Polynomial approximate discretization of geometric centers in high-dimensional Euclidean space

Abstract

Many geometric optimization problems can be reduced to choosing points in space (centers) minimizing some objective function which continuously depends on the distances from the chosen centers to given input points. We prove that, for any fixed $\varepsilon >0$, every finite set of points in any-dimensional real space admits a polynomial-size set of candidate centers which can be computed in polynomial time and which contains a $(1+\varepsilon )$-approximation of each point of space with respect to the Euclidean distances to all the given points. It provides a universal approximation-preserving reduction of any geometric center-based problems whose objective function satisfies a natural continuity-type condition to their discrete versions where the desired centers are selected from a polynomial-size set of candidates. The obtained polynomial upper bound for the size of a universal centers set is supplemented by a theoretical lower bound for this size in the worst case.

Appendix

Here, we prove Statements 1 and 2, which contain estimations of the functions

$$\begin{aligned} a(\varepsilon )=\frac{\zeta +1}{\lceil \frac{1}{\varepsilon }\log \frac{2}{\varepsilon }\rceil } \hbox { and } b(\varepsilon )=\frac{\ell (\varepsilon )}{(\frac{1}{\varepsilon }\log \frac{2}{\varepsilon })^{\lceil \frac{1}{\varepsilon }\log \frac{2}{\varepsilon }\rceil }}, \end{aligned}$$

where

$$\begin{aligned}&\varepsilon \in (0,1),\zeta =\Big \lceil \frac{\log \frac{1}{0.87\varepsilon }}{\log (1+0.87\varepsilon )}\Big \rceil ,\hbox { and}\\&\ell (\varepsilon )=\left( \frac{\sqrt{\zeta }}{0.26\varepsilon \,(0.87 \varepsilon )^{1/\zeta }}+\frac{1}{0.87\varepsilon }\right) ^\zeta (1+0.87\varepsilon )^{-\zeta (\zeta -1)/2}\zeta . \end{aligned}$$

Statement 1

If $\varepsilon \in (0,1)$, then $a(\varepsilon )\le 1$.

Proof

Case 1: $\varepsilon \in (0,0.4)$. In this case, by using Taylor’s theorem, we obtain that

$$\begin{aligned} \log (1+0.87\varepsilon )=\frac{\ln (1+0.87\varepsilon )}{\ln 2}\ge \frac{0.87\varepsilon -(0.87\varepsilon )^2/2}{\ln 2}>\frac{0.71\varepsilon }{\ln 2}>\varepsilon ,\hbox { so} \\ \frac{\log \frac{1}{0.87\varepsilon }}{\log (1+0.87\varepsilon )}+1< \frac{\log \frac{1+0.87\varepsilon }{0.87\varepsilon }}{\varepsilon }<{\textstyle \frac{1}{\varepsilon }\log \frac{2}{\varepsilon }}. \end{aligned}$$

It follows that $a(\varepsilon )<1$.

Case 2: $\varepsilon \in [0.4,1)$. The interval [0.4, 1) can be divided into 8 subintervals with constant values of both $\zeta$ and $\lceil \frac{1}{\varepsilon }\log \frac{2}{\varepsilon }\rceil$. For each of these subintervals, the inequality $a(\varepsilon )\le 1$ is verified directly. $\square$

The proof of Statement 2 is more complicated. To improve understanding, we first give a short sketch for the justification of the weaker statement:

Simplified stimate

If $\varepsilon \in (0,1)$, then $\displaystyle \ell (\varepsilon )=\big (\,{\mathcal {O}}({\textstyle \frac{1}{\varepsilon }})\log {\textstyle \frac{2}{\varepsilon }}\,\big )^{\lceil \frac{1}{\varepsilon }\log \frac{2}{\varepsilon }\rceil }$.

Proof

By using the inequality $\ln (1+x)\le x$, we obtain that $\displaystyle 1/\zeta \le \frac{0.87\varepsilon }{\ln \frac{1}{0.87\varepsilon }}={\mathcal {O}}(\varepsilon )$, so $(0.87\varepsilon )^{1/\zeta }=\Omega (1)$. Next, by Statement 1, we have $\zeta <\frac{1}{\varepsilon }\log \frac{2}{\varepsilon }$, therefore, the expression $\displaystyle \frac{\sqrt{\zeta }}{0.26\varepsilon \,(0.87\varepsilon )^{1/\zeta }}+\frac{1}{0.87\varepsilon }$ in the definition of $\ell (\varepsilon )$ is $\displaystyle \frac{{\mathcal {O}}(1)}{\varepsilon ^{1.5}}\sqrt{\log {\textstyle \frac{2}{\varepsilon }}}$. On the other hand, the definition of $\zeta$ gives the inequality $\displaystyle (1+0.87\varepsilon )^\zeta \ge \frac{1}{0.87\varepsilon }$, which can be written as $(1+0.87\varepsilon )^{-(\zeta -1)/2}\le \sqrt{0.87\varepsilon \,(1+0.87\varepsilon )}$. It follows that

$$\begin{aligned} \ell (\varepsilon )= \left( \,\frac{{\mathcal {O}}(1)\sqrt{\varepsilon }}{\varepsilon ^{1.5}}{ \textstyle \sqrt{(1+0.87\varepsilon )\log \frac{2}{\varepsilon }}}\ \right) ^\zeta \zeta . \end{aligned}$$

Finally, we recall that $\zeta \le \lceil \frac{1}{\varepsilon }\log \frac{2}{\varepsilon }\rceil -1$, as shown in Statement 1, so we obtain the estimate $\ell (\varepsilon )= \big (\,{\mathcal {O}}({\textstyle \frac{1}{\varepsilon }})\log {\textstyle \frac{2}{\varepsilon }}\,\big )^{\lceil \frac{1}{\varepsilon }\log \frac{2}{\varepsilon }\rceil }$. $\square$

Now, let us prove the required stronger statement.

Statement 2

If $\varepsilon \in (0,1)$, then $b(\varepsilon )< 37$.

Proof

Case 1: $\varepsilon \in (0,2^{-14})$. The inequality $\ln (1+x)\le x$ implies that $\displaystyle \zeta \ge \frac{\ln \frac{1}{0.87\varepsilon }}{0.87\varepsilon }$. Hence, for small $\varepsilon$, the second term in the sum

$$\begin{aligned} \frac{\sqrt{\zeta }}{0.26\varepsilon \,(0.87\varepsilon )^{1/\zeta }}+\frac{1}{0.87\varepsilon } \end{aligned}$$

is much less than the first. It follows that this sum can be estimated, e.g., as the value $\displaystyle \frac{\sqrt{\zeta }}{0.25\varepsilon \,(0.87\varepsilon )^{1/\zeta }}$. Next, by the definition of $\zeta$, we have $\displaystyle (1+0.87\varepsilon )^\zeta \ge \frac{1}{0.87\varepsilon }$. Then

$$\begin{aligned} \ell (\varepsilon )< \left( \frac{\sqrt{\zeta }}{0.25\varepsilon \,(0.87\varepsilon )^{1/\zeta }}\right) ^\zeta (0.87\varepsilon )^{(\zeta -1)/2}\zeta = \left( \frac{\sqrt{0.87\zeta }}{0.25\sqrt{\varepsilon }}\,\right) ^\zeta (0.87\varepsilon )^{-1.5}\zeta . \end{aligned}$$

But $\displaystyle \zeta \le \frac{\log \frac{1}{0.87\varepsilon }}{\log (1+0.87\varepsilon )}+1$ and, for small $\varepsilon$, it is less than $\frac{1}{\varepsilon }\log \frac{2}{\varepsilon }$, as shown in the proof of Statement 1. So

$$\begin{aligned} \ell (\varepsilon )<\left( \frac{\sqrt{0.87}}{0.25\varepsilon }{ \textstyle \sqrt{\log \frac{2}{\varepsilon }}}\,\right) ^{\frac{1}{\varepsilon }\log \frac{2}{\varepsilon }}\frac{\log \frac{2}{\varepsilon }}{0.87^{1.5}\varepsilon ^{2.5}}. \end{aligned}$$

It remains to note that the obtained expression is less than $\big (\frac{1}{\varepsilon }\log \frac{2}{\varepsilon }\big )^{\frac{1}{\varepsilon }\log \frac{2}{\varepsilon }}$ since

$$\begin{aligned} \left( \frac{\sqrt{0.87}}{0.25}\,\right) ^{\frac{1}{\varepsilon } \log \frac{2}{\varepsilon }}\frac{\log \frac{2}{\varepsilon }}{0.87^{1.5} \varepsilon ^{2.5}}<\left( {\textstyle \sqrt{\log \frac{2}{\varepsilon }}}\,\right) ^{\frac{1}{\varepsilon }\log \frac{2}{\varepsilon }} \end{aligned}$$

for $\varepsilon =2^{-14}$ and, therefore, for any $\varepsilon <2^{-14}$. Thus, we have $b(\varepsilon )<1$.

Case 2: $\varepsilon \in [2^{-14},1)$. First, let us consider the expression

$$\begin{aligned} \displaystyle \ell (z,\varepsilon )=\left( \frac{\sqrt{z}}{0.26\varepsilon \,(0.87\varepsilon )^{1/z}}+\frac{1}{0.87\varepsilon }\right) ^z(1+0.87\varepsilon )^{-z(z-1)/2}z \end{aligned}$$

as the function of $\varepsilon \in [2^{-14},1)$ and $\displaystyle z\in \{\zeta -1,\,\zeta \}$. For any fixed positive integer z, if we decrease $\varepsilon$, then the value of this function increases. Similarly, if we fix $\varepsilon$ and increase z from $\zeta -1$ to $\zeta$, then we obtain that $\ell (\zeta ,\varepsilon )>\ell (\zeta -1,\varepsilon )$ since the terms

$$\begin{aligned} \Big (\frac{\sqrt{z}}{0.26\varepsilon \,(0.87\varepsilon )^{1/z}}(1+0.87\varepsilon )^{-(z-1)/2}\Big )^z \hbox { and } \Big (\frac{1}{0.87\varepsilon }(1+0.87\varepsilon )^{-(z-1)/2}\Big )^z \end{aligned}$$

increase at least in $\displaystyle \frac{(1+0.87\varepsilon )^{1-\zeta }}{0.26\varepsilon }>\frac{0.87\varepsilon }{0.26\varepsilon }>1$ and $\displaystyle \frac{(1+0.87\varepsilon )^{1-\zeta }}{0.87\varepsilon }>\frac{0.87\varepsilon }{0.87\varepsilon }=1$ times, respectively. Taking into account that the value of $\zeta$ is an integer-value function of $\varepsilon$, increasing when $\varepsilon$ decreases, the above observations imply that the function $\ell (\varepsilon )=\ell (\zeta ,\varepsilon )$ increases with decreasing $\varepsilon$.

On the other hand, the function $L(\varepsilon )=\big (\frac{1}{\varepsilon }\log \frac{2}{\varepsilon }\big )^{\lceil \frac{1}{\varepsilon }\log \frac{2}{\varepsilon }\rceil }$, the denominator in the expression for $b(\varepsilon )$, also increases with decreasing $\varepsilon$. Hence, for any positive integer J and each $j=1,\dots ,J$, we obtain the inequality $\displaystyle \max _{\varepsilon \in [\varepsilon _{j-1},\varepsilon _j]}b(\varepsilon )\le \frac{\ell (\varepsilon _{j-1})}{L(\varepsilon _j)}$, where $\varepsilon _0=2^{-14}$ and $\varepsilon _j=\varepsilon _0+(1-\varepsilon _0)\,j/J$. It follows that

$$\begin{aligned} \max _{\varepsilon \in [2^{-14},1]}b(\varepsilon )=\max _{j=1,\dots ,J}\max _{\varepsilon \in [\varepsilon _{j-1},\varepsilon _j]}b(\varepsilon )\le \max _{j=1,\dots ,J}\frac{\ell (\varepsilon _{j-1})}{L(\varepsilon _j)}. \end{aligned}$$

To finish the proof, we choose $J=2^{15}$ and, by using computer calculations, verify that $\displaystyle \frac{\ell (\varepsilon _{j-1})}{L(\varepsilon _j)}<37$ for all $j=1,\dots ,J$. $\square$