Group Divisible Designs with Two Associate Classes and 𝜆₂=1

Abstract

The original classiffcation of PBIBDs defined group divisible designs GDD(𝑣=𝑣₁+𝑣₂+⋯+𝑣_𝑔,𝑔,𝑘,𝜆₁,𝜆₂) with 𝜆₁≠0. In this paper, we prove that the necessary conditions are suffcient for the existence of the group divisible designs with two groups of unequal sizes and block size three with 𝜆₂=1.

1. Introduction

A pairwise balanced design is an ordered pair (𝑆,ℬ), denoted PBD(𝑆,ℬ), where 𝑆 is a finite set of symbols, and ℬ is a collection of subsets of 𝑆 called blocks, such that each pair of distinct elements of 𝑆 occurs together in exactly one block of ℬ. Here |𝑆|=𝑣 is called the order of the PBD. Note that there is no condition on the size of the blocks in ℬ. If all blocks are of the same size 𝑘, then we have a Steiner system S(𝑣,𝑘). A PBD with index 𝜆 can be defined similarly; each pair of distinct elements occurs in 𝜆 blocks. If all blocks are same size, say 𝑘, then we get a balanced incomplete block design BIBD(𝑣,𝑏,𝑟,𝑘,𝜆). In other words, a BIBD(𝑣,𝑏,𝑟,𝑘,𝜆) is a set 𝑆 of 𝑣 elements together with a collection of 𝑏  𝑘-subsets of 𝑆, called blocks, where each point occurs in 𝑟 blocks, and each pair of distinct elements occurs in exactly 𝜆 blocks (see [1–3]).

Note that in a BIBD(𝑣,𝑏,𝑟,𝑘,𝜆), the parameters must satisfy the necessary conditions(1)𝑣𝑟=𝑏𝑘 and(2)𝜆(𝑣−1)=𝑟(𝑘−1).

With these conditions, a BIBD(𝑣,𝑏,𝑟,𝑘,𝜆) is usually written as BIBD(𝑣,𝑘,𝜆).

A group divisible design GDD(𝑣=𝑣₁+𝑣₂+⋯+𝑣_𝑔,𝑔,𝑘,𝜆₁,𝜆₂) is an ordered triple (𝑉,𝐺,ℬ), where 𝑉 is a 𝑣-set of symbols, 𝐺 is a partition of 𝑉 into 𝑔 sets of size 𝑣₁,𝑣₂,…,𝑣_𝑔, each set being called group, and ℬ is a collection of 𝑘-subsets (called blocks) of 𝑉, such that each pair of symbols from the same group occurs in exactly 𝜆₁ blocks, and each pair of symbols from different groups occurs in exactly 𝜆₂ blocks (see [1, 2, 4]). Elements occurring together in the same group are called first associates, and elements occurring in different groups we called second associates. We say that the GDD is defined on the set 𝑉. The existence of such GDDs has been of interest over the years, going back to at least the work of Bose and Shimamoto in 1952 who began classifying such designs [5]. More recently, much work has been done on the existence of such designs when 𝜆₁=0 (see [6] for a summary), and the designs here are called partially balanced incomplete block designs (PBIBDs) of group divisible type in [6]. The existence question for 𝑘=3 has been solved by Fu and Rodger [1, 2] when all groups are the same size.

In this paper, we continue to focus on blocks of size 3, solving the problem when the required designs having two groups of unequal size, namely, we consider the problem of determining necessary conditions for an existence of GDD(𝑣=𝑚+𝑛,2,3,𝜆₁,𝜆₂) and prove that the conditions are sufficient for some infinite families. Since we are dealing on GDDs with two groups and block size 3, we will use GDD(𝑚,𝑛;𝜆₁,𝜆₂) for GDD(𝑣=𝑚+𝑛,2,3,𝜆₁,𝜆₂) from now on, and we refer to the blocks as triples. We denote (𝑋,𝑌;ℬ) for a GDD(𝑚,𝑛;𝜆₁,𝜆₂) if 𝑋 and 𝑌 are 𝑚-set and 𝑛-set, respectively. Chaiyasena, et al. [7] have written a paper in this direction. In particular, they have completely solved the problem of determining all pairs of integers (𝑛,𝜆) in which a GDD(1,𝑛;1,𝜆) exists. We continue to investigate in this paper all triples of integers (𝑚,𝑛,𝜆) in which a GDD(𝑚,𝑛;𝜆,1) exists. We will see that necessary conditions on the existence of a GDD(𝑚,𝑛;𝜆₁,𝜆₂) can be easily obtained by describing it graphically as follows.

Let 𝜆𝐾_𝑣 denote the graph on 𝑣 vertices in which each pair of vertices is joined by 𝜆 edges. Let 𝐺₁ and 𝐺₂ be graphs. The graph 𝐺₁∨_𝜆𝐺₂ is formed from the union of 𝐺₁ and 𝐺₂ by joining each vertex in 𝐺₁ to each vertex in 𝐺₂ with 𝜆 edges. A 𝐺-decomposition of a graph 𝐻 is a partition of the edges of 𝐻 such that each element of the partition induces a copy of 𝐺. Thus the existence of a GDD(𝑚,𝑛;𝜆₁,𝜆₂) is easily seen to be equivalent to the existence of a 𝐾₃-decomposition of 𝜆₁𝐾_𝑚∨_𝜆2𝜆₁𝐾_𝑛. The graph 𝜆₁𝐾_𝑚∨_𝜆2𝜆₁𝐾_𝑛 is of order 𝑚+𝑛 and size 𝜆₁[(^𝑚₂)+(^𝑛₂)]+𝜆₂𝑚𝑛. It contains 𝑚 vertices of degree 𝜆₁(𝑚−1)+𝜆₂𝑛 and 𝑛 vertices of degree 𝜆₁(𝑛−1)+𝜆₂𝑚. Thus the existence of a 𝐾₃-decomposition of 𝜆₁𝐾_𝑚∨_𝜆2𝜆₁𝐾_𝑛 implies(1)3∣𝜆₁[(^𝑚₂)+(^𝑛₂)]+𝜆₂𝑚𝑛, and (2)2∣𝜆₁(𝑚−1)+𝜆₂𝑛 and 2∣𝜆₁(𝑛−1)+𝜆₂𝑚.

2. Preliminary Results

We will review some known results concerning triple designs that will be used in the sequel, most of which are taken from [3].

Theorem 2.1. Let 𝑣 be a positive integer. Then there exists a BIBD(𝑣,3,1) if and only if 𝑣≡1𝑜𝑟3(mod6).

A BIBD(𝑣,3,1) is usually called Steiner triple system and is denoted by STS(𝑣). Let (𝑉,ℬ) be an STS(𝑣). Then the number of triples 𝑏=|ℬ|=𝑣(𝑣−1)/6. A parallel class in an STS(𝑣) is a set of disjoint triples whose union is the set 𝑉. A parallel class contains 𝑣/3 triples, and, hence, an STS(𝑣) having a parallel class can exist only when 𝑣≡3(mod6). When the set ℬ can be partitioned into parallel classes, such a partition ℛ is called a resolution of the STS(𝑣), and the STS(𝑣) is called resolvable. If (𝑉,ℬ) is an STS(𝑣), and ℛ is a resolution of it, then (𝑉,ℬ,ℛ) is called a Kirkman triple system, denoted by KTS(𝑣), with (𝑉,ℬ) as its underlying STS. It is well known that a KTS(𝑣) exists if and only if 𝑣≡3(mod6). Thus if (𝑉,ℬ,ℛ) is a KTS(𝑣), then ℛ contains (𝑣−1)/2 parallel classes.

Theorem 2.2. There exists a PBD(6𝑘+5) with one block of size 5 and 6𝑘²+9𝑘 blocks of size 3.

Example 2.3. Let 𝑆={1,2,3,…,11}. Then PBD(11) is an ordered pair (𝑆,ℬ), where ℬ contains the following blocks: {}{}{}{}.1,2,3,4,5}{2,6,9}{3,7,8}{4,8,111,6,7}{2,7,11}{3,9,10}{5,6,81,8,9}{2,8,10}{4,6,10}{5,7,101,10,11}{3,6,11}{4,7,9}{5,9,11 (2.1)

A factor of a graph 𝐺 is a spanning subgraph. An 𝑟-factor of a graph is a spanning 𝑟-regular subgraph, and an 𝑟-factorization is a partition of the edges of the graph into disjoint 𝑟-factors. A graph 𝐺 is said to be 𝑟-factorable if it admits an 𝑟-factorization. In particular, a 1-factor is a perfect matching, and a 1-factorization of an 𝑟-regular graph 𝐺 is a set of 1-factors which partition the egde set of 𝐺. The following results are well known.

Theorem 2.4. The complete graph 𝐾_2𝑛 is 1-factorable, 𝐾_2𝑛+1 is 2-factorable, and 𝐾_3𝑛+1 is 3-factorable.

The following results on existence of 𝜆-fold triple systems are well known (see, e.g., [3]).

Theorem 2.5. Let 𝑛 be a positive integer. Then a BIBD(𝑛,3,𝜆) exists if and only if 𝜆 and 𝑛 are in one of the following cases:
(a)𝜆≡0(mod6) and 𝑛≠2, (b)𝜆≡1𝑜𝑟5(mod6) and 𝑛≡1𝑜𝑟3(mod6), (c)𝜆≡2𝑜𝑟4(mod6) and 𝑛≡0𝑜𝑟1(mod3), and (d)𝜆≡3(mod6) and 𝑛 is odd.

The results of Chaiyasena, et al. [7] will be useful, and we will state their results as follows.

Theorem 2.6. Let 𝑣 be a positive integer with 𝑣≥3. The spectrum of 𝜆, denoted 𝑆_1,𝑣 is defined as 𝑆_1,𝑣={𝜆∶aGDD(1,𝑣;1,𝜆)exists}. (2.2) Then

(a)𝑆_1,𝑣={1,3,5,…,𝑣−1} if 𝑣≡0(mod6),(b)𝑆_1,𝑣={6,12,18,…,𝑣−1} if 𝑣≡1(mod6),(c)𝑆_1,𝑣={1,7,13,…,𝑣−1} if 𝑣≡2(mod6),(d)𝑆_1,𝑣={2,4,6,…,𝑣−1} if 𝑣≡3(mod6),(e)𝑆_1,𝑣={3,9,15,…,𝑣−1} if 𝑣≡4(mod6), and(f)𝑆_1,𝑣={4,10,16,…,𝑣−1} if 𝑣≡5(mod6).

The following notations will be used throughout the paper for our constructions. (1)Let 𝑇={𝑥,𝑦,𝑧} be a triple and 𝑎∉𝑇. We use 𝑎∗𝑇 for three triples of the form {𝑎,𝑥,𝑦},{𝑎,𝑥,𝑧},{𝑎,𝑦,𝑧}. If 𝒯 is a set of triples, then 𝑎∗𝒯 is defined as {𝑎∗𝑇∶𝑇∈𝒯}. (2)Let 𝐺=⟨𝑉(𝐺),𝐸(𝐺)⟩ be a graph. If 𝑢,𝑣∈𝑉(𝐺), 𝑒=𝑢𝑣∈𝐸(𝐺), and 𝑎∉𝑉(𝐺), then we use 𝑎+𝑒 for the triple {𝑎,𝑢,𝑣}. We further use 𝑎+𝐸(𝐺) for the collection of triples 𝑎+𝑒 for all 𝑒∈𝐸(𝐺). In other words, 𝑎+𝐸(𝐺)∶={𝑎+𝑒∶𝑒∈𝐸(𝐺)}. (2.3) In particular, if ℱ={𝑥₁𝑦₁,𝑥₂𝑦₂,…,𝑥_𝑛𝑦_𝑛} is a 1-factor of 𝐾_2𝑛 and 𝑎 is not in the vertex set of 𝐾_2𝑛, then 𝑎+ℱ=𝑎,𝑥₁,𝑦₁,𝑎,𝑥₂,𝑦₂,…,𝑎,𝑥_𝑛,𝑦_𝑛. (2.4) If 𝐶_𝑚∶𝑥₁,𝑥₂,…,𝑥_𝑚+1=𝑥₁ is a cycle in 𝐾_𝑛, then 𝑎+𝐶_𝑚=𝑎,𝑥₁,𝑥₂,𝑎,𝑥₂,𝑥₃,…,𝑎,𝑥_𝑚−1,𝑥_𝑚,𝑎,𝑥_𝑚,𝑥₁. (2.5) Also if 𝐺 is a 2-regular graph and 𝑎∉𝑉(𝐺), then 𝑎+𝐸(𝐺) forms a collection of triples such that for each 𝑢∈𝑉(𝐺), there are exactly two triples in 𝑎+𝐸(𝐺) containing 𝑎 and 𝑢. In general if 𝐺 is an 𝑟-regular graph and 𝑎∉𝑉(𝐺), then 𝑎+𝐸(𝐺) forms a collection of triples such that for each 𝑢∈𝑉(𝐺), there are exactly 𝑟 triples in 𝑎+𝐸(𝐺) containing 𝑎 and 𝑢. (3)Let 𝑉 be a 𝑣-set. We use 𝐾(𝑉) for the complete graph 𝐾_𝑣 on the vertex set 𝑉. (4)Let 𝑉 be a 𝑣-set. Let STS(𝑉) be defined as STS(𝑉)={ℬ∶(𝑉,ℬ)isanSTS(𝑣)}. (2.6) KTS(𝑉) and BIBD(𝑉,3,𝜆) can be defined similarly, that is, KTS(𝑉)={ℬ∶(𝑉,ℬ)isaKTS(𝑣)},BIBD(𝑉,3,𝜆)={ℬ∶(𝑉,ℬ)isaBIBD(𝑣,3,𝜆)}. (2.7) Let 𝑋 and 𝑌 be disjoint sets of cardinality 𝑚 and 𝑛, respectively. We define GDD(𝑋,𝑌;𝜆₁,𝜆₂) as GDD𝑋,𝑌;𝜆₁,𝜆₂=ℬ∶(𝑋,𝑌;ℬ)isaGDD𝑚,𝑛;𝜆₁,𝜆₂. (2.8)(5)When we say that ℬ is a collection of subsets (blocks) of a 𝑣-set 𝑉, ℬ may contain repeated blocks. Thus, “∪” in our construction will be used for the union of multisets.

3. GDD(𝑚,𝑛;𝜆,1)

Let 𝜆 be a positive integer. We consider in this section the problem of determining all pairs of integers (𝑚,𝑛) in which a GDD(𝑚,𝑛;𝜆,1) exists. Recall that the existence of GDD(𝑚,𝑛;𝜆,1) implies 3∣𝜆[𝑚(𝑚−1)+𝑛(𝑛−1)]+2𝑚𝑛, 2∣𝜆(𝑚−1)+𝑛 and 2∣𝜆(𝑛−1)+𝑚. Let 𝑆(𝜆)∶={(𝑚,𝑛)∶aGDD(𝑚,𝑛;𝜆,1)exists}. (3.1)

By solving systems of linear congruences, we obtain the following necessary conditions.

Lemma 3.1. Let 𝑡 be a nonnegative integer.
(a)If (𝑚,𝑛)∈𝑆(6𝑡+1), then there exist nonnegative integers ℎ and 𝑘 such that {𝑚,𝑛}∈{{6𝑘+1,6ℎ+2},{6𝑘+1,6ℎ+6},{6𝑘+3,6ℎ+4},{6𝑘+3,6ℎ+6},{6𝑘+5,6ℎ+2},{6𝑘+5,6ℎ+4}}.(b)If (𝑚,𝑛)∈𝑆(6𝑡+2), then there exist nonnegative integers ℎ and 𝑘 such that {𝑚,𝑛}∈{{6𝑘+6,6ℎ+4},{6𝑘+6,6ℎ+6}}.(c)If (𝑚,𝑛)∈𝑆(6𝑡+3), then there exist nonnegative integers ℎ and 𝑘 such that {𝑚,𝑛}∈{{6𝑘+1,6ℎ+6},{6𝑘+3,6ℎ+2},{6𝑘+3,6ℎ+4},{6𝑘+3,6ℎ+6},{6𝑘+5,6ℎ+6}}.(d)If (𝑚,𝑛)∈𝑆(6𝑡+4), then there exist nonnegative integers ℎ and 𝑘 such that {𝑚,𝑛}∈{{6𝑘+2,6ℎ+2},{6𝑘+2,6ℎ+4},{6𝑘+6,6ℎ+4},{6𝑘+6,6ℎ+6}}.(e)If (𝑚,𝑛)∈𝑆(6𝑡+5), then there exist nonnegative integers ℎ and 𝑘 such that {𝑚,𝑛}∈{{6𝑘+1,6ℎ+6},{6𝑘+3,6ℎ+4},{6𝑘+3,6ℎ+6}}.(f)If (𝑚,𝑛)∈𝑆(6𝑡+6), then there exist nonnegative integers ℎ and 𝑘 such that {𝑚,𝑛}∈{{6𝑘+6,6ℎ+2},{6𝑘+6,6ℎ+4},{6𝑘+6,6ℎ+6}}.

In order to obtain sufficient conditions on an existence of GDD(𝑚,𝑛;𝜆,1), we first observe the following facts. (1)Let 𝑋 and 𝑌 be two disjoint sets of size 𝑚 and 𝑛, respectively. Then STS(𝑋∪𝑌)≠∅ if and only if GDD(𝑋,𝑌;1,1)≠∅. (2)Let 𝑋 and 𝑌 be two disjoint sets of size 𝑚 and 𝑛; respectively, and let 𝜆∈{2,3,4,5,6}. Then GDD(𝑋,𝑌;𝜆,1)≠∅ if STS(𝑋∪𝑌)≠∅, BIBD(𝑋,3,𝜆−1)≠∅, and BIBD(𝑌,3,𝜆−1)≠∅.

Thus, we have the following results.

Lemma 3.2. Let ℎ and 𝑘 be nonnegative integers. Then
(a)(6𝑘+1,6ℎ+6),(6𝑘+6,6ℎ+1),(6𝑘+3,6ℎ+6),(6𝑘+6,6ℎ+3),(6𝑘+3,6ℎ+4),(6𝑘+4,6ℎ+3), (6𝑘+1,6ℎ+2),(6𝑘+2,6ℎ+1),(6𝑘+5,6ℎ+2),(6𝑘+2,6ℎ+5),(6𝑘+5,6ℎ+4), (6𝑘+4,6ℎ+5)∈𝑆(1),(b)(6𝑘+1,6ℎ+6),(6𝑘+6,6ℎ+1),(6𝑘+3,6ℎ+6),(6𝑘+6,6ℎ+3),(6𝑘+3,6ℎ+4),(6𝑘+4,6ℎ+3)∈𝑆(3), and(c)(6𝑘+1,6ℎ+6),(6𝑘+6,6ℎ+1),(6𝑘+3,6ℎ+6),(6𝑘+6,6ℎ+3),(6𝑘+3,6ℎ+4), (6𝑘+4,6ℎ+3)∈𝑆(5).

Lemma 3.3. Let ℎ and 𝑘 be nonnegative integers. Then,
(a)(6𝑘+6,6ℎ+6)∈𝑆(2) and(b)(6𝑘+6,6ℎ+4),(6𝑘+4,6ℎ+6)∈𝑆(2).

Proof. (a) We first consider an existence of GDD(6,6;2,1), where the groups are 𝑋={1,2,3,4,5,6} and 𝑌={𝑎₁,𝑎₂,…,𝑎₆}. Let ℱ={ℱ₁,ℱ₂,…,ℱ₅} be a 1-factorization of 𝐾(𝑋). Let ℬ₁=⋃⁵_𝑖=1(𝑎_𝑖+ℱ_𝑖), ℬ₂∈STS(𝑋∪{𝑎₆}), and ℬ₃∈BIBD(𝑌,3,2). Then (𝑋,𝑌;ℬ) forms a GDD(6,6;2,1), where ℬ=ℬ₁∪ℬ₂∪ℬ₃. Thus (6,6)∈𝑆(2).
Let 𝑋 and 𝑌 be two sets of size 6𝑘+6 and 6ℎ+6, respectively. Suppose that 𝑘≤ℎ and ℎ≥1. Let 𝑎₁,𝑎₂,𝑎₃∈𝑌 and let 𝑌′=𝑌−{𝑎₁,𝑎₂,𝑎₃}. Thus, KTS(𝑌′)≠∅. Let ℬ₁∈KTS(𝑌′) with 𝒫₁,𝒫₂,…,𝒫_3ℎ+1 as its parallel classes. Since STS(𝑋∪𝑌′) and STS(𝑋∪{𝑎₁,𝑎₂,𝑎₃}) are not empty, there exist ℬ₂∈STS(𝑋∪𝑌′) and ℬ₃∈STS(𝑋∪{𝑎₁,𝑎₂,𝑎₃}). We now let ℬ as ³_𝑖=1𝑎_𝑖∗𝒫_𝑖∪^3ℎ+1_𝑖=4𝒫_𝑖∪ℬ₂∪ℬ₃∪𝑎₁,𝑎₂,𝑎₃. (3.2) Thus, (𝑋,𝑌;ℬ) forms a GDD(6𝑘+6,6ℎ+6;2,1) and (6𝑘+6,6ℎ+6)∈𝑆(2).
(b) Let 𝑋 and 𝑌 be two sets of size 6𝑘+6 and 6ℎ+4, respectively, 𝑎∈𝑌 and let 𝑌′=𝑌−{𝑎}. Choose ℬ₁∈KTS(𝑌′) with 𝒫₁,𝒫₂,…,𝒫_3ℎ+1 as its parallel classes. Since STS(𝑋∪𝑌′) and STS(𝑋∪{𝑎}) are not empty, there exist ℬ₂∈STS(𝑋∪𝑌′) and ℬ₃∈STS(𝑋∪{𝑎}). We now let ℬ as ℬ₂∪ℬ₃∪𝑎∗𝒫₁∪^3ℎ+1_𝑖=2𝒫_𝑖. (3.3) Thus, (𝑋,𝑌;ℬ) forms a GDD(6𝑘+6,6ℎ+4;2,1) and (6𝑘+6,6ℎ+4)∈𝑆(2).
Therefore, the proof is complete.

Part of the proof of the following lemma is based on an existence of GDD(4,4;2,3) which we now construct. Let 𝐴={𝑎,𝑏,𝑐,𝑑} and 𝐵={1,2,3,4}. Then it is easy to check that 𝐹∈GDD(𝐴,𝐵;2,3), where 𝐹={{1,𝑎,𝑏},{1,𝑎,𝑐},{1,𝑎,𝑑},{2,𝑏,𝑐},{2,𝑏,𝑑},{2,𝑏,𝑎},{3,𝑐,𝑑},{3,𝑐,𝑎},{3,𝑐,𝑏},{4,𝑑,𝑎}, {4,𝑑,𝑏},{4,𝑑,𝑐},{𝑎,2,3},{𝑎,2,4},{𝑎,3,4},{𝑏,1,3},{𝑏,1,4},{𝑏,3,4},{𝑐,1,2},{𝑐,1,4}, {𝑐,2,4},{𝑑,1,2},{𝑑,1,3},{𝑑,2,3}}.

Lemma 3.4. Let ℎ and 𝑘 be nonnegative integers. Then(6𝑘+2,6ℎ+3),(6𝑘+3,6ℎ+2),(6𝑘+5,6ℎ+6),(6𝑘+6,6ℎ+5)∈𝑆(3) (3.4)

Proof. Case 1. Let 𝑋_𝑘 be a (6𝑘+2)-set containing 𝑎₁,𝑎₂, and 𝑌_ℎ be a (6ℎ+3)-set containing 1,2,3.
Subcase 1 (k = 0). Let ℬ₀={{1,2,3},{1,2,3},{1,2,3},{1,𝑎₁,𝑎₂},{2,𝑎₁,𝑎₂},{3,𝑎₁,𝑎₂}}, and we can see that ℬ₀∈GDD(𝑋₀,𝑌₀;3,1). Suppose that ℎ≥1. Since 𝑋₀∪𝑌_ℎ is a set of size 6ℎ+5, it follows by Theorem 2.2, that there exists a PBD(6ℎ+5), (𝑋₀∪𝑌_ℎ,ℬ₁), in which {1,2,3,𝑎₁,𝑎₂}∈ℬ₁ and 6ℎ²+9ℎ triples in ℬ₁. Let ℬ^′₁=ℬ₁−{{1,2,3,𝑎,𝑏}}. Since 𝑌_ℎ is a (6ℎ+3)-set, it follows, by Theorem 2.5(c), that BIBD(𝑌_ℎ,3,2)≠∅. Let ℬ₂∈BIBD(𝑌_ℎ,3,2). It is easy to see that (𝑋₀,𝑌_ℎ;ℬ) forms a GDD(2,6ℎ+3;3,1), where ℬ=ℬ₀∪ℬ^′₁∪ℬ₂. (3.5)
Subcase 2 (k = 1). A GDD(8,6ℎ+3;3,1) can be constructed as follows. Let 𝑋=𝐴∪𝐵, where 𝐴 and 𝐵 are sets of size four. It is clear that STS(𝑌_ℎ), STS(𝐴∪𝑌_ℎ), and STS(𝐵∪𝑌_ℎ) are not empty. It has been shown above that GDD(𝐴,𝐵;2,3) is not empty. We now choose ℬ₁∈STS(𝑌_ℎ), ℬ₂∈STS(𝐴∪𝑌_ℎ), ℬ₃∈STS(𝐵∪𝑌_ℎ), and ℬ₄∈GDD(𝐴,𝐵;2,3), and let ℬ=ℬ₁∪ℬ₂∪ℬ₃∪ℬ₄. Then, (𝑋,𝑌_ℎ;ℬ) form a GDD(8,6ℎ+3;3,1).
Subcase 3 (k = 2). We first consider the existence of GDD(4,10;2,3) with 𝐴={0,1,2,…,9} and 𝐵={𝑎₀,𝑎₁,𝑎₂,𝑎₃}. Let 𝐾(𝐴) be the complete graph of order 10 with 𝐴 as its vertex set. It is well known that 𝐾₁₀ is 1-factorable. In other words, 𝐾₁₀ can be decomposed as a union of nine edge-disjoint 1-factors. Consequently, 𝐾₁₀ can be decomposed as a union of three edge-disjoint 3-factors. Also, 𝐾₁₀ can be decomposed as a union of 10𝐶₃ and a 3-factor: ten triples {{𝑥,𝑥+1,𝑥+3}∶𝑥=0,1,…,9} and a 3-factor ℱ₀ of 𝐾₁₀, where 𝐸ℱ₀=⁹_𝑖=0{{𝑖,𝑖+4},{𝑖,𝑖+5},{𝑖,𝑖+6}}, (3.6) reducing arithmetic operations (mod10). Therefore, 2𝐾₁₀ can be decomposed as a union of 10𝐶₃ and four 3-factors.
Let ℱ₁,ℱ₂,ℱ₃ be a 3-factorization of 𝐾₁₀ and 10𝐶₃ and ℱ₀ as described above. Then (𝐴,𝐵;ℬ) forms a GDD(10,4;2,3), where the collection ℬ=10𝐶₃∪³_𝑖=0𝑎_𝑖+ℱ_𝑖∪ℬ₁ (3.7) with ℬ₁={{𝑎₀,𝑎₁,𝑎₂},{𝑎₁,𝑎₂,𝑎₃},{𝑎₂,𝑎₃,𝑎₀},{𝑎₃,𝑎₀,𝑎₁}}.
A GDD(14,6ℎ+3;3,1) can be constructed as follows. Let 𝑋=𝐴∪𝐵, where 𝐴 and 𝐵 are sets of size ten and four, respectively. It is clear that STS(𝑌_ℎ), STS(𝐴∪𝑌_ℎ), and STS(𝐵∪𝑌_ℎ) are not empty. It has been shown above that GDD(𝐴,𝐵;2,3) is not empty. We now choose ℬ₁∈STS(𝑌_ℎ), ℬ₂∈STS(𝐴∪𝑌_ℎ), ℬ₃∈STS(𝐵∪𝑌_ℎ), and ℬ₄∈GDD(𝐴,𝐵;2,3) and let ℬ=ℬ₁∪ℬ₂∪ℬ₃∪ℬ₄. Then (𝑋,𝑌_ℎ;ℬ) form a GDD(14,6ℎ+3;3,1).
Subcase 4 (k ≥ 3). Let 𝐴={𝑎₁,𝑎₂,𝑎₃,𝑎₄,𝑎₅}. Suppose that 𝐴⊆𝑋_𝑘 and 𝑋′=𝑋_𝑘−𝐴. Since 𝑋′ is a (6𝑘−3)-set, it follows that STS(𝑋′)≠∅ and KTS(𝑋′)≠∅. Choose ℬ₁∈ STS(𝑋′) and let 𝒦∈KTS(𝑋′) with 𝒫₁,𝒫₂,…,𝒫_3𝑘−2 as its parallel classes. Let ℬ₂=⋃⁵_𝑖=1(𝑎_𝑖∗𝒫_𝑖)∪⋃^3𝑘−1_𝑖=6𝒫_𝑖. Since 𝑋_𝑘∪𝑌_ℎ is a set of size 6(𝑘+ℎ)+5, we choose a PBD(6(𝑘+ℎ)+5), (𝑋_𝑘∪𝑌_ℎ,ℬ₃), as in Theorem 2.2 in which 𝐴∈ℬ₃. Let ℬ^′₃=𝐵₃−{𝐴}. Since 𝐴 is a 5-set and 𝑌_ℎ is a (6ℎ+3)-set, it follows, by Theorem 2.5(c) and (d), that there exist ℬ₄∈BIBD(5,3,3) and ℬ₅∈BIBD(𝑌_ℎ,3,2). Thus, we can see that (𝑋_𝑘,𝑌_ℎ;ℬ) forms a GDD(6𝑘+2,6ℎ+3;3,1), where ℬ=ℬ₁∪ℬ₂∪ℬ^′₃∪ℬ₄∪ℬ₅. (3.8)
Case 2. We now suppose that 𝑋 and 𝑌 be sets of size 6𝑘+5 and 6ℎ+6, respectively. We suppose further that 𝑎∈𝑋 and 𝑋′=𝑋−{𝑎}. By Lemma 3.3(b), we have GDD(𝑋′,𝑌;2,1)≠∅. Choose ℬ₁∈GDD(𝑋′,𝑌;2,1) and ℬ₂∈STS(𝑌∪{𝑎}). By Theorem 2.6(e) that GDD({𝑎},𝑋′;1,3)≠∅. Choose ℬ₃∈GDD({𝑎},𝑋′;1,3). Let ℬ=ℬ₁∪ℬ₂∪ℬ₃, and it is easy to see that ℬ∈GDD(𝑋,𝑌;3,1).
Thus, (6𝑘+5,6ℎ+6)∈𝑆(3).

Lemma 3.5. Let ℎ and 𝑘 be nonnegative integers. Then
(a)(6𝑘+6,6ℎ+6)∈𝑆(4) and (6𝑘+6,6ℎ+6)∈𝑆(6),(b)(6𝑘+6,6ℎ+4),(6𝑘+4,6ℎ+6)∈𝑆(4) and (6𝑘+6,6ℎ+4),(6𝑘+4,6ℎ+6)∈𝑆(6),(c)(6𝑘+2,6ℎ+2) withℎ,𝑘not both zero, (6𝑘+2,6ℎ+4),(6𝑘+4,6ℎ+2)∈𝑆(4), and(d)(6𝑘+6,6ℎ+2),(6𝑘+2,6ℎ+6)∈𝑆(6).

Proof. The proofs of (a) and (b) follow from the results of Lemma 3.3(a), and (b), respectively, and Theorem 2.5(c).
(c) We have the following cases.
Case 1 (6k + 2, 6h + 2). Let 𝑋_𝑘 be a (6𝑘+2)-set and 𝑌_ℎ be a (6ℎ+2)-set. It is clear that GDD(𝑋₀,𝑌₀;4,1)=∅. We now construct a GDD(2,8;4,1), (𝑋₀,𝑌₁;ℬ), with 𝑋₀={𝑥,𝑦}, 𝑌₁={𝑎₁,𝑎₂,…,𝑎₈}, 𝐴={𝑎₁,𝑎₂,𝑎₃}, 𝑌^′₁=𝑌₁−𝐴, and ℬ=ℬ₁∪ℬ₂∪ℬ₃∪ℬ₄, where ℬ₁∈BIBD(𝐴,3,4), ℬ₂∈STS(𝑋∪𝑌^′₁), ℬ₃=⋃³_𝑖=1(𝑎_𝑖+𝐸(𝐾(𝑌^′₁))), and ℬ₄={{𝑎_𝑖,𝑥,𝑦}∶𝑖=1,2,3}. We now construct a GDD(6𝑘+2,6ℎ+2;4,1), (𝑋_𝑘,𝑌_ℎ;ℬ), in general case, where 𝑘≥0 and ℎ≥1. We first let 𝐴={𝑎₁,𝑎₂,𝑎₃}⊆𝑌_ℎ, 𝑌^′_ℎ=𝑌_ℎ−𝐴, and we will use a result on the existence of GDD(1,6ℎ−1;1,4) which has been shown in Theorem 2.6(f), namely, GDD({𝑎},𝑌^′_ℎ;1,4)≠∅. Therefore, we can choose ℬ_𝑖∈GDD({𝑎_𝑖},𝑌^′_ℎ;1,4),ℬ₄∈BIBD(𝐴,3,4),ℬ₅∈STS(𝑋_𝑘∪𝑌^′_ℎ), and ℬ₆=⋃³_𝑖=1𝐹_𝑖, where 𝐹_𝑖∈STS(𝑋_𝑘∪{𝑎_𝑖}). We can see that ℬ∈GDD(𝑋_𝑘,𝑌_ℎ;4,1), where ℬ=ℬ₁∪ℬ₂∪ℬ₃∪ℬ₄∪ℬ₅∪ℬ₆.
Case 2 (6k + 2, 6h + 4). Let 𝑋_𝑘 be a (6𝑘+2)-set and 𝑌_ℎ be a (6ℎ+4)-set. It is easy to see that GDD(𝑋₀,𝑌₀;4,1)≠∅ by constructing (𝑋₀,𝑌₀;ℬ) as follows. Let 𝑋₀={𝑎,𝑏}, 𝑌₀={1,2,3,4}, and ℬ=⋃⁴_𝑖=1{{𝑖,𝑎,𝑏}}∪𝒫₂∪𝒫₂, where 𝒫₂={{1,2,3},{1,2,4},{1,3,4},{2,3,4}}. We now turn to more general cases. Suppose that 𝑎∈𝑌_ℎ and 𝑌^′_ℎ=𝑌_ℎ−{𝑎}. Since 𝑌^′_ℎ is a (6ℎ+3)-set, it follows that KTS(𝑌^′_ℎ)≠∅. Choose ℬ₁∈KTS(𝑌^′_ℎ) with parallel classes 𝒫₁,𝒫₂,…,𝒫_3ℎ+1. Let ℬ₂=(𝑎∗𝒫₁)∪(𝑎∗𝒫₂)∪(⋃^3ℎ+1_𝑖=3𝒫_𝑖). We have shown in Lemma 3.4(d) that GDD(𝑋_𝑘,𝑌^′_ℎ;3,1)≠∅. Choose ℬ₃∈GDD(𝑋_𝑘,𝑌^′_ℎ;3,1) and ℬ₄∈STS(𝑋_𝑘∪{𝑎}). We can see that ℬ∈GDD(𝑋_𝑘,𝑌_ℎ;4,1), where ℬ=ℬ₂∪ℬ₃∪ℬ₄.
(d) Let 𝑋_𝑘 be a (6𝑘+6)-set and 𝑌_ℎ be a (6ℎ+2)-set. Let 𝑋₀={𝑎₁,𝑎₂,…,𝑎₆} and 𝑌₀={𝑎,𝑏}. Let ℬ₁={{𝑎_𝑖,𝑎,𝑏}∶𝑖=1,2,…,6}, ℬ₂∈BIBD(𝑋₀,3,6). Then ℬ₁∪ℬ₂∈GDD(𝑋₀,𝑌₀;6,1).
Next we will show that GDD(𝑋_𝑘,𝑌₀;6,1)≠∅ by letting 𝑋_𝑘={𝑎₁,𝑎₂,…,𝑎_6𝑘+6}, 𝑌₀={𝑎,𝑏}, 𝐴={𝑎₁,𝑎₂,…,𝑎₅} and 𝑋^′_𝑘=𝑋_𝑘−𝐴. Let ℬ₁={{𝑎_𝑖,𝑎,𝑏}∶𝑖=1,2,…,5}, ℬ₂∈STS({𝑎,𝑏}∪𝑋^′_𝑘), and ℬ₃∈BIBD(𝐴,3,6). Theorem 2.6(b) shows an existence of a GDD(1,6ℎ+1;1,6). Let ℬ₄=⋃⁵_𝑖=1ℬ^′_𝑖, where ℬ^′_𝑖∈GDD({𝑎_𝑖},𝑋^′_𝑘;1,6). It is easy to check that ℬ∈GDD(𝑋_𝑘,𝑌₀;6,1), where ℬ=ℬ₁∪ℬ₂∪ℬ₃∪ℬ₄.
Finally, let ℎ≥1, 𝑎∈𝑌 and 𝑌^′_ℎ=𝑌_ℎ−{𝑎}. We now choose ℬ₁∈BIBD(𝑋_𝑘,3,4), ℬ₂∈BIBD(𝑌^′_ℎ,3,4), ℬ₃∈STS(𝑋_𝑘∪𝑌^′_ℎ), and ℬ₄∈STS(𝑋_𝑘∪{𝑎}). By Theorem 2.6(b) that GDD({𝑎},𝑌^′_ℎ;1,6)≠∅. Choose ℬ₅∈GDD({𝑎},𝑌^′_ℎ;1,6). Thus, we can check that ℬ∈GDD(𝑋_𝑘,𝑌_ℎ;6,1), where ℬ=ℬ₁∪ℬ₂∪ℬ₃∪ℬ₄∪ℬ₅. Thus (6𝑘+6,6ℎ+2)∈𝑆(6) for all positive integers ℎ,𝑘.

Now we have an existence of a GDD(𝑚,𝑛;𝑟,1) for 𝑟=1,2,…,6 whenever 𝑚 and 𝑛 are not equal to 2, so we can readily extend to any 6𝑡+𝑟 by the following lemma.

Lemma 3.6. Let 𝑚 and 𝑛 be positive integers with 𝑚≠2 and 𝑛≠2. If there exists a GDD(𝑚,𝑛;𝑟,1) with 𝑟≥1, then a GDD(𝑚,𝑛;6𝑡+𝑟,1),𝑡≥0, exists

Proof. Let 𝑋 be an 𝑚-set and 𝑌 be an 𝑛-set. By assumption we have GDD(𝑋,𝑌;𝑟,1)≠∅. Choose ℬ₁∈GDD(𝑋,𝑌;𝑟,1). Since 𝑚 and 𝑛 are not equal to 2, by Theorem 2.5(a) there exist ℬ₂∈BIBD(𝑋,3,6𝑡) and ℬ₃∈BIBD(𝑌,3,6𝑡). It is easy to see that (𝑋,𝑌;ℬ) forms a GDD(𝑚,𝑛;6𝑡+𝑟,1), where ℬ=ℬ₁∪ℬ₂∪ℬ₃. Thus (𝑚,𝑛)∈𝑆(6𝑡+𝑟) with 𝑟≥1.

Finally, we have the main result as in the following.

Theorem 3.7. Let 𝑚 and 𝑛 be positive integers with 𝑚≠2 and 𝑛≠2. There exists a GDD(𝑚,𝑛;𝜆,1),𝜆≥1 if and only if
(1)3∣𝜆[𝑚(𝑚−1)+𝑛(𝑛−1)]+2𝑚𝑛 and(2)2∣𝜆(𝑚−1)+𝑛 and 2∣𝜆(𝑛−1)+𝑚.

Proof. The proof follows from Lemmas 3.1–3.6.

Achnowledgment

N. Pabhapote is supported by the University of the Thai Chamber of Commerce, and N. Punnim is supported by the Thailand Research Fund.