Maximin distance designs based on densest packings

Abstract

Computer experiments play a crucial role when physical experiments are expensive or difficult to be carried out. As a kind of designs for computer experiments, maximin distance designs have been widely studied. Many existing methods for obtaining maximin distance designs are based on stochastic algorithms, and these methods will be infeasible when the run size or number of factors is large. In this paper, we propose some deterministic construction methods for maximin $L_2$-distance designs in two to five dimensions based on densest packings. The resulting designs have large $L_2$-distances and are mirror-symmetric. Some of them have the same $L_2$-distances as the existing optimal maximin distance designs, and some of the others are completely new. Especially, the resulting 2-dimensional designs possess a good projection property.

Appendix

The points of $P_{0i},i=1,\ldots ,4$, in Algorithm 1.

$$\begin{aligned} \begin{aligned} P_{01}&=\Big \{(\sqrt{3}j\sin \theta ,-\sqrt{3}j\cos \theta )|j=1,2,\ldots ,\lfloor K/\sqrt{3}\rfloor \Big \},\\ P_{02}&=\Bigg \{\Bigg (\sqrt{3}\left[ (2i+1)\Big \lfloor \frac{K}{\sqrt{3}} \Big \rfloor +2i\Big \lfloor \frac{K}{\sqrt{3}}-\frac{1}{2}\Big \rfloor +3i+j\right] \sin \theta ,\Big (i+\frac{1}{2}\Big )\sin \theta \\&\quad +\sqrt{3}\Big (\Big \lfloor \frac{K}{\sqrt{3}} -\frac{1}{2}\Big \rfloor -j+\frac{3}{2}\Big )\cos \theta \Bigg )\\&\quad \Bigg |i=0,1,\ldots ,\lfloor K\rfloor -1,\ j=1,2,\ldots ,2\left( \Big \lfloor \frac{K}{\sqrt{3}} -\frac{1}{2}\Big \rfloor +1\right) \Bigg \},\\ P_{03}&=\Bigg \{\Bigg (\sqrt{3}\left[ (2i+1)\Big \lfloor \frac{K}{\sqrt{3}} \Big \rfloor +2(i+1)\Big \lfloor \frac{K}{\sqrt{3}}-\frac{1}{2} \Big \rfloor +3i+j+2\right] \sin \theta ,\\&\quad (i+1)\sin \theta +\sqrt{3}\Big (\Big \lfloor \frac{K}{\sqrt{3}} -\frac{1}{2}\Big \rfloor -j+2\Big )\cos \theta \Bigg )\Bigg |i=0,1,\ldots , \Big \lfloor K-\frac{1}{2}\Big \rfloor -1,\\&\quad j=1,2,\ldots , 2\Big \lfloor \frac{K}{\sqrt{3}}\Big \rfloor +1\Bigg \},\\ P_{04}&=\left\{ \begin{array}{l}\Bigg \{\Bigg (K\cos \theta -\sqrt{3} \Big (\left\lfloor \frac{K}{\sqrt{3}}\right\rfloor -j+1\Big ) \sin \theta ,K\sin \theta +\sqrt{3}\Big (\left\lfloor \frac{K}{\sqrt{3}} \right\rfloor -j+1\Big )\cos \theta \Bigg )\\ ~~~\Bigg |j=1,2,\ldots ,\left\lfloor \frac{K}{\sqrt{3}}\right\rfloor -t+1\Bigg \},\qquad \ \ \text {if}\ K\in \ {\mathbb {N}},\\ \\ \Bigg \{\Bigg (K\cos \theta -\sqrt{3}\left( \left\lfloor \frac{K}{\sqrt{3}}-\frac{1}{2}\right\rfloor -j+\frac{3}{2}\right) \sin \theta ,K\sin \theta +\sqrt{3}\left( \left\lfloor \frac{K}{\sqrt{3}}-\frac{1}{2}\right\rfloor -j+\frac{3}{2}\right) \cos \theta \Bigg )\\ ~~~\Bigg |j=1,2,\ldots ,\left\lfloor \frac{K}{\sqrt{3}} \right\rfloor -t+\frac{3}{2}\Bigg \},\qquad \ \ \text {if}\ K+\frac{1}{2}\in {\mathbb {N}}.\end{array}\right. \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}&t=\min \{\max \{t_0,-t_b\},t_b\},\\&t_0=\left\{ \begin{array}{l} \left\lfloor (3+4\lfloor \frac{K}{\sqrt{3}}\rfloor )\left( K+\sqrt{3}\left( \lfloor \frac{K}{\sqrt{3}}\rfloor +\frac{1}{2}\right) \right) -\frac{2K-1}{2\sqrt{3}}\right\rfloor , \quad \text {if}\ K\in {\mathbb {N}},\frac{K}{\sqrt{3}}-\lfloor \frac{K}{\sqrt{3}} \rfloor \ge \frac{1}{2},\\ \left\lfloor (3+4\lfloor \frac{K}{\sqrt{3}}\rfloor )\left( K+\sqrt{3} \left( \lfloor \frac{K}{\sqrt{3}}\rfloor +\frac{1}{2}\right) \right) -\frac{K}{\sqrt{3}} +\frac{1}{2}\right\rfloor -\frac{1}{2},\quad \text {if}\ K+\frac{1}{2}\in {\mathbb {N}}, \frac{K}{\sqrt{3}}-\lfloor \frac{K}{\sqrt{3}}\rfloor \ge \frac{1}{2},\\ \left\lfloor \left( 1+4\lfloor \frac{K}{\sqrt{3}}\rfloor \right) \left( K+\sqrt{3} \lfloor \frac{K}{\sqrt{3}}\rfloor \right) -\frac{K}{\sqrt{3}}\right\rfloor , \quad \text {if}\ K\in {\mathbb {N}},\frac{K}{\sqrt{3}} -\lfloor \frac{K}{\sqrt{3}}\rfloor<\frac{1}{2},\\ \left\lfloor \left( 1+4\lfloor \frac{K}{\sqrt{3}}\rfloor \right) \left( K+\sqrt{3} \lfloor \frac{K}{\sqrt{3}}\rfloor \right) -\frac{2K-1}{2\sqrt{3}} +\frac{1}{2}\right\rfloor -\frac{1}{2},\quad \text {if}\ K+\frac{1}{2}\in {\mathbb {N}},\frac{K}{\sqrt{3}} -\lfloor \frac{K}{\sqrt{3}}\rfloor <\frac{1}{2},\\ \end{array}\right. \\&t_b=\left\{ \begin{array}{l}\lfloor \frac{K}{\sqrt{3}}\rfloor , \quad \text {if}\ K\in {\mathbb {N}},\\ \frac{1}{2}+\lfloor \frac{K}{\sqrt{3}}-\frac{1}{2}\rfloor ,\quad \text {if}\ K+1/2\in {\mathbb {N}}.\\ \end{array}\right. \end{aligned} \end{aligned}$$

Proof of Proposition 1

According to the definition of C, we have

$$\begin{aligned} \begin{aligned} C&=\{z_1{\varvec{\nu }}_1+z_2{\varvec{\nu }}_2:\ z_1,z_2\in {\mathbb {Z}}\}=\{(z_1,0)+(z_2/2,\sqrt{3}z_2/2):\ z_1,z_2\in {\mathbb {Z}}\}\\&=\{(z_1+z_2/2,\sqrt{3}z_2/2):\ z_1,z_2\in {\mathbb {Z}}\}. \end{aligned} \end{aligned}$$

Thus, the points in C can be divided into two categories, which are points on the lines $x=i$ for $i\in {\mathbb {Z}}$ and points on the lines $x=i+1/2$ for $i\in {\mathbb {Z}}$. Combining with the relationship between two coordinates, it is easy to obtain the expressions of $C_1$ and $C_2$ in Proposition 1. Moreover, for any point in C, ${\mathbf{p}}_0=(x_0,y_0)$, according to the structure of C, it is obvious that the neighboring points of ${\mathbf{p}}_0$ are

$$\begin{aligned} \begin{aligned}&{\mathbf{p}}_1=(x_0-1,y_0),\ {\mathbf{p}}_2=\left( x_0-\frac{1}{2},y_0+\frac{\sqrt{3}}{2}\right) ,\ {\mathbf{p}}_3=\left( x_0-\frac{1}{2},y_0-\frac{\sqrt{3}}{2}\right) ,\quad \text {and}\\&{\mathbf{p}}_4=(x_0+1,y_0),\ {\mathbf{p}}_5=\left( x_0+\frac{1}{2},y_0+\frac{\sqrt{3}}{2}\right) ,\ {\mathbf{p}}_6=\left( x_0+\frac{1}{2},y_0-\frac{\sqrt{3}}{2}\right) . \end{aligned} \end{aligned}$$

Then, the $L_2$-distances between ${\mathbf{p}}_0$ and these points are always one. The proof is completed. $\square$

Proof of Corollary 1

We have $X=C\cap [-K,K]^2=(C_1\cup C_2)\cap [-K,K]^2=X_1\cup X_2$, where

$$\begin{aligned} \begin{aligned} X_1&=C_1\cap [-K,K]^2=\left\{ (\pm i,\pm \sqrt{3}j)\big |\ i\le K; \sqrt{3}j\le K; i,j\in {\mathbb {N}}\right\} ,\\&=\left\{ (\pm i,\pm \sqrt{3}j)\big |\ i=0,1,\ldots ,\lfloor K\rfloor ;j=0,1,\ldots ,\lfloor K/\sqrt{3}\rfloor \right\} ,\quad \text {and}\\ X_2&=C_2\cap [-K,K]^2=\left\{ (\pm i,\pm \sqrt{3}j)\big |\ i\le K; \sqrt{3}j\le K; i-\frac{1}{2},j-\frac{1}{2}\in {\mathbb {N}}\right\} \\&=\left\{ (\pm i,\pm \sqrt{3}j)\big |\ i=\frac{1}{2},\ldots ,\frac{1}{2}+\lceil K-1\rceil ;j=\frac{1}{2},\ldots ,\frac{1}{2}+\lfloor K/\sqrt{3}-\frac{1}{2}\rfloor \right\} . \end{aligned} \end{aligned}$$

Thus, the number of points in X is equal to the sum of the numbers of points in $X_1$ and $X_2$, i.e.

$$\begin{aligned} \begin{aligned} n_0&=(2\lfloor K\rfloor +1)(1+2\lfloor K/\sqrt{3}\rfloor )+2(\lceil K-1\rceil +1)\times 2\left( 1+\lfloor K/\sqrt{3}-{1/2}\rfloor \right) \\&=(2\lfloor K\rfloor +1)(1+2\lfloor K/\sqrt{3}\rfloor )+4\lceil K\rceil \left( 1+\lfloor K/\sqrt{3}-{1/2}\rfloor \right) . \end{aligned} \end{aligned}$$

According to the coordinates of points in X, it is obvious that both $X_1$ and $X_2$ are symmetric against two axes and the origin. Thus, if there is a point $\mathbf{a }=(x_1,x_2)\in X_1$ then the points $\mathbf{b }=(x_1,-x_2)$, $\mathbf{c }=(-x_1,x_2)$ and $\mathbf{d }=(-x_1,-x_2)$ all belong to $X_1$, which implies that $X_1$ is column-orthogonal and mirror-symmetric. Similarly, $X_2$ is column-orthogonal and mirror-symmetric. Then, we have that $X=X_1\cup X_2$ is column-orthogonal and mirror-symmetric. $\square$

Proof of Proposition 2

Note that

$$\begin{aligned} \begin{aligned} C^*&=C\cdot R=\{(z_1{\varvec{\nu }}_1+z_2{\varvec{\nu }}_2)\cdot R:\ z_i\in {\mathbb {Z}},i=1,2\}\\&=\{(z_1,z_2)\cdot G_0\cdot R:\ z_i\in {\mathbb {Z}},i=1,2\}\\&=\{(z_1,z_2)\cdot G_0^*:\ z_i\in {\mathbb {Z}},i=1,2\}, \end{aligned} \end{aligned}$$

where $G_0^*=G_0\cdot R$. In other words, obtaining $C^*$ by rotating C is equivalent to generating $C^*$ by the rotated generator matrix $G_0^*$. Thus, we only need to consider the property about the rotation of generator matrix $G_0$.

(1) If $\theta =\pi /6$, we have

$$\begin{aligned} G_0^*=\begin{pmatrix}1 &{}0\\ \frac{1}{2} &{}\frac{\sqrt{3}}{2}\end{pmatrix}\cdot \begin{pmatrix}\frac{\sqrt{3}}{2} &{}\frac{1}{2}\\ -\frac{1}{2} &{}\frac{\sqrt{3}}{2}\end{pmatrix} =\begin{pmatrix}\frac{\sqrt{3}}{2} &{}\frac{1}{2}\\ 0 &{}1\end{pmatrix}. \end{aligned}$$

Then,

$$\begin{aligned} C^*=\left\{ \left( \frac{\sqrt{3}}{2}z_1,\frac{1}{2}z_1+z_2\right) :\ z_1,z_2\in {\mathbb {Z}}\right\} =\left\{ \left( \frac{\sqrt{3}}{2}z_2^*,z_1^* +\frac{1}{2}z_2^*\right) :\ z_1^*,z_2^*\in {\mathbb {Z}}\right\} , \end{aligned}$$

where $z_1^*=z_2$ and $z_2^*=z_1$. It implies that $C^*$ is equivalent to exchanging the two columns of C in the sense of reordering the rows.

(2) If $\theta =\pi /3$, we have

$$\begin{aligned} G_0^*=\begin{pmatrix}1 &{}0\\ \frac{1}{2} &{}\frac{\sqrt{3}}{2}\end{pmatrix}\cdot \begin{pmatrix}\frac{1}{2} &{}\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} &{}\frac{1}{2}\end{pmatrix} =\begin{pmatrix}\frac{1}{2} &{}\frac{\sqrt{3}}{2}\\ -\frac{1}{2} &{}\frac{\sqrt{3}}{2}\end{pmatrix}. \end{aligned}$$

Then,

$$\begin{aligned} C^*= & {} \left\{ \left( \frac{1}{2}z_1-\frac{1}{2}z_2,\frac{\sqrt{3}}{2}z_1 +\frac{\sqrt{3}}{2}z_2\right) :\ z_1,z_2\in {\mathbb {Z}}\right\} \\= & {} \left\{ \left( z_1^*+\frac{1}{2}z_2^*, \frac{\sqrt{3}}{2}z_2^*\right) :\ z_1^*,z_2^*\in {\mathbb {Z}}\right\} , \end{aligned}$$

where $z_1^*=-z_2$ and $z_2^*=z_1+z_2$. It implies that $C^*$ is equivalent to C in the sense of reordering the rows. In these two cases, $C^*$ and C are isomorphic. The proof is completed. $\square$

Proof of Theorem 1

The mirror-symmetrical property of P can be easily obtained from the property of $P_0$. Thus, we only prove the distance property of P here. Combining with Proposition 1 and the relationship between P and $P_0$, we have $d(P)=d(P_0)/(2l)=1/(2l)$ and ${\mathrm{mpd}}(P)={\mathrm{mpd}}(P_0)/(2l)$. So, we only need to obtain ${\mathrm{mpd}}(P_0)$. Since there is a slight difference in the points for the cases of $K\in {\mathbb {N}}$ and $K+1/2 \in {\mathbb {N}}$, the proofs need to be considered separately. Here we only discuss the first case, and the other case follows similarly.

To prove the first case, we define a way of rotation, called non-crossed rotation. Given an $n\times k$ matrix $X=(x_{il})$, let $X^*=(x_{il}^*)=XR$ with rotation angle $\theta$. If for any i, j, l satisfying $x_{il}<x_{jl}$, there is always $x_{il}^*<x_{jl}^*$, then we call this rotation is non-crossed. The rotation used in Algorithm 1 is a non-crossed rotation. According to Corollary 1 and the definition of non-crossed rotation, there are two kinds of projection distances on the first dimension and two kinds on the second dimension between any two neighboring points in $X^*$, which are

$$\begin{aligned} \begin{aligned}&d_{1,1}=\sqrt{3}\sin \theta ,\quad d_{1,2}=\frac{1}{2}\cos \theta -\sqrt{3}\left( \frac{1}{2} +\left\lfloor \frac{K}{\sqrt{3}}-\frac{1}{2}\right\rfloor +\left\lfloor \frac{K}{\sqrt{3}}\right\rfloor \right) \sin \theta ,\quad \text {and}\\&d_{2,1}=\sin \theta ,\quad d_{2,2}=\left( \frac{1}{2}-2K\right) \sin \theta +\frac{\sqrt{3}}{2}\cos \theta , \end{aligned} \end{aligned}$$

respectively. To show these projection distances more clearly, see Fig. 2. Points with the same shape have the same horizontal coordinates before rotating. In this example, there are two kinds of projection distances on the first dimension, which are

$$\begin{aligned} d_1(O,A)=d_1(C,D)=d_{1,1}\quad \text {and}\quad d_1(A,C)=d_1(D,E)=d_{1,2}. \end{aligned}$$

And there are two kinds of projection distances on the second dimension, which are

$$\begin{aligned} d_2(O,B)=d_2(D,H)=d_{2,1}\quad \text {and}\quad d_2(F,G)=d_2(H,I)=d_{2,2}. \end{aligned}$$

Fig. 2

An example of non-crossed rotation

Thus, to achieve equal spacing on the first dimension, let $d_{1,1}=d_{1,2}$. Therefore, we have the value of $\theta$ shown in Algorithm 1, i.e, $\theta =\arctan \left[ \sqrt{3}\left( 3+2\left\lfloor \frac{K}{\sqrt{3}} -\frac{1}{2}\right\rfloor +2\left\lfloor \frac{K}{\sqrt{3}} \right\rfloor \right) \right] ^{-1}$. Based on values of $n,\theta$ and $l=K\cos \theta -\sqrt{3}t\sin \theta$, we have $l=(\sqrt{3}/2)(n-1)\sin \theta$. At this time, it is easy to see that ${\mathrm{mpd}}(P_0)=\min \{d_{1,1},d_{2,1},d_{2,2}\}=\sin \theta$. Thus, we have

$$\begin{aligned} d(P)=\frac{1}{2l}=\frac{1}{\sqrt{3}(n-1)\sin \theta }\quad \text {and}\quad {\mathrm{mpd}}(P)=\frac{{\mathrm{mpd}}(P_0)}{2l}=\frac{1}{\sqrt{3}(n-1)}. \end{aligned}$$

$\square$

Proof of Proposition 3

The coordinates of points in $P_0$ can be obtained easily. Now, let us calculate the number of points in $P_0$. According to the coordinates of points, there are only two cases of points: (1) $x_1,x_2,x_3$ are all even, (2) only one of $x_1,x_2,x_3$ is even and the others are odd. Thus, we have

$$\begin{aligned} \begin{aligned} n&=(1+s/2)^3+{3\atopwithdelims ()1}(1+s/2)(s/2)^2=\frac{1}{2}[(1+s)^3+1]\ \text {for an even}\ s,\ \text {and}\\ n&=[(1+s)/2]^3+{3\atopwithdelims ()1}[(1+s)/2]\cdot [(1+s)/2]^2=\frac{1}{2}(1+s)^3\ \text {for an odd}\ s. \end{aligned} \end{aligned}$$

$\square$

Proof of Corollary 2

It is easy to obtain the coordinates and the number of points of $P_0^c$, so we omit it here. Conway and Sloane (1998) showed that the packing radius, the radius of sphere, of lattice $D_3$ is $r=1/\sqrt{2}$, and the kissing number is $\tau = 12$. To get the $L_2$-distance of $P_0^c$, we only need to illustrate that the structure of $D_3^c$ is the same as that of $D_3$. Based on the coordinates of $D_3$, it can be found that the points in $D_3$ can be divided into two parts $D_{31}$ and $D_{32}$, whose coordinates are

$$\begin{aligned} \begin{aligned}&D_{31}=\{(x_1,x_2,x_3)\in {\mathbb {Z}}^3:\ x_1+x_2=0\ (\text {mod}\ 2)\ \text {and}\ x_3=0\ (\text {mod}\ 2)\},\quad \text {and}\\&D_{32}=\{(x_1,x_2,x_3)\in {\mathbb {Z}}^3:\ x_1+x_2=1\ (\text {mod}\ 2)\ \text {and}\ x_3=1\ (\text {mod}\ 2)\}. \end{aligned} \end{aligned}$$

Obviously, $D_3^c$ has a similar interleaved structure as follows:

$$\begin{aligned} \begin{aligned}&D_{31}^c=\{(x_1,x_2,x_3)\in {\mathbb {Z}}^3:\ x_1+x_2=0\ (\text {mod}\ 2)\ \text {and}\ x_3=1\ (\text {mod}\ 2)\},\quad \text {and}\\&D_{32}^c=\{(x_1,x_2,x_3)\in {\mathbb {Z}}^3:\ x_1+x_2=1\ (\text {mod}\ 2)\ \text {and}\ x_3=0\ (\text {mod}\ 2)\}, \end{aligned} \end{aligned}$$

where $D_3^c=D_{31}^c\cup D_{32}^c$. Thus, the points in $D_3^c$ can be regarded as being obtained from $D_3$ with shifting, which means that the radius of sphere and kissing number of $D_3^c$ are equal to those of $D_3$. So, the $L_2$-distance of $P_0^c$ is $d=2r=\sqrt{2}$, which is the same as that of $P_0$. $\square$

Proof of Corollary 3

According to Corollary 2, we have that the $L_2$-distance is $d_0=\sqrt{2}$ for both $P_0$ and $P_0^c$. Thus, it is easy to obtain that $d(P)=d_0/s=\sqrt{2}/s$ since $P=P_0/s$. Moreover, since the orthogonality property of P is the same as that of $P_0$, we only need to prove that $P_0$ is a column-orthogonal and mirror-symmetric design. Let $P_0^*=P_0+ (-s/2)$. According to the coordinates of points in $P_0^*$, it is easy to find that if there is a point $\mathbf{a }=(x_1,x_2,x_3)\in P_0^*$ then the points $\mathbf{b }=(-x_1,x_2,x_3)$, $\mathbf{c }=(x_1,-x_2,x_3)$, $\mathbf{d }=(x_1,x_2,-x_3)$, $\mathbf{e }=(-x_1,-x_2,x_3)$, $\mathbf{f }=(-x_1,x_2,-x_3)$, $\mathbf{g }=(x_1,-x_2,-x_3)$ and $\mathbf{h }=(-x_1,-x_2,-x_3)$ all belong to $P_0^*$. With a simple calculation, it can be seen that the inner product of any two distinct columns in $P_0^*$ is zero, which means that $P_0$ is column-orthogonal. Moreover, $P_0^*$ can be written as $P_0^*=(\mathbf{X }^T,-\mathbf{X }^T)^T$, which indicates that $P_0$ is a mirror-symmetric design. $\square$