Group Divisible Designs with Two Associate Classes and
The original classiffcation of PBIBDs defined group divisible designs GDD() with . 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 .
A pairwise balanced design is an ordered pair , denoted PBD, where is a finite set of symbols, and is a collection of subsets of called blocks, such that each pair of distinct elements of occurs together in exactly one block of . Here is called the order of the PBD. Note that there is no condition on the size of the blocks in . If all blocks are of the same size , then we have a Steiner system S. A PBD with index can be defined similarly; each pair of distinct elements occurs in blocks. If all blocks are same size, say , then we get a balanced incomplete block design BIBD. In other words, a BIBD is a set of elements together with a collection of -subsets of , called blocks, where each point occurs in blocks, and each pair of distinct elements occurs in exactly blocks (see [1–3]).
Note that in a BIBD, the parameters must satisfy the necessary conditions(1) and(2).
With these conditions, a BIBD is usually written as BIBD.
A group divisible design GDD is an ordered triple , where is a -set of symbols, is a partition of into sets of size , each set being called group, and is a collection of -subsets (called blocks) of , such that each pair of symbols from the same group occurs in exactly blocks, and each pair of symbols from different groups occurs in exactly 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 . 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 . More recently, much work has been done on the existence of such designs when (see  for a summary), and the designs here are called partially balanced incomplete block designs (PBIBDs) of group divisible type in . The existence question for 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 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 for GDD from now on, and we refer to the blocks as triples. We denote for a GDD if and are -set and -set, respectively. Chaiyasena, et al.  have written a paper in this direction. In particular, they have completely solved the problem of determining all pairs of integers in which a GDD exists. We continue to investigate in this paper all triples of integers in which a GDD exists. We will see that necessary conditions on the existence of a GDD can be easily obtained by describing it graphically as follows.
Let denote the graph on vertices in which each pair of vertices is joined by edges. Let and be graphs. The graph is formed from the union of and by joining each vertex in to each vertex in with edges. A -decomposition of a graph is a partition of the edges of such that each element of the partition induces a copy of . Thus the existence of a GDD is easily seen to be equivalent to the existence of a -decomposition of . The graph is of order and size . It contains vertices of degree and vertices of degree . Thus the existence of a -decomposition of implies(1), and (2) and .
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 .
Theorem 2.1. Let be a positive integer. Then there exists a BIBD if and only if .
A BIBD is usually called Steiner triple system and is denoted by STS. Let be an STS. Then the number of triples . A parallel class in an STS is a set of disjoint triples whose union is the set . A parallel class contains triples, and, hence, an STS having a parallel class can exist only when . When the set can be partitioned into parallel classes, such a partition is called a resolution of the STS, and the STS is called resolvable. If is an STS, and is a resolution of it, then is called a Kirkman triple system, denoted by KTS, with as its underlying STS. It is well known that a KTS exists if and only if . Thus if is a KTS, then contains parallel classes.
Theorem 2.2. There exists a PBD with one block of size 5 and blocks of size 3.
Example 2.3. Let . Then PBD(11) is an ordered pair , where contains the following blocks:
A factor of a graph is a spanning subgraph. An -factor of a graph is a spanning -regular subgraph, and an -factorization is a partition of the edges of the graph into disjoint -factors. A graph is said to be -factorable if it admits an -factorization. In particular, a 1-factor is a perfect matching, and a 1-factorization of an -regular graph is a set of 1-factors which partition the egde set of . The following results are well known.
Theorem 2.4. The complete graph is 1-factorable, is 2-factorable, and is 3-factorable.
The following results on existence of -fold triple systems are well known (see, e.g., ).
Theorem 2.5. Let be a positive integer. Then a BIBD exists if and only if and are in one of the following cases:
(a) and , (b) and , (c) and , and (d) and is odd.
The results of Chaiyasena, et al.  will be useful, and we will state their results as follows.
Theorem 2.6. Let be a positive integer with . The spectrum of , denoted is defined as Then(a) if ,(b) if ,(c) if ,(d) if ,(e) if , and(f) if .
The following notations will be used throughout the paper for our constructions. (1)Let be a triple and . We use for three triples of the form . If is a set of triples, then is defined as . (2)Let be a graph. If , , and , then we use for the triple . We further use for the collection of triples for all . In other words, In particular, if is a 1-factor of and is not in the vertex set of , then If is a cycle in , then Also if is a 2-regular graph and , then forms a collection of triples such that for each , there are exactly two triples in containing and . In general if is an -regular graph and , then forms a collection of triples such that for each , there are exactly triples in containing and . (3)Let be a -set. We use for the complete graph on the vertex set . (4)Let be a -set. Let STS be defined as KTS and BIBD can be defined similarly, that is, Let and be disjoint sets of cardinality and , respectively. We define GDD as (5)When we say that is a collection of subsets (blocks) of a -set , may contain repeated blocks. Thus, “” in our construction will be used for the union of multisets.
Let be a positive integer. We consider in this section the problem of determining all pairs of integers in which a GDD exists. Recall that the existence of GDD implies , and . Let
By solving systems of linear congruences, we obtain the following necessary conditions.
Lemma 3.1. Let be a nonnegative integer.
(a)If , then there exist nonnegative integers and such that .(b)If , then there exist nonnegative integers and such that .(c)If , then there exist nonnegative integers and such that .(d)If , then there exist nonnegative integers and such that .(e)If , then there exist nonnegative integers and such that .(f)If , then there exist nonnegative integers and such that .
In order to obtain sufficient conditions on an existence of GDD, we first observe the following facts. (1)Let and be two disjoint sets of size and , respectively. Then STS if and only if GDD. (2)Let and be two disjoint sets of size and ; respectively, and let . Then GDD if STS, BIBD, and BIBD.
Thus, we have the following results.
Lemma 3.2. Let and be nonnegative integers. Then
(a), , (b),, and(c),
Lemma 3.3. Let and be nonnegative integers. Then,
Proof. (a) We first consider an existence of GDD, where the groups are and . Let be a 1-factorization of . Let ), , and . Then forms a GDD, where . Thus .
Let and be two sets of size and , respectively. Suppose that and . Let and let . Thus, KTS. Let with as its parallel classes. Since STS and STS are not empty, there exist and . We now let as Thus, forms a GDD and .
(b) Let and be two sets of size and , respectively, and let . Choose with as its parallel classes. Since STS and STS are not empty, there exist and . We now let as Thus, forms a GDD and .
Therefore, the proof is complete.
Part of the proof of the following lemma is based on an existence of GDD which we now construct. Let and . Then it is easy to check that , where , , .
Lemma 3.4. Let and be nonnegative integers. Then
Proof. Case 1. Let be a -set containing , and be a -set containing .
Subcase 1 (k = 0). Let , and we can see that . Suppose that . Since is a set of size , it follows by Theorem 2.2, that there exists a PBD, , in which and triples in . Let . Since is a -set, it follows, by Theorem 2.5(c), that BIBD. Let . It is easy to see that forms a GDD, where
Subcase 2 (k = 1). A GDD can be constructed as follows. Let , where and are sets of size four. It is clear that STS, STS, and STS are not empty. It has been shown above that GDD is not empty. We now choose , , , and , and let . Then, form a GDD.
Subcase 3 (k = 2). We first consider the existence of GDD with and . Let be the complete graph of order 10 with as its vertex set. It is well known that is 1-factorable. In other words, can be decomposed as a union of nine edge-disjoint 1-factors. Consequently, can be decomposed as a union of three edge-disjoint 3-factors. Also, can be decomposed as a union of and a 3-factor: ten triples and a 3-factor of , where reducing arithmetic operations . Therefore, can be decomposed as a union of and four 3-factors.
Let be a 3-factorization of and and as described above. Then forms a GDD, where the collection with .
A GDD can be constructed as follows. Let , where and are sets of size ten and four, respectively. It is clear that STS, STS, and STS are not empty. It has been shown above that GDD is not empty. We now choose , , , and and let . Then form a GDD.
Subcase 4 (k ≥ 3). Let . Suppose that and . Since is a -set, it follows that STS and KTS. Choose STS and let with as its parallel classes. Let . Since is a set of size , we choose a PBD, , as in Theorem 2.2 in which . Let . Since is a 5-set and is a -set, it follows, by Theorem 2.5(c) and (d), that there exist and . Thus, we can see that forms a GDD, where
Case 2. We now suppose that and be sets of size and , respectively. We suppose further that and . By Lemma 3.3(b), we have GDD. Choose and . By Theorem 2.6(e) that GDD. Choose . Let , and it is easy to see that .
Lemma 3.5. Let and be nonnegative integers. Then
(a) and ,(b) and ,(c) withnot both zero, , and(d).
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 be a -set and be a -set. It is clear that GDD. We now construct a GDD, , with , , , , and , where , , , and . We now construct a GDD, , in general case, where and . We first let , , and we will use a result on the existence of GDD which has been shown in Theorem 2.6(f), namely, GDD. Therefore, we can choose , and , where . We can see that , where .
Case 2 (6k + 2, 6h + 4). Let be a -set and be a -set. It is easy to see that GDD by constructing as follows. Let , , and , where . We now turn to more general cases. Suppose that and . Since is a -set, it follows that KTS. Choose with parallel classes . Let . We have shown in Lemma 3.4(d) that GDD. Choose and . We can see that , where .
(d) Let be a -set and be a -set. Let and . Let , . Then .
Next we will show that GDD by letting , , and . Let , , and . Theorem 2.6(b) shows an existence of a GDD. Let , where . It is easy to check that , where .
Finally, let , and . We now choose , , , and . By Theorem 2.6(b) that GDD. Choose . Thus, we can check that , where . Thus for all positive integers .
Now we have an existence of a GDD for whenever and are not equal to 2, so we can readily extend to any by the following lemma.
Lemma 3.6. Let and be positive integers with and . If there exists a GDD with , then a GDD, exists
Proof. Let be an -set and be an -set. By assumption we have GDD. Choose . Since and are not equal to 2, by Theorem 2.5(a) there exist and . It is easy to see that forms a GDD, where . Thus with .
Finally, we have the main result as in the following.
Theorem 3.7. Let and be positive integers with and . There exists a GDD if and only if
(1) and(2) and .
N. Pabhapote is supported by the University of the Thai Chamber of Commerce, and N. Punnim is supported by the Thailand Research Fund.
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