Asymptotically optimal maximin distance Latin hypercube designs

Abstract

Maximin distance designs and orthogonal designs have become increasingly popular in computer and physical experiments. The construction of such designs is challenging, especially under the maximin distance criterion. This paper studies a class of Latin hypercube designs by calculating the minimum distances between design points. We derive a general formula for the minimum intersite distance of this kind of design. The row pairs with the minimum intersite distance are also specified. The results show that such kind of Latin hypercube design is asymptotically optimal under both the maximin distance criterion and the orthogonality criterion.

Appendix

A.1. Proof of Lemma 4

First, based on Property 2, we can easily obtain the $\phi$ values for the following three row pairs

$$\begin{aligned} \begin{aligned}&\phi (m_1,m_{2^c+1}) = \sum _{r=1}^{2^c} ( m_{1,r}-m_{2^c+1,r} ) \\&\quad = 2 \times ( 1 + 3 + \dots + 2^{c+1}-1 )/2 = 2^{2c}, \\&\phi (m_1, m_{2^c+2^{c-1}+1})\\&\quad = \,( 1 + 3 + \dots + (2^{c+1}\!-\!1) )/2 - (1 + 3 \\&\qquad + \dots + (2^{c}\!-\!1) - (2^{c}\!+\!1) - (2^{c}\!+\!3) - \dots - (2^{c+1}\!-\!1) )/2 \\&\quad = \,2 \times ((2^{c}+1) + (2^{c}+3) + \dots + (2^{c+1}-1))/2 = \frac{3}{4} \times 2^{2c}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\phi (m_{2^{c-1}+1}, m_{2^c+1})\\&\quad =\sum _{r=1}^{2^c} (m_{2^{c-1}+1,r}-m_{2^c+1,r}) = \sum _{r=1}^{2^c} ((-m_{2^c+2^{c-1}+1,r})-(-m_{1,r})) \\&\quad = \sum _{r=1}^{2^c} (m_{1,r} - m_{2^c+2^{c-1}+1,r}) \\&\quad = \phi (m_1, m_{2^c+2^{c-1}+1}) = \frac{3}{4} \times 2^{2c}. \end{aligned} \end{aligned}$$

Next, we confirm that the value $\phi (m_1,m_{2^c+1})$ is the maximum $\phi$ value and the values $\phi (m_1, m_{2^c+2^{c-1}+1})$, $\phi (m_{2^{c-1}+1}, m_{2^c+1})$ are the second-largest $\phi$ value. According to Properties 1 and 2 , it is easy to see that the maximum $\phi$ value is the $\phi$ value of the first row and the ($2^c+1$)th row. We will focus on the second-largest $\phi$ value.

For any $i,j \ne 1,2^c+1$, since the row vector $m_i$ has a half of elements positive and others negative, so half of the $2^{c+1}$ elements of the sum formula $\phi (m_i, m_j)$ are negative. Then, if possible, the maximum value of $\phi (m_i,m_j)$ is

$$\begin{aligned} \begin{aligned}&2 \times ( ((2^{c}+1) + (2^{c}+3) + \dots + (2^{c+1}-1))/2 - (1 + 3 + \dots + (2^{c}-1))/2 ) \\&\quad =\, 2^{c} \times 2^{c-1} = \frac{1}{2} \times 2^{2c} < \frac{3}{4} \times 2^{2c}. \end{aligned} \end{aligned}$$

By Property 1, we can know the sum formula $\phi (m_1,m_i)$ has $2^c+2^{c-1}$ positive elements. For the values $\phi (m_{2^c+1},m_i)$, $\phi (m_i,m_1)$, and $\phi (m_i,m_{2^c+1})$, we have a similar conclusion. By the (i) and (iii) of Property 2, we have

$$\begin{aligned}&\mathrm{max} \{ \phi (m_1, m_{i}), \phi (m_{2^c+1}, m_{i}), \phi (m_i,m_1), \phi (m_i,m_{2^c+1}) \} \\&\quad \leqslant \,(( 1 + 3 + \dots + (2^{c+1}\!-\!1))\!+\! ((2^{c}\!+\!1) + (2^{c}\!+\!3) + \cdots \\&\qquad + (2^{c+1}\!-\!1))\!-\!(1 + 3 + \dots + (2^{c}\!-\!1)) )/2 \\&\quad = \,\frac{3}{4} \times 2^{2c}, \end{aligned}$$

where the equality achieves at $\phi (m_1, m_{2^c+2^{c-1}+1})$ and $\phi (m_{2^{c-1}+1}, m_{2^c+1})$. Also, we note that

$$\begin{aligned} \phi (m_{2^c+1},m_1) = -\phi (m_1,m_{2^c+1}) = -2^{2c} < \frac{3}{4} \times 2^{2c} . \end{aligned}$$

Summarizing the above results, we can see that the second-largest $\phi$ value can only be achieved at $\phi (m_1, m_{2^c+2^{c-1}+1})$ and $\phi (m_{2^{c-1}+1}, m_{2^c+1})$. This completes the proof. $\square$

A.2. Proof of Theorem 1

For $c=2$, it is easy to show that the minimum $L_2$-distance between any two distinct rows in $H_2$ is $d_2(h_i,h_j)=2$, where $\left| i-j \right| =2^{c}$; and the second-smallest $L_2$-distance is $d_2(h_1,h_8) = d_2(h_2,h_7) = \frac{1}{3} \times 2^6 - \frac{3}{4} \times 2^4 + \frac{1}{6} \times 2^2 = 10$. In the following, we only consider the case of $c > 2$.

For convenience, we always assume $i < j$ and denote ${\mathscr {A}}=\{ 1,2,\dots , 2^{c} \}$ and ${\mathscr {B}}=\{ 2^{c}+1,2^{c}+2,\dots , 2^{c+1} \}$. We calculate the $L_2$-distances between any two distinct rows in $H_c$ by dividing i and j into the following four cases.

• (i) If $i,j \in {\mathscr {A}} \,\mathrm{or}\, i,j \in {\mathscr {B}}, \,\mathrm{and}\, \vert i-j \vert =2^{c-1}$, by Lemma 3, we have

  $$\begin{aligned} d_2(h_i,h_j) = \sum _{r=1}^{2^{c-1}} \vert h_{ir}-h_{jr} \vert ^2 = 4 \times \frac{2^{3(c-1)+2}-2^{c-1}}{3} = \frac{2}{3}\times 2^{3c} - \frac{2}{3}\times 2^{c}. \end{aligned}$$

• (ii) If $i,j \in {\mathscr {A}} \,\mathrm{or}\, i,j \in {\mathscr {B}}, \,\mathrm{and}\, \vert i-j \vert \ne 2^{c-1}$, by Lemma 3, we have

  $$\begin{aligned} d_2(h_i,h_j) =4 \times \frac{2^{3(c-1)+2}-2^{c-1}}{6} =\frac{1}{3}\times 2^{3c} - \frac{1}{3}\times 2^{c}. \end{aligned}$$

• (iii) If $i \in {\mathscr {A}}, j \in {\mathscr {B}} \,\mathrm{and}\, \vert i-j \vert =2^{c}$, we have

  $$\begin{aligned} d_2(h_i,h_j) = \sum _{r=1}^{2^{c-1}} \vert h_{ir}-h_{jr} \vert ^2 = \sum _{r=1}^{2^{c-1}} \left| (2m_{ir}-\frac{1}{2}) - (2m_{ir}+\frac{1}{2}) \right| ^2 = 2^{c-1}. \end{aligned}$$

• (iv) If $i \in {\mathscr {A}}, j \in {\mathscr {B}} \,\mathrm{and}\, \vert i-j \vert \ne 2^{c}$, we have

  $$\begin{aligned} \begin{aligned} d_2(h_i,h_j)&= \sum _{r=1}^{2^{c-1}} \vert h_{ir}-h_{jr} \vert ^2\\&= \sum _{r=1}^{2^{c-1}} \left| (2m_{ir}-\frac{1}{2}) - (2m_{j-2^{c},r}+\frac{1}{2}) \right| ^2\\&= 4d_2(m_i,m_{j-2^{c}}) - 4\phi (m_i,m_{j-2^{c}}) + 2^{c-1}. \end{aligned} \end{aligned}$$

  If $\left| i-(j-2^{c}) \right| = 2^{c-1}$, then we have

  $$\begin{aligned} d_2(h_i,h_j)&=\; 4 \times \frac{2^{3(c-1)+2}-2^{c-1}}{3} -4\phi (m_i,m_{j-2^{c}}) + 2^{c-1} \nonumber \\&\geqslant \, 4 \times \frac{2^{3(c-1)+2}-2^{c-1}}{3} -4\phi (m_1,m_{1+2^{c-1}}) + 2^{c-1} \nonumber \\&=\; 4 \times \frac{2^{3(c-1)+2}-2^{c-1}}{3} -4\times 2^{2(c-1)} + 2^{c-1} \nonumber \\&=\; \frac{2}{3}\times 2^{3c} - 2^{2c} -\frac{1}{6} \times 2^c. \end{aligned}$$

  (6)

  The equality (6) holds if and only if $i = 1$ and $j = 1+2^{c-1}+2^c$.

  If $\left| i-(j-2^{c}) \right| \ne 2^{c-1}$, then we have

  $$\begin{aligned} d_2(h_i,h_j)&=\; 4 \times \frac{2^{3(c-1)+2}-2^{c-1}}{6} -4\phi (m_i,m_{j-2^{c}}) + 2^{c-1} \nonumber \\&\geqslant \, 4 \times \frac{2^{3(c-1)+2}-2^{c-1}}{6} -4\times \frac{3}{4}\times 2^{2(c-1)} + 2^{c-1} \nonumber \\&=\; \frac{1}{3}\times 2^{3c} - \frac{3}{4} \times 2^{2c} +\frac{1}{6} \times 2^c. \end{aligned}$$

  (7)

  The equality (7) holds if and only if one of the following two conditions holds: (a) $i = 1$ and $j = 2^c+2^{c-1}+2^{c-2}+1$; and (b) $i = 2^{c-2}+1$ and $j = 2^c+2^{c-1}+1$.

By summarizing the above results, when $c \geqslant 2$, we can see that the minimum $L_2$-distance between any two distinct rows in $H_c$ is $d_2(h_i,h_j) = 2^{c-1}$, where $\left| i-j \right| =2^{c}$; and the second-smallest $L_2$-distance is $d_2(h_1,h_{2^{c}+2^{c-1}+2^{c-2}+1}) = d_2(h_{2^{c-2}+1},h_{2^{c}+2^{c-1}+1}) = \frac{1}{3}\times 2^{3c} - \frac{3}{4} \times 2^{2c} +\frac{1}{6} \times 2^c$. All other $L_2$-distances between row pairs are larger than $\frac{1}{3}\times 2^{3c} - \frac{3}{4} \times 2^{2c} +\frac{1}{6} \times 2^c$. This completes the proof. $\square$

A.3. Proof of Theorem 2

By Lemma 3, we know that, for any two rows of $M_c$, their $L_2$-distance is

$$\begin{aligned} d_2(m_i,m_j) = \left\{ \begin{array}{ll} \frac{1}{3} \times (2^{3c+2}-2^c), &{} \text{ if } \, \vert i-j \vert = 2^c;\\ \frac{1}{6} \times (2^{3c+2}-2^c), &{} \text{ otherwise }. \end{array}\right. \end{aligned}$$

By Theorem 1, we know that the minimum $L_2$-distance between two distinct rows in $H_{c}$ is $2^{c-1}$. This value is the $L_2$-distance between $h_i$ and $h_j$ with $\vert i-j \vert = 2^c$. So when $i,j \in \{ 1,2,\dots , 2^{c+1} \} \,\mathrm{and}\, \vert i-j \vert = 2^c$, the $L_2$-distance between two rows $p_i$ and $p_j$ is

$$\begin{aligned} d_2(p_i,p_j) = \frac{1}{3}\times (2^{3c+2}-2^c) + 2^{c-1} = \frac{4}{3} \times 2^{3c} + \frac{1}{6} \times 2^c. \end{aligned}$$

By Theorem 1, we know that the second-smallest $L_2$-distance between two distinct rows of $H_{c}$ is $d_2(h_1,h_{2^{c}+2^{c-1}+2^{c-2}+1}) = d_2(h_{2^{c-2}+1},h_{2^{c}+2^{c-1}+1}) = \frac{1}{3}\times 2^{3c} - \frac{3}{4} \times 2^{2c} +\frac{1}{6} \times 2^c.$ So when $i,j \in \{ 1,2,\dots , 2^{c+1} \} \,\mathrm{and}\, \vert i-j \vert \ne 2^c$, the minimum $L_2$-distance between two rows $p_i$ and $p_j$ is

$$\begin{aligned} \begin{aligned}&d_2(p_1,p_{2^{c}+2^{c-1}+2^{c-2}+1}) = d_2(p_{2^{c-2}+1},p_{2^{c}+2^{c-1}+1})\\&\quad =\, \frac{1}{6}\times (2^{3c+2}-2^c) + \frac{1}{3} \times 2^{3c} - \frac{3}{4} \times 2^{2c} + \frac{1}{6} \times 2^{c}\\&\quad =\, 2^{3c} - \frac{3}{4} \times 2^{2c}. \end{aligned} \end{aligned}$$

By summarizing above results, we know that the $L_2$-distance of the design $P_c$ is $2^{3c} - \frac{3}{4} \times 2^{2c}$, which is either $d_2(p_1, p_{2^{c}+2^{c-1}+2^{c-2}+1})$ or $d_2(p_{2^{c-2}+1}, p_{2^{c}+2^{c-1}+1})$. This completes the proof. $\square$

A.4. Proof of Theorem 3

According to Theorem 2, the $L_2$-distance efficiency of the design $P_c$ is

$$\begin{aligned} d_{eff}(P_c) = \dfrac{2^{3c}-\dfrac{3}{4}\times 2^{2c}}{\dfrac{1}{6} \times 2^{c+1}(2^{c+1}+1)(2^c+2^{c-1})} = 1- \dfrac{5}{2^{c+2}+2}, \end{aligned}$$

which clearly tends to one as c increases infinitely. $\square$