Research Article | Open Access

# Holey Perfect Mendelsohn Designs of Type with Block Size Four

**Academic Editor:**M. Gionfriddo

#### Abstract

Let 4-HPMD denote a *holey perfect Mendelsohn design with block size four*. The existence of 4-HPMDs with holes of size 2 and one hole of size 3, that is, of type , was established by Bennett et al. in 1997. In this paper, we investigate the existence of 4-HPMDs of type for : a 4-HPMD() exists if and only if , except possibly for , (7, 6), (11, 9), (11, 10). We also investigate the existence of 4-HPMD() for general and prove that there exists a 4-HPMD() for all . Moreover, if , then a 4-HPMD() exists for all ; if , then a 4-HPMD() exists for all .

#### 1. Introduction

Let be positive integers, a -*Mendelsohn design*, briefly -MD, is a pair , where is a -set (of points) and is a collection of cyclically ordered -subset of (called *blocks*) such that every ordered pair of points of are consecutive in exactly blocks of . If for all , every ordered pair of points of is -apart in exactly blocks of , then the -MD is called *perfect* and is denoted by -PMD.

The existence of a -PMD is equivalent to the existence of an idempotent quasigroup satisfying the identity, called *Stein's third law*,
for all . Let . Then is a -PMD.

For the existence of -PMD, we have the following theorem from [1–3].

Theorem 1. *A -PMD exists if and only if and . *

In this paper, we are only interested in PMDs where . A *holey perfect Mendelsohn design* (briefly HPMD) is a triple which satisfies the following properties. (1) is a partition of into subsets called *holes*. (2) is a family of cyclically ordered -subset of (called *blocks*) such that a hole and a block contain at most one common point.(3)Every ordered pair of points from distinct holes are -apart in exactly one block for . The *type* of the HPMD is the multiset and is described by an exponential notation. Throughout this paper, we shall use HPMD to describe a PMD of the type in which occurs times, .

In graph theoretic terminology, an HPMD is a decomposition of a complete multipartite directed graph DK into -circuits such that for any two vertices and from different components, there is one circuit along which the directed distance from to is , where .

Another class of designs related to HPMDs is *group divisible design* (GDD). A GDD is a 4-tuple ) which satisfies the following properties. (1) is a partition of into subsets called *groups*. (2) is a family of subsets of (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 exactly blocks. The *type* of the GDD is the multiset and we will also use an “exponential” notation for the type of GDD. We also use the notation GDD() to denote the GDD when its block sizes belong to and group sizes belong to . In particular, a GDD, where there is only one group of size , is denoted by -GDD of type .

Theorem 2 (see [4, 5]). *There exists a -GDD of type for each , and with , except for and possibly excepting . *

If , then the GDD becomes a *pairwise balanced design* (*PBD*) [6]. If , and the type is , then the GDD becomes a *transversal design*, TD. The following results are well known (see [7, 8], e.g.).

Theorem 3. *
(a) There exists a TD for any positive integer .**
(b) There exists a TD for every positive integer .**
(c) There exists a TD for and ,**
(d) There exists a TD for and . *

It is well known that the existence of a TD is equivalent to the existence of mutually orthogonal Latin squares, denoted by -MOLS. It is easy to see that if we ignore the direction of elements in the blocks, an HPMD becomes a GDD with . But the converse may be not true. It is proved in [9] that a -GDD of type and index exists if and only if . However, for the existence of HPMDs with block size four (briefly 4-HPMDs), the following results are known ([8, 10–14]).

Theorem 4. *
(a) A 4-HPMD exists if and only if with the exception of types , , and for odd .**
(b) For , a 4-HPMD exists if and only if and , except and except possibly .**
(c) A 4-HPMD exists if and only if .**
(d) There exist HPMDs of type for and , except .**
(e) For , a 4-HPMD exists if and only if , except possibly for , and and .*

In this paper, we consider the existence of a 4-HPMD and extend the results in Theorems 4(c) and 4(d). It is fairly well known that the Latin square corresponding to a quasigroup satisfying Stein's third law is a self-orthogonal Latin square (briefly SOLS). So the existence of a 4-HPMD will imply the existence of a holey SOLS of the same type. Accordingly, the necessary condition for the existence of holey SOLS is also a necessary condition for the existence of 4-HPMD, which can be stated as follows (see, e.g., [15, 16]).

Lemma 5. *If a 4-HPMD exists, then . *

The main result of our investigation is the following theorem.

Theorem 6. *
(a) For , a 4-HPMD exists if and only if , , with the possible exception of .**
(b) For , a 4-HPMD exists if . Moreover, if , then a 4-HPMD exists for all ; if , then a 4-HPMD exists for all . *

For convenience in what follows, we shall simply refer to a 4-HPMD as an HPMD, where it is to be understood that that blocks are of size four.

#### 2. Construction Tools

To construct HPMDs directly, sometimes we use *starter blocks*. Suppose the block set of an HPMD is closed under the action of some Abelian group , then we are able to list only part of the blocks (starter or base blocks) which determines the structure of the HPMD. We can also attach some infinite points to an Abelian group . When the group acts on the blocks, the infinite points remain fixed. Formally, let be the block set of an HPMD over the point set , where is a group, is a set of *infinite points*, . The addition is extended over as follows: for any and . A set is called *starter blocks* of if is a minimum subset of satisfying the following property: for any and any , , and for any , there exist and such that , where when . In the following example , are infinite points.

*Example 7. *An HPMD points: holes: starter blocks : , , , , , .

In this example, the entire set of 108 blocks are developed from the starter blocks by adding to each point of the starter blocks. To check the starter blocks, we need only to calculate whether the differences from all pairs , , in each of the starter blocks are precisely , where is the set of the differences of the holes. For the above example, the set of differences from the six blocks is exactly . This is also true for the set of the differences .

The above idea of starter blocks can be generalized: instead of adding to each point of the starter blocks, we may have to add , where , to develop the block set; we refer to this as the method. In this case, for a set to be starter blocks, we require that for any and any , . For quasigroups, we require that for all if and only if for any [17, 18].

*Example 8. *An HPMD points: holes: starter blocks: : , , , , , , , , , , , , , , , , , , , .

By adding , where , to the 20 starter blocks, we obtain a set of 60 blocks.

Next, we state several recursive constructions of HPMDs, which are commonly used in other block designs [8]. The following construction comes from the weighting construction of GDDs [6].

*Construction 1 (Weighting). *Suppose is a GDD with and let . Suppose there exist HPMDs of type for every . Then there exists an HPMD of type .

Lemma 9. *There exists an HPMD for each , and with , except for and possibly excepting . *

*Proof. *From Theorem 2, there exist 4-GDDs of the same type. So we can give all points of this GDD weight one to get the desired HPMD by Construction 1.

Using Theorem 3, if we give every point of an HPMD weight and input TD to each block of the HPMD, we can obtain the following construction.

*Construction 2. *Suppose there exists an HPMD of type , then there exists an HPMD, where .

The next construction may be called “filling in holes,” which is used commonly in constructing designs.

*Construction 3. *Suppose there exist an HPMD of type and HPMDs of type , where and , then there exists an HPMD of type .

Lemma 10. *If there exists an HPMD, there exists an HPMD. *

*Proof. *By Theorem 4, there exists an HPMD for any . We adjoin points to this HPMD and fill three holes of size with an HPMD, leaving one hole of size . The result is the desired HPMD.

Lemma 11. *If there exists HPMD, then there exists an HPMD for . *

*Proof. *Applying Construction 2 with to the HPMD, we have an HPMD of type . Add points, , to this HPMD and fill an HPMD, which exists by Theorem 4, into the holes of size 8, we have an HPMD, where .

Lemma 12. *If an HPMD exists, then an HPMD exists for . *

*Proof. *Applying Construction 2 with to the HPMD, we obtain an HPMD of type . To this HPMD we adjoin points, where , and fill the holes of size 10 with an HPMD, which exists by Theorem 4, we obtain an HPMD.

Lemma 13. *If there exists a TD, then there exists an HPMD, where . *

*Proof. *Give weight 2 to each point of first four groups of a TD. Give weight 1, 2 or 3 to the points of the fifth group so that the total is . Since there exist HPMDs of type , , from Theorem 4, we obtain the desired HPMD by Construction 1.

Lemma 14. *If there exists a TD, then there exist*(a)*HPMD, where and ,*(b)*HPMD for and . *

*Proof. *(a) We will use Construction 1: give weight 2 to each point of first four groups of a TD; give weight 2 to the points of the fifth group and give weight 1, 2, or 3 to the points of the sixth group so that the total is . Since there exist HPMDs of type for and from Theorem 4, we obtain the HPMD of type .

(b) Take the HPMD from (a), where , we fill the holes of sizes and with an HPMD and an HPMD, to obtain an HPMD, where .

Lemma 15. *If there exist a TD, an HPMD, and an HPMD, where , then there exists an HPMD for . *

*Proof. *Take the HPMD from Lemma 14, where and , we first adjoin points to the HPMD and then fill the holes of sizes and with an HPMD and an HPMD, we obtain an HPMD where .

Lemma 16 (see [8, Lemma 6.3]). *If there exists a TD, then there exist*(a)*HPMD for and ,*(b)*HPMD for and . *

*Proof. *(a) Give weight 2 to each points of five groups of the TD, give weight 2 to points of the sixth group and weight 0 to the remainder points of this group, and give weight , where , to the points of the seventh group such that the sum of these weights is equal to . As there exists an HPMD for and , we obtain an HPMD.

(b) Take the HPMD from (a), we obtain an HPMD by filling HPMDs of types and into the holes of sizes and .

Lemma 17. *If there exist a TD, an HPMD, and an HPMD, where , then there exists HPMD for . *

*Proof. *We adjoin points to the HPMD from Lemma 16, where and fill the holes of sizes and with an HPMD and an HPMD, we obtain an HPMD, where .

Lemma 18. *An HPMD exists for . *

*Proof. *We start with a TD. In the first four groups of the TD, we give all the points weight 3. In the last group we give the points a weight of 1, 2, 3, or 4 for a total weight of . We need HPMDs of type , where . For , the design is given in Theorem 4; for , they are given in Appendix J. The resulting design is an HPMD by Construction 1.

#### 3. HPMD() for Some Specific

From the results in the previous section, we can obtain the following results.

Lemma 19. *There exists an HPMD for any . *

*Proof. *For , we have Theorem 4. For except , please see Appendix A. For , we can obtain the design from an HPMD given in Appendix G, by filling the hole of size 9 with an HPMD. For , we take an HPMD from Theorem 4, adjoin one point to this HPMD, and fill each hole of size 8 with an HPMD.

For , let , where and . For , we apply Construction 2 with on an HPMD (Theorem 4) to get an HPMD. Adjoin one point to this HPMD and fill the holes of size 8 with an HPMD, we obtain an HPMD. If we adjoin 3 points to the HPMD and fill holes of size 8 with an HPMD and the remaining hole with an HPMD, we obtain an HPMD.

For , because an HPMD exists for any by Theorem 4, we have an HPMD by Construction 2 with . Adjoin points, where , to this HPMD, and fill the holes of size 8 with an HPMD, the hole of size 12 with an HPMD if and an HPMD if , we obtain an HPMD, where .

Combining Theorems 4 (with ) and 4, and the above lemma, we have the following.

Lemma 20. *An HPMD exists for all and . *

Now we show the existence of some HPMD for .

Lemma 21. *An HPMD exists for . *

*Proof. *For , we apply Lemma 20. For please see Appendices B–F. For , we first get an HPMD by applying Lemma 13 with and . Adjoin points, where , to this HPMD, and fill three holes of size 8 with an HPMD and the hole of size with an HPMD, we obtain an HPMD for . For , please see Appendix I, except , which is given in [8].

Lemma 22. *An HPMD exists for . *

*Proof. *For , we apply Lemma 20. For , please see Appendices B, D, and E. For , we apply Lemma 9.

For , we first get an HPMD by applying Lemma 13 with . Adjoin points, where , to this HPMD, and fill three holes of size 8 with an HPMD and the hole of size with an HPMD, we obtain an HPMD for .

For , we obtain an HPMD by Lemma 10 with , because an HPMD exists for by Theorem 4 and an HPMD is given in Example 8.

Lemma 23. *An HPMD exists for . *

*Proof. *For , we apply Lemma 20. For , please see Appendices B–G. For , we first get an HPMD by applying Lemma 13 with , . Adjoin points, where , to the HPMD, and fill three holes of size 10 with an HPMD and the hole of size 8 with an HPMD, we obtain an HPMD for . For or , please see Appendix I.

Lemma 24. *An HPMD exists for . *

*Proof. *For , we apply Lemma 20. For , we first obtain an HPMD from HPMD by Construction 2 with . Add two points to this HPMD and fill the holes of size 8 with an HPMD and the hole of 12 with an HPMD, we obtain an HPMD.

For , from Lemma 14 with , , we have HPMD. For , we apply Lemma 10 with and . For , please see Appendix I. For , we apply Lemma 9.

Lemma 25. *An HPMD exists for . *

*Proof. *For , we apply Lemma 20. For , we take the HPMD from Lemma 21 and fill the hole of size 17 with an HPMD. For , please see Appendices C–F.

For , we generate an HPMD from an HPMD (Theorem 4) by Construction 2 with . To this HPMD, we adjoin one point, fill four holes of size 8 by an HPMD and one hole of size 12 by an HPMD, we obtain an HPMD.

For , we first get an HPMD by applying Lemma 13 with . Adjoin points, where , to this HPMD, and fill three holes of size 10 with an HPMD and the hole of size 14 with an HPMD, we obtain an HPMD for .

For , we take an HPMD from Lemma 18. Adjoin three points to this HPMD and fill three holes of size 12 with an HPMD and one hole of size 8 with an HPMD, we obtain an HPMD.

Finally for , please see Appendix I.

Lemma 26. *An HPMD exists for . *

*Proof. *For , we apply Lemma 20. For , we add points to an HPMD and fill in three holes with an HPMD and one hole with an HPMD, to obtain an HPMD. For , please see Appendices C–F.

For , we take the HPMD from Appendix J, adjoin points, , to the HPMD, and fill four holes of size 8 with an HPMD and one hole of size 14 with an HPMD, we obtain an HPMD.

For , we take an HPMD from Lemma 18. Adjoin points, where , to this HPMD, and fill three holes of size 12 with an HPMD and the hole of size 10 with an HPMD (Theorem 4), we obtain an HPMD for .

For , please see Appendix I.

Lemma 27. *An HPMD exists for . *

*Proof. *For , we first apply Lemma 14 with and to get an HPMD, where . Adjoin points, where , to this HPMD, and fill the holes of sizes 10 and 12 with an HPMD and an HPMD, respectively. The result is an HPMD, where .

For , we first get an HPMD by applying Lemma 13 with , . Adjoin points, where , to this HPMD, and fill three holes of size 14 with an HPMD and the hole of size 10 with an HPMD, we obtain an HPMD for .

Lemma 28. *An HPMD exists for . *

*Proof. *For , we first apply Lemma 14 with and to get an HPMD, where , then adjoin points, where , to this HPMD, and fill the holes of sizes 10 and 14 with an HPMD and an HPMD, respectively. The result is an HPMD, where .

For , we first get an HPMD by applying Lemma 13 with , . Adjoin points, where , to this HPMD, and fill three holes of size 14 with an HPMD and the hole of size 12 with an HPMD, we obtain an HPMD for .

For , we start with an HPMD, adjoin points to this HPMD, and fill in the first three holes of size , where , with an HPMD. The resulting design is an HPMD of type , where , and this completes the proof of the lemma.

Lemma 29. *An HPMD exists for . *

*Proof. *For , we apply Lemma 20. For , we take an HPMD from Lemma 18 and fill the holes of sizes 12 and 14 with an HPMD and an HPMD, respectively. For , we take an HPMD from Lemma 26 and fill the hole of size 17 with an HPMD. For , please see Appendix D. For , we can get the designs from Lemma 14 with and .

Lemma 30. *An HPMD exists for . *

*Proof. *For , we can get an HPMD from Lemma 12 with , and , and and . Fill the hole of size with an HPMD, we have an HPMD.

For , we form a -GDD of type by deleting one block from a TD. In the first five groups of this GDD, we give all of the points weight . In the fifth group, we give one point of weight and the other points weight . In the last group, we give the points a weight of 1, 2, or 3 for a total weight of where . Since there are HPMDs of types for and for and by Theorem 4, we get an HPMD of type for . To this HPMD, we fill the holes of size 12 with an HPMD and obtain an HPMD for .

For , we use a TD: in the first four groups of the TD, we give all of the points a weight of two. In the fifth, sixth, and seventh groups, we give one point weight two and the other points weight zero. In the last group, we give the points a weight of 1, 2, or 3, for a total weight of . Since we have HPMDs of types for and , we get an HPMD of type for . By filling in the holes of size with an HPMD, the resulting design is an HPMD for .

Lemma 31. *An HPMD exists for . *

*Proof. *We will use a TD: in the first five groups of the TD, we give all of the points a weight of two. In the sixth and seventh groups, we give one point weight two and the other points weight zero. In the last group, we give the points a weight of 0, 1, 2, 3, or 4, for a total weight of . Since we have HPMDs of types for and , we get an HPMD of type for . By filling in the holes of size with an HPMD, the resulting design is an HPMD for .

#### 4. HPMD() for

Lemma 32. *An HPMD exists for and , except possibly . *

*Proof. *For , we apply Lemma 20.

For , let us consider first. For and , an HPMD exists by Theorem 4. For and , we apply Lemma 9 when . For and , we obtain an HPMD by Lemma 10 with , because an HPMD exists for . For and , we obtain an HPMD by Lemma 10 with , because an HPMD exists for . Besides an HPMD in Example 8 and an HPMD in Example 7, an HPMD for can be found in [8]. The other designs for are given in Appendices B–H, except .

Now let us consider and . For , , and , by Lemma 11, we have an HPMD. For and , let , we obtain first an HPMD by Lemma 13 with and when ; for , we have an HPMD in Appendix J. To the HPMD, we adjoin 2 points and fill the holes of size 8 with an HPMD, to obtain an HPMD for . For and , we adjoin points to an HPMD and fill the holes of size 12 with an HPMD. For and , we get them from Lemma 15 with and . The cases of are covered by Lemmas 21, 22, 23, 24, 25, and 26, respectively. For and , we can get an HPMD from Lemma 12 with , and .

Lemma 33. *An HPMD exists for and . *

*Proof. *For , and , we have Lemmas 27, 28, 29, and 30 to cover these cases, respectively.

For and , the designs are provided by Lemma 11 with . For and , we get these designs from Lemma 14 with and .

For , we apply Lemma 12 with , , and .

Lemma 34. *An HPMD exists for and . *

*Proof. *For and , we obtain at first an HPMD by Lemma 12 with , , and , . Fill an HPMD and an HPMD into the holes of sizes and , we obtain an HPMD and an HPMD, respectively. For and , an HPMD exists by Lemma 15 with , , .

For , the designs come from Lemma 16 with . For , the designs are provided by Lemma 31.

For and , we first apply Lemma 12 with , , and , to get an HPMD. Fill the hole of size by an HPMD, we obtain an HPMD. For and , we have an HPMD by Lemma 14 with and . Finally for , we apply Lemma 17 with , , .

Lemma 35. *An HPMD exists for and . *

*Proof. *For and , an HPMD exists by Lemma 17 with and , and and , and .

For and , there exists an HPMD by Lemma 12 with , , and . Fill an HPMD into the hole of size , we obtain an HPMD. For and , because we have an HPMD, by Construction 2 with , we have an HPMD. Add two points to this design and fill an HPMD into the holes of size and an HPMD into the hole of size , we obtain an HPMD. For and , an HPMD exists by Lemma 14 with and . For and , an HPMD exists by Lemma 15 with , , and .

For and , the designs come from Lemma 17 with , and .

For and , because an HPMD exists for , we get an HPMD from Construction 2 with . Add points to this design, , and fill an HPMD into the holes of size , we obtain an HPMD for .

Lemma 36. *An HPMD exists for and . *

*Proof. *For and , we apply Lemma 17 with , and . For and , we apply Lemma 16 with and .

For and , by Lemma 16 with and , there exist HPMD for and . Fill in the hole of size with an HPMD, we obtain an HPMD for . For and , an HPMD exists from Lemma 15 with and .

For and , an HPMD exists by Lemma 17 with , , and .

Lemma 37. *An HPMD exists for and . *

*Proof. *The cases of are covered by Lemmas 24, 29, and 31, respectively. The cases of are covered by Lemmas 34, 35, and 36, respectively. For , we apply Lemma 12 with and because HPMDs of type exist for all .

Lemma 38. *An HPMD exists for and . *

*Proof. *For and , an HPMD exists by by Lemma 17 with and , and and , and .

For and , by Lemma 17 with and , and and , and . there exist HPMD for . Fill in the hole of size with HPMD in HPMD, we obtain an HPMD for . For and , we apply Lemma 15 with and , and . For and , we apply Lemma 14 with , and , where , to obtain an HPMD. Fill the holes of sizes 36 and with an HPMD of types , , respectively, we obtain an HPMD for . For and , we apply Lemma 15 with , , .

For and , we apply Lemma 16 with and . Finally for and , we apply Lemma 17 with , , and .

The required HPMDs of type for the application of Lemmas 15 and 17 in the above proof as well as in the proofs of the next two lemmas come from either Lemma 32 or Lemma 37.

Lemma 39. *An HPMD exists for and . *

*Proof. *Let .

For , , let , where and . Because an TD exists, using Lemma 17 with , , , we obtain an HPMD for and .

For , , let , where and . using Lemma 17 with , , , we obtain an HPMD for and .

For , , let , where and . using Lemma 17 with , , , we obtain an HPMD for and .

For , , we apply Lemma 16 with and to obtain an HPMD for .

Lemma 40. *An HPMD exists for and . *

*Proof. *Let , where . Since , we have . If there exists a TD, then using Lemma 17 with , , , we obtain an HPMD for . Because and , we have .

For