Construction of column-orthogonal strong orthogonal arrays

Abstract

Strong orthogonal arrays were recently introduced as a new class of space-filling designs for computer experiments due to their better stratifications than orthogonal arrays. To further improve the space-filling properties in low dimensions while possessing the column orthogonality, we propose column-orthogonal strong orthogonal arrays of strength two star and three. Construction methods and characterizations of such designs are provided. The resulting strong orthogonal arrays, with the numbers of levels being increased, have their space-filling properties in one and two dimensions being strengthened. They can accommodate comparable or even larger numbers of factors than those in the existing literature, enjoy flexible run sizes, and possess the column orthogonality. The construction methods are convenient and flexible, and the resulting designs are good choices for computer experiments.

Appendix: Proofs of theorems

Proof of Theorem 2

From (3) and (4), we have $D^*= C^*R$, where $C^*=(C_1^*, \dots ,$ $C_p^*)$ and $R=\mathrm{diag}\{V,\dots ,V\}$ with V repeating p times. Since $(a_i, a_j, b_j)$ is an OA of strength 3 for any $i \ne j$, $C^*$ is an OA of strength 2 with levels from $\Omega (s)$. By noting that R is column-orthogonal, we have ${D^*}^T D^*=(C^* R)^T C^* R = R^T ({C^*}^T C^*) R = c_1 R^T R=c_2I_{2p}$, where $c_1$ and $c_2$ are two constants, and $I_{2p}$ is the identity matrix of order 2p. This shows that $D^*$ in (3) is column-orthogonal. So is D in (5).

Before showing the stratifications of D, we first derive the form of any column d of D in (5). Let $a_i^*=a_i- (s-1)/2$ and $b_i^*=b_i- (s-1)/2$ for $i=1,\dots , m$. Note that the column $d^*$ of $D^*$ in (3) corresponding to d has the form $d^*=c_1^*s^2 + c_2^*s \pm c_3^*,$ where $(c_1^*, c_2^*, c_3^*)$ is a copy of some $( a_j^*, b_j^*,a_i^*)$ with $i \ne j$. From (5), the column d of D has the form

$$\begin{aligned} \begin{aligned} d&= d^* + (s^3-1)/2 \\&= c_1^*s^2 + c_2^*s \pm c_3^* + (s^3-1)/2 \\&= r_1 s^2 + r_2 s + r_3, \end{aligned} \end{aligned}$$

where $r_1=c_1^* +(s-1)/2$, $r_2= c_2^* + (s-1)/2$ and $r_3=\pm c_3^* + (s-1)/2$. Since $(c_1^*, c_2^*, c_3^*)$ is a copy of some $( a_j^*, b_j^*, a_i^*)$ with $i \ne j$, then $(r_1, r_2, r_3)$ is a copy of $( a_j, b_j, a_i)$ or $( a_j, b_j, s-1-a_i)$. Thus, all entries of $(r_1, r_2, r_3)$ take values in $\{0, 1, \dots , s-1\} \times \{0, 1, \dots , s-1\} \times \{0, 1, \dots , s-1\}$. Therefore, we have that for $i \ne j$, any column d of D in (5) has the form $d=a_j s^2 + b_j s + a$ where $a=a_{i}$ or $a=s-1 - a_{i}$.

We now show the stratifications of D. For any two columns $d_i=a_i s^2 + b_i s + a_{i'}$ and $d_j=a_j s^2 + b_j s + a_{j'}$, let us show that the array $(d_i, d_{j})$ can be collapsed into an OA$(n,2,s^2 \times s, 2)$ and an OA$(n,2,s \times s^2, 2)$, that is to say, $(\lfloor d_i/s \rfloor , \lfloor d_j/s^2 \rfloor )=(a_i s +b_i, a_j)$ and $(\lfloor d_i/s^2 \rfloor , \lfloor d_j/s \rfloor )=(a_i, a_j s + b_j )$ are an OA$(n,2,s^2 \times s, 2)$ and an OA$(n,2,s \times s^2, 2)$, respectively. In fact, this is true by noting the following three facts: (i) $(a_i, a_j, b_j)$ is an OA(n, 3, s, 3) for any $i\ne j$; (ii) $x_1 s+x_2$ establishes a one-to-one correspondence between the $s^2$ levels in $\{0, 1, \dots , s^2-1\}$ and the $s^2$ pairs $(x_1, x_2)$, where $x_1, x_2 \in \{0, 1, \dots , s-1\}$; (iii) $x_1 s^2 + x_2 s + x_3$ establishes a one-to-one correspondence between the $s^3$ levels in $\{0, 1, \dots , s^3-1\}$ and the $s^3$ pairs $(x_1, x_2, x_3)$, where $x_1, x_2, x_3 \in \{0, 1, \dots , s-1\}$. Therefore, D satisfies the property of stratifications in Definition 1. This completes the proof. $\square$

Proof of Theorem 4

From the definition of two-level OA and Construction 1, m must be a multiple of 4 and $p= \lfloor (m-1)/2 \rfloor =(m-2)/2$. Therefore the number of columns for the constructed OSOA is $m-2$. According to the structures of A and B in (6), A is an OA$(2m, m-1, 2, 3)$ and B is an OA$(2m, m-1, 2, 2)$. Since $b_j$ is orthogonal to the interaction column $a_i a_j$ for any $i \ne j$, $(a_i, a_j, b_j)$ is an OA(2m, 3, 2, 3). Then from Theorem 3, the resulting design is an OSOA$(2m, m-2, 8, 3)$. The proof is completed. $\square$

Proof of Theorem 5

Note that $A=(a_1, \dots ,a_m)$ and $B=(b_1, \dots , b_m)$ with $m=(s^{k-1})/(s-1)$. From Theorem 2, we only need to show $(a_i, a_j, b_j)$ is an OA(n, 3, s, 3) for any $i \ne j$. To do so, we assume that $b_j$ is the interaction column of $a_i$ and $a_j$, i.e.,

$$\begin{aligned} a_i a_j^l=b_j^t \end{aligned}$$

(8)

holds for some $l, t \in \{1, \dots ,s-1 \}$. Note that $a_i=b_i e_k$ and $a_j=b_j e_k$. So Equation (8) is possible to hold only when $l=s-1$. However, if $l=s-1$, Equation (8) implies that $b_i b_j^{s-1}=b_j^t$ for some $t\in \{1, \dots ,s-1 \}$. This is impossible because $b_i$ and $b_j$ are two different columns of B. This completes the proof. $\square$