2; 2) coming from Lemma 2.4 to 8ll in holes, we obtain the desired designs. For n = 58, start from a TD(6; 5) and truncate one group to 3 points to obtain a {5; 6}-GDD of type 55 31 . Applying Construction 2.6 with weight 12, adjoining 12 points and 8lling in holes with a 4-RGDD(68 ) and a 4-IRGDD of type 6(10+2; 2) , we obtain the desired design. For n = 74, start from a TD(6; 7), remove a block and use one of the deleted point to rede8ne the groups, we obtain a {5; 6}-GDD of type 56 61 . Applying Construction 2.6 with weight 12, adjoining 12 points and 8lling in holes with a 4-RGDD(614 ) and a 4-IRGDD of type 6(10+2; 2) , we obtain the desired design. For n = 114, start from an RTD(6; 11), remove 10 points from the 8rst group and take one truncated parallel class to rede8ne the groups, we obtain a {5; 6; 11}-GDD of type 510 61 . Applying Construction 2.6 with weight 12, adjoining 12 points and 8lling in holes with a 4-RGDD(614 ) and a 4-IRGDD of type 6(10+2; 2) , we obtain the desired design. For n = 124, start from a TD(6; 11), truncate one group to size 6 and use the deleted point to rede8ne the groups, we obtain a {5; 6; 11}-GDD of type 511 61 . Applying Construction 2.6 with weight 12, adjoining 12 points and 8lling in holes with a 4-RGDD(614 ) and a 4-IRGDD of type 6(10+2; 2) , we obtain the desired design. In the remainder of this section, we shall focus mainly on the existence of 4-RGDDs of type 18n . First, we establish a preliminary bound and then reduce it. Denote N18 = {n: a 4-RGDD of type 18n exists}. We need the following working lemma. Lemma 6.2. Let x and s be nonnegative integers. Suppose a TD(6 + x; m) exists. Suppose also that there exists a 4-RGDD(182b=3+s ) and 4-IRGDDs of types 18(2m=3+s; s) and 18(2ai =3+s; s) when x ¿ 0, where 0 6 b 6 m and 0 6 ai 6 m for i ∈ {1; 2; : : : ; x}. If n = (10m + 2(a1 + a2 + · · · + ax ))=3 + (2b=3 + s), then there exists a 4-RGDD of type 18n . Proof. Truncate any x + 1 groups in the TD(6 + x; m) to sizes a1 ; a2 ; : : : ; ax and b. This gives a {5; 6; : : : ; 6 + x}-GDD with groups of sizes m; a1 ; a2 ; : : : ; ax and b. Apply Construction 2.6 with weight 12, adjoin 18s in8nite points and apply Construction 2.9 with a 4-RGDD(182b=3+s ) and 4-IRGDDs of types 18(2m=3+s; s) and 18(2ai =3+s; s) , we then obtain the design as required. Here we need 4-frames of types 12u for u ∈ {5; 6; : : : ; 6 + x} as input designs, which all come from Theorem 2.1. The proof is complete. Lemma 6.3. Let T = {t: t ¿ 2; t = 11; 13}, x be an integer and x ¿ 0. Suppose a TD(6 + x; 9t) exists. Suppose also that there exists a 4-RGDD(182c+4 ), where 0 6 c 6 3t. If bi = 0 or bi 6 t and bi ∈ T for i ∈ {1; 2; : : : ; x}, then 30t + 6(b1 + b2 + · · · + bx ) + (2c + 4) ∈ N18 .
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Proof. First, for each t ∈ T apply Construction 2.9 with a 4-frame of type 542t+1 coming from Theorem 2.1 and a 4-RGDD of type 184 coming from Lemma 2.3, which is also a 4-IRGDD of type 18(4; 1) , we get a 4-IRGDD of type 18(6t+4; 4) . Then, apply Lemma 6.2 with m = 9t, s = 4, b = 3c and ai = 9bi for i ∈ {1; 2; : : : ; x}, we obtain the 4-RGDDs as required. Lemma 6.4. If n ¿ 256, then n ∈ N18 . Proof. From Lemma 2.3, we have a TD(8; 9t) for any t ¿ 8. Apply Lemma 6.3 with x = 2, t ¿ 8, c ∈ {0; 1; 2}, 0 6 bi 6 t and bi ∈ {1; 11; 13} for i ∈ {1; 2}, we then have [30t + 6 × 2 + 4; 30t + 6 × (2t − 4) + 4] = [30t + 16; 42t − 20] ⊂ N18 noting that [2; 2t − 4] ⊂ {b1 + b2 : 0 6 bi 6 t; bi = 1; 11; 13; i = 1; 2}. Here, we need a 4-RGDD of type 182c+4 for c ∈ {0; 1; 2} whose existence can be found in Lemma 2.3 or Lemma 3.3. It is not diIcult to check that the above intervals [30t + 16; 42t − 20] overlap when t runs over [8; ∞) \ {11; 13; 19}. This completes the proof. Lemma 6.5. If 118 6 n 6 254, then n ∈ N18 . Proof. The proof is similar to that of Lemma 6.4. Here, starting from a TD(15; 27) and applying Lemma 6.3 with x =9, t =3, c ∈ {0; 1; 2}, and bi ∈ {0; 2; 3} for i ∈ {1; 2; : : : ; 9}, we then have [18 · 5 + 6 × 4 + 4; 18 · 5 + 6 × 27 + 4] = [118; 256] ⊂ N18 noting that [4; 27] ⊂ {b1 +b2 +· · ·+b9 : bi ∈ {0; 2; 3}; i =1; 2; : : : ; 9}. This completes the proof. Now, we treat the small orders of n. Lemma 6.6. If n ∈{4,6,8,10,12,14,16,24,32,40,48}, then n ∈ N18 . Proof. For n=4, the design comes from Lemma 2.3. For n ∈{6,8,10,12,14}, the designs come from Lemma 3.3. For other values of n, from Lemma 2.3, we have a 4-RGDD of type (18m)4 for each m ∈ {4; 6; 8; 10; 12}. Start from these 4-RGDDs and apply Construction 2.5 with a 4-RGDD of type 18m , we obtain the 4-RGDDs of type 184m as desired. Lemma 6.7. If n ∈{22,26,34,46,50,52,56,58}, then n ∈ N18 . Proof. Start from a 4-frame of type (18s)t where s ∈ {3; 5; 7; 9} and t ∈ {5; 7; 11; 17; 19} coming from Theorem 2.1 and apply Construction 2.8. Adjoining 18 in8nite points and using 4-RGDDs of type 18u for u ∈ {4; 6; 8; 10} coming from Lemma 6.6 to 8ll in the holes, we obtain the required 4-RGDDs. The suitable parameters such that n = st + 1 are listed below: n = 22: s = 3, t = 7; n = 50: s = 7, n = 26: s = 5, t = 5; n = 52: s = 3, n = 34: s = 3, t = 11; n = 56: s = 5, n = 46: s = 9, t = 5; n = 58: s = 3,
t = 7; t = 17; t = 11; t = 19.
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Lemma 6.8. n ∈{20,28,30,36,42,44,54,60}, then n ∈ N18 . Proof. Apply Construction 2.10 with g = 18, u ∈ {4; 6; 12}, m = 1 and v ∈ {5; 7; 9; 11}. Here, we need 4-RGDDs of type 18u , 4-frames of type 18v and an RTD(4; v), which come from Lemma 6.6, Theorem 2.1 and Lemma 2.3, respectively. It is easy to see that u ¿ 4 ¿ 2 = m + 1, so we can get the designs as desired. The suitable parameters such that n = uv are listed below: n = 20: u = 4, v = 5; n = 42: u = 6, v = 7; n = 28: u = 4, v = 7; n = 44: u = 4, v = 11; n = 30: u = 6, v = 5; n = 54: u = 6, v = 9; n = 36: u = 4, v = 9; n = 60: u = 12, t = 5. Lemma 6.9. [64; 88] ⊂ N18 . Proof. First, starting from a 4-RGDD of type 724 coming from Lemma 2.3 and applying Construction 2.5 with a 4-RGDD of type 184 coming from Lemma 2.3, we obtain a 4-IRGDD of type 18(12+4; 4) . Then, take a TD(7; 9) and apply Construction 2.6 with weight 24 to the points of the 8rst 5 groups, weight 0 or 24 to all the points of the sixth group and weight 0; 12 or 24 to the points in the last group. We obtain a 4-frame of type 2165 (18u)1 or 2166 (18u)1 with u ∈ {0; 2; 4; : : : ; 12}. Adjoin 72 in8nite points and apply Construction 2.9 with a 4-RGDD of type 18u+4 and a 4-IRGDD of type 18(12+4; 4) , we obtain a 4-RGDD of type 1860+u+4 or 1872+u+4 . Here, we need the input 4-frames of types 245 , 245 121 , 246 , 246 121 and 247 , which come from Theorem 2.1 or Lemma 2.2. Lemma 6.10. [90; 116] ⊂ N18 . Proof. First, starting from a 4-RGDD of type 725 coming from Theorem 1.1 and applying Construction 2.5 with a 4-RGDD of type 184 coming from Lemma 2.3, we obtain a 4-IRGDD of type 18(16+4; 4) . Similarly, we can get a 4-IRGDD of type 18(12+4; 4) from 4-RGDDs of types 724 and 184 . Then, take a TD(8; 24) and apply Construction 2.6 with weight 12 to the points of the 8rst 5 groups, weight 12 to 18 or 24 points of both the sixth and the seventh group and weight 0 to the remaining points in these two groups, and weight 0 or 12 to the points in the last group. We obtain a 4-frame of type 2885 (18u)1 , 2885 2161 (18u)1 , 2885 2162 (18u)1 , 2886 2161 (18u)1 , or 2167 (18u)1 with u ∈ {0; 2; 4; : : : ; 12}. Adjoin 72 in8nite points and apply Construction 2.9 with a 4-RGDD of type 18u+4 and 4-IRGDDs of types 18(12+4; 4) and 18(16+4; 4) , we obtain the designs as desired. Theorem 6.11. If h ≡ 6 (mod 12), then the necessary conditions for the existence of a 4-RGDD(hn ), namely, n ¿ 4 and n ≡ 0 (mod 2) are also su
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Construction 2.7 with a 4-RGDD of type 6n and an RTD(4; h=6) coming from Lemma 2.3. For h ¿ 30 and n ∈ {4; 6}, the designs come from Lemma 2.3 and Theorem 4.6. This leaves h ¿ 30 and n ∈ {54; 68} to be considered. For h ¿ 30 and n = 54, starting from a 4-RGDD of type 129 coming from Theorem 1.1 and applying Construction 2.7 with an RTD(4; h=2), we obtain a 4-RGDD of type (6h)9 . Applying Construction 2.5 with a 4-RGDD of type h6 gives the desired designs. For h ¿ 30 and n = 68, starting from a 4-RGDD of type (4h)4 and applying Construction 2.5 with a 4-RGDD of type h4 , we obtain a 4-IRGDD of type h(12+4; 4) . Then, take a TD(6; h) and truncate one group into size h=3. Applying Construction 2.6 with weight 12 gives a 4-frame of type (12h)5 (4h)1 . Adjoining 4h in8nite points and applying Construction 2.9 with a 4-RGDD of type h8 and a 4-IRGDD of type h(12+4; 4) , we obtain the designs as desired. 7. 4-RGDD(hn ) with h ≡ 0 (mod 12) In this section, we discuss the existence of 4-RGDD(hn ) with h ≡ 0 (mod 12). From Construction 2.7, we need mainly work on the cases when h = 24; 36 or 72. Lemma 7.1. Let h ∈ {24; 36; 72}. If n ∈{4,5,6,7,8,10}, then there exists a 4-RGDD of type hn . Proof. For n ∈ {4; 5}, the required designs come from Theorem 1.1. For n = 6 and h ∈ {24; 36}, the required designs come from Lemma 3.5 or Lemma 3.7. For n = 6 and h = 72, apply Construction 2.7 with a 4-RGDD of type 186 coming from Lemma 3.3 and an RTD(4; 4). For n=7 and h=24, the required design comes from Lemma 3.6. For n = 7 and h ∈ {36; 72}, apply Construction 2.7 with a 4-RGDD of type 47 coming from Theorem 1.1. For n = 8 and h ∈ {24; 36; 72}, apply Construction 2.7 with a 4-RGDD of type 38 coming from Theorem 1.1. For n = 10 and h = 24, apply Construction 2.7 with a 4-RGDD of type 610 coming from Theorem 1.1. For n = 10 and h ∈ {36; 72}, apply Construction 2.7 with a 4-RGDD of type 410 coming from Theorem 1.1. Lemma 7.2. Let h ∈ {24; 36; 72}. Suppose a TD(7; m) exists. Suppose also there exist 4-frames of type ht for t = m, a2 and a 4-RGDD(ha1 +1 ), where 3 6 a1 ¡ m and 5 6 a2 6 m. If n = 5m + a1 + a2 + 1, then there exists a 4-RGDD of type hn . Proof. Truncate 2 groups in the TD(7; m) to sizes a1 and a2 . Take a deleted point from the group of size a1 to rede8ne groups. This gives a {5; 6; 7; m; a2 }-GDD with groups of sizes 5; 6 and a1 . Apply Construction 2.6 with weight h, add h in8nite points and apply Construction 2.8 with 4-RGDDs of types h6 , h7 coming from Lemma 7.1, and type ha1 +1 , we then get the design as desired. Here we also need 4-frames of types hu for u ∈ {5; 6; 7} as input designs, which all come from Theorem 2.1. The proof is complete. Lemma 7.3. Let h ∈ {24; 36; 72}. If n ¿ 84, then there exists a 4-RGDD of type hn .
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Proof. From Lemma 2.3, we have a TD(7; m) or a TD(7; m + 1) for any m ¿ 15. Apply Lemma 7.2 with m ¿ 15, a1 = 3 and a2 ∈ [5; 11] ∪ [13; m] or a1 = 4 and a2 = 11 noting that we do not have a 4-frame of type 3612 , we then obtain a 4-RGDD of type hn for each n ∈ [5m + 9; 6m + 4]. Here, we need a 4-RGDD of type h4 or h5 to 8ll in holes, which comes from Lemma 7.1. It is not diIcult to check that the intervals [5m + 9; 6m + 4] overlap when m runs over [15; ∞). Lemma 7.4. Let h ∈ {24; 36; 72}. If 49 6 n 6 83, then there exists a 4-RGDD of type hn . Proof. Start with a TD(10; 9) and truncate 5 groups to size ai , where ai ∈ {0; 3; 4; 5; 6; 7; 9} and 1 6 i 6 5. This gives a {5; 6; 7; 8; 9; 10}-GDD of type 95 a11 a12 · · · a15 . Give weight h to each point of this GDD to obtain a 4-frame of type (9h)5 (ha1 )1 (ha2 )1 · · · (ha5 )1 . Adding h in8nite points and using 4-RGDDs of types hu for u ∈ {4; 5; 6; 7; 8; 10} coming from Lemma 7.1 to 8ll in the holes gives the desired 4-RGDDs. Lemma 7.5. Let h ∈ {24; 36; 72}. If 39 6 n 6 48, then there exists a 4-RGDD of type hn . Proof. The proof is similar to that of Lemma 7.4. Here, we start with a TD(7; 7) and truncate 2 groups to size ai , where ai ∈ {0; 3; 4; : : : ; 7} and 1 6 i 6 2. Lemma 7.6. There exists a 4-RGDD of type h38 for h ∈ {24; 36; 72}. Proof. Start with a TD(6; 7) and delete 5 points from a block so as to form a {5; 6}-GDD of type 65 71 . Give weight h to the points of this GDD to obtain a 4-frame of type (6h)5 (7h)1 . Adding h in8nite points and using 4-RGDDs of types hu for u=7; 8 coming from Lemma 7.1 to 8ll in the holes gives the desired 4-RGDD of type h38 . Lemma 7.7. Let h ∈ {24; 36; 72}. If n ∈{19,21,22,25,26,29,31,33,34,37}, then there exists a 4-RGDD of type hn . Proof. Start from a 4-frame of type (hs)t where s ∈ {3; 4; 5} and t ∈ {5; 6; 7; 8; 9; 10; 11} coming from Theorem 2.1 and apply Construction 2.8. Adding h in8nite points and using 4-RGDDs of types hu for u = 4; 5; 6 coming from Lemma 7.1 to 8ll in the holes gives the desired 4-RGDDs. We list the suitable parameters such that n = st + 1 below: n = 19: n = 21: n = 22: n = 25: n = 26:
s = 3, s = 4, s = 3, s = 4, s = 5,
t = 6; t = 5; t = 7; t = 6; t = 5;
n = 29: n = 31: n = 33: n = 34: n = 37:
s = 4, s = 3, s = 4, s = 3, s = 4,
t = 7; t = 10; t = 8; t = 11; t = 9.
Lemma 7.8. Let h ∈ {24; 36; 72}. If n ∈{12,16,20,24,28,32,36}, then there exists a 4-RGDD of type hn .
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Proof. For each given n, a 4-RGDD of type 3n exists from Theorem 1.1. Apply Construction 2.7 to these 4-RGDDs with an RTD(4; h=3), we get the results as desired. Lemma 7.9. Let h ∈ {24; 36; 72}. If n ∈ {30; 35}, then there exists a 4-RGDD of type hn . Proof. Starting from a 4-RGDD of type (6h)5 or (7h)5 coming from Theorem 1.1 and apply Construction 2.5 with a 4-RGDD of type h6 or h7 coming from Lemma 7.1, we obtain the desired designs. Combining Lemma 7.1 and Lemmas 7.3–7.9 and applying Construction 2.7, we have the following result. Lemma 7.10. Let h ∈ {24; 36; 72; 120}. If n ¿ 4 and n ∈ {9; 11; 13; 14; 15; 17; 18; 23; 27}, then there exists a 4-RGDD of type hn . Lemma 7.11. If h ≡ 0 (mod 12) and h ∈ {12; 60}, then there exists a 4-RGDD of type h27 . Proof. For h = 180, take a TD(7; h=3) and truncate 2 groups to size h=4. This gives a {5; 6; 7}-GDD of type (h=3)5 (h=4)2 . Give weight 12 to each point of this GDD to obtain a 4-frame of type (4h)5 (3h)2 . Adding h in8nite points and using 4-RGDDs of types hu for u ∈ {4; 5} coming from Theorem 1.1 to 8ll in the holes gives the desired 4-RGDDs. For a 4-RGDD of type 18027 , apply Construction 2.7 with a 4-RGDD of type 3627 . Lemma 7.12. If h ≡ 24; 72 (mod 96), then there exists a 4-RGDD of type h15 . Proof. From [21], we have an 8-RGDD of type 815 , which can be extended to a 9-GDD of type 815 161 . Similarly, we can obtain a 5-GDD of type (h=8)8 (7h=24)1 from a 4-RGDD of type (h=8)8 coming from Theorem 1.1. Now, in the 9-GDD of type 815 161 we give weight h=8 to the points of groups of size 8 and weight 7h=24 to the points of the group of size 16, to obtain a 5-GDD of type h15 (14h=3)1 . Removing the distinguished group of this 5-GDD gives the 4-RGDD of type h15 as desired. Lemma 7.13. Let h ∈ {72; 120}. If n ∈{11,13,17,23}, then there exists a 4-RGDD of type hn . Proof. Start with a TD(h=12 + 1; n) and adjoin an in8nite point ∞ to the groups, then delete a 8nite point so as to form a {h=12 + 1; n + 1}-GDD of type (h=12)n n1 . Note that each block of size n + 1 intersects the group of size n in the in8nite point ∞ and each block of size h=12 + 1 intersects the group of size n, but certainly not in ∞. Now, in the group of size n, we give ∞ weight 4(n − 1) and give the remaining points weight h=3 − 4 for a total weight of h(n − 1)=3. Give all other points in the GDD weight 12. The result is a 5-GDD of type hn (h(n − 1)=3)1 , where we make use of input designs
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from 5-GDDs of types 12h=12 (h=3 − 4)1 (from a 4-RGGD(12h=12 )) and 12n (4(n − 1))1 (from a 4-RGDD(12n )). Removing the distinguished group of this 5-GDD gives the 4-RGDD of type hn as desired. Lemma 7.14. There exist 4-RGDDs of types 2414 , 2418 , 3613 , 7214 , 7218 , 12014 and 12018 . Proof. Starting from 4-RGDDs of types 614 , 618 , 413 and applying Construction 2.7, we obtain the designs as desired. Combining Theorem 1.1, Lemmas 3.4 and 3.6, Theorem 4.7 and Lemmas 7.10–7.14 and applying Construction 2.7, we have the following result. Theorem 7.15. If h ≡ 0 (mod 12), then the necessary conditions for the existence of a 4-RGDD(hn ), namely, n ¿ 4 are also su
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