#### 4.1. Methods for Constructing $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$

In this subsection, we modify some constructions for MCAs to the case of $(\overline{1},t)$-LAs. The next two lemmas provide the “truncation” and “derivation” constructions, which were first used to construct mixed CAs.

**Lemma** **5.** (Truncation) Let $2\le {v}_{1}\le {v}_{2}\le \cdots \le {v}_{i-1}\le {v}_{i}\le {v}_{i+1}\le \cdots \le {v}_{k}$. Then, $(\overline{1},t)$-LAN$(k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))\le $$(\overline{1},t)$-LAN$(k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i},{v}_{i+1},\cdots ,{v}_{k}))$.

**Proof.** Let A be a $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i},{v}_{i+1},\cdots ,{v}_{k}))$ with $N=(\overline{1},t)$-LAN$(k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i},{v}_{i+1},\cdots ,{v}_{k}))$. Delete the ith column from A to obtain a $(\overline{1},t)$-LA$(N;k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))$. Thus, $(\overline{1},t)$-LAN$(k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))\le N=(\overline{1},t)$-LAN$(k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i},{v}_{i+1},\cdots ,{v}_{k}))$. □

**Lemma** **6.** (Derivation) Let $2\le {v}_{1}\le {v}_{2}\le \cdots \le {v}_{i-1}\le {v}_{i}\le {v}_{i+1}\le \cdots \le {v}_{k}$. Then ${v}_{i}\xb7(\overline{1},t-1)$-LAN$(k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))\le $$(\overline{1},t)$-LAN$(k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i},{v}_{i+1},\cdots ,{v}_{k}))$, where $t\ge 2$.

**Proof.** Let A be a $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ with $N=(\overline{1},t)$-LAN$(k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$. By Lemma 1, A is an MCA and a $(1,t)$-LA. For each $x\in \{0,1,\cdots ,{v}_{i}-1\}$, taking the rows in A that involve the symbol x in the ith columns and omitting the column yields an MCA$({N}_{x};t-1,k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))$. We use $A\left(x\right)$ to denote the derived array. Next, we prove that $A\left(x\right)$ is a $(1,t-1)$-LA$({N}_{x};k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))$. In fact, for any $(t-1)$-way interactions ${T}_{1}$ and ${T}_{2}$ with ${T}_{1}\ne {T}_{2}$, if $\rho (A\left(x\right),{T}_{1})=\rho (A\left(x\right),{T}_{2})$, we can form two t-way interactions ${T}_{1}^{\prime}$ and ${T}_{2}^{\prime}$ by inserting $(i,x)$ into ${T}_{1}$ and ${T}_{2}$, respectively. Hence, $\rho (A,{T}_{1}^{\prime})=\rho (A,{T}_{2}^{\prime})$, where $|\rho (A,{T}_{1}^{\prime})|=|\rho (A\left(x\right),{T}_{1})|$ but ${T}_{1}^{\prime}\ne {T}_{2}^{\prime}$. Consequently, A is not a $(1,t)$-LA. It is clear that ${N}_{i}\ge (\overline{1},t-1)$-LAN$(k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))$ for $0\le i\le {v}_{i}-1$. Thus, $N={N}_{0}+{N}_{1}+\cdots +{N}_{{v}_{i}-1}\ge {v}_{i}\xb7(\overline{1},t-1)$-LAN$(k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))$. □

The following product construction can be used to produce a new LA from old LAs, which is a typical construction in combinatorial design.

**Construction** **1.** (Product Construction) If both a $(\overline{1},t)$-LA$({N}_{1};k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ and an MCA$({N}_{2};t,k,({s}_{1},{s}_{2},\cdots ,{s}_{k}))$ exist, then a $(\overline{1},t)$-LA$({N}_{1}{N}_{2};k,({v}_{1}{s}_{1},{v}_{2}{s}_{2},\cdots ,{v}_{k}{s}_{k}))$ exists, where $t<k$. In particular, if both a $(\overline{1},t)$-LA$({N}_{1};k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ and a $(\overline{1},t)$-LA$({N}_{2};k,({s}_{1},{s}_{2},\cdots ,{s}_{k}))$ exist, then a $(\overline{1},t)$-LA$({N}_{1}{N}_{2};k,({v}_{1}{s}_{1},{v}_{2}{s}_{2},\cdots ,{v}_{k}{s}_{k}))$ also exists, where $t<k$.

**Proof.** Let $A=\left({a}_{ij}\right)\phantom{\rule{4pt}{0ex}}(i\in {I}_{{N}_{1}},j\in {I}_{k})$ and $B=\left({b}_{ij}\right)\phantom{\rule{4pt}{0ex}}(i\in {I}_{{N}_{2}},j\in {I}_{k})$ be the given $(\overline{1},t)$-LA$({N}_{1};k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ and MCA$({N}_{2};t,k,({s}_{1},{s}_{2},\cdots ,{s}_{k}))$, respectively. We form an ${N}_{1}{N}_{2}\times k$ array as follows. For each row $({a}_{i1},{a}_{i2},\cdots ,{a}_{ik})$ of A and each row $({b}_{h1},{b}_{h2},\cdots ,{b}_{hk})$ of B, include the row $(({a}_{i1},{b}_{h1}),({a}_{i2},{b}_{h2}),\cdots ,({a}_{ik},{b}_{hk}))$ as a row of $\overline{A}$, where $1\le i\le {N}_{1},1\le h\le {N}_{2}$.

From the typical method in design theory, the resultant array $\overline{A}$ is an MCA$({N}_{1}{N}_{2};t,k,({v}_{1}{s}_{1},{v}_{2}{s}_{2},\cdots ,{v}_{k}{s}_{k}))$, as both A and B are MCAs. By Lemma 1, we only need to prove that $\overline{A}$ is a $(1,t)$-LA. Suppose that $\rho (\overline{A},{T}_{1})=\rho (\overline{A},{T}_{2})$, where ${T}_{1}=\{(i,({a}_{hi},{b}_{ci})):i\in I,\left|I\right|=t,I\subset \{1,2,\cdots ,k\},h\in {I}_{{N}_{1}},c\in {I}_{{N}_{2}}\}$ and ${T}_{2}=\{(j,({a}_{{h}^{\prime}j},{b}_{{c}^{\prime}j})):j\in {I}^{\prime},|{I}^{\prime}|=t,{I}^{\prime}\subset \{1,2,\cdots ,k\},{h}^{\prime}\in {I}_{{N}_{1}},{c}^{\prime}\in {I}_{{N}_{2}}\}$ with ${T}_{1}\ne {T}_{2}$. It is noteworthy that the projection on the first component of ${T}_{1}$ and ${T}_{2}$ 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 $(1,t)$-LA. The first assertion is then proved because a $(\overline{1},t)$-LA$({N}_{2};k,({s}_{1},{s}_{2},\cdots ,{s}_{k}))$ is an MCA$({N}_{2};t,k,({s}_{1},{s}_{2},\cdots ,{s}_{k}))$. 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.

**Construction** **2.** If a $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ exists, then a $(\overline{1},t)$-LA$(2N;k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},a,{v}_{i+1},\cdots ,{v}_{k}))$ exists, where $i\in \{1,2,3,\cdots ,k\}$ and ${v}_{i}<a\le 2{v}_{i}$.

**Proof.** Let $A=\left({a}_{ij}\right),(i\in {I}_{N},j\in {I}_{k})$ be the given $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ with entries in the ith column from a set ${V}_{i}$ of size ${v}_{i}$. For a certain $i\in {I}_{k}$, we replace the symbols $0,1,\cdots ,a-{v}_{i}-1$ in the ith column of A by ${v}_{i},{v}_{i}+1,\cdots ,a-1$, respectively. We denote the resultant array by ${A}^{\prime}$. Clearly, permuting the symbols in a certain column does not affect the property of $(\overline{1},t)$-LAs. Thus, ${A}^{\prime}$ is also a $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$, where entries in the ith column of ${A}^{\prime}$ from the set $\{a-{v}_{i},a-{v}_{i}+1,\cdots ,{v}_{i}-1,{v}_{i},{v}_{i}+1,\cdots ,a-1\}$. Subsequently, write $M={\left({A}^{T}\right|{\left({A}^{\prime}\right)}^{T})}^{T}$. It is easy to prove that M is a $(1,t)$-LA$(2N;k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},a,{v}_{i+1},\cdots ,{v}_{k}))$ and an MCA$(2N;t,k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},a,{v}_{i+1},\cdots ,{v}_{k}))$. By Lemma 1, M is the desired array. □

The following example illustrates the idea in Construction 2.

**Example** **1.** The transpose of the following array is a

$(\overline{1},2)$-LA

$(12;5,(2,2,2,2,3\left)\right)$0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |

0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |

0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 |

0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 |

0 | 2 | 1 | 0 | 1 | 1 | 2 | 0 | 1 | 0 | 2 | 2 |

Replace the symbols

$0,1$ by

$2,3$ 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.

0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |

0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |

0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 2 | 2 | 2 | 3 | 2 | 3 | 3 | 3 | 3 | 2 | 2 | 3 |

0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 |

0 | 2 | 1 | 0 | 1 | 1 | 2 | 0 | 1 | 0 | 2 | 2 | 0 | 2 | 1 | 0 | 1 | 1 | 2 | 0 | 1 | 0 | 2 | 2 |

It is easy to verify that M is a $(\overline{1},2)$-LA$(24;5,(2,2,4,2,3\left)\right)$.

Replace the symbol 0 by 2 in the 3th column. Juxtapose two such arrays from top to bottom to obtain the following array

${M}^{\prime}$; we list it as its transpose to conserve space.

0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |

0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |

0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 2 | 2 | 2 | 1 | 2 | 1 | 1 | 1 | 1 | 2 | 2 | 1 |

0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 |

0 | 2 | 1 | 0 | 1 | 1 | 2 | 0 | 1 | 0 | 2 | 2 | 0 | 2 | 1 | 0 | 1 | 1 | 2 | 0 | 1 | 0 | 2 | 2 |

It is easy to verify that ${M}^{\prime}$ is a $(\overline{1},2)$-LA$(24;5,(2,2,3,2,3\left)\right)$. □

**Remark** **1.** Construction 2 may produce an optimal $(\overline{1},t)$-LA. For example, a $(\overline{1},2)$-LA$(16;(2,2,3,4\left)\right)$ is shown in Table 1. By Construction 2, we can obtain a $(\overline{1},2)$-LA$(32;(2,2,3,8\left)\right)$, which is optimal by Lemma 4. Fusion is an effective construction for MCAs from CAs, for example, see [

29]. As with CAs, fusion for

$(\overline{1},t)$-LAs guarantees the extension of uniform constructions to mixed cases; however, fusion for a

$(\overline{1},t)$-LA

$(N;k,v)$ may not produce mixed-level

$(\overline{1},t)$-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

$\mathcal{T}\subseteq {\mathcal{I}}_{t}$ with

$\left|\mathcal{T}\right|=d$ and any

$T\in {\mathcal{I}}_{t}$, we have

$\rho (A,T)\subseteq \rho (A,\mathcal{T})\iff T\in \mathcal{T},$ then the array

A is called a

$(d,t)$-DA or a

$(d,t)$-DA

$(N;k,v)$.

**Construction** **3.** (Fusion) Suppose that A is a $(1,t)$-DA$(N;k,v)$ with $t\ge 2$. If A is also a $(\lceil \frac{v}{{v}_{i}}\rceil ,t)$-LA$(N;k,v)$, then a $(\overline{1},t)$-LA$(N;k,(v,\cdots ,v,{v}_{i},v,\cdots ,v))$ exists, where $2\le {v}_{i}<v$.

**Proof.** Let A be a $(1,t)$-DA$(N;k,v)$ over the symbol set V of size v. Suppose that ${a}_{i}(i=1,2,\cdots ,{v}_{i})$ are positive integers with ${a}_{1}+{a}_{2}+\cdots +{a}_{{v}_{i}}=v$. It is clear that there exists a certain ${a}_{i}$ such that ${a}_{i}=\lceil \frac{v}{{v}_{i}}\rceil $ and ${a}_{i}\ge {a}_{j}$, where $1\le i\ne j\le {v}_{i}$. We select ${a}_{1},{a}_{2},\cdots ,{a}_{{v}_{i}}$ elements from V in the ith column of A to form the element sets ${A}_{i}(1\le i\le {v}_{i})$, respectively. The elements in ${A}_{i}(1\le i\le {v}_{i})$ are identical with $1,2,\cdots ,{v}_{i}$, respectively. Then, we obtain an $N\times k$ array ${A}^{\prime}$. Clearly, ${A}^{\prime}$ is an MCA. We only need to prove that ${A}^{\prime}$ is a $(1,t)$-LA by Lemma 1, i.e., for any two distinct t-way interactions ${T}_{1}=\{({a}_{1},{u}_{{a}_{1}}),\cdots ,({a}_{t},{u}_{{a}_{t}})\}$ and ${T}_{2}=\{({b}_{1},{s}_{{b}_{1}}),\cdots ,({b}_{t},{s}_{{b}_{t}})\}$, we have $\rho ({A}^{\prime},{T}_{1})\ne \rho ({A}^{\prime},{T}_{2})$. It is clear that $\rho (A,{T}_{1})=\rho ({A}^{\prime},{T}_{1})$ and $\rho ({A}^{\prime},{T}_{2})=\rho (A,{T}_{2})$ when $i\notin \{{a}_{1},\cdots ,{a}_{t}\}$ and $i\notin \{{b}_{1},\cdots ,{b}_{t}\}$. Hence, $\rho ({A}^{\prime},{T}_{1})\ne \rho ({A}^{\prime},{T}_{2})$.

When $i\in \{{a}_{1},\cdots ,{a}_{t}\}$ and $i\notin \{{b}_{1},\cdots ,{b}_{t}\}$, we can obtain a t-way interaction ${T}_{1}^{\prime}=\{({a}_{1},{u}_{{a}_{1}}),\cdots ,(i,a),\cdots ,({a}_{t},{u}_{{a}_{t}})\}$ of A, where $a\in {A}_{{u}_{i}}$. If $\rho ({A}^{\prime},{T}_{1})=\rho ({A}^{\prime},{T}_{2})$, then $\rho (A,{T}_{1}^{\prime})\subset \rho ({A}^{\prime},{T}_{1})=\rho ({A}^{\prime},{T}_{2})=\rho (A,{T}_{2})$; however, ${T}_{1}^{\prime}\ne {T}_{2}$; as such, it is a contradiction that A is a $(1,t)$-DA$(N;k,v)$. If $i\notin \{{a}_{1},\cdots ,{a}_{t}\}$ and $i\in \{{b}_{1},\cdots ,{b}_{t}\}$, then the similar argument can prove the conclusion.

When

$i\in \{{a}_{1},\cdots ,{a}_{t}\}$ and

$i\in \{{b}_{1},\cdots ,{b}_{t}\}$, it is clear that

$\rho ({A}^{\prime},{T}_{1})\ne \rho ({A}^{\prime},{T}_{2})$ if

${u}_{i}\ne {s}_{i}$. The case

${u}_{i}={s}_{i}$ remains to be considered. Without loss of generality, suppose that

${a}_{j}$ elements are identical with

${u}_{i}$. It is clear that

${T}_{1}$ and

${T}_{2}$ can be obtained from

${\mathcal{T}}_{1}$ and

${\mathcal{T}}_{2}$ by fusion, respectively, where

${\mathcal{T}}_{1}$ and

${\mathcal{T}}_{2}$ are sets of

t-way interactions with

$|{\mathcal{T}}_{1}|=|{\mathcal{T}}_{2}|={a}_{j}$. If

$\rho ({A}^{\prime},{T}_{1})=\rho ({A}^{\prime},{T}_{2})$, then

$\rho ({A}^{\prime},{T}_{1})=\rho (A,{\mathcal{T}}_{1})=\rho ({A}^{\prime},{T}_{2})=\rho (A,{\mathcal{T}}_{2})$. It is a contradiction that

A is a

$(\lceil \frac{v}{{v}_{i}}\rceil ,t)$-LA

$(N;k,v)$ because the existence of

$(\lceil \frac{v}{{v}_{i}}\rceil ,t)$-LA

$(N;k,v)$ implies the existence of

$({a}_{j},t)$-LA

$(N;k,v)$ [

12]. □

Constructions 2 and 3 provide an effective and efficient method to construct a mixed-level

$(\overline{1},t)$-LA from a

$(1,t)$-LA

$(N;k,v)$. The existence of

$(d,t)$-DA

$(N;k,v)$ with

$d\ge 1$ implies the existence of

$(d,t)$-LA

$(N;k,v)$ [

12]. Hence, the array

A in Construction 3 can be obtained by a

$(d,t)$-DA

$(N;k,v)$, which is characterized in terms of super-simple OAs. The existence of super-simple OAs can be found in [

17,

30,

31,

32,

33,

34]. It is noteworthy that the derived array is not optimal. In the remainder of this section, we present two “Roux-type” recursive constructions [

35].

**Construction** **4.** If both a $(\overline{1},t)$-LA$({N}_{1};k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ and a $(\overline{1},t-1)$-LA$({N}_{2};k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))$ exist, then a $(\overline{1},t)$-LA$({N}_{1}+e{N}_{2};k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i}+e,{v}_{i+1},{v}_{i+2}\cdots ,{v}_{k}))$ exists, where $e\ge 0$.

**Proof.** Let

A and

B be the given

$(\overline{1},t)$-LA

$({N}_{1};k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ and

$(\overline{1},t-1)$-LA

$({N}_{2};k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))$, respectively. Clearly, if

$e=0$, then

A is the required array. Now, suppose that

$e\ge 1$. Insert a column vector

$(j,j,\cdots ,j)$ of length

${N}_{2}$ to the front of the

ith column of

B to form an

${N}_{2}\times k$ array

${B}_{j}$, where

$j\in \{{v}_{i},{v}_{i}+1,{v}_{i}+2,\cdots ,{v}_{i}+e-1\}$. Let

$M=({A}^{T}|{B}_{{v}_{i}}^{T}|{B}_{{v}_{i}+1}^{T}|\cdots |{B}_{{v}_{i}+e-1}^{T}{)}^{T}$. Clearly,

M is an MCA

$({N}_{1}+e{N}_{2};t,k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i}+e,{v}_{i+1},{v}_{i+2}\cdots ,{v}_{k}))$ [

9]. By Lemma 1, we only need to prove that

M is a

$(1,t)$-LA, i.e.,

$\rho (M,{T}_{1})\ne \rho (M,{T}_{2})$ for any two distinct

t-way interactions

${T}_{1}$ and

${T}_{2}$, where

${T}_{1}=\{({a}_{1},{u}_{{a}_{1}}),\cdots ,({a}_{t},{u}_{{a}_{t}})\}$ and

${T}_{2}=\{({b}_{1},{s}_{{b}_{1}}),\cdots ,({b}_{t},{s}_{{b}_{t}})\}$. Next, we distinguish the following cases.

Case 1. $i\notin \{{a}_{1},\cdots ,{a}_{t}\}$ and $i\notin \{{b}_{1},\cdots ,{b}_{t}\}$

In this case, because A is a $(\overline{1},t)$-LA, $\rho (A,{T}_{1})\ne \rho (A,{T}_{2})$, $\rho (M,{T}_{1})\ne \rho (M,{T}_{2})$ as A is part of M.

Case 2. $i\notin \{{a}_{1},\cdots ,{a}_{t}\}$ and $i\in \{{b}_{1},\cdots ,{b}_{t}\}$ or $i\in \{{a}_{1},\cdots ,{a}_{t}\}$ and $i\notin \{{b}_{1},\cdots ,{b}_{t}\}$

When $i\notin \{{a}_{1},\cdots ,{a}_{t}\}$ and $i\in \{{b}_{1},\cdots ,{b}_{t}\}$, if ${s}_{i}\notin \{{v}_{i},{v}_{i}+1,\cdots ,{v}_{i}+e-1\}$, then $\rho (A,{T}_{1})\ne \rho (A,{T}_{2})$. Thus, $\rho (M,{T}_{1})\ne \rho (M,{T}_{2})$. If ${s}_{i}\in \{{v}_{i},{v}_{i}+1,\cdots ,{v}_{i}+e-1\}$, then ${T}_{2}$ must be included by rows of ${B}_{i}$, where $i\in \{{v}_{i},{v}_{i}+1,\cdots ,{v}_{i}+e-1\}$; however, it must not be included by any row of A. Clearly, ${T}_{1}$ must be included by some rows of A. Consequently, $\rho (M,{T}_{1})\ne \rho (M,{T}_{2})$. When $i\in \{{a}_{1},\cdots ,{a}_{t}\}$ and $i\notin \{{b}_{1},\cdots ,{b}_{t}\}$, the same argument can prove the conclusion.

Case 3. $i\in \{{a}_{1},\cdots ,{a}_{t}\}$ and $i\in \{{b}_{1},\cdots ,{b}_{t}\}$

Clearly, $\rho (M,{T}_{1})\ne \rho (M,{T}_{2})$ holds whenever ${u}_{i}\ne {s}_{i}$. If ${u}_{i}={s}_{i}\notin \{{v}_{i},{v}_{i}+1,\cdots ,{v}_{i}+e-1\}$, then $\rho (A,{T}_{1})\ne \rho (A,{T}_{2})$, which implies that $\rho (M,{T}_{1})\ne \rho (M,{T}_{2})$. If ${u}_{i}={s}_{i}\in \{{v}_{i},{v}_{i}+1,\cdots ,{v}_{i}+e-1\}$, then ${T}_{1}$ and ${T}_{2}$ must be included by some rows for a certain ${B}_{i}$, where $i\in \{{v}_{i},{v}_{i}+1,\cdots ,{v}_{i}+e-1\}$. Because B is a $(\overline{1},t-1)$-LA, $\rho ({B}_{i},{T}_{1})\ne \rho ({B}_{i},{T}_{2})$, which implies $\rho (M,{T}_{1})\ne \rho (M,{T}_{2})$. □

More generally, we have the following construction.

**Construction** **5.** Let $p\ge 0,q\ge 0$ and $1\le i<j\le k$. If a $(\overline{1},t)$-LA$({N}_{1};k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i},{v}_{i+1},\cdots ,{v}_{j-1},{v}_{j},{v}_{j+1},\cdots ,{v}_{k}))$, $(\overline{1},t-1)$-LA$({N}_{2};k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))$, a $(\overline{1},t-1)$-LA$({N}_{3};k-1,({v}_{1},{v}_{2},\cdots ,{v}_{j-1},{v}_{j+1},\cdots ,{v}_{k}))$ and $(\overline{1},t-2)$-LA$({N}_{4};k-2,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{j-1},{v}_{j+1},\cdots ,{v}_{k}))$ exist, then a $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i}+p,{v}_{i+1},\cdots ,{v}_{j-1},{v}_{j}+q,{v}_{j+1},\cdots ,{v}_{k}))$ exists, where $N={N}_{1}+p{N}_{2}+q{N}_{3}+pq{N}_{4}$.

**Proof.** We begin with a $(\overline{1},t)$-LA$({N}_{1};k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i},{v}_{i+1},\cdots ,{v}_{j-1},{v}_{j},{v}_{j+1},\cdots ,{v}_{k}))$, an ${N}_{1}\times k$ array A that is on ${V}_{1}\times \cdots \times {V}_{i-1}\times {V}_{i}^{\prime}\times {V}_{i+1}\times \cdots \times {V}_{j-1}\times {V}_{j}^{\prime}\times {V}_{j+1}\times \cdots \times {V}_{k}$. Let ${H}_{1}$ and ${H}_{2}$ be two sets with $|{H}_{1}|=p$ and $|{H}_{2}|=q$ such that ${H}_{1}\bigcap {V}_{i}^{\prime}=\varnothing $ and ${H}_{2}\bigcap {V}_{j}^{\prime}=\varnothing $, respectively. Suppose that ${B}^{\prime}$, an ${N}_{2}\times (k-1)$ array, is a $(\overline{1},t-1)$-LA$({N}_{2};k-1,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{k}))$, which is on ${V}_{1}\times \cdots \times {V}_{i-1}\times {V}_{i+1}\times \cdots \times {V}_{k}$. For each row $({a}_{1}$, ${a}_{2},\cdots ,{a}_{i-1},{a}_{i+1},\cdots ,{a}_{k})$ of ${B}^{\prime}$, add $x\in {H}_{1}$ to obtain a k-tuple $({a}_{1}$, ${a}_{2},\cdots ,{a}_{i-1},x,{a}_{i+1},\cdots ,{a}_{k})$. Then, we obtain a $p{N}_{2}\times k$ array from ${B}^{\prime}$, denoted by B. Similarly, from a $(\overline{1},t-1)$-LA$({N}_{3};k-1,({v}_{1},{v}_{2},\cdots ,{v}_{j-1},{v}_{j+1},\cdots ,{v}_{k}))$, we obtain a $q{N}_{3}\times k$ array, denoted by C. For each pair $(x,y)\in {H}_{1}\times {H}_{2}$, we construct k-tuple $({a}_{1},{a}_{2},\cdots ,{a}_{i-1},x,{a}_{i+1},\cdots ,{a}_{j-1},y,{a}_{j+1},\cdots ,{a}_{k})$ for each row of the given $(\overline{1},t-2)$-LA$({N}_{4};k-2,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{j-1},{v}_{j+1},\cdots ,{v}_{k}))$. These tuples result in a $pq{N}_{4}\times k$ array, denoted by D.

Denote ${V}_{i}^{\prime}\cup {H}_{1}={V}_{i}$, ${V}_{j}^{\prime}\cup {H}_{2}={V}_{j}$ and $F=\left(\begin{array}{c}A\hfill \\ B\hfill \\ C\hfill \\ D\hfill \end{array}\right)$. We claim that F, an $({N}_{1}+p{N}_{2}+q{N}_{3}+pq{N}_{4})\times k$ array, is a $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i}+p,{v}_{i+1},\cdots ,{v}_{j-1},{v}_{j}+q,{v}_{j+1},\cdots ,{v}_{k})$ which is on ${V}_{1}\times \cdots \times {V}_{i-1}\times {V}_{i}\times {V}_{i+1}\times \cdots \times {V}_{j-1}\times {V}_{j}\times {V}_{j+1}\times \cdots \times {V}_{k}$.

Clearly, F is an MCA $(N;t,k,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i}+p,{v}_{i+1},\cdots ,{v}_{j-1},{v}_{j}+q,{v}_{j+1},\cdots ,{v}_{k})$. To prove this assertion, we only need to demonstrate that $\rho (F,{T}_{a})\ne \rho (F,{T}_{b})$ for any two distinct t-way interactions ${T}_{a}=\{({a}_{1},{u}_{{a}_{1}}),\cdots ,({a}_{t},{u}_{{a}_{t}})\}$ and ${T}_{b}=\{({b}_{1},{v}_{{b}_{1}}),\cdots ,({b}_{t},{v}_{{b}_{t}})\}$. By similar argument as the proof of Construction 4, we can prove the conclusion except for the case where $i,j\in \{{a}_{1},{a}_{2},\cdots ,{a}_{t}\}$ and $i,j\in \{{b}_{1},{b}_{2},\cdots ,{b}_{t}\}$, ${u}_{i}={v}_{i}\in {H}_{1}$, and ${u}_{j}={v}_{j}\in {H}_{2}$. In this case, ${T}_{a}$ and ${T}_{b}$ are only included by some rows of D. If $\rho (F,{T}_{a})=\rho (F,{T}_{b})$, then $\rho (D,{T}_{a})=\rho (D,{T}_{b})=\rho (F,{T}_{a})=\rho (F,{T}_{b})$. Consequently, $\rho (D,{T}_{a}\setminus \{(i,{u}_{i}),(j,{u}_{j})\})=\rho (D,{T}_{b}\setminus \{(i,{u}_{i}),(j,{u}_{j})\})$, which implies that $\rho ({D}^{\prime},{T}_{a}\setminus \{(i,{u}_{i}),(j,{u}_{j})\})=\rho ({D}^{\prime},{T}_{b}\setminus \{(i,{u}_{i}),(j,{u}_{j})\})$ by the construction of D. It is a contradiction with ${D}^{\prime}$ being a $(\overline{1},t-2)$-LA$({N}_{4};k-2,({v}_{1},{v}_{2},\cdots ,{v}_{i-1},{v}_{i+1},\cdots ,{v}_{j-1},{v}_{j+1},\cdots ,{v}_{k}))$. The proof is completed. □

#### 4.2. Constructions and Existence of Optimal $(\overline{1},t)$-LA$({\prod}_{i=k-t+1}^{k}{v}_{i};k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$

Let $2\le {v}_{1}\le {v}_{2}\le \cdots \le {v}_{k}$. An $N\times k$ array A is called ${\mathrm{MCA}}_{2}^{\ast}({\prod}_{i=k-t+1}^{k}{v}_{i};t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ if $\left|\rho \right(A,T\left)\right|=1$ for any t-way interaction $T\in \mathcal{T}=\{\{(k-t+1,{v}_{k-t+1}),\cdots ,(k,{v}_{k})\}:{v}_{i}\in {V}_{i}\phantom{\rule{4pt}{0ex}}(k-t+1\le i\le k)\}$ and $\left|\rho \right(A,{T}^{\prime}\left)\right|\ge 2$ for any t-way interaction ${T}^{\prime}\notin \mathcal{T}$. If an optimal $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ with $N={\prod}_{i=k-t+1}^{k}{v}_{i}$ exists, then the following condition must be satisfied.

**Lemma** **7.** Let $2\le {v}_{1}\le {v}_{2}\le \cdots \le {v}_{k-t},2{v}_{k-t}\le {v}_{k-t+1}\le {v}_{k-t+2}\le \cdots \le {v}_{k}$. If A is an optimal $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ with $N={\prod}_{i=k-t+1}^{k}{v}_{i}$. Then, A is an ${\mathit{MCA}}_{2}^{\ast}(N;t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$.

**Proof.** Let A be the given optimal $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ with $N={\prod}_{t=k-t+1}^{k}{v}_{i}$. Then, A is an MCA$(N;t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k})$ by Lemma 1. Because $N={\prod}_{t=k-t+1}^{k}{v}_{i}$, we have $\left|\rho \right(A,T\left)\right|=1$ for any t-way interaction $T\in \mathcal{T}$. It follows that $\left|\rho \right(A,{T}^{\prime}\left)\right|\ge 2$ for any t-way interaction ${T}^{\prime}$ of A from the definition of $(\overline{1},t)$-LA, where ${T}^{\prime}\notin \mathcal{T}$. Hence, A is an ${\mathrm{MCA}}_{2}^{\ast}({\prod}_{i=k-t+1}^{k}{v}_{i};t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$, as desired. □

Clearly, an ${\mathrm{MCA}}_{2}^{\ast}(N;t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ is not always a $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$. Next, we present a special case of ${\mathrm{MCA}}_{2}^{\ast}$, which produces optimal $(\overline{1},t)$-LAs. First, we introduce the notion of mixed orthogonal arrays (MOAs).

An MOA, or MOA$(N;t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ is an $N\times k$ array with entries in the ith column from a set ${V}_{i}$ of size ${v}_{i}$ such that each $N\times t$ sub-array contains each t-tuple occurring an equal number of times as a row. When ${v}_{1}={v}_{2}=\cdots ={v}_{k}=v$, an MOA is merely an orthogonal array, denoted by OA$(N;t,k,v)$.

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

${v}_{1}\le {v}_{2}\le \cdots \le {v}_{k}$. By definition of MOA, all

t-tuples occur in the same number of rows for any

$N\times t$ sub-array of an MOA. This number of rows is called

index. It is obvious that

$\left(\genfrac{}{}{0pt}{}{k}{t}\right)$ indices exist. We denote it by

${\lambda}_{1},{\lambda}_{2},\cdots ,{\lambda}_{\left(\genfrac{}{}{0pt}{}{k}{t}\right)}$. If

${\lambda}_{i}\ne {\lambda}_{j}$ for any

$i\ne j$, then an MOA is termed as a

pairwise distinct index mixed orthogonal array, denoted by PDIMOA

$(N;t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$. Moreover, if

${\lambda}_{i}=1$ for a certain

$i\in \{1,2,\cdots ,\left(\genfrac{}{}{0pt}{}{k}{t}\right)\}$ holds, then it is termed as

${\mathrm{PDIMOA}}^{\ast}(N;t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$. It is clear that

$N={\prod}_{i=k-t+1}^{k}{v}_{i}$ in the definition of

${\mathrm{PDIMOA}}^{\ast}$.

**Example** **2.** The transpose of the following array is a ${\mathit{PDIMOA}}^{\ast}(24;2,3,(2,4,6))$. □

The following lemma can be easily obtained by the definition of ${\mathrm{PDIMOA}}^{\ast}$; therefore, we omit the proof herein.

**Lemma** **8.** Suppose that ${v}_{1}\le {v}_{2}\le \cdots \le {v}_{k}$. If A is a ${\mathit{PDIMOA}}^{\ast}({\prod}_{i=k-t+1}^{k}{v}_{i};t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$, then ${v}_{1}<{v}_{2}<\cdots <{v}_{k}$ and ${v}_{i}|{v}_{j}$, where $1\le i\le k-t$ and $k-t+1\le j\le k$.

**Lemma** **9.** Let $2<{v}_{1}<{v}_{2}<\cdots <{v}_{k}$. If a PDIMOA$(N;t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ exists, then a $(\overline{1},t)$-LA$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ exists. Moreover, if $N={\prod}_{i=k-t+1}^{k}{v}_{i}$, then the derived $(\overline{1},t)$-LA is optimal.

**Proof.** Let A be a PDIMOA$(N;t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$. Clearly, A is an MCA. By Lemma 1, we only need to prove that ${T}_{1}\ne {T}_{2}$ implies $\rho (A,{T}_{1})\ne \rho (A,{T}_{2})$, where ${T}_{1}$ and ${T}_{2}$ are two t-way interactions. In fact, if $\rho (A,{T}_{1})=\rho (A,{T}_{2})$, then $|\rho (A,{T}_{1})|=|\rho (A,{T}_{2})|$, which contradicts the definition of a PDIMOA. The optimality can be obtained by Theorem 1. □

We construct an optimal

$(\overline{1},t)$-LA

$(N;k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ with

$N={\prod}_{i=k-t+1}^{k}{v}_{i}$ in terms of

${\mathrm{PDIMOA}}^{\ast}$. First, we have the following simple and useful construction for

${\mathrm{PDIMOA}}^{\ast}$. A similar construction for MOAs was first stated in [

37].

**Construction** **6.** Let $b={r}_{1}{r}_{2}\cdots {r}_{m}<{v}_{2}<\cdots <{v}_{k}$ and ${r}_{1}<{r}_{2}<\cdots <{r}_{m}$. If a ${\mathit{PDIMOA}}^{\ast}({\prod}_{i=k-t+1}^{k}{v}_{i};t,k,({r}_{1}{r}_{2}\cdots {r}_{m},{v}_{2},{v}_{3},\cdots {v}_{k}))$ exists, then a ${\mathit{PDIMOA}}^{\ast}({\prod}_{i=k-t+1}^{k}{v}_{i};t,k+m-1,({r}_{1},{r}_{2},\cdots ,{r}_{m},{v}_{2},{v}_{3},\cdots ,{v}_{k}))$ also exists.

**Proof.** Let A be ${\mathrm{PDIMOA}}^{\ast}(N;t,k,(b,{v}_{2},{v}_{3},\cdots {v}_{k}))$ with $b={r}_{1}{r}_{2},\cdots {r}_{m}$. We can form an $N\times (k+m-1)$ array ${A}^{\prime}$ by replacing the symbols in ${V}_{b}$ by those of ${V}_{{r}_{1}}\times {V}_{{r}_{2}}\times \cdots \times {V}_{{r}_{m}}$. It is easily verified that ${A}^{\prime}$ is the required ${\mathrm{PDIMOA}}^{\ast}$. □

The following construction can be obtained easily; thus, we omit its proof.

**Construction** **7.** Let ${a}_{1}<{a}_{2}<\cdots <{a}_{k}$ and ${b}_{1}<{b}_{2}<\cdots <{b}_{k}$. If both a ${\mathit{PDIMOA}}^{\ast}({\prod}_{i=k-t+1}^{k}{a}_{i};t,k,({a}_{1},{a}_{2},\cdots ,{a}_{k}))$ and a ${\mathit{PDIMOA}}^{\ast}({\prod}_{i=k-t+1}^{k}{b}_{i};t,k,({b}_{1},{b}_{2},\cdots ,{b}_{k}))$ exist, then a ${\mathit{PDIMOA}}^{\ast}({\prod}_{i=k-t+1}^{k}{a}_{i}{b}_{i};t,k,({a}_{1}{b}_{1},{a}_{2}{b}_{2},\cdots ,{a}_{k}{b}_{k}))$ exists. In particular, if both a ${\mathit{PDIMOA}}^{\ast}({\prod}_{i=k-t+1}^{k}{a}_{i};t,k,({a}_{1},{a}_{2},\cdots ,{a}_{k}))$ and an OA$(t,k,v)$ exist, then a ${\mathit{PDIMOA}}^{\ast}({\prod}_{i=k-t+1}^{k}{a}_{i}{v}^{t};t,k,({a}_{1}v,{a}_{2}v,\cdots ,{a}_{k}v))$ exists.

Next, some series of optimal mixed-level $(\overline{1},t)$-LAs are presented. First, we list some known results for later use.

**Lemma** **10.** [38] An OA$({v}^{t};t,t+1,v)$ exists for any integer $v\ge 2,t\ge 2$. The existence of ${\mathrm{PDIMOA}}^{\ast}{(t,t+1,({v}_{1},{v}_{2},\cdots ,{v}_{t},{v}_{t+1}))}^{\prime}$s is determined completely by the following theorem.

**Theorem** **2.** Let ${v}_{1}<{v}_{2}<\cdots <{v}_{t+1}$. A ${\mathit{PDIMOA}}^{\ast}({\prod}_{i=2}^{t+1}{v}_{i};t,t+1,({v}_{1},{v}_{2},\cdots ,{v}_{t},{v}_{t+1}))$ exists if and only if ${v}_{1}|{v}_{i}$ for $2\le i\le t+1$.

**Proof.** The necessity can be easily obtained by Lemma 8. For sufficiency, we write ${v}_{i}={v}_{1}{r}_{i}$ for $i=2,3,\cdots ,t+1$. Clearly, ${r}_{i}\ge 2$ and ${r}_{i}\ne {r}_{j}$ for $2\le i\ne j\le t+1$. We list all t-tuples from ${Z}_{{r}_{2}}\times {Z}_{{r}_{3}}\times \cdots \times {Z}_{{r}_{t+1}}$ to form an MOA$({\prod}_{i=2}^{t+1}{r}_{i};t,t,({r}_{2},{r}_{3},\cdots ,{r}_{t},{r}_{t+1}))$, which is also a ${\mathrm{PDIMOA}}^{\ast}({\prod}_{i=2}^{t+1}{r}_{i};t,t+1,(1,{r}_{2},{r}_{3},\cdots ,{r}_{t},{r}_{t+1})$. Apply Construction 7 with an OA$({v}_{1}^{t};t,t+1,{v}_{1})$ given by Lemma 10 to obtain the required ${\mathrm{PDIMOA}}^{\ast}$. □

More generally, we have the following results.

**Theorem** **3.** Let ${v}_{1}<{v}_{2}<\cdots <{v}_{k}$ and ${v}_{i}={k}_{i}{v}_{1}{v}_{2}\cdots {v}_{k-t}$, where ${k}_{i}\ge 2$, $i=k-t+1,k-t+2,\cdots ,k$. Then, a ${\mathit{PDIMOA}}^{\ast}({\prod}_{i=k-t+1}^{k}{v}_{i};t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ exists.

**Proof.** Let $M={v}_{1}{v}_{2}\cdots {v}_{k-t}$. Then, ${v}_{i}=M{k}_{i}$, where $i=k-t+1,\cdots ,k$. By Theorem 2, a ${\mathrm{PDIMOA}}^{\ast}(N;t,t+1,(M,{v}_{k-t+1},\cdots ,{v}_{k}))$ with $N={\prod}_{i=k-t+1}^{k}{v}_{i}$ exists. Apply Construction 6 to obtain a ${\mathrm{PDIMOA}}^{\ast}({\prod}_{i=k-t+1}^{k}{v}_{i},t,k,({v}_{1},{v}_{2},\cdots ,{v}_{k}))$ as desired. □

**Theorem** **4.** Let ${v}_{1}\le {v}_{2}\le {v}_{3}$ with ${v}_{2}\ge 2{v}_{1}$. Then, an optimal $(\overline{1},2)$-LA$({v}_{2}{v}_{3};3,({v}_{1},{v}_{2},{v}_{3}))$ exists.

**Proof.** First, we construct a ${v}_{2}{v}_{3}\times 3$ array $A=\left({a}_{ij}\right)$ : ${a}_{i+r{v}_{3},1}=(i-1+r)\phantom{\rule{3.33333pt}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}{v}_{1}$, where $i=1,2,\cdots ,{v}_{3}$ and $r=0,1,\cdots ,{v}_{2}-1$; ${a}_{i,2}=\u230a\frac{i-1}{{v}_{3}}\u230b$ and ${a}_{i,3}=(i-1)\phantom{\rule{3.33333pt}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}{v}_{3}$ for $i=1,2,\cdots ,{v}_{2}{v}_{3}$.

We prove that A is an optimal $(\overline{1},2)$-LA. Optimality is guaranteed by Theorem 1. It is clear that A is ${\mathrm{MCA}}_{2}^{\ast}({v}_{2}{v}_{3},({v}_{1},{v}_{2},{v}_{3}))$. Consequently, $\left|\rho \right(A,\left\{\right(1,a),(2,b\left)\right\}\left)\right|\ge 2,\left|\rho \right(A,\left\{\right(1,c),(3,d\left)\right\}\left)\right|\ge 2$ and $\left|\rho \right(A,\left\{\right(2,e),(3,f\left)\right\}\left)\right|=1$, where $a,c\in {V}_{1},b,e\in {V}_{2},d,f\in {V}_{3}$. It is clear that $\rho (A,\{(1,a),(2,b)\left\}\right)\ne \rho (A,\{(2,e),(3,f)\left\}\right)$ and $\rho (A,\{(1,c),(3,d)\left\}\right)\ne \rho (A,\{(2,e),(3,f)\left\}\right)$. We only need to prove $\rho (A,\{(1,a),(2,b)\left\}\right)\ne \rho (A,\{(1,c),(3,d)\left\}\right)$. In fact, by construction, $\rho (A,\{(1,a),(2,b)\})\subset \{r{v}_{3}+1,r{v}_{3}+2,\cdots ,(r+1){v}_{3}\}$ for a certain $r\in \{0,1,2,\cdots ,{v}_{2}-1\}$ but $\{i,i+{v}_{1}{v}_{3}\}\subset \rho (A,\{(1,c),(3,d)\})$, where $i\in \{1,2,\cdots ,{v}_{1}{v}_{3}\}$, which implies $\rho (A,\{(1,a),(2,b)\left\}\right)\ne \rho (A,\{(1,c),(3,d)\left\}\right)$. Thus, A is a $(\overline{1},t)$-LA by Lemma 1. □

The following example illustrates the idea in Theorem 4.

**Example** **3.** The transpose of the following array is an optimal $(\overline{1},2)$-LA$(42;3,(3,6,7\left)\right)$0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 |

0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 5 | 5 | 5 | 5 | 5 | 5 | 5 |

0 | 1 | 2 | 3 | 4 | 5 | 6 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

**Theorem** **5.** Let $2<w<v$ with $v\ge 2w$. Then, an optimal $(\overline{1},1)$-LA$(v;w+1,(w,w,\cdots ,w,v\left)\right)$ exists.

**Proof.** First, we construct a

$2w\times (w+1)$ array

$A=\left({a}_{ij}\right)$ as follows:

When $v>2w$, let $C=\left({c}_{ij}\right)$ be a $(v-2w)\times (w+1)$ array with ${c}_{i,(w+1)}=i-1$ for $i=2w+1,2w+2,\cdots ,v$ and ${c}_{i,j}$ be an arbitrary element for $\{0,1,\cdots ,w-1\}$ with $i=2w+1,2w+2,\cdots ,v,j=1,2,\cdots ,w$. Let $M=A$ and $N={\left({A}^{T}\right|{C}^{T})}^{T}$. It is easy to prove that M and N are the required arrays if $v=2w$ and $v>2w$, respectively. □

The following results need the notion of a Latin square. A

Latin square of order

n is an

$n\times n$ 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

$n\ne 2,6$, there exists a Latin square of order

n with

n distinct transversals [

39].

**Theorem** **6.** Let $2<w\ne 6$ and $v\ge 3w$. Then, an optimal $(\overline{1},1)$-LA$(v;2{w}^{\u230a\frac{v-2w}{w}\u230b}+1,(w,w,\cdots ,w,v))$ exists.

**Proof.** Let ${L}_{0}={\left({a}_{ij}\right)}_{w\times w}$ with ${a}_{ij}=i-1$, where $i=1,2,\cdots ,w,j=1,2,3,\cdots ,w$. For $2<w\ne 6$, a latin square of order w with w disjoint transversals, denoted by ${L}_{1}$, exists. We take each of the w disjoint transversals from ${L}_{1}$ as a column to form a $w\times w$ array ${L}_{2}$, which, clearly, is also a Latin square. Let $\pi =(1,2,3,\cdots ,w)$. The permutation $\pi $ is applied to the columns of L to obtain a new array denoted by ${L}^{\left(\pi \right)}$. If L is a Latin square of w, then ${L}^{\left(\pi \right)},{L}^{\left({\pi}^{2}\right)},\cdots ,{L}^{\left({\pi}^{w-1}\right)}$ are also Latin squares of order w. The corresponding $1,2,\cdots ,w$ columns of ${L}^{\left({\pi}^{i}\right)}$ and ${L}^{\left({\pi}^{j}\right)}$ for $0\le i\ne j\le w-1$ have no common symbols.

Let $A=\left(\begin{array}{c}{L}_{0}\\ {L}_{1}\\ {L}_{1}\end{array}\right|\left.\begin{array}{c}{L}_{0}\\ {L}_{2}\\ {L}_{2}^{\left(\pi \right)}\end{array}\right)$ be a $3w\times 2w$ array. It is easy to verify that A is a $(\overline{1},1)$-LA$(3w;2w,w)$. For each part $\left(\begin{array}{c}{L}_{x}\\ {L}_{y}\\ {L}_{z}\end{array}\right)$ of A, we can construct a $4w\times {w}^{2}$ array of the form $\left(\begin{array}{c}{L}_{x}\\ {L}_{y}\\ {L}_{z}\\ {L}_{z}\end{array}\right|\left.\begin{array}{c}{L}_{x}\\ {L}_{y}\\ {L}_{z}\\ {L}_{z}^{\left(\pi \right)}\end{array}\right|\left.\begin{array}{c}\cdots \\ \cdots \\ \cdots \\ \cdots \end{array}\right|\left.\begin{array}{c}{L}_{x}\\ {L}_{y}\\ {L}_{z}\\ {L}_{z}^{\left({\pi}^{w-1}\right)}\end{array}\right)$. Next, juxtapose these resultant arrays to obtain a $4w\times 2{w}^{2}$ array ${A}^{\prime}$, which is easily verified to be a $(\overline{1},1)$-LA$(4w;2{w}^{2},w)$. Continue this process until the $(\u230a\frac{v-2w}{w}\u230b+2)w\times 2{w}^{\u230a\frac{v-2w}{w}\u230b}$ array B can be obtained. Clearly, B is a $(\overline{1},1)$-LA$((\u230a\frac{v-2w}{w}\u230b+2)w;2{w}^{\u230a\frac{v-2w}{w}\u230b},w)$.

Let $V={(0,1,2,\cdots ,v-1)}^{T}$. Suppose C is a $(v-(\u230a\frac{v-2w}{w}\u230b+2)w)\times 2{w}^{\u230a\frac{v-2w}{w}\u230b}$ array with entries from $\{0,1,\cdots ,w-1\}$, where $v\ne kw$. Write ${B}^{\prime}=B$; if $v=kw$, ${B}^{\prime}={\left({B}^{T}\right|{C}^{T})}^{T}$ if $v\ne kw$. It is easy to verify that $M=\left({B}^{\prime}\right|V)$ is the required array. □

The following theorem considers the case $w=2$.

**Theorem** **7.** Let $v\ge 4$ be a positive integer. Then an optimal $(\overline{1},1)$-LA$(v;{2}^{v}-2v-1,(2,2,\cdots ,2,v))$ exists.

**Proof.** Let $V={(0,1,2,\cdots ,v-1)}^{T}$. We only need to construct a $(\overline{1},1)$-LA$(v;{2}^{v}-2v-2,2)$, A, because $\left(A\right|V)$ is the required array. As $v\ge 4$, 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 $\{0,1\}$ form the $(\overline{1},1)$-LA$(v;k,2)$ 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, $x+y=v$ and $x\ge 2,y\ge 2$. Because there exist x positions with 0s, the number of different column vectors is $\left(\genfrac{}{}{0pt}{}{v}{x}\right)$. Consequently, the number of all the different column vectors is $\left(\genfrac{}{}{0pt}{}{v}{2}\right)+\cdots +\left(\genfrac{}{}{0pt}{}{v}{v-2}\right)={2}^{v}-2v-2$. □