4.1. Methods for Constructing -LA
In this subsection, we modify some constructions for MCAs to the case of -LAs. The next two lemmas provide the “truncation” and “derivation” constructions, which were first used to construct mixed CAs.
(Truncation) Let . Then, -LAN-LAN.
Let A be a -LA with -LAN. Delete the ith column from A to obtain a -LA. Thus, -LAN-LAN. □
(Derivation) Let . Then -LAN-LAN, where .
Let A be a -LA with -LAN. By Lemma 1, A is an MCA and a -LA. For each , taking the rows in A that involve the symbol x in the ith columns and omitting the column yields an MCA. We use to denote the derived array. Next, we prove that is a -LA. In fact, for any -way interactions and with , if , we can form two t-way interactions and by inserting into and , respectively. Hence, , where but . Consequently, A is not a -LA. It is clear that -LAN for . Thus, -LAN. □
The following product construction can be used to produce a new LA from old LAs, which is a typical construction in combinatorial design.
(Product Construction) If both a -LA and an MCA exist, then a -LA exists, where . In particular, if both a -LA and a -LA exist, then a -LA also exists, where .
Let and be the given -LA and MCA, respectively. We form an array as follows. For each row of A and each row of B, include the row as a row of , where .
From the typical method in design theory, the resultant array is an MCA, as both A and B are MCAs. By Lemma 1, we only need to prove that is a -LA. Suppose that , where and with . It is noteworthy that the projection on the first component of and is the corresponding t-way interaction of A, while the projection on the second component is the corresponding t-way interaction of B. Therefore, A is not a -LA. The first assertion is then proved because a -LA is an MCA. The second assertion can be proven by the first assertion. □
The following construction can be used to increase the number of levels for a certain factor.
If a -LA exists, then a -LA exists, where and .
Let be the given -LA with entries in the ith column from a set of size . For a certain , we replace the symbols in the ith column of A by , respectively. We denote the resultant array by . Clearly, permuting the symbols in a certain column does not affect the property of -LAs. Thus, is also a -LA, where entries in the ith column of from the set . Subsequently, write . It is easy to prove that M is a -LA and an MCA. By Lemma 1, M is the desired array. □
The following example illustrates the idea in Construction 2.
The transpose of the following array is a
Replace the symbols
in the 3th column, respectively. Juxtapose two such arrays from top to bottom to obtain the following array M
; we list it as its transpose to conserve space.
It is easy to verify that M is a -LA.
Replace the symbol 0 by 2 in the 3th column. Juxtapose two such arrays from top to bottom to obtain the following array
; we list it as its transpose to conserve space.
It is easy to verify that is a -LA. □
Remark 1. Construction 2 may produce an optimal -LA. For example, a -LA is shown in Table 1. By Construction 2, we can obtain a -LA, which is optimal by Lemma 4.
Fusion is an effective construction for MCAs from CAs, for example, see [29
]. As with CAs, fusion for
-LAs guarantees the extension of uniform constructions to mixed cases; however, fusion for a
may not produce mixed-level
-LAs. This problem can be circumvented by introducing the notion of detecting arrays (DAs). Suppose that all the factors have the same levels. If, for any
, we have
then the array A
is called a
-DA or a
(Fusion) Suppose that A is a -DA with . If A is also a -LA, then a -LA exists, where .
Let A be a -DA over the symbol set V of size v. Suppose that are positive integers with . It is clear that there exists a certain such that and , where . We select elements from V in the ith column of A to form the element sets , respectively. The elements in are identical with , respectively. Then, we obtain an array . Clearly, is an MCA. We only need to prove that is a -LA by Lemma 1, i.e., for any two distinct t-way interactions and , we have . It is clear that and when and . Hence, .
When and , we can obtain a t-way interaction of A, where . If , then ; however, ; as such, it is a contradiction that A is a -DA. If and , then the similar argument can prove the conclusion.
, it is clear that
. The case
remains to be considered. Without loss of generality, suppose that
elements are identical with
. It is clear that
can be obtained from
by fusion, respectively, where
are sets of t
-way interactions with
. It is a contradiction that A
because the existence of
implies the existence of
Constructions 2 and 3 provide an effective and efficient method to construct a mixed-level
-LA from a
. The existence of
implies the existence of
]. Hence, the array A
in Construction 3 can be obtained by a
, which is characterized in terms of super-simple OAs. The existence of super-simple OAs can be found in [17
]. It is noteworthy that the derived array is not optimal. In the remainder of this section, we present two “Roux-type” recursive constructions [35
If both a -LA and a -LA exist, then a -LA exists, where .
be the given
, respectively. Clearly, if
, then A
is the required array. Now, suppose that
. Insert a column vector
to the front of the i
th column of B
to form an
. Clearly, M
is an MCA
]. By Lemma 1, we only need to prove that M
for any two distinct t
. Next, we distinguish the following cases.
Case 1. and
In this case, because A is a -LA, , as A is part of M.
Case 2. and or and
When and , if , then . Thus, . If , then must be included by rows of , where ; however, it must not be included by any row of A. Clearly, must be included by some rows of A. Consequently, . When and , the same argument can prove the conclusion.
Case 3. and
Clearly, holds whenever . If , then , which implies that . If , then and must be included by some rows for a certain , where . Because B is a -LA, , which implies . □
More generally, we have the following construction.
Let and . If a -LA, -LA, a -LA and -LA exist, then a -LA exists, where .
We begin with a -LA, an array A that is on . Let and be two sets with and such that and , respectively. Suppose that , an array, is a -LA, which is on . For each row , of , add to obtain a k-tuple , . Then, we obtain a array from , denoted by B. Similarly, from a -LA, we obtain a array, denoted by C. For each pair , we construct k-tuple for each row of the given -LA. These tuples result in a array, denoted by D.
Denote , and . We claim that F, an array, is a -LA which is on .
Clearly, F is an MCA . To prove this assertion, we only need to demonstrate that for any two distinct t-way interactions and . By similar argument as the proof of Construction 4, we can prove the conclusion except for the case where and , , and . In this case, and are only included by some rows of D. If , then . Consequently, , which implies that by the construction of D. It is a contradiction with being a -LA. The proof is completed. □
4.2. Constructions and Existence of Optimal -LA
Let . An array A is called if for any t-way interaction and for any t-way interaction . If an optimal -LA with exists, then the following condition must be satisfied.
Let . If A is an optimal -LA with . Then, A is an .
Let A be the given optimal -LA with . Then, A is an MCA by Lemma 1. Because , we have for any t-way interaction . It follows that for any t-way interaction of A from the definition of -LA, where . Hence, A is an , as desired. □
Clearly, an is not always a -LA. Next, we present a special case of , which produces optimal -LAs. First, we introduce the notion of mixed orthogonal arrays (MOAs).
An MOA, or MOA is an array with entries in the ith column from a set of size such that each sub-array contains each t-tuple occurring an equal number of times as a row. When , an MOA is merely an orthogonal array, denoted by OA.
The notion of mixed or asymmetric orthogonal arrays, introduced by Rao [36
], have received significant attention in recent years. These arrays are important in experimental designs as universally optimal fractions of asymmetric factorials. Without loss of generality, we assume that
. By definition of MOA, all t
-tuples occur in the same number of rows for any
sub-array of an MOA. This number of rows is called index
. It is obvious that
indices exist. We denote it by
, then an MOA is termed as a pairwise distinct index mixed orthogonal array
, denoted by PDIMOA
. Moreover, if
for a certain
holds, then it is termed as
. It is clear that
in the definition of
Example 2. The transpose of the following array is a .
The following lemma can be easily obtained by the definition of ; therefore, we omit the proof herein.
Suppose that . If A is a , then and , where and .
Let . If a PDIMOA exists, then a -LA exists. Moreover, if , then the derived -LA is optimal.
Let A be a PDIMOA. Clearly, A is an MCA. By Lemma 1, we only need to prove that implies , where and are two t-way interactions. In fact, if , then , which contradicts the definition of a PDIMOA. The optimality can be obtained by Theorem 1. □
We construct an optimal
in terms of
. First, we have the following simple and useful construction for
. A similar construction for MOAs was first stated in [37
Let and . If a exists, then a also exists.
Let A be with . We can form an array by replacing the symbols in by those of . It is easily verified that is the required . □
The following construction can be obtained easily; thus, we omit its proof.
Let and . If both a and a exist, then a exists. In particular, if both a and an OA exist, then a exists.
Next, some series of optimal mixed-level -LAs are presented. First, we list some known results for later use.
Lemma 10.  An OA exists for any integer .
The existence of s is determined completely by the following theorem.
Let . A exists if and only if for .
The necessity can be easily obtained by Lemma 8. For sufficiency, we write for . Clearly, and for . We list all t-tuples from to form an MOA, which is also a . Apply Construction 7 with an OA given by Lemma 10 to obtain the required . □
More generally, we have the following results.
Let and , where , . Then, a exists.
Let . Then, , where . By Theorem 2, a with exists. Apply Construction 6 to obtain a as desired. □
Let with . Then, an optimal -LA exists.
First, we construct a array : , where and ; and for .
We prove that A is an optimal -LA. Optimality is guaranteed by Theorem 1. It is clear that A is . Consequently, and , where . It is clear that and . We only need to prove . In fact, by construction, for a certain but , where , which implies . Thus, A is a -LA by Lemma 1. □
The following example illustrates the idea in Theorem 4.
Example 3. The transpose of the following array is an optimal -LA
Let with . Then, an optimal -LA exists.
First, we construct a
When , let be a array with for and be an arbitrary element for with . Let and . It is easy to prove that M and N are the required arrays if and , respectively. □
The following results need the notion of a Latin square. A Latin square
of order n
array of n
symbols in which each symbol occurs exactly once in each row and in each column. The diagonal
of such a square is a set of entries that contains exactly one representative of each row and column, respectively. A transversal
is a diagonal in which none of the symbols are repeated. For
, there exists a Latin square of order n
distinct transversals [39
Let and . Then, an optimal -LA exists.
Let with , where . For , a latin square of order w with w disjoint transversals, denoted by , exists. We take each of the w disjoint transversals from as a column to form a array , which, clearly, is also a Latin square. Let . The permutation is applied to the columns of L to obtain a new array denoted by . If L is a Latin square of w, then are also Latin squares of order w. The corresponding columns of and for have no common symbols.
Let be a array. It is easy to verify that A is a -LA. For each part of A, we can construct a array of the form . Next, juxtapose these resultant arrays to obtain a array , which is easily verified to be a -LA. Continue this process until the array B can be obtained. Clearly, B is a -LA.
Let . Suppose C is a array with entries from , where . Write ; if , if . It is easy to verify that is the required array. □
The following theorem considers the case .
Let be a positive integer. Then an optimal -LA exists.
Let . We only need to construct a -LA, A, because is the required array. As , the number of occurrences of 0, 1 should be at least 2. It is easy to prove that all the different column vectors of length v with entries from form the -LA as desired. Thus, all that remains is to calculate the number of all the different column vectors. Write x and y as the number of 0s and 1s in a column vector of length v, respectively. Clearly, and . Because there exist x positions with 0s, the number of different column vectors is . Consequently, the number of all the different column vectors is . □