The best approximation problems between the least-squares solution manifolds of two matrix equations

1.   Introduction

Computer experiments, as a widely used method in scientific research, simulate complex real-world problems through complex computer codes ^[1,2,3]. It is very important to plan computer experiments efficiently. Latin hypercube designs (LHDs) introduced by McKay et al. ^[4] are very suitable to plan computer experiments involving only quanlitative factors. Numerous methods have been proposed to construct LHDs with good properties, such as low-dimensional projection property, orthogonality, and other uniform criteria (such as uniform discrepancies, maximin distance, etc.). Computer experiments with both qualitative and quantitative factors have also received a lot of attention (see, for example, ^[5,6,7,8,9]). Sliced space-filling designs and sliced LHDs (SLHDs) are efficient choices when both quantitative and qualitative factors are included in computer experiments ^[10,11]. However, such two types of designs are inefficient due to the increase in the number of runs as the number of level combinations of the qualitative factors increases. To solve this problem, Deng et al. ^[12] first proposed marginally coupled design (MCD), which is more cost-effective in terms of the number of runs, and possesses excellent space-filling properties, i.e., in which the design for the quantitative factors is an LHD, and such quantitative factor design is also an SLHD with respect to each qualitative factor. Some researchers have worked on improving the low-dimensional stratification of the design for the quantitative factors in MCDs; see, among others, ^[13,14] and ^[15]. Other researchers have constructed orthogonal MCDs in which the designs for the quantitative factors are orthogonal ^[16]. In order to improve the stratification between qualitative and quantitative factors, Yang et al. ^[17] proposed doubly coupled design (DCD) which has the following attractive space-filling properties: (1) the whole design is an MCD, and (2) the design points for the quantitative factors form an SLHD with respect to the level combinations of any two qualitative factors. In the above improved MCDs and DCDs, the designs for the qualitative factors are all equal-level orthogonal arrays (OAs). However, there exist qualitative factors being mixed-level in real-world problems, and in MCDs the designs of the qualitative factors are often mixed-level OAs. In this paper, we aim to construct MCDs in which the designs for qualitative factors are mixed-level OAs.

For an MCD $(D_1, D_2)$ where $D_1$ and $D_2$ are the designs for qualitative and quantitative factors, respectively, Deng et al. ^[12] investigated the existence and construction of an MCD for mixed-level qualitative factors. They gave the existence of an MCD $(D_1, D_2)$ with $D_1$ being an $OA\left(n, s_{1}^{k_{1}} s_{2}^{k_{2}}, 2\right)$, $s_{1} = \beta s_{2}$, in terms of the structure of $D_1$. The existence is somewhat limited by the restriction $s_{1} = \beta s_{2}$. To overcome this limitation, we provide a necessary and sufficient condition on both $D_1$ and $D_2$ to ensure the existence of an MCD with $D_1$ being an $OA\left(n, s_{1}^{k_{1}} s_{2}^{k_{2}}, 2\right)$, $s_{1} = \beta s_{2}$, or $s_{1} \neq \beta s_{2}$. Given a small initial MCD with mixed-level qualitative factors, a large MCD with mixed-level qualitative factors can be constructed by Construction 3 of Deng et al. ^[12]. However, Deng et al. ^[12] did not address the question of how to construct the initial MCDs. Fortunately, the MCDs constructed in this paper can be used as initial MCDs for Construction 3 of ^[12]. Therefore, for the MCDs obtained in this paper, the run sizes are more flexible than for the MCDs constructed in Construction 3 of Deng et al. ^[12]. Based on the existence result of ^[12], for $s_{1} = \beta s_{2}$, we give an algorithm to construct MCDs for $D_2$ with a large number of columns. By the necessary and sufficient condition in this paper, two algorithms are proposed to construct MCDs with $D_1$ being an $OA\left(n, 2^{k_{1}} s^{k_{2}}, 2\right)$, $s = 2\beta$, or $s \neq 2\beta$. For the $D_2$ constructed by Construction 3 of Deng et al. ^[12], the $D_2$ only has stratification property in one-dimensional projections. To enhance the space-filling property of $D_2$, we present two algorithms to construct MCDs with $D_2$ possessing stratification properties in two-, three-, or four- dimensional projections.

The paper is organized as follows: Section 2 introduces the basic definitions and notation. Section 3 gives five methods for constructing MCDs with mixed-level qualitative factors. Section 4 provides the conclusions. All proofs are deferred to Appendix A. Some tables are listed in Appendix B.

2.   Definitions and notation

Let $GF(s) = \left \{ \alpha _{0}, \alpha _{1}, \dots, \alpha _{s-1} \right \}$, $\alpha _{0} = 0$, $\alpha _{1} = 1$, denote a Galois field of order $s$, which is simplified as $GF(s) = \left \{ 0, 1, \dots, s-1 \right \}$ if $s$ is a prime. An $n\times p$ matrix is called a Latin hypercube design of $n$ runs and $p$ factors, denoted by $LHD\left(n, p\right)$, if each of its columns is a random permutation of $\{0, 1, \ldots, n-1\}$. An $n \times k$ array $A$ is said to be a mixed-level OA of strength $2$, denoted by $OA\left(n, {s_{1} }^{k_{1}} {s_{2} }^{k_{2}}, 2\right)$, if any $n \times2$ sub-array of $A$ contains all possible level combinations with equal frequency, where the entries in the first $k_{1}$ columns and the last $k_{2}$ columns are taken from $\left \{ 0, 1, \dots, s_{1}-1 \right \}$ and $\left \{ 0, 1, \dots, s_{2}-1 \right \}$, respectively. When $s_{1} = s_{2} = s$ and $k_{1}+k_{2} = k$, the orthogonal array $A$ is equal-level, denoted by $OA\left(n, {s }^{k}, 2\right)$. An $OA\left(s^{t}, s^{k}, 2\right)$ with $v = (s^{t}-1)/(s-1)$ can be constructed using the Rao-Hamming construction, the details of which are described in Section 3.4 of ^[18]. For a prime power $s$, let $\eta _{1}$ and $\eta _{2}$ be two $s$-level independent columns of length $s^2$, where the entries of both $\eta _{1}$ and $\eta _{2}$ are taken from $GF(s) = \left \{ \alpha _{0}, \alpha _{1}, \dots, \alpha _{s-1} \right \}$, $\alpha _{0} = 0$, $\alpha _{1} = 1$. We apply the Rao-Hamming construction to create an $OA\left(s^{2}, s^{s+1}, 2\right)$ $\Phi$ as

where the addition and multiplication operations are based on $GF(s)$. An $OA\left(n, s_{1}^{k_1}s_{2}^{k_2}, 2\right)$ $A$ is said to be a $(\beta _1\mbox{ $\times$ } \beta _2)$ - resolvable OA, denoted by $(\beta_1\mbox{ $\times$ } \beta _2)\mbox{-} ROA\left(n, s_{1}^{k_1}s_{2}^{k_2}, 2\right)$, if for $i = 1, 2$, its rows can be divided into $n/(\beta_{i}s_{i})$ sub-arrays $A_{1}, \dots, A_{n/(\beta_{i}s_{i})}$ of $\beta_{i}s_{i}$ rows each, where $A_{i}$ is an $OA\left(\beta_{i}s_{i}, {s_{1} }^{k_{1}} {s_{2} }^{k_{2}}, 1\right)$ for $i = 1, 2$. In particular, when $s_1\mbox{ = }s_2\mbox{ = }s, \beta_1\mbox{ = } \beta_2\mbox{ = } \beta$, and $k_{1}+k_{2} = k$, then the array reduces to $\beta \mbox{-} ROA\left(n, s^{k}, 2\right)$. If $\beta \mbox{ = }1$, the array $A$ is called a completely resolvable OA (CROA), denoted by $CROA\left(n, s^{k}, 2\right)$.

Suppose $D_{1}$ is an $OA\left(n, s_{1}^{k_1}s_{2}^{k_2}, 2\right)$ and $D_{2}$ is an $LHD\left(n, p\right)$, then the design $D = \left(D_{1}, D_{2}\right)$ is called a MCD, denoted by $MCD\left(n, s_{1}^{k_1}s_{2}^{k_2}, p\right)$, where $D_{1}$ and $D_{2}$ are sub-designs for qualitative factors and quantitative factors, respectively, if for every level of any factor of $D_{1}$, the corresponding rows in $D_{2}$ form a small LHD. When $s_{1} = s_{2} = s$ and $k_{1}+k_{2} = k$, the MCD is denoted by $MCD\left(n, {s }^{k}, p\right)$.

Let $\mathbf{1}_s$ be an $s$ - dimensional column vector whose entries are all ones. An $u \times r$ matrix $A$ with entries from $GF(s)$ is called a difference scheme of strength 2 based on $GF(s)$, denoted by D$\left(u, r, s\right)$, if for all $i$ and $k$ with $1\le i$, $k\le r$, $i \ne k$, the vector difference between the $i$th and $k$th columns contains each element of $GF(s)$ exactly $u/s$ times. Throughout, $D(u, r, s)$ is a $u$-row, $r$-column, and $s$-level difference schme (of strength 2). For an $n\times m$ matrix $X$ and an $s \times p$ matrix $Y$, their Kronecker sum and Kronecker product are defined as $X \oplus Y$ = $\left(x_{ij}+Y\right)$ and $X \otimes Y$ = $\left(x_{ij}Y\right)$, respectively, where x$_{ij}$ is the $(i, j)$th entry of $X$. For a matrix $X = \left(x_{ij} \right)_{n\times m}$, define an $n\times m$ matrix $f\left(X, s \right)$ as

He et al. ^[13] demonstrated a necessary and sufficient condition for the design $\left(D_1, D_2\right)$ being an $MCD\left(n, s^{k}, p\right)$, as stated in Lemma 1.

Lemma 1 (^[13]). Suppose $D_1$ is an $OA\left(n, s^{k}, 2\right)$ and $D_2$ is an $LHD\left(n, p\right)$. Let $d_i$ be the $i$th column of $D_2$ for $i = 1, 2, \ldots, p$. Then, $D = \left(D_1, D_2\right)$ is an $MCD\left(n, s^{k}, p\right)$ if, and only if, for $i = 1, 2, \ldots, p$, the $(D_1, f(d_i, s))$ is an $OA\left(n, s^{k}(n/s), 2\right)$, where $f(\ast, \cdot)$ can be obtained from Equation $(2.1)$.

Lemma 2 given by Deng et al. ^[12] presents a necessary and sufficient condition for the existence of an $MCD\left(n, s_{1}^{k_{1}} s_{2}^{k_{2}}, p\right)$ with $s_{1} = \beta s_{2}$.

Lemma 2 (^[12]). Given that $D_{1}$ is an $OA\left(n, s_{1}^{k_{1}} s_{2}^{k_{2}}, 2\right)$ with $s_{1} = \beta s_{2}$, an $MCD\left(n, s_{1}^{k_{1}} s_{2}^{k_{2}}, p\right)$ $D = \left(D_{1}, D_{2}\right)$ exists if, and only if, $D_{1}$ is a $(1 \times\beta)$-$ROA\left(n, s_{1}^{k_{1}} s_{2}^{k_{2}}, p\right)$ that can be expressed as

such that $\left(A_{i 1}, A_{i 2}\right)$ is an $O A\left(s_{1}, s_{1}^{k_{1}} s_{2}^{k_{2}}, 1\right)$, where $m = n / s_{1}$, and the $A_{i 2}$ is a $CRO A\left(s_{1}, s_{2}^{k_{2}}, 2\right)$, for $i = 1, 2, \ldots, m$.

The necessary and sufficient condition given by Lemma 2 is rather restricted by the restriction $s_{1} = \beta s_{2}$. Similar to Lemma 1, we provide directly a necessary and sufficient condition to break this restriction, as shown in the following lemma.

Lemma 3. Suppose $D_{1} = (\Omega, \Lambda)$ is an $OA\left(n, s_1^{k_1}s_2^{k_2}, 2\right)$, where $\Omega$ and $\Lambda$ are an $OA\left(n, s_1^{k_1}, 2\right)$ and an $OA\left(n, s_2^{k_2}, 2\right)$, respectively, and $D_{2}$ is an $LHD\left(n, p\right)$. Let $d_{i}$ be the $i$th column of $D_{2}$ for $i = 1, 2 \dots, p$. Then, $D$ = $\left(D_{1}, D_{2}\right)$ is an $MCD\left(n, s_{1}^{k_{1}} s_{2}^{k_{2}}, p\right)$ if, and only if, for $i = 1, 2, \ldots, p$, $\left(\Omega, f\left(d_i, s_1\right)\right)$ and $\left(\Lambda, f\left(d_i, s_2\right)\right)$ are an $OA\left(n, s_1^{k_1}(n/s_1), 2\right)$ and an $OA\left(n, s_2^{k_2}(n/s_2), 2\right)$, respectively, where $f\left(\ast, \cdot\right)$ can be obtained from Equation $(2.1)$.

3.   Construction of MCDs

3.1.   Construction of MCDs for $D_1$ being an $OA\left(N, s_1^{k_1}s_2^{k_2}, 2\right)$ with $N=s_1^{2}$ ($s_1= \beta s_2$), $N=2s$ ($s\geq 2$), $N=2\lambda s^{2}$ ($s\geq 2$)

This section presents three construction algorithms to construct MCDs. First, we construct an MCD $D = (D_1, D_2)$ via an $OA\left(s_1^2, s_1^{k_1+2}, 2\right)$ and a $CROA\left(s_1, s_2^{k_2}, 2\right)$ with $s_1 = \beta s_2$. Motivated by Lemma 2, we present the following algorithm.

Theorem 1. For $D = \left(D_1, D_2\right)$ constructed by Algorithm 1, we have

$\rm(i)$ $D_1 = \left(A, B^*\right)$ is an $OA\left(s_1^2, s_1^{k_1} s_2^{k_2}, 2\right)$ with $s_1 = \beta s_2$;

$\rm(ii)$ $D_2$ is an $LHD\left(s_1^2, p\right)$;

$\rm(iii)$ $D = \left(D_1, D_2\right)$ is an $MCD\left(s_1^2, s_{1}^{k_{1}} s_{2}^{k_{2}}, p\right)$.

If we take $G$ in Step 1 of Algorithm 1 to be a regular saturated $OA\left(s_1^2, s_1^{s_1+1}, 2\right)$, then $D = \left(D_1, D_2\right)$ is an $MCD\left(s_1^2, s_1^{s_1-1}s_2^{k_2}, p\right)$; alternatively, the $G$ can also be a non-regular $s_1$-level OA. Hence, $s_1$ may or may not be a prime power. Furthermore, if $B$ is a $CROA\left(s_1, s_2^{s_2}, 2\right)$ and $s_1 = {s_2}^2$ in Step 2, then $D = \left(D_1, D_2\right)$ is an $MCD\left(s_1^2, s_1^{s_1-1}s_2^{s_2}, p\right)$.

In Theorem 1, the MCDs with at most $\beta!s_1!s_2!$ distinct quantitative columns can be constructed from Steps 4 and 5 of Algorithm 1. Thus, Theorem 1 provides the MCD with a large number of quantitative factors. Let $\eta = \beta!s_1!s_2!$, and there can be as many as $(\eta!)/ (p!(\eta-p)!)$ different MCDs. Thus, an optimal $D_2$ under maximin distance criterion ^[19] (or the centered $L_2$-discrepancy criterion ^[20,21]) can be found by ranking the $(\eta!)/(p!(\eta-p)!)$ candidate MCDs or via the simulated annealing ^[22] or the threshold accepting algorithms ^[23] when the number of candidate MCDs is very large.

Example 1. Applying the Rao-Hamming construction to generate an $OA\left(16, 4^{5}, 2\right)$ $G$, then the $A$ is obtained and listed in Table 1. The $CROA\left(4, 2^{2}, 2\right)$ $B$ is obtained as $B = \left(\begin{array}{cccc} 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{array}\right)^{T}$, then $B^{*} = \left(B^{T}, B^{T}, B^{T}, B^{T}\right)^{T}$. Thus, $D_{1}$ is constructed as $D_{1} = \left(A, B^{*}\right)$. Consider the case $p = 3$. In Step 4, $\mu_{1}$, $\mu_{2}$, and $\mu_{3}$ are obtained as $\mu_{1} = (0, 1, 2, 3)^{T}$, $\mu_{2} = (1, 0, 2, 3)^{T}$, $\mu_{3} = (1, 2, 0, 3)^{T}$, then $E$ can be obtained and listed in Table 1. In Step 5, $w_{1}$, $w_{2}$, and $w_{3}$ are obtained as $w_{1} = (0, 1, 2, 3)^{T}$ and $w_{2} = w_{3} = (2, 3, 0, 1)^{T}$, then $C$ can be obtained and listed in Table 1. By the matrix operation of $s_{1}E+C$ in Step 6, $D_{2}$ can be generated. It is easy to check that $D = \left(D_{1}, D_{2}\right)$ is an $MCD\left(16, 4^{3} 2^{2}, 3\right)$, which is provided in Table 2.

Table 1.  Matrices $A$, $E$, and $C$ in Example 1.

$Run$	$A$			$E$			$C$			$Run$	$A$			$E$			$C$
1	0	0	0	0	1	1	0	2	2	9	2	3	1	2	2	0	0	2	2
2	1	1	1	0	1	1	1	3	3	10	3	2	0	2	2	0	1	3	3
3	2	2	2	0	1	1	2	0	0	11	0	1	3	2	2	0	2	0	0
4	3	3	3	0	1	1	3	1	1	12	1	0	2	2	2	0	3	1	1
5	1	2	3	1	0	2	0	2	2	13	3	1	2	3	3	3	0	2	2
6	0	3	2	1	0	2	1	3	3	14	2	0	3	3	3	3	1	3	3
7	3	0	1	1	0	2	2	0	0	15	1	3	0	3	3	3	2	0	0
8	2	1	0	1	0	2	3	1	1	16	0	2	1	3	3	3	3	1	1

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Table 2.  $D = \left(D_{1}, D_{2}\right)$ in Example 1.

$Run$	$MCD(D_1, D_2)$								$Run$	$MCD(D_1, D_2)$
	$D_{1}$					$D_{2}$				$D_{1}$					$D_{2}$
1	0	0	0	0	0	0	6	6	9	2	3	1	0	0	8	10	2
2	1	1	1	1	1	1	7	7	10	3	2	0	1	1	9	11	3
3	2	2	2	0	1	2	4	4	11	0	1	3	0	1	10	8	0
4	3	3	3	1	0	3	5	5	12	1	0	2	1	0	11	9	1
5	1	2	3	0	0	4	2	10	13	3	1	2	0	0	12	14	14
6	0	3	2	1	1	5	3	11	14	2	0	3	1	1	13	15	15
7	3	0	1	0	1	6	0	8	15	1	3	0	0	1	14	12	12
8	2	1	0	1	0	7	1	9	16	0	2	1	1	0	15	13	13

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For the $MCD\left(s_{1}^{2}, s_{1}^{k_1}s_{2}^{k_2}, p\right)$ constructed by Algorithm 1, the $s_1$ can be a non-prime power. The following example provides an illustration of the non-prime power case in Algorithm 1.

Example 2. The $OA\left(144, 12^{7}, 2\right)$ $G$ listed in Table 17 of Appendix B and the $CROA\left(12, 2^{2}, 2\right)$ $B$ listed in Table 3 are obtained from the library of orthogonal arrays maintained by Sloane (http://neilsloane.com/oadir/index.html). Then, divide $G$ into $G = ({l_1}, l_{2}, A)$ listed in Table 17 of Appendix B. Moreover, $B^{*} = \left(B^{T}, B^{T}, B^{T}, B^{T}, B^{T}, B^{T}\right)^{T}$ is obtained and listed in Table 18. Thus, $D_{1}$ is constructed as $D_{1} = \left(A, B^{*}\right)$. Consider the case $p = 3$. In Step 4, $\mu_{1}$, $\mu_{2}$, and $\mu_{3}$ are obtained, which are listed in Table 3, and then the $E$ can be obtained, which is listed in Table 17 of Appendix B. In Step 5, $w_{1}$, $w_{2}$, and $w_{3}$ are obtained, which are listed in Table 3, then the $C$ can be obtained and listed in Table 17 of Appendix B. By the matrix operation of $s_{1}E+C$ in Step 6, $D_{2}$ can be generated. It is easy to check that $D = \left(D_{1}, D_{2}\right)$ is an $MCD\left(144, 12^{5} 2^{2}, 3\right)$, which is provided in Table 18.

Table 3.  Matrix $B$, vectors $u_i$ and $w_i$ in Example 2, where $i = 1, 2, 3$.

$Run$	$B$		$u_1$	$u_2$	$u_3$	$w_1$	$w_2$	$w_3$	$Run$	$B$		$u_1$	$u_2$	$u_3$	$w_1$	$w_2$	$w_3$	$Run$		$B$	$u_1$	$u_2$	$u_3$	$w_1$	$w_2$	$w_3$
1	1	1	0	1	1	0	2	0	5	1	1	4	4	4	4	4	8	9	0	1	8	8	8	8	8	4
2	0	0	1	0	0	1	3	1	6	0	0	5	5	5	5	5	9	10	1	0	9	9	9	9	9	5
3	0	0	2	2	2	2	0	2	7	0	1	6	6	6	6	6	10	11	1	0	10	10	10	10	10	6
4	1	1	3	3	3	3	1	3	8	1	0	7	7	7	7	7	11	12	0	1	11	11	11	11	11	7

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Algorithm 1 can produce some MCDs based on the above Theorem 1, as shown in Table 4.

Table 4.  Some MCDs from Algorithm 1.

D$_1$	D$_2$	MCDs
$OA\left(16, 4^{3}2^{2}, 2\right)$	$LHD\left(16, 2!4!2!\right)$	$MCD\left(16, 4^{3}2^{2}, 2!4!2!\right)$
$OA\left(36, 6^{1}2^{2}, 2\right)$	$LHD\left(36, 3!6!2!\right)$	$MCD\left(36, 6^{1}2^{2}, 3!6!2!\right)$
$OA\left(64, 8^{7}2^{2}, 2\right)$	$LHD\left(64, 4!8!2!\right)$	$MCD\left(64, 8^{7}2^{2}, 4!8!2!\right)$
$OA\left(36, 6^{1}3^{3}, 2\right)$	$LHD\left(36, 2!6!3!\right)$	$MCD\left(36, 6^{1}3^{3}, 2!6!3!\right)$
$OA\left(81, 9^{7}3^{3}, 2\right)$	$LHD\left(81, 3!9!3!\right)$	$MCD\left(81, 9^{7}3^{3}, 3!9!3!\right)$
$OA\left(144, 12^{1}3^{3}, 2\right)$	$LHD\left(144, 4!12!3!\right)$	$MCD\left(144, 12^{1}3^{3}, 4!12!3!\right)$
$OA\left(64, 8^{7}4^{4}, 2\right)$	$LHD\left(64, 2!8!4!\right)$	$MCD\left(64, 8^{7}4^{4}, 2!8!4!\right)$
$OA\left(144, 12^{1}4^{4}, 2\right)$	$LHD\left(144, 3!12!4!\right)$	$MCD\left(144, 12^{1}4^{4}, 3!12!4!\right)$
$OA\left(256, 16^{15}4^{4}, 2\right)$	$LHD\left(256, 4!16!4!\right)$	$MCD\left(256, 16^{15}4^{4}, 4!16!4!\right)$

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To employ Algorithm 1, we need several CROAs. Theorem 3 of He et al. ^[13] gives four types of CROAs, as shown in Lemma 4.

Lemma 4 (^[13]). For a prime $h$ and three positive integers $k$, $t$ ($t\geq2$), and $w$, if $s = h^k$, the following four CROAs exist: $\rm (i)$ $CROA\left(s^{t}, s^{s^{t-1}}, 2\right)$; $\rm (ii)$ $CROA\left(2s^{t}, s^{2s^{t-1}}, 2\right)$; $\rm (iii)$ $CROA\left(4s^{t}, s^{4s^{t-1}}, 2\right)$; and $\rm (iv)$ $CROA\left(h^{w}s^{2}, s^{h^{w}s}, 2\right)$.

According to Theorem 1 and Lemma 4, we can obtain a wealth of MCDs for mixed-level qualitative factors as follows.

Corollary 1. For a prime $h$ and three positive integers $k$, $t$ ($t\geq2$), and $w$, let $s = h^k$, then by Algorithm 1,

$\rm(i)$ if an $OA\left(s^{2t}, (s^{t})^{k_1+2}, 2\right)$ exists, an $MCD\left(s^{2t}, (s^{t})^{k_1}(s)^{s^{t-1}}, p\right)$ can be obtained;

$\rm(ii)$ if an $OA\left(4s^{2t}, (2s^{t})^{k_1+2}, 2\right)$ exists, an $MCD\left(4s^{2t}, (2s^{t})^{k_1}(s)^{2s^{t-1}}, p\right)$ can be obtained;

$\rm(iii)$ if an $OA\left(16s^{2t}, (4s^{t})^{k_1+2}, 2\right)$ exists, an $MCD\left(16s^{2t}, (4s^{t})^{k_1}(s)^{4s^{t-1}}, p\right)$ can be obtained;

$\rm(iv)$ if an $OA\left(h^{2w}s^{4}, (h^{w}s^{2})^{k_1+2}, 2\right)$ exists, an $MCD\left(h^{2w}s^{4}, (h^{w}s^{2})^{k_1}(s)^{h^{w}s}, p\right)$ can be obtained.

If there exists a small initial MCD for mixed-level qualitative factors, then a series of large MCDs for mixed-level qualitative factors can be constructed by Construction 3 of Deng et al. ^[12], as shown in Lemma 5.

Lemma 5 (^[12]). Let $D_1^{(0)} = \left(\Phi, \Psi\right)$ and $D_2^{(0)}$ be an $OA\left(n, s_{1}^{k_{1}} s_{2}^{k_{2}}, 2\right)$ and an $LHD(n, p)$, respectively, where $\Phi$ and $\Psi$ are an $OA\left(n, s_{1}^{k_{1}}, 2\right)$ and an $OA\left(n, s_{2}^{k_{2}}, 2\right)$, respectively. For some $u$, there are two difference schemes $D\left(u, r_{1}, s_{1}\right)$ and $D\left(u, r_{2}, s_{2}\right)$ (of strength 2), denoted by $D(i)$ for $i = 1, 2$, respectively. Let $\mathrm{C} = \left(c_{i j}\right)$ be an $u \times f$ matrix with $c_{i j} = 1$ and $H$ be an $LHD\left(u, pf\right)$. Construct $D_1 = \left(D(1) \oplus \Phi, D(2) \oplus \Psi\right)$ and $D_2 = C \otimes D_2^{(0)}+n H \otimes\textbf{1}_{n}$. If $D^{(0)} = \left(D_1^{(0)}, D_2^{(0)} \right)$ is an MCD, then $D = \left(D_1, D_2 \right)$ is also an MCD, where $D_1$ and $D_2$ are an $OA\left(nu, s_{1}^{k_{1}r_{1}} s_{2}^{k_{2}r_{2}}, 2\right)$ and an $LHD\left(nu, pf\right)$, respectively.

The key to constructing MCDs, $D = \left(D_1, D_2\right)$, using Lemma 5 is the existence of the initial MCD $D^{(0)} = \left(D_1^{(0)}, D_2^{(0)}\right)$. However, the construction method of $D^{(0)} = \left(D_1^{(0)}, D_2^{(0)}\right)$ is not mentioned in ^[12]. Excitingly, the MCDs obtained by Theorem 1 can be used as the initial MCDs. Based on Lemma 5, a large number of MCDs with more columns can be constructed from the initial MCDs obtained by Theorem 1 as follows.

Corollary 2. For $D = \left(D_1, D_2\right)$ constructed by Algorithm 1 and Theorem 1, if there exist two difference schemes $D\left(u, r_{1}, s_{1}\right)$ and $D\left(u, r_{2}, s_{2}\right)$ (of strength 2) for some $u$, then for any integer $f$, an $MCD\left(us_{1}^{2}, (s_{1})^{k_1r_1}(s_{2})^{k_2r_2}, pf\right)$ can be obtained by Lemma 5.

Based on Algorithm 1, Theorem 1 and Corollary 2 can generate a series of MCDs with $D_1$ being an $OA\left(n, s_{1}^{k_{1}} s_{2}^{k_{2}}, 2\right)$ with $s_{1} = \beta s_{2}$, but they can be criticized for the $s_{1} = \beta s_{2}$ restriction. However, when $s_{1}\neq \beta s_{2}$, an MCD also exists, as in the following example.

Example 3. Given $D_{1}$ is an $OA\left(6, 2^{1}3^{1}, 2\right)$ and $D_{2}$ is an $LHD\left(6, 6\right)$ as listed in Table 5, it is easy to verify that $D = \left(D_{1}, D_{2} \right)$ is an $MCD\left(6, 2^{1}3^{1}, 6\right)$ according to Lemma 3.

Table 5.  $D = \left(D_{1}, D_{2}\right)$ in Example 3.

$Run$	$MCD(D_1, D_2)$								$Run$	$MCD(D_1, D_2)$
	$D_{1}$		$D_{2}$							$D_{1}$		$D_{2}$
1	0	0	0	0	2	2	4	4	4	1	0	5	5	3	3	1	1
2	0	1	2	4	0	4	0	2	5	1	1	3	1	5	1	5	3
3	0	2	4	2	4	0	2	0	6	1	2	1	3	1	5	3	5

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Obviously, the $MCD\left(6, 2^{1}3^{1}, 6\right)$ listed in Table 5 cannot be constructed by Algorithm 1. Next, we propose a new algorithm for constructing $MCDs\left(2s, 2^{1}s^{1}, s!\right)$.

Theorem 2. The design $D = (D_{1}, D_{2})$ constructed by Algorithm 2 is an $MCD\left(2s, 2^{1}s^{1}, s!\right)$, where $D_{1}$ is an $OA\left(2s, 2^{1}s^{1}, 2\right)$ and $D_{2}$ is an $LHD\left(2s, s!\right)$.

If $p < s!$, there can be as many as $(s!)/(p!(s-p)!)$ different MCDs from Algorithm 2. Similar to Algorithm 1, an optimal $D_2$ under the maximin distance criterion or the centered $L_2$-discrepancy criterion can be obtained Hickernell ^[19,20]. Next, we provide an example to illustrate Algorithm 2 and Theorem 2.

Example 4. Let $s = 4$, and an $8 \times2$ matrix $D_{1} = \left(L_{1}, L_{2}\right)$ is obtained from Step 1, as shown in Table 6. For $1\le i\le 24$, $D_{2} = \left(d_{1}, d_{2}, \cdots, d_{24}\right)$ is constructed according to Step 2, as shown in Table 6. It is easy to verify that $D = \left(D_{1}, D_{2}\right)$ is an $MCD\left(8, 2^{1}4^{1}, 24\right)$ from Lemma 3, which is provided in Table 6.

Table 6.  $D = (D_{1}, D_{2})$ in Example 4.

$Run$	$MCD(D_1, D_2)$
	$D_1$		$D_2$
1	0	0	0	0	0	0	0	0	2	2	2	2	2	2	4	4	4	4	4	4	6	6	6	6	6	6
2	0	1	2	2	4	4	6	6	0	0	4	4	6	6	0	0	2	2	6	6	0	0	2	2	4	4
3	0	2	4	6	2	6	2	4	4	6	0	6	0	4	2	6	0	6	0	2	2	4	0	4	0	2
4	0	3	6	4	6	2	4	2	6	4	6	0	4	0	6	2	6	0	2	0	4	2	4	0	2	0
5	1	0	7	7	7	7	7	7	5	5	5	5	5	5	3	3	3	3	3	3	1	1	1	1	1	1
6	1	1	5	5	3	3	1	1	7	7	3	3	1	1	7	7	5	5	1	1	7	7	5	5	3	3
7	1	2	3	1	5	1	5	3	3	1	7	1	7	3	5	1	7	1	7	5	5	3	7	3	7	5
8	1	3	1	3	1	5	3	5	1	3	1	7	3	7	1	5	1	7	5	7	3	5	3	7	5	7

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Algorithm 2 can produce some MCDs based on the above Theorem 2, as shown in Table 7.

Table 7.  Some MCDs from Algorithm 2.

D$_1$	D$_2$	MCDs
$OA\left(6, 2^{1}3^{1}, 2\right)$	$LHD\left(6, 3!\right)$	$MCD\left(6, 2^{1}3^{1}, 3!\right)$
$OA\left(8, 2^{1}4^{1}, 2\right)$	$LHD\left(8, 4!\right)$	$MCD\left(8, 2^{1}4^{1}, 4!\right)$
$OA\left(10, 2^{1}5^{1}, 2\right)$	$LHD\left(10, 5!\right)$	$MCD\left(10, 2^{1}5^{1}, 5!\right)$
$OA\left(12, 2^{1}6^{1}, 2\right)$	$LHD\left(12, 6!\right)$	$MCD\left(12, 2^{1}6^{1}, 6!\right)$
$OA\left(14, 2^{1}7^{1}, 2\right)$	$LHD\left(14, 7!\right)$	$MCD\left(14, 2^{1}7^{1}, 7!\right)$
$OA\left(16, 2^{1}8^{1}, 2\right)$	$LHD\left(16, 8!\right)$	$MCD\left(16, 2^{1}8^{1}, 8!\right)$
$OA\left(18, 2^{1}9^{1}, 2\right)$	$LHD\left(18, 9!\right)$	$MCD\left(18, 2^{1}9^{1}, 9!\right)$
$OA\left(20, 2^{1}10^{1}, 2\right)$	$LHD\left(20, 10!\right)$	$MCD\left(20, 2^{1}10^{1}, 10!\right)$
$OA\left(22, 2^{1}11^{1}, 2\right)$	$LHD\left(22, 11!\right)$	$MCD\left(22, 2^{1}11^{1}, 11!\right)$

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In Lemma 5, the MCDs constructed by Theorem 2 can also be used as the initial MCDs for Construction 3 of ^[12]. Based on Lemma 5, a large number of MCDs with $D_1$ being an $OA\left(2us, 2^{r_{1}}s^{r_{2}}, 2 \right)$ can be obtained from the initial MCDs constructed by Theorem 2 as follows.

Corollary 3. For $D = \left(D_1, D_2\right)$ constructed by Algorithm 2 and Theorem 2, if there exist two difference schemes $D\left(u, r_{1}, 2\right)$ and $D\left(u, r_{2}, s\right)$ (of strength 2) for some $u$, then for any integer $f$, an $MCD\left(2us, 2^{r_1}s^{r_2}, pf\right)$ can be obtained by Lemma 5.

In the MCD ($D_1$, $D_2$) constructed by Algorithm 2 and Theorem 2, the $D_1$ has only two columns. In order to construct $D_1$ that can accommodate more qualitative factors, we present Algorithm 3 as follows.

Theorem 3. For $D_{1}^{(0)}$ and $D_{2}^{(0)}$ in Algorithm 3, if $D^{(0)} = \left(D_{1}^{(0)}, D_{2}^{(0)}\right)$ is an $MCD\left(n, s^{m}, p\right)$, then the design $D = \left(D_{1}, D_{2}\right)$ constructed by Algorithm 3 is an $MCD\left(2n, 2^{1}s^{m}, p\right)$, where $D_{1}$ is an $OA\left(2n, 2^{1}s^{m}, 2\right)$, and $D_{2}$ is an $LHD\left(2n, p\right)$.

Remark 1. Note that Algorithm 2 and Algorithm 3 can construct MCDs with $D_1$ being an $OA\left(N, 2^1 s^k, 2\right)$, $s = 2\beta$, or $s \neq 2\beta$, but the values of $N$ in the two Algorithms are different. Algorithm 2 works for $k = 1$ and $N = 2s$, while Algorithm 3 works for $k \geq 2$ and $N = 2\lambda s^{2}$, where $\lambda$ is a positive integer. Thus, Algorithm 3 is able to construct MCDs with more columns in $D_1$ than Algorithm 2, and Algorithm 2 is not a special case of Algorithm 3. For example, for $s = 3$, Algorithm 2 constructs an $MCD\left(6, 2^{1}3^{1}, 6\right)$, where $D_{1}$ and $D_{2}$ are an $OA\left(6, 2^{1}3^{1}, 2\right)$ and an $LHD\left(6, 6\right)$, respectively, while Algorithm 3 constructs an $MCD\left(18, 2^{1}3^{2}, 2\right)$, where $D_{1}$ and $D_{2}$ are an $OA\left(18, 2^{1}3^{2}, 2\right)$ and an $LHD\left(18, 2\right)$, respectively. This shows that Algorithm 2 and Algorithm 3 cannot be replaced by each other.

Next, we provide an example to illustrate Algorithm 3 and Theorem 3.

Example 5. Table 8 gives an $MCD\left(9, 3^{2}, 2\right)$ $D^{(0)} = \left(D_{1}^{(0)}, D_{2}^{(0)}\right)$, where $D_{1}^{(0)}$ is an $OA\left(9, 3^{2}, 2\right)$ and $D_{2}^{(0)}$ is an $LHD\left(9, 2\right)$. Then, an $18 \times3$ matrix $D_{1} = \left(L_{1}, L_{2}\right)$ is obtained by the operations $L_{1} = (0, 1)^{T}\otimes \textbf{1}_{n}$ and $L_{2} = \textbf{1}_{2}\otimes D_{1}^{(0)}$ in Step 2, as shown in Table 9. An $18 \times 2$ matrix $D_{2}$ is obtained by the operations $D_{2} = \left(\left(2D_{2}^{(0)}\right)^{T}, \left((2n-1)\textbf{1}_{n}- 2D_{2}^{(0)}\right)^{T} \right)^{T}$ in Step 3, as shown in Table 9. It is easy to verify that $D = (D_{1}, D_{2})$ is an $MCD\left(18, 2^{1}3^{2}, 2\right)$ from Lemma 3, which is provided in Table 9.

Table 8.  $D^{(0)} = \left(D_{1}^{(0)}, D_{2}^{(0)}\right)$ in Example 5.

$Run$	$MCD(D_{1}^{(0)}, D_{2}^{(0)})$				$Run$	$MCD(D_{1}^{(0)}, D_{2}^{(0)})$				$Run$	$MCD(D_{1}^{(0)}, D_{2}^{(0)})$
	$D_1$		$D_2$			$D_1$		$D_2$			$D_1$		$D_2$
1	0	0	0	2	4	1	0	4	4	7	2	0	8	7
2	0	1	3	8	5	1	1	7	0	8	2	1	2	3
3	0	2	6	5	6	1	2	1	6	9	2	2	5	1

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Table 9.  $D = (D_{1}, D_{2})$ in Example 5.

$Run$	$MCD(D_1, D_2)$					$Run$	$MCD(D_1, D_2)$					$Run$	$MCD(D_1, D_2)$
	$D_1$			$D_2$			$D_1$			$D_2$			$D_1$			$D_2$
1	0	0	0	0	4	7	0	2	0	16	14	13	1	1	0	9	9
2	0	0	1	6	16	8	0	2	1	4	6	14	1	1	1	3	17
3	0	0	2	12	10	9	0	2	2	10	2	15	1	1	2	15	5
4	0	1	0	8	8	10	1	0	0	17	13	16	1	2	0	1	3
5	0	1	1	14	0	11	1	0	1	11	1	17	1	2	1	13	11
6	0	1	2	2	12	12	1	0	2	5	7	18	1	2	2	7	15

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Algorithm 3 can produce some MCDs based on the above Theorem 3, as shown in Table 10.

Table 10.  Some MCDs from Algorithm 3.

$MCD\left({D_1}^{(0)}, {D_2}^{(0)} \right)$				$MCD\left(D_1, D_2 \right)$
Source	${D_1}^{(0)}$	${D_2}^{(0)}$		$D_1$	$D_2$	MCDs
Table 5	$OA\left(9, 3^{2}, 2\right)$	$LHD\left(9, 2 \right)$		$OA\left(18, 2^{1}3^{2}, 2\right)$	$LHD\left(18, 2 \right)$	$MCD\left(18, 2^{1}3^{2}, 2 \right)$
Table B1	$OA\left(27, 3^{9}, 2\right)$	$LHD\left(27, 4 \right)$		$OA\left(54, 2^{1}3^{9}, 2\right)$	$LHD\left(54, 4 \right)$	$MCD\left(54, 2^{1}3^{9}, 4 \right)$
	$OA\left(32, 4^{8}, 2\right)$	$LHD\left(32, 7 \right)$		$OA\left(64, 2^{1}4^{8}, 2\right)$	$LHD\left(64, 7\right)$	$MCD\left(64, 2^{1}4^{8}, 7 \right)$
	$OA\left(32, 4^{8}, 2\right)$	$LHD\left(32, 7 \right)$		$OA\left(64, 2^{1}4^{8}, 2\right)$	$LHD\left(64, 7\right)$	$MCD\left(64, 2^{1}4^{8}, 7 \right)$
	$OA\left(100, 5^{20}, 2\right)$	$LHD\left(200, 19 \right)$		$OA\left(100, 2^{1}5^{20}, 2\right)$	$LHD\left(200, 19\right)$	$MCD\left(200, 2^{1}5^{20}, 19 \right)$
Example 2	$OA\left(49, 7^{5}, 2\right)$	$LHD\left(49, 3 \right)$		$OA\left(98, 2^{1}7^{5}, 2\right)$	$LHD\left(98, 3\right)$	$MCD\left(98, 2^{1}7^{5}, 3 \right)$
	$OA\left(64, 8^{7}, 2\right)$	$LHD\left(64, 2 \right)$		$OA\left(128, 2^{1}8^{7}, 2\right)$	$LHD\left(128, 2 \right)$	$MCD\left(128, 2^{1}8^{7}, 2 \right)$
	$OA\left(81, 9^{8}, 2\right)$	$LHD\left(81, 2 \right)$		$OA\left(162, 2^{1}9^{8}, 2\right)$	$LHD\left(162, 2 \right)$	$MCD\left(162, 2^{1}9^{8}, 2 \right)$
1Table 5, Table B1 and Example 2 come from [15], [13] and [12], respectively.

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Similar to Corollary 2 and Corollary 3, we can obtain the following Corollary 4 for the initial MCDs constructed by Algorithm 3 and Theorem 3.

Corollary 4. For $D = \left(D_1, D_2\right)$ constructed by Algorithm 3 and Theorem 3, if there exist two difference schemes $D\left(u, r_{1}, 2\right)$ and $D\left(u, r_{2}, s\right)$ (of strength 2) for some $u$, then for any integer $f$, an $MCD\left(2un, 2^{r_{1}}s^{r_{2}m}, pf\right)$ can be obtained by Lemma 5.

Table 11 presents some designs $D_1$ for mixed-level qualitative factors in MCDs constructed via Algorithms 1, 2, and 3. In the fourth column of Table 11, the $D_1$'s are obtained by Construction 3 of Deng et al. ^[12] from the initial designs listed in the first three columns.

Table 11.  Some designs $D_1$ constructed by different algorithms.

Algorithm 1	Algorithm 2	Algorithm 3	Corollaries
D$_1$	D$_1$	D$_1$	D$_1$	source
$OA\left(16, 4^{3}2^{2}, 2\right)$	$OA\left(6, 2^{1}3^{1}, 2\right)$	$OA\left(18, 2^{1}3^{3}, 2\right)$	$OA\left(64, 4^{12}2^{8}, 2\right)$	corollary 2
$OA\left(36, 6^{1}2^{2}, 2\right)$	$OA\left(8, 2^{1}4^{1}, 2\right)$	$OA\left(32, 2^{1}4^{4}, 2\right)$	$OA\left(512, 8^{56}2^{16}, 2\right)$	corollary 2
$OA\left(64, 8^{7}2^{2}, 2\right)$	$OA\left(10, 2^{1}5^{1}, 2\right)$	$OA\left(50, 2^{1}5^{5}, 2\right)$	$OA\left(729, 9^{63}3^{15}, 2\right)$	corollary 2
$OA\left(36, 6^{1}3^{3}, 2\right)$	$OA\left(12, 2^{1}6^{1}, 2\right)$	$OA\left(72, 2^{1}6^{2}, 2\right)$	$OA\left(36, 2^{2}3^{3}, 2\right)$	corollary 3
$OA\left(81, 9^{7}3^{3}, 2\right)$	$OA\left(14, 2^{1}7^{1}, 2\right)$	$OA\left(98, 2^{1}7^{7}, 2\right)$	$OA\left(32, 2^{4}4^{4}, 2\right)$	corollary 3
$OA\left(144, 12^{1}3^{3}, 2\right)$	$OA\left(16, 2^{1}8^{1}, 2\right)$	$OA\left(128, 2^{1}8^{8}, 2\right)$	$OA\left(64, 2^{8}4^{4}, 2\right)$	corollary 3
$OA\left(64, 8^{7}4^{4}, 2\right)$	$OA\left(18, 2^{1}9^{1}, 2\right)$	$OA\left(162, 2^{1}9^{9}, 2\right)$	$OA\left(108, 2^{2}3^{3}, 2\right)$	corollary 4
$OA\left(144, 12^{1}4^{4}, 2\right)$	$OA\left(20, 2^{1}10^{1}, 2\right)$	$OA\left(200, 2^{1}10^{2}, 2\right)$	$OA\left(128, 2^{4}4^{16}, 2\right)$	corollary 4
$OA\left(256, 16^{15}4^{4}, 2\right)$	$OA\left(22, 2^{1}11^{1}, 2\right)$	$OA\left(242, 2^{1}11^{11}, 2\right)$	$OA\left(500, 2^{2}5^{25}, 2\right)$	corollary 4

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For the $MCD\left(s_{1}^{2}, s_{1}^{k_1}s_{2}^{k_2}, p\right)$ constructed by Algorithm 1, the relation $s_{1} = \beta s_{2}$ is indispensable. When $s = 2\beta$, the $MCD\left(s^{2}, 2^{2}s^{s-1}, p\right)$, $MCD\left(2s, 2^{1}s^{1}, s!\right)$, and $MCD\left(2\lambda s^{2}, 2^{1}s^{m}, p\right)$ ($\lambda \ge1$) can be constructed by Algorithms 1, 2, and 3, respectively. Clearly, the three MCDs have different numbers of run sizes. For $s\neq 2\beta$, the $MCD\left(2s, 2^{1}s^{1}, s!\right)$ and $MCD\left(2\lambda s^{2}, 2^{1}s^{m}, p\right)$ ($\lambda \ge1$) can also be obtained using Algorithms 2 and 3, respectively. Algorithm 2 is not a special case of Algorithm 3 due to the different number of run sizes for the constructed MCDs.

3.2.   Construction of MCDs for $D_1$ being an $OA\left(s_1^{2}, s_1^{k_1}s_2^{k_2}, 2\right)$ with $s_1=s_2^{2}$ and $D_2$ with the better space-filling property

In the MCDs ($D_1, D_2$) constructed by the above three algorithms, the space-filling property of $D_2$ is not considered. The space-filling property is very important for the quantitative factor design $D_2$. In this section, we introduce another algorithm to construct MCDs $D = (D_1, D_2)$ for $D_2$ with the better space-filling property.

Theorem 4. For $s_{1} = s_{2}^{2}$, $D_{1}$, and $D_{2}$ obtained in Algorithm 4, we have

$\rm(i)$ $D_1$ is an $OA\left(s_{1}^{2}, s_{1}^{s_{1}-1}s_{2}^{s_2}, 2\right)$;

$\rm(ii)$ $D_2$ is an $LHD\left(s_{1}^{2}, 2k\right)$, where if $s_{2}$ is odd, $k = (s_{2}+1)/2$; if $s_{2}$ is even, $k = s_{2}/2$;

$\rm(iii)$ $(D_1, D_2)$ is an $MCD\left(s_{1}^{2}, s_{1}^{s_{1}-1}s_{2}^{s_2}, 2k\right)$;

$\rm(iv)$ any two distinct columns of $D_2$ achieve $s_2 \times s_2$ grids stratification.

Theorem 4 (iv) tells us that $D_2$ has two-dimensional projection property without considering $D_1$. For each level of any factor in $D_1$, and for each level combination of any two factors in some columns of $D_1$, the corresponding rows in $D_2$ can also achieve the two-dimensional space-filling property, as stated in the following corollary.

Corollary 5. For $D = \left(D_1, D_2\right)$ ($D_1 = (F_{0}, U_{0}, u_{1})$) constructed by Algorithm 4 and Theorem 4, we have

$\rm(i)$ the rows in $D_2$ corresponding to each level of any factor in $D_1$ can achieve stratification on the $s_2 \times s_2$ grids in any two-dimensional projection;

$\rm(ii)$ the rows in $D_2$ corresponding to each level combination of any two factors in $(U_{0}, u_{1})$ can achieve stratification on the $s_2 \times s_2$ grids in any two-dimensional projection.

Next, we provide an example to illustrate Algorithm 4 and Theorem 4.

Example 6. Consider the case $s_1 = 4$ and $s_2 = 2$. An $OA\left(16, 4^{5}, 2\right)$ $F$ and an $OA\left(4, 2^{3}, 2\right)$ $H$ are obtained from the Rao-Hamming construction. Divide $F$ as $F = (F_{0}, f_{1}, f_{2})$ listed in Table 12. For the $H$ listed in Table 12, we obtain an $16\times 3$ matrix $U$ by replacing the levels $0, 1, 2, 3$ of the $f_{1}$ with the $1$st, $2$nd, $3$rd, and the $4$th row of the $H$, respectively. Then partition $U$ as $U = \left(U_{0}, u_{1}, u_{2}\right)$ listed in Table 12. In Step 3 and Step 4, $H^{*}$ is the first 2 columns of $H$ and $k = s_{2}/2 = 1$, after replacing the levels $0, 1, 2, 3$ of the $f_{2}$ by the $1$st, 2nd, 3rd, and the $4$th row of the $H^{*}$, respectively. Then, $V$ is obtained, and denote $V$ as $V = (v_{1}, v_{2})$ listed in Table 12.

Table 12.  Matrices $H$, $F$, $U$, and $V$ in Example 6.

$Run$				$F$						$U$				$V$		$Run$	$F$						$U$				$V$
	${H}$			$F_1$			$f_1$	$f_2$		$U_0$	$u_1$	$u_2$		$v_1$	$v_2$		$F_1$			$f_1$	$f_2$		$U_0$	$u_1$	$u_2$		$v_1$	$v_2$
1	0	0	0	0	0	0	0	0		0	0	0		0	0	9	0	2	3	1	2		0	1	1		1	0
2	0	1	1	1	1	1	1	0		0	1	1		0	0	10	1	3	2	0	2		0	0	0		1	0
3	1	0	1	2	2	2	2	0		1	0	1		0	0	11	2	0	1	3	2		1	1	0		1	0
4	1	1	0	3	3	3	3	0		1	1	0		0	0	12	3	1	0	2	2		1	0	1		1	0
5				0	1	2	3	1		1	1	0		0	1	13	0	3	1	2	3		1	0	1		1	1
6				1	0	3	2	1		1	0	1		0	1	14	1	2	0	3	3		1	1	0		1	1
7				2	3	0	1	1		0	1	1		0	1	15	2	1	3	0	3		0	0	0		1	1
8				3	2	1	0	1		0	0	0		0	1	16	3	0	2	1	3		0	1	1		1	1

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From Step 5, $D_1 = (F_{0}, U_{0}, u_{1})$, and it is easy to check that $D_{1}$ is an $O A\left(16, 4^{3} 2^{2}, 2\right)$. In Step 6, let $W_1 = V$, $W_2 = (v_{2}, v_{1})$, $W_3 = (u_{2}, u_{2})$, $W_4 = (u_{1}, u_{1})$, then by matrix operation of $s_{2}^{3}W_{1}+s_{2}^{2}W_{2}+s_{2}W_{3}+W_{4}$, $D_{2}$ can be generated. It is easy to verify that $(D_1, D_2)$ is an $MCD\left(16, 4^{3} 2^{2}, 2\right)$, which is provided in Table 13.

Table 13.  $D = \left(D_{1}, D_{2}\right)$ in Example 6.

$Run$	$MCD(D_1, D_2)$							$Run$	$MCD(D_1, D_2)$
	$D_{1}$					$D_{2}$			$D_{1}$					$D_{2}$
1	0	0	0	0	0	0	0	9	0	2	3	0	1	11	7
2	1	1	1	0	1	3	3	10	1	3	2	0	0	8	4
3	2	2	2	1	0	2	2	11	2	0	1	1	1	9	5
4	3	3	3	1	1	1	1	12	3	1	0	1	0	10	6
5	0	1	2	1	1	5	9	13	0	3	1	1	0	14	14
6	1	0	3	1	0	6	10	14	1	2	0	1	1	13	13
7	2	3	0	0	1	7	11	15	2	1	3	0	0	12	12
8	3	2	1	0	0	4	8	16	3	0	2	0	1	15	15

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Next, let $D_2 = (d_1, d_2)$. It is easy to see that $d_1$ and $d_2$ achieve stratification on $2 \times 2$ grids, as shown in Figure 1.

Figure 1.  Stratification on $2\times 2$ grids.

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Algorithm 4 can produce some MCDs based on the above Theorem 4, as shown in Table 14.

Table 14.  Some MCDs from Algorithm 4.

D$_1$	D$_2$	MCDs
$OA\left(16, 4^{3}2^{2}, 2\right)$	$LHD\left(6, 2\right)$	$MCD\left(6, 4^{3}2^{2}, 2 \right)$
$OA\left(81, 9^{8}3^{3}, 2\right)$	$LHD\left(8, 4\right)$	$MCD\left(8, 9^{8}3^{3}, 4 \right)$
$OA\left(256, 16^{15}4^{4}, 2\right)$	$LHD\left(10, 4\right)$	$MCD\left(10, 16^{15}4^{4}, 4 \right)$
$OA\left(625, 25^{24}5^{5}, 2\right)$	$LHD\left(12, 6 \right)$	$MCD\left(12, 25^{24}5^{5}, 6 \right)$
$OA\left(1296, 36^{35}6^{6}, 2\right)$	$LHD\left(14, 6 \right)$	$MCD\left(14, 36^{35}6^{6}, 6 \right)$

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Next, we introduce the following algorithm to generate MCDs by modifying Algorithm 4, that is, we rearrange the columns of F in Algorithm 4 and apply the idea of Step 4 in Algorithm 4 twice.

Theorem 5. For $s_{1} = s_{2}^{2}$, $D_{1}$, and $D_{2}$ obtained in Algorithm 5, we have

$\rm(i)$ $D_1$ is an $OA\left(s_{1}^{2}, s_{1}^{s_{1}-2}s_{2}^{s_2}, 2\right)$;

$\rm(ii)$ $D_2$ is an $LHD\left(s_{1}^{2}, 4k\right)$, where if $s_{2}$ is odd, $k = (s_{2}+1)/2$; if $s_{2}$ is even, $k = s_{2}/2$;

$\rm(iii)$ $(D_1, D_2)$ is an $MCD\left(s_{1}^{2}, s_{1}^{s_{1}-2}s_{2}^{s_2}, 4k\right)$.

Theorem 6. For $D_1$ and $D_2$ constructed by Algorithm 5 and Theorem 5, $D_2$ can be partitioned into two disjoint groups of $2k$ columns, i.e., $D_2 = (D_{21}, D_{22})$. For $i = 1, 2, \ldots, 2k$, let $d_{1}^{i}$ and $d_{2}^{i}$ be the $i$th columns of $D_{21}$ and $D_{22}$, respectively. Then,

$\rm(i)$ any two distinct columns of $D_2$ achieve $s_2 \times s_2$ grids stratification;

$\rm(ii)$ any two columns from different groups, $d_1^{j}$ and $d_2^{j'}$, achieve $s_2^2 \times s_2^2$ grids stratification, where $j, j' = 1, 2, \ldots, k$;

$\rm(iii)$ any three columns from two different groups, $d_i^{j}$, $d_{i'}^{t}$ and $d_{i'}^{h}$, achieve $s_2^2 \times s_2 \times s_2$ grids stratification, where $i, i' = 1, 2$, $i \neq i'$, $j, t, h = 1, 2, \ldots, 2k$, $t \neq h$;

$\rm(iv)$ any four columns from two different groups, $d_i^{j}$, $d_i^{r}$, $d_{i'}^{t}$ $d_{i'}^{h}$, achieve $s_2 \times s_2 \times s_2 \times s_2$ grids stratification, where $i, i' = 1, 2$, $i\neq i^{'}$, $j, r, t, h = 1, 2, \ldots, 2k$, , $j\neq r$, and $t \neq h$.

According to Theorem 6, there are $4k^2$ two-column groups achieving stratifications on $s_2^2 \times s_2^2$ grids, $2k^2(2k-1)$ three-column groups achieving stratifications on $s_2^2 \times s_2 \times s_2$ grids, and $k^2(2k-1)^2$ four-column groups achieving stratifications on $s_2 \times s_2 \times s_2 \times s_2$ grids, respectively. Theorem 6 shows that a large number of columns in $D_2$ have good two-, three-, or four-dimensional projections. Next, we provide an example to illustrate Algorithm 5, Theorem 5, and Theorem 6.

Example 7. Consider the case $s_1=9$ and $s_2=3$. An an $OA\left(9, 3^{4}, 2\right)$ $H$ listed in Table 19 of Appendix B and an $OA\left(81, 9^{10}, 2\right)$ $F$ listed in Table 20 of Appendix B are obtained from the library of orthogonal arrays maintained by Sloane (http://neilsloane.com/oadir/index.html). Divide $F$ as $F=({F_0}^{*}, f_{0}, f_{1}, f_{2})$ listed in Table 20 of Appendix B. For the $H$, obtain an $81\times 4$ matrix $U$ by replacing the levels $0, 1, \ldots, 8$ of the $f_{1}$ with the $1$st, $2$nd, $3$rd, $\ldots$, the $9$th row of the $H$ according to Step 2 of Algorithm 4, respectively. Then, partition $U$ as $U=(U_{0}, u_{1}, u_{2})$ listed in Table 20 of Appendix B. In Step 3 and Step 4 of Algorithm 4, due to $s_2=3$, let $H^{*}=H$ and $k=(s_{2}+1)/2 =2$, after replacing the levels $0, 1, \ldots, 8$ of the $f_{2}$ by the $1$st, 2nd, 3rd, $\ldots$, the $9$th row of the $H^{*}$, respectively. Then, $V$ is obtained, and denote $V$ as $V=(v_{1}, v_{2}, v_{3}, v_{4})$ listed in Table 20 of Appendix B. From Step 3, $D_1=({F_0}^{*}, U_{0}, u_{1})$, and it is easy to check that $D_{1}$ is an $O A\left(81, 9^{7} 3^{3}, 2\right)$. In Step 4, obtain an $81\times 4$ matrix $Z$ by replacing the levels $0, 1, \ldots, 8$ of the $f_{0}$ with the $1$st, $2$nd, $3$rd, $\ldots$, the $9$th row of the $H^{*}$, respectively. Then, $Z$ is obtained, and denote $Z$ as $Z=(z_{1}, z_{2}, z_{3}, z_{4})$ listed in Table 20 of Appendix B. In Step 5, let $W_1=V$, $W_2=(v_{2}, v_{1}, v_{4}, v_{3})$, $W_3=(u_{2}, u_{2}, u_{2}, u_{2})$, $W_4=(u_{1}, u_{1}, u_{1}, u_{1})$ according to Step 6 of Algorithm 4 and let $X_1=Z$, $X_2=(z_{2}, z_{1}, z_{4}, z_{3})$, then by matrix operation of $s_{2}^{3}W_{1}+s_{2}^{2}W_{2}+s_{2}W_{3}+W_{4}$ and $s_{2}^{3}X_{1}+s_{2}^{2}X_{2}+s_{2}W_{3}+W_{4}$, $D_{21}$ and $D_{22}$ can be generated, respectively. Then, $D_2=(D_{21}, D_{22})$. It is easy to verify that $(D_{1}, D_{2})$ is an $MCD\left(81, 9^{7} 3^{3}, 8\right)$ listed in Table 21 of Appendix B. Next, let the first two columns of $D_{21}$ be $d_1$ and $d_2$, and the first two columns of $D_{22}$ be $d_3$, $d_4$. After collapsing the levels of $d_1$, $d_2$, $d_3$, $d_4$, it is easy to see that the $d_1$, $d_2$, $d_3$, $d_4$ satisfies the stratifications of (i) and (ii) in Theorem 6, as shown in Figure 2 and Figure 3.

Figure 2.  Stratification on $3\times 3$ grids.

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Figure 3.  Stratification on $9\times 9$ grids.

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Inspired by Corollary 5, Corollary 6 is given as follows.

Corollary 6. For $D = \left(D_1, D_2\right)$ ($D_1 = ({F_0}^{*}, U_{0}, u_{1})$, $D_2 = (D_{21}, D_{22})$) constructed by Algorithm 5 and Theorem 5, we have

$\rm(i)$ the rows in $D_{2i}$, $i = 1, 2$, corresponding to each level of any factor in $D_1$ can achieve stratification on the $s_2 \times s_2$ grids in any two-dimensional projection;

$\rm(ii)$ the rows in $D_{2i}$, $i = 1, 2$, corresponding to each level combination of any two factors in $(U_{0}, u_{1})$ can achieve stratification on the $s_2 \times s_2$ grids in any two-dimensional projection.

Algorithm 5 can produce some MCDs based on the above Theorem 5, as shown in Table 15.

Table 15.  Some MCDs from Algorithm 5.

D$_1$	D$_2$	MCDs
$OA\left(16, 4^{2}2^{2}, 2\right)$	$LHD\left(6, 4 \right)$	$MCD\left(6, 4^{3}2^{2}, 4 \right)$
$OA\left(81, 9^{7}3^{3}, 2\right)$	$LHD\left(8, 8\right)$	$MCD\left(8, 9^{8}3^{3}, 8 \right)$
$OA\left(256, 16^{14}4^{4}, 2\right)$	$LHD\left(10, 8 \right)$	$MCD\left(10, 16^{15}4^{4}, 8 \right)$
$OA\left(625, 25^{23}5^{5}, 2\right)$	$LHD\left(12, 12 \right)$	$MCD\left(12, 25^{24}5^{5}, 12 \right)$
$OA\left(1296, 36^{33}6^{6}, 2\right)$	$LHD\left(14, 12 \right)$	$MCD\left(14, 36^{35}6^{6}, 12 \right)$

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4.   Conclusions

Many researchers have constructed MCDs for equal-level qualitative factors. However, there has been less research on MCDs when the qualitative factors are mixed-level. Construction 3 of Deng et al. ^[12] generates large MCDs for mixed-level qualitative factors from small initial MCDs for mixed-level qualitative factors. Obviously, such a construction is not valid when the initial MCD does not exist. The key to Construction 3 of Deng et al. ^[12] is how to obtain a small initial MCD. However, they did not answer the question. Fortunately, the constructed MCDs in this paper can be considered as the initial MCDs for Construction 3 of ^[12].

In this paper, we propose five algorithms to construct MCDs where the designs for the qualitative factors are mixed-level. The construction of the first algorithm is characterized by the fact that it is based on an $OA\left(s_1^2, s_1^{k_1+2}, 2\right)$ and a $CROA\left(s_1, s_2^{k_2}, 2\right)$ with $s_1 = \beta s_2$. Clearly, its constructed MCD is limited by $s_{1} = \beta s_{2}$. To break this limitation, Algorithms 2 and 3 employ a mirror-symmetric structure to construct $D_2$. Moreover, the $D_1$ constructed by Algorithm 3 can accommodate more columns than the one constructed by Algorithm 2, and the two algorithms construct different numbers of run sizes. The fourth and fifth algorithms construct the MCD using the level replacement method and the rotation method, where $D_2$ has stratification in two- or higher-dimensional projection. Finally, Table 16 lists some types and features of MCDs that can be constructed using our five algorithms. Obviously, compared to the MCDs constructed by Construction 3 of Deng et al. ^[12], our constructed MCDs have more flexible run sizes, and the more flexible fixed level $D_1$ -$D_1$ is an $OA\left(n, s_{1}^{k_{1}} s_{2}^{k_{2}}, 2\right)$, $s_{1} = \beta s_{2}$, or $s_{1} \neq \beta s_{2}$. Moreover, in contrast to Construction 3 of Deng et al. ^[12], which does not consider the space-filling property of $D_2$, Algorithm 4 and Algorithm 5 construct $D_2$ with the space-filling property.

Table 16.  Some of the MCDs $( D_{1},D_{2})$ results.

Source	$D_{1}$	Constraints
Theorem 1	$OA\left(s_{1}^{2}, s_{1}^{k_{1}} s_{2}^{k_{2}}, 2\right)$	$s_1=\beta s_{2}$, an $OA\left(s_{1}^{2}, s_{1}^{k_{1}+2}, 2\right)$ and
		a $CROA\left(s_{1}, s_{2}^{k_{2}}, 2\right)$ exist.
Corollary 1	$OA\left(s^{2t}, (s^{t})^{k_1}(s)^{s^{t-1}}, 2\right)$	an $OA\left(s^{2t}, (s^{t})^{k_{1}+2}, 2\right)$ exists.
	$OA\left(4s^{2t}, (2s^{t})^{k_1}(s)^{2s^{t-1}}, 2\right)$	an $OA\left(4s^{2t}, (2s^{t})^{k_{1}+2}, 2\right)$ exists.
	$OA\left(16s^{2t}, (4s^{t})^{k_1}(s)^{4s^{t-1}}, 2\right)$	an $OA\left(16s^{2t}, (4s^{t})^{k_{1}+2}, 2\right)$ exists.
	$OA\left(h^{2w}s^{4}, (h^{w}s^{2})^{k_1}(s)^{h^{w}s}, 2\right)$	an $OA\left(h^{2w}s^{4}, (h^{w}s^{2})^{k_{1}+2}, 2\right)$ exists.
Corollary 2	$OA\left(us_{1}^{2}, s_{1}^{r_{1}k_{1}}s_{2}^{r_{2}k_{2}}, 2\right)$	$s_1=\beta s_{2}$, D$\left(u, r_1, s_1\right)$ and D$(u, r_2, s_2)$ exist.
Theorem 2	$OA\left(2s, 2^{1} s^{1}, 2\right)$	$s\ge 2$.
Corollary 3	$OA\left(2su, 2^{r_{1}} s^{r_{2}}, 2\right)$	$s\ge 2$, D$(u, r_1, 2)$ and D$(u, r_2, s)$ exist.
Theorem 3	$OA\left(2n, 2^{1} s^{m}, 2\right)$	$s\ge 2$.
Corollary 4	$OA\left(2nu, 2^{r_{1}} s^{r_{2}m}, 2\right)$	$s\ge 2$, D$(u, r_1, 2)$ and D$(u, r_2, s)$ exist.
Theorems 4	$OA\left(s_{1}^{2}, s_{1}^{s_{1}-1}s_{2}^{s_2}, 2\right)$	$s_{1}=s_{2}^{2}$, $s_{2}$ is a prime or prime power.
Theorems 5	$OA\left(s_{1}^{2}, s_{1}^{s_{1}-2}s_{2}^{s_2}, 2\right)$	$s_{1}=s_{2}^{2}$, $s_{2}$ is a prime or prime power.

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For future work, a direction is to introduce methods that can produce MCDs with three or more mixed-level qualitative factors, which deserves further investigation.

Author contributions

Weiping Zhou: Algorithm, methodology, validation, investigation, resources, data curation, writing—original draft preparation, writing—review and editing; Wan He: Algorithm, software, validation, writing—original draft preparation; Wei Wang: Methodology, writing—review and editing, visualization, supervision, project administration; Shigui Huang: Software, writing—original draft preparation, writing—review and editing. All authors have read and agreed to the published version of the manuscript.

Acknowledgments

This work was supported by the Special Fund for Scientific and Technological Bases and Talents of Guangxi (Grant No. Guike AD21075008), the Guangxi Young Teachers Basic Ability Improvement Project (Nos. 2021KY0203), and Science and Technology Project of Guangxi (Grant No. Guike AD23023002).

Conflict of interest

The authors declare no conflict of interest in this paper.

Appendix A.   PROOFS

Proof of Lemma ${\rm3}$. From the definition of an MCD, it is clear that $D = \left(D_1, D_2\right)$ is an $MCD\left(n, {s_1}^{k_1} {s_2}^{k_2}, p\right)$ if, and only if, $(\Omega, D_2)$ and $(\Lambda, D_2)$ are an $MCD\left(n, {s_1}^{k_1}, p\right)$ and an $MCD\left(n, {s_2}^{k_2}, p\right)$, respectively. Let $d_i$ be the $i$th column of $D_2$, for $i = 1, \ldots, p$. From Lemma 1, we have (i) $(\Omega, D_2)$ is an $MCD\left(n, {s_1}^{k_1}, p\right)$ if, and only if, $\left(\Omega, f\left(d_i, s_1\right)\right)$ is an $OA\left(n, {s_1}^{k_1}(n/{s_1}), 2\right)$; (ii) $\left(\Lambda, D_2\right)$ is an $MCD\left(n, {s_2}^{k_2}, p\right)$ if, and only if, $\left(\Lambda, f\left(d_i, s_2\right)\right)$ is an $OA\left(n, {s_2}^{k_2}(n/{s_2}), 2\right)$ for $i = 1, \ldots, p$. □

Proof of Theorem ${\rm1}$. (i) In the design $\left(l_2, A, B^*\right)$, the levels $0$, $1$, $\ldots$, $s_1-1$ of the $l_2$ correspond to the $1$st, $2$nd, $\ldots$, $s_1$th rows of the $B$, respectively, where $B^{*} = 1_{s_{1}} \otimes B$. Thus, $D_1 = \left(A, B^*\right)$ is an $OA\left(s_1^2, s_1^{k_1} s_2^{k_2}, 2\right)$ with $s_1 = \beta s_2$.

(ii) From Steps 4 and 5 of Algorithm 1, it is clear that $\left(e_i, c_i\right)$ is an $OA\left(s_1^2, s_1^{2}, 2\right)$ for $i = 1, 2, \ldots, p$, where $s_1 = \beta s_2$. Thus, $D_2$ is an ${LHD}\left(s_1^2, p\right)$ from Step 6 of Algorithm 1.

(iii) Let $a$, $b$, and $d$ be any columns of $A$, $B^*$, and $D_2$, respectively. From Steps 3, 4, 5, and 6, let $e$ and $c$ be the columns corresponding to $d$ in $E$ and $C$, respectively. From Step 6 and Equation (2.1), $\left(a, f\left(D_2, s \right)\right) = \left(a, e\right)$, thus $\left(a, f\left(D_2, s_1 \right)\right)$ is an $OA\left(s_1^2, s_1^{2}, 2\right)$. From Steps 4, 5, and 6 of Algorithm 1, it is clear that $f\left(D_2, s_2 \right) = \beta e + c^{*}$, where $c^{*} = 1_{s_{1}} \otimes w$, $w$ is a random permutation of $\left((c_{i, 1}^{*})^{T}, (c_{i, 2}^{*})^{T}, \cdots, (c_{i, \beta }^{*})^{T}\right)^{T}$ with $c_{i, j}^{*} = (j-1) \textbf{1}_{s_{2}}$. Since $\left(b, e, c^{*} \right)$ is an $OA\left(s_1^2, s_2^{1}s_1^{1}\beta ^{1}, 3\right)$, $\left(b, f\left(D_2, s_2 \right)\right)$ is an $OA\left(s_1^2, s_2^{1}(\beta s_1)^{1}, 2\right)$. Thus, $D = \left(D_1, D_2\right)$ is an $MCD\left(s_1^2, s_1^{k_1}s_2^{k_2}, p\right)$ from Lemma 3. □

Proof of Theorem ${\rm2}$. From Steps 1 and 2 of Algorithm 2, it is easy to check that $D_{1}$ is an $OA\left(2s, 2^{1}s^{1}, 2\right)$ and $D_{2}$ is an $LHD\left(2s, s!\right)$. By Step 2, we can see that $d_i$ is the $i$th column of $D_2$ for $i = 1, 2, \ldots, s!$. Since $f\left(d_i, 2\right)$ = $\left((u_{i})^{T}, \left((s-1)\textbf{1}_{s}- u_{i} \right)^{T} \right)^{T}$, $(L_1, f(d_i, 2))$ is an $OA\left(2s, 2^{1}s^{1}, 2\right)$, where $u_{i}$ is a random permutation of $(0, 1, 2, \dots, s-1)^{T}$, $i = 1, 2, \ldots, s!$. For $1\le i\le s!$, let $\xi_{1} = 2u_{i}$ and $\xi_{2} = (2s-1)\textbf{1}_{s}- 2u_{i}$, then

where $e = (0, 1, \dots, s-1)^{T}$. Obviously, the elements of $f(\xi_{1}, s)$ and $f(\xi_{2}, s)$ are all taken from $\{0, 1\}$. Since $f(\xi_{2}, s) = \textbf{1}_{s}-f(\xi_{1}, s)$, $(L_2, f(d_i, s))$ is an $OA\left(2s, s^{1}2^{1}, 2\right)$, $i = 1, 2, \ldots, s!$. From Lemma 3, the design $D = (D_{1}, D_{2})$ is an $MCD\left(2s, 2^{1}s^{1}, s!\right)$ □

Proof of Theorem ${\rm3}$. The proof of Theorem 3 is similar to that of Theorem 2 and is therefore omitted here. □

Proof of Theorem ${\rm4}$. For $i = 1, 2, \ldots, s_{1}-1$, $j = 1, 2, \ldots, s_{2}-1$, let $f_{0i}$ and $u_{0j}$ be the $i$th and $j$th columns of $F_{0}$ and $U_{0}$, respectively.

(i) Since $F = (F_{0}, f_{1}, f_{2})$ is an $OA\left(s_{1}^{2}, s_{1}^{s_{1}+1}, 2\right)$, $U = (U_{0}, u_{1}, u_{2})$ is an $OA\left(s_{1}^{2}, s_{2}^{s_{2}+1}, 2\right)$, $(f_{0i}, u_{0j})$ is an $OA\left(s_{1}^{2}, s_{1}^{1}s_{2}^{1}, 2\right)$, and $(f_{0i}, u_{1})$ is an $OA\left(s_{1}^{2}, s_{1}^{1}s_{2}^{1}, 2\right)$, $i = 1, 2, \ldots, s_{1}-1$, $j = 1, 2, \ldots, s_{2}-1$, thus $D_1$ is an $OA\left(s_{1}^{2}, s_{1}^{s_{1}-1}s_{2}^{s_2}, 2\right)$.

(ii) According to Proposition 1 of ^[24], we can obtain that $(v_{i}, v_{j}, u_{1}, u_{2})$ is an $OA\left(s_{1}^{2}, s_{2}^{4}, 4\right)$, where $s_{1} = s_{2}^{2}$, $i\neq j$, $i, j = 1, 2, \ldots, 2k$. Thus, $D_2$ is an $LHD\left(s_{1}^{2}, 2k\right)$, where if $s_{2}$ is odd, $k = (s_{2}+1)/2$; if $s_{2}$ is even, $k = s_{2}/2$.

(iii) For $h = 1, 2, \ldots, k$, let $d_{2h-1}$ and $d_{2h}$ be the $(2h-1)$th and $2h$th columns of $D_{2}$, respectively, then $d_{2h-1} = s_{2}^{3}v_{2h-1}+s_{2}^{2}v_{2h}+s_{2}u_{2}+u_{1}$ and $d_{2h} = s_{2}^{3}v_{2h}+s_{2}^{2}v_{2h-1}+s_{2}u_{2}+u_{1}$. Obviously, for $i = 1, 2, \ldots, s_{1}-1$, $j = 1, 2, \ldots, s_{2}-1$, $h = 1, 2, \ldots, k$, $(f_{0i}, f(d_{2h-1}, s_{1})) = (f_{0i}, s_{2}v_{2h-1}+v_{2h})$, $(f_{0i}, f(d_{2h}, s_{1})) = (f_{0i}, s_{2}v_{2h}+v_{2h-1})$, $(u_{0j}, f(d_{2h-1}, s_{1})) = (u_{0j}, s_{2}^{2}v_{2h-1}+s_{2}v_{2h}+u_{2})$, $(u_{0j}, f(d_{2h}, s_{1})) = (u_{0j}, s_{2}^{2}v_{2h}+s_{2}v_{2h-1}+u_{2})$, $(u_{1}, f(d_{2h-1}, s_{1})) = (u_{1}, s_{2}^{2}v_{2h-1}+s_{2}v_{2h}+u_{2})$, and $(u_{1}, f(d_{2h}, s_{1})) = (u_{0j}, s_{2}^{2}v_{2h}+s_{2}v_{2h-1}+u_{2})$, where $s_{1} = s_{2}^{2}$. According to Proposition 1 of ^[24], for $s_{1} = s_{2}^{2}$, it is easy to obtain that $(f_{0i}, v_{2h-1}, v_{2h})$ is an $OA\left(s_{1}^{2}, s_{1}^{1}s_{2}^{2}, 3\right)$, and both $(u_{1}, u_{2}, v_{2h-1}, v_{2h})$ and $(u_{2}, u_{0j}, v_{2h-1}, v_{2h})$ are $OA\left(s_{2}^{4}, s_{2}^{4}, 4\right)$'s, where $i = 1, 2, \ldots, s_{1}-1$, $j = 1, 2, \ldots, s_{2}-1$, $h = 1, 2, \ldots, k$. Therefore, both $(f_{0i}, f(d_{2h-1}, s_{1}))$ and $(f_{0i}, f(d_{2h}, s_{1}))$ are $OA\left(s_{1}^{2}, s_{1}^{2}, 2\right)$'s, and $(u_{0j}, f(d_{2h-1}, s_{2}))$, $(u_{0j}, f(d_{2h}, s_{2}))$, $(u_{1}, f(d_{2h-1}, s_{2}))$, and $(u_{1}, f(d_{2h}, s_{2}))$ are all $OA\left(s_{2}^{4}, s_{2}^{1}(s_{2}^{3})^{1}, 2\right)$'s, where $s_{1} = s_{2}^{2}$, $i = 1, 2, \ldots, s_{1}-1$, $j = 1, 2, \ldots, s_{2}-1$, $h = 1, 2, \ldots, k$. From Lemma 3, the design $(D_1, D_2)$ is an $MCD\left(s_{1}^{2}, s_{1}^{s_{1}-1}s_{2}^{s_2}, 2k\right)$.

(iv) Since $f\left (D_2, {s_2}^3 \right) = W_1$ and $W_1$ is an $OA\left(s_{1}^{2}, s_{2}^{2k}, 2\right)$ with $s_1 = {s_2}^2$, thus any two distinct columns of $D_2$ achieve $s_2 \times s_2$ grids stratification. □

Proof of Theorem ${\rm5}$. The proof of Theorem 5 is similar to that of Theorem 4 and is therefore omitted here. □

Proof of Theorem ${\rm6}$. (i) Since $f(D_{2}, s_{2}^{3}) = (W_{1}, X_{1})$ and $(W_{1}, X_{1})$ is an $OA\left(s_{1}^{2}, s_{2}^{4k}, 2\right)$, thus Theorem 6 (i) is true.

(ii) For $j, j' = 1, 2, \ldots, 2k$, it is easy to see that $f(d_1^{j}, s_{2}^{2}) = s_{2}v_{2j-1}+v_{2j}$ or $f(d_1^{j}, s_{2}^{2}) = s_{2}v_{2j}+v_{2j-1}$, and $f(d_2^{j'}, s_{2}^{2}) = s_{2}z_{2j'-1}+z_{2j'}$ or $f(d_2^{j'}, s_{2}^{2}) = s_{2}z_{2j'}+z_{2j'-1}$. According to Proposition 1 of ^[24], it is easy to obtain that $(v_{2j-1}, v_{2j}, z_{2j'-1}, z_{2j'})$ is an $OA\left(s_{2}^{4}, s_{2}^{4}, 4\right)$. Thus, Theorem 6 (ii) is true.

(iii-iv) From Proposition 1 of ^[24], it is known that any two columns of $V$ in Algorithm 4 and any two columns of $Z$ in Algorithm 5 form an $OA\left(s_{1}^{4}, s_{2}^{4}, 4\right)$ with $s_1 = {s_2}^2$. Similar to the proof of (ii), thus (iii) and (iv) are true. □

Appendix B.   Tables

Table 17.  Matrices $G$, $E$, and $C$ in Example 2.

$Run$	$G$								$E$				$C$			$Run$	$G$								$E$				$C$			$Run$	$G$								$E$				$C$
	$l_1$	$l_2$	$A$						$e_1$	$e_2$	$e_3$		$c_1$	$c_2$	$c_3$		$l_1$	$l_2$	$A$						$e_1$	$e_2$	$e_3$		$c_1$	$c_2$	$c_3$		$l_1$	$l_2$	$A$						$e_1$	$e_2$	$e_3$		$c_1$	$c_2$	$c_3$
1	0	0	0	0	0	0	0		0	1	1		0	2	0	49	4	0	4	4	4	4	4		4	4	4		0	2	0	97	8	0	8	8	8	8	8		8	8	8		0	2	0
2	0	1	1	6	3	8	4		0	1	1		1	3	1	50	4	1	5	10	1	6	2		4	4	4		1	3	1	98	8	1	9	2	11	4	6		8	8	8		1	3	1
3	0	2	2	8	6	1	11		0	1	1		2	0	2	51	4	2	0	6	10	5	9		4	4	4		2	0	2	99	8	2	10	4	2	9	1		8	8	8		2	0	2
4	0	3	3	2	1	11	10		0	1	1		3	1	3	52	4	3	1	0	5	9	8		4	4	4		3	1	3	100	8	3	11	10	9	1	0		8	8	8		3	1	3
5	0	4	4	7	9	5	2		0	1	1		4	4	8	53	4	4	2	11	7	3	0		4	4	4		4	4	8	101	8	4	6	3	5	7	10		8	8	8		4	4	8
6	0	5	5	1	11	9	7		0	1	1		5	5	9	54	4	5	3	5	9	7	11		4	4	4		5	5	9	102	8	5	7	9	1	5	3		8	8	8		5	5	9
7	0	6	6	9	2	3	8		0	1	1		6	6	10	55	4	6	10	7	0	1	6		4	4	4		6	6	10	103	8	6	2	5	10	11	4		8	8	8		6	6	10
8	0	7	7	11	8	10	6		0	1	1		7	7	11	56	4	7	11	9	6	8	10		4	4	4		7	7	11	104	8	7	3	1	4	0	2		8	8	8		7	7	11
9	0	8	8	4	5	2	9		0	1	1		8	8	4	57	4	8	6	2	3	0	7		4	4	4		8	8	4	105	8	8	4	6	7	10	5		8	8	8		8	8	4
10	0	9	9	10	4	7	1		0	1	1		9	9	5	58	4	9	7	8	2	11	5		4	4	4		9	9	5	106	8	9	5	0	6	3	9		8	8	8		9	9	5
11	0	10	10	5	7	6	3		0	1	1		10	10	6	59	4	10	8	3	11	10	1		4	4	4		10	10	6	107	8	10	0	7	3	2	11		8	8	8		10	10	6
12	0	11	11	3	10	4	5		0	1	1		11	11	7	60	4	11	9	1	8	2	3		4	4	4		11	11	7	108	8	11	1	11	0	6	7		8	8	8		11	11	7
13	1	0	1	1	1	1	1		1	0	0		0	2	0	61	5	0	5	5	5	5	5		5	5	5		0	2	0	109	9	0	9	9	9	9	9		9	9	9		0	2	0
14	1	1	2	7	4	9	5		1	0	0		1	3	1	62	5	1	0	11	2	7	3		5	5	5		1	3	1	110	9	1	10	3	6	5	7		9	9	9		1	3	1
15	1	2	3	9	7	2	6		1	0	0		2	0	2	63	5	2	1	7	11	0	10		5	5	5		2	0	2	111	9	2	11	5	3	10	2		9	9	9		2	0	2
16	1	3	4	3	2	6	11		1	0	0		3	1	3	64	5	3	2	1	0	10	9		5	5	5		3	1	3	112	9	3	6	11	10	2	1		9	9	9		3	1	3
17	1	4	5	8	10	0	3		1	0	0		4	4	8	65	5	4	3	6	8	4	1		5	5	5		4	4	8	113	9	4	7	4	0	8	11		9	9	9		4	4	8
18	1	5	0	2	6	10	8		1	0	0		5	5	9	66	5	5	4	0	10	8	6		5	5	5		5	5	9	114	9	5	8	10	2	0	4		9	9	9		5	5	9
19	1	6	7	10	3	4	9		1	0	0		6	6	10	67	5	6	11	8	1	2	7		5	5	5		6	6	10	115	9	6	3	0	11	6	5		9	9	9		6	6	10
20	1	7	8	6	9	11	7		1	0	0		7	7	11	68	5	7	6	10	7	9	11		5	5	5		7	7	11	116	9	7	4	2	5	1	3		9	9	9		7	7	11
21	1	8	9	5	0	3	10		1	0	0		8	8	4	69	5	8	7	3	4	1	8		5	5	5		8	8	4	117	9	8	5	7	8	11	0		9	9	9		8	8	4
22	1	9	10	11	5	8	2		1	0	0		9	9	5	70	5	9	8	9	3	6	0		5	5	5		9	9	5	118	9	9	0	1	7	4	10		9	9	9		9	9	5
23	1	10	11	0	8	7	4		1	0	0		10	10	6	71	5	10	9	4	6	11	2		5	5	5		10	10	6	119	9	10	1	8	4	3	6		9	9	9		10	10	6
24	1	11	6	4	11	5	0		1	0	0		11	11	7	72	5	11	10	2	9	3	4		5	5	5		11	11	7	120	9	11	2	6	1	7	8		9	9	9		11	11	7
25	2	0	2	2	2	2	2		2	2	2		0	2	0	73	6	0	6	6	6	6	6		6	6	6		0	2	0	121	10	0	10	10	10	10	10		10	10	10		0	2	0
26	2	1	3	8	5	10	0		2	2	2		1	3	1	74	6	1	7	0	9	2	10		6	6	6		1	3	1	122	10	1	11	4	7	0	8		10	10	10		1	3	1
27	2	2	4	10	8	3	7		2	2	2		2	0	2	75	6	2	8	2	0	7	5		6	6	6		2	0	2	123	10	2	6	0	4	11	3		10	10	10		2	0	2
28	2	3	5	4	3	7	6		2	2	2		3	1	3	76	6	3	9	8	7	5	4		6	6	6		3	1	3	124	10	3	7	6	11	3	2		10	10	10		3	1	3
29	2	4	0	9	11	1	4		2	2	2		4	4	8	77	6	4	10	1	3	11	8		6	6	6		4	4	8	125	10	4	8	5	1	9	6		10	10	10		4	4	8
30	2	5	1	3	7	11	9		2	2	2		5	5	9	78	6	5	11	7	5	3	1		6	6	6		5	5	9	126	10	5	9	11	3	1	5		10	10	10		5	5	9
31	2	6	8	11	4	5	10		2	2	2		6	6	10	79	6	6	0	3	8	9	2		6	6	6		6	6	10	127	10	6	4	1	6	7	0		10	10	10		6	6	10
32	2	7	9	7	10	6	8		2	2	2		7	7	11	80	6	7	1	5	2	4	0		6	6	6		7	7	11	128	10	7	5	3	0	2	4		10	10	10		7	7	11
33	2	8	10	0	1	4	11		2	2	2		8	8	4	81	6	8	2	10	11	8	3		6	6	6		8	8	4	129	10	8	0	8	9	6	1		10	10	10		8	8	4
34	2	9	11	6	0	9	3		2	2	2		9	9	5	82	6	9	3	4	10	1	7		6	6	6		9	9	5	130	10	9	1	2	8	5	11		10	10	10		9	9	5
35	2	10	6	1	9	8	5		2	2	2		10	10	6	83	6	10	4	11	1	0	9		6	6	6		10	10	6	131	10	10	2	9	5	4	7		10	10	10		10	10	6
36	2	11	7	5	6	0	1		2	2	2		11	11	7	84	6	11	5	9	4	10	11		6	6	6		11	11	7	132	10	11	3	7	2	8	9		10	10	10		11	11	7
37	3	0	3	3	3	3	3		3	3	3		0	2	0	85	7	0	7	7	7	7	7		7	7	7		0	2	0	133	11	0	11	11	11	11	11		11	11	11		0	2	0
38	3	1	4	9	0	11	1		3	3	3		1	3	1	86	7	1	8	1	10	3	11		7	7	7		1	3	1	134	11	1	6	5	8	1	9		11	11	11		1	3	1
39	3	2	5	11	9	4	8		3	3	3		2	0	2	87	7	2	9	3	1	8	0		7	7	7		2	0	2	135	11	2	7	1	5	6	4		11	11	11		2	0	2
40	3	3	0	5	4	8	7		3	3	3		3	1	3	88	7	3	10	9	8	0	5		7	7	7		3	1	3	136	11	3	8	7	6	4	3		11	11	11		3	1	3
41	3	4	1	10	6	2	5		3	3	3		4	4	8	89	7	4	11	2	4	6	9		7	7	7		4	4	8	137	11	4	9	0	2	10	7		11	11	11		4	4	8
42	3	5	2	4	8	6	10		3	3	3		5	5	9	90	7	5	6	8	0	4	2		7	7	7		5	5	9	138	11	5	10	6	4	2	0		11	11	11		5	5	9
43	3	6	9	6	5	0	11		3	3	3		6	6	10	91	7	6	1	4	9	10	3		7	7	7		6	6	10	139	11	6	5	2	7	8	1		11	11	11		6	6	10
44	3	7	10	8	11	7	9		3	3	3		7	7	11	92	7	7	2	0	3	5	1		7	7	7		7	7	11	140	11	7	0	4	1	3	5		11	11	11		7	7	11
45	3	8	11	1	2	5	6		3	3	3		8	8	4	93	7	8	3	11	6	9	4		7	7	7		8	8	4	141	11	8	1	9	10	7	2		11	11	11		8	8	4
46	3	9	6	7	1	10	4		3	3	3		9	9	5	94	7	9	4	5	11	2	8		7	7	7		9	9	5	142	11	9	2	3	9	0	6		11	11	11		9	9	5
47	3	10	7	2	10	9	0		3	3	3		10	10	6	95	7	10	5	6	2	1	10		7	7	7		10	10	6	143	11	10	3	10	0	5	8		11	11	11		10	10	6
48	3	11	8	0	7	1	2		3	3	3		11	11	7	96	7	11	0	10	5	11	6		7	7	7		11	11	7	144	11	11	4	8	3	9	10		11	11	11		11	11	7

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Table 18.  $D = \left(D_{1}, D_{2}\right)$ in Example 2.

$Run$	$MCD(D_1, D_2)$										$Run$	$MCD(D_1, D_2)$										$Run$	$MCD(D_1, D_2)$
	$D_1$							$D_2$				$D_1$							$D_2$				$D_1$							$D_2$
	$A$					$B^*$						$A$					$B^*$						$A$					$B^*$
1	0	0	0	0	0	1	1	0	14	12	49	4	4	4	4	4	1	1	48	50	48	97	8	8	8	8	8	1	1	96	98	96
2	1	6	3	8	4	0	0	1	15	13	50	5	10	1	6	2	0	0	49	51	49	98	9	2	11	4	6	0	0	97	99	97
3	2	8	6	1	11	0	0	2	12	14	51	0	6	10	5	9	0	0	50	48	50	99	10	4	2	9	1	0	0	98	96	98
4	3	2	1	11	10	1	1	3	13	15	52	1	0	5	9	8	1	1	51	49	51	100	11	10	9	1	0	1	1	99	97	99
5	4	7	9	5	2	1	1	4	16	20	53	2	11	7	3	0	1	1	52	52	56	101	6	3	5	7	10	1	1	100	100	104
6	5	1	11	9	7	0	0	5	17	21	54	3	5	9	7	11	0	0	53	53	57	102	7	9	1	5	3	0	0	101	101	105
7	6	9	2	3	8	0	1	6	18	22	55	10	7	0	1	6	0	1	54	54	58	103	2	5	10	11	4	0	1	102	102	106
8	7	11	8	10	6	1	0	7	19	23	56	11	9	6	8	10	1	0	55	55	59	104	3	1	4	0	2	1	0	103	103	107
9	8	4	5	2	9	0	1	8	20	16	57	6	2	3	0	7	0	1	56	56	52	105	4	6	7	10	5	0	1	104	104	100
10	9	10	4	7	1	1	0	9	21	17	58	7	8	2	11	5	1	0	57	57	53	106	5	0	6	3	9	1	0	105	105	101
11	10	5	7	6	3	1	0	10	22	18	59	8	3	11	10	1	1	0	58	58	54	107	0	7	3	2	11	1	0	106	106	102
12	11	3	10	4	5	0	1	11	23	19	60	9	1	8	2	3	0	1	59	59	55	108	1	11	0	6	7	0	1	107	107	103
13	1	1	1	1	1	1	1	12	2	0	61	5	5	5	5	5	1	1	60	62	60	109	9	9	9	9	9	1	1	108	110	108
14	2	7	4	9	5	0	0	13	3	1	62	0	11	2	7	3	0	0	61	63	61	110	10	3	6	5	7	0	0	109	111	109
15	3	9	7	2	6	0	0	14	0	2	63	1	7	11	0	10	0	0	62	60	62	111	11	5	3	10	2	0	0	110	108	110
16	4	3	2	6	11	1	1	15	1	3	64	2	1	0	10	9	1	1	63	61	63	112	6	11	10	2	1	1	1	111	109	111
17	5	8	10	0	3	1	1	16	4	8	65	3	6	8	4	1	1	1	64	64	68	113	7	4	0	8	11	1	1	112	112	116
18	0	2	6	10	8	0	0	17	5	9	66	4	0	10	8	6	0	0	65	65	69	114	8	10	2	0	4	0	0	113	113	117
19	7	10	3	4	9	0	1	18	6	10	67	11	8	1	2	7	0	1	66	66	70	115	3	0	11	6	5	0	1	114	114	118
20	8	6	9	11	7	1	0	19	7	11	68	6	10	7	9	11	1	0	67	67	71	116	4	2	5	1	3	1	0	115	115	119
21	9	5	0	3	10	0	1	20	8	4	69	7	3	4	1	8	0	1	68	68	64	117	5	7	8	11	0	0	1	116	116	112
22	10	11	5	8	2	1	0	21	9	5	70	8	9	3	6	0	1	0	69	69	65	118	0	1	7	4	10	1	0	117	117	113
23	11	0	8	7	4	1	0	22	10	6	71	9	4	6	11	2	1	0	70	70	66	119	1	8	4	3	6	1	0	118	118	114
24	6	4	11	5	0	0	1	23	11	7	72	10	2	9	3	4	0	1	71	71	67	120	2	6	1	7	8	0	1	119	119	115
25	2	2	2	2	2	1	1	24	26	24	73	6	6	6	6	6	1	1	72	74	72	121	10	10	10	10	10	1	1	120	122	120
26	3	8	5	10	0	0	0	25	27	25	74	7	0	9	2	10	0	0	73	75	73	122	11	4	7	0	8	0	0	121	123	121
27	4	10	8	3	7	0	0	26	24	26	75	8	2	0	7	5	0	0	74	72	74	123	6	0	4	11	3	0	0	122	120	122
28	5	4	3	7	6	1	1	27	25	27	76	9	8	7	5	4	1	1	75	73	75	124	7	6	11	3	2	1	1	123	121	123
29	0	9	11	1	4	1	1	28	28	32	77	10	1	3	11	8	1	1	76	76	80	125	8	5	1	9	6	1	1	124	124	128
30	1	3	7	11	9	0	0	29	29	33	78	11	7	5	3	1	0	0	77	77	81	126	9	11	3	1	5	0	0	125	125	129
31	8	11	4	5	10	0	1	30	30	34	79	0	3	8	9	2	0	1	78	78	82	127	4	1	6	7	0	0	1	126	126	130
32	9	7	10	6	8	1	0	31	31	35	80	1	5	2	4	0	1	0	79	79	83	128	5	3	0	2	4	1	0	127	127	131
33	10	0	1	4	11	0	1	32	32	28	81	2	10	11	8	3	0	1	80	80	76	129	0	8	9	6	1	0	1	128	128	124
34	11	6	0	9	3	1	0	33	33	29	82	3	4	10	1	7	1	0	81	81	77	130	1	2	8	5	11	1	0	129	129	125
35	6	1	9	8	5	1	0	34	34	30	83	4	11	1	0	9	1	0	82	82	78	131	2	9	5	4	7	1	0	130	130	126
36	7	5	6	0	1	0	1	35	35	31	84	5	9	4	10	11	0	1	83	83	79	132	3	7	2	8	9	0	1	131	131	127
37	3	3	3	3	3	1	1	36	38	36	85	7	7	7	7	7	1	1	84	86	84	133	11	11	11	11	11	1	1	132	134	132
38	4	9	0	11	1	0	0	37	39	37	86	8	1	10	3	11	0	0	85	87	85	134	6	5	8	1	9	0	0	133	135	133
39	5	11	9	4	8	0	0	38	36	38	87	9	3	1	8	0	0	0	86	84	86	135	7	1	5	6	4	0	0	134	132	134
40	0	5	4	8	7	1	1	39	37	39	88	10	9	8	0	5	1	1	87	85	87	136	8	7	6	4	3	1	1	135	133	135
41	1	10	6	2	5	1	1	40	40	44	89	11	2	4	6	9	1	1	88	88	92	137	9	0	2	10	7	1	1	136	136	140
42	2	4	8	6	10	0	0	41	41	45	90	6	8	0	4	2	0	0	89	89	93	138	10	6	4	2	0	0	0	137	137	141
43	9	6	5	0	11	0	1	42	42	46	91	1	4	9	10	3	0	1	90	90	94	139	5	2	7	8	1	0	1	138	138	142
44	10	8	11	7	9	1	0	43	43	47	92	2	0	3	5	1	1	0	91	91	95	140	0	4	1	3	5	1	0	139	139	143
45	11	1	2	5	6	0	1	44	44	40	93	3	11	6	9	4	0	1	92	92	88	141	1	9	10	7	2	0	1	140	140	136
46	6	7	1	10	4	1	0	45	45	41	94	4	5	11	2	8	1	0	93	93	89	142	2	3	9	0	6	1	0	141	141	137
47	7	2	10	9	0	1	0	46	46	42	95	5	6	2	1	10	1	0	94	94	90	143	3	10	0	5	8	1	0	142	142	138
48	8	0	7	1	2	0	1	47	47	43	96	0	10	5	11	6	0	1	95	95	91	144	4	8	3	9	10	0	1	143	143	139

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Table 19.  Matrix $H$ in Example 7.

$Run$	$H$				$Run$	$H$				$Run$	$H$
1	0	0	0	0	4	1	0	1	1	7	2	0	2	2
2	0	1	1	2	5	1	1	2	0	8	2	1	0	1
3	0	2	2	1	6	1	2	0	2	9	2	2	1	0

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