A Las Vegas approximation algorithm for metric $1$-median selection

TL;DR

This paper introduces a randomized approximation algorithm for the metric 1-median problem that guarantees a solution within a factor of (2+ε) in expected linear time, with a high-probability near-optimal solution.

Contribution

It presents a novel Las Vegas approximation algorithm for the metric 1-median problem with improved expected runtime and approximation guarantees.

Findings

Always outputs a (2+ε)-approximate solution in expected O(n/ε^2) time.

Inherits properties from Indyk's algorithm to produce a (1+ε)-approximate median with constant probability.

Provides a practical randomized approach for efficient median approximation in metric spaces.

Abstract

Given an $n$-point metric space, consider the problem of finding a point with the minimum sum of distances to all points. We show that this problem has a randomized algorithm that {\em always} outputs a $(2+\epsilon)$-approximate solution in an expected $O(n/\epsilon^2)$ time for each constant $\epsilon>0$. Inheriting Indyk's algorithm, our algorithm outputs a $(1+\epsilon)$-approximate $1$-median in $O(n/\epsilon^2)$ time with probability $\Omega(1)$.

Equations

$$\sum _{x\in [n]}d\left(p,x\right)\le \alpha \cdot \underset{y\in [n]}{\min }\sum _{x\in [n]}d\left(y,x\right).$$

$$\sum _{x\in [n]}d\left(z,x\right)\le \underset{y\in [n]\setminus B(z,8r)}{\inf }\sum _{x\in [n]}d\left(y,x\right).$$

$$\underset{y\in B(z,8r)}{\min }\sum _{x\in B(z,\delta nr)}d\left(y,x\right)$$

$$\ge$$

$$\sum _{x\in B(z,\delta nr)}d\left(z,x\right)\le \left(2+ϵ\right)\cdot \sum _{i=1}^{\lfloor |B(z,\delta nr)|/2\rfloor }d\left(\pi \left(2i-1\right),\pi \left(2i\right)\right).$$

$$\sum _{x\in B(z,\delta nr)}d\left(z^{\prime },x\right)\le \left(2+o(1)\right)\cdot \mathrm{E}\left[\sum _{i=1}^{\lfloor |B(z,\delta nr)|/2\rfloor }d\left(\pi \left(2i-1\right),\pi \left(2i\right)\right)\right].$$

$$\sum _{x\in [n]}d\left(z,x\right)\le \left(1+\frac{ϵ}{10^{10}}\right)\cdot \sum _{x\in [n]}d\left(z^{\prime },x\right).$$

$$B\left(x,R\right)\equiv \left\{y\in [n]\mid d\left(x,y\right)<R\right\}$$

$$B^2\left(x,R\right)\equiv B\left(x,R\right)\times B\left(x,R\right).$$

$$\mathrm{Pr}\left[\left|X-\mathrm{E}[X]\right|\ge k\sqrt{\mathrm{var}(X)}\right]\le \frac{1}{k^2}.$$

$$\underset{y\in [n]\setminus B(z,8r)}{\inf }\sum _{x\in [n]}d(y,x)\ge 7\cdot \sum _{x\in [n]}d(z,x).$$

$$\sum _{x\in [n]}d(y,x)$$

$$\underset{y\in B(z,8r)}{\min }\sum _{x\in [n]}d(y,x)\ge \frac{1-100\sqrt{\delta }}{2}\cdot \sum _{x\in [n]}d(z,x).$$

$$\sum _{x\in B(z,\delta nr)}d(y,x)$$

$$\sum _{x\in [n]\setminus B(z,\delta nr)}d(y,x)$$

$$\sum _{x\in [n]}d(y,x)\ge \sum _{i=1}^{\lfloor |B(z,\delta nr)|/2\rfloor }d\left(\pi \left(2i-1\right),\pi \left(2i\right)\right)+\sum _{x\in [n]\setminus B(z,\delta nr)}\left(d(z,x)-8r\right).$$

$$\sum _{x\in [n]}d(y,x)\ge \frac{1-100\sqrt{\delta }}{2}\cdot nr$$

$$\left(2+ϵ\right)\cdot \underset{y\in [n]}{\min }\sum _{x\in [n]}d(y,x)\ge \sum _{x\in [n]}d(z,x),$$

$$\left|[n]\setminus B\left(z,\delta nr\right)\right|\le \frac{1}{\delta }$$

$$\left|B\left(z,\delta nr\right)\right|\ge n-\frac{1}{\delta }=\left(1-o(1)\right)n.$$

$$\sum _{x\in [n]}d\left(z,x\right)\ge \sum _{x\in [n]\setminus B(z,\delta nr)}d\left(z,x\right)\ge \sum _{x\in [n]\setminus B(z,\delta nr)}\delta nr=\left|[n]\setminus B\left(z,\delta nr\right)\right|\cdot \delta nr.$$

$$r^{\prime }\equiv \frac{1}{|B(z,\delta nr){|}^2}\cdot \sum _{u,v\in B(z,\delta nr)}d\left(u,v\right)$$

$$r^{\prime }$$

$$\frac{1}{|B(z,\delta nr)|}\cdot \sum _{u\in B(z,\delta nr)}d\left(z,u\right)\le \frac{1}{n}\cdot \sum _{u\in [n]}d\left(z,u\right).$$

$$\sum _{i=1}^{\lfloor |B(z,\delta nr)|/2\rfloor }d\left(\pi \left(2i-1\right),\pi \left(2i\right)\right),$$

$$\mathrm{E}\left[\sum _{i=1}^{\lfloor |B(z,\delta nr)|/2\rfloor }d\left(\pi \left(2i-1\right),\pi \left(2i\right)\right)\right]=\frac{1}{2}\cdot \left(1\pm o(1)\right)n{r}^{\prime }.$$

$$\mathrm{E}\left[d\left(\pi \left(2i-1\right),\pi \left(2i\right)\right)\right]$$

$$=$$

$$\mathrm{E}\left[{\left(\sum _{i=1}^{\lfloor |B(z,\delta nr)|/2\rfloor }d\left(\pi \left(2i-1\right),\pi \left(2i\right)\right)\right)}^2\right]$$

$$+$$

$$\sum _{\text{distinct i,j=1}}^{\lfloor |B(z,\delta nr)|/2\rfloor }\mathrm{E}\left[d\left(\pi \left(2i-1\right),\pi \left(2i\right)\right)\cdot d\left(\pi \left(2j-1\right),\pi \left(2j\right)\right)\right]\le \frac{1}{4}\cdot \left(1+o(1)\right)n^2{\left(r^{\prime }\right)}^2.$$

$$\left\{\pi \left(2i-1\right),\pi \left(2i\right),\pi \left(2j-1\right),\pi \left(2j\right)\right\}$$

Full text

1 Introduction

A metric space is a nonempty set $M$ endowed with a metric, i.e., a function $d\colon M\times M\to[\,0,\infty\,)$ such that

• •

$d(x,y)=0$ if and only if $x=y$ (identity of indiscernibles),

• •

$d(x,y)=d(y,x)$ (symmetry), and

• •

$d(x,y)+d(y,z)\geq d(x,z)$ (triangle inequality)

for all $x$, $y$, $z\in M$ [13].

For all $n\in\mathbb{Z}^{+}$, define $[n]\equiv\{1,2,\ldots,n\}$. Given $n\in\mathbb{Z}^{+}$ and oracle access to a metric $d\colon[n]\times[n]\to[\,0,\infty\,)$, metric $1$-median asks for $\mathop{\mathrm{argmin}}_{y\in[n]}\,\sum_{x\in[n]}\,d(y,x)$, breaking ties arbitrarily. It generalizes the classical median selection on the real line and has a brute-force $\Theta(n^{2})$-time algorithm. More generally, metric $k$-median asks for $c_{1}$, $c_{2}$, $\ldots$, $c_{k}\in[n]$ minimizing $\sum_{x\in[n]}\,\min_{i=1}^{k}\,d(x,c_{i})$. Because $d(\cdot,\cdot)$ defines $\binom{n}{2}=\Theta(n^{2})$ nonzero distances, only $o(n^{2})$-time algorithms are said to run in sublinear time [8]. For all $\alpha\geq 1$, an $\alpha$-approximate $1$-median is a point $p\in[n]$ satisfying

[TABLE]

For all $\epsilon>0$, metric $1$-median has a Monte Carlo $(1+\epsilon)$-approximation $O(n/\epsilon^{2})$-time algorithm [8, 9]. Guha et al. [7] show that metric $k$-median has a Monte Carlo, $O(\exp(O(1/\epsilon)))$-approximation, $O(nk\log n)$-time, $O(n^{\epsilon})$-space and one-pass algorithm for all small $k$ as well as a deterministic, $O(\exp(O(1/\epsilon)))$-approximation, $O(n^{1+\epsilon})$-time, $O(n^{\epsilon})$-space and one-pass algorithm. Given $n$ points in $\mathbb{R}^{D}$ with $D\geq 1$, the Monte Carlo algorithms of Kumar et al. [10] find a $(1+\epsilon)$-approximate $1$-median in $O(D\cdot\exp(1/\epsilon^{O(1)}))$ time and a $(1+\epsilon)$-approximate solution to metric $k$-median in $O(Dn\cdot\exp((k/\epsilon)^{O(1)}))$ time. All randomized $O(1)$-approximation algorithms for metric $k$-median take $\Omega(nk)$ time [11, 7]. Chang [2] shows that metric $1$-median has a deterministic, $(2h)$-approximation, $O(hn^{1+1/h})$-time and nonadaptive algorithm for all constants $h\in\mathbb{Z}^{+}\setminus\{1\}$, generalizing the results of Chang [1] and Wu [15]. On the other hand, he disproves the existence of deterministic $(2h-\epsilon)$-approximation $O(n^{1+1/(h-1)}/h)$-time algorithms for all constants $h\in\mathbb{Z}^{+}\setminus\{1\}$ and $\epsilon>0$ [3, 4].

In social network analysis, the closeness centrality of a point $v$ is the reciprocal of the average distance from $v$ to all points [14]. So metric $1$-median asks for a point with the maximum closeness centrality. Given oracle access to a graph metric, the Monte-Carlo algorithms of Goldreich and Ron [6] and Eppstein and Wang [5] estimate the closeness centrality of a given point and those of all points, respectively.

All known sublinear-time algorithms for metric $1$-median are either deterministic or Monte Carlo, the latter having a positive probability of failure. For example, Indyk’s Monte Carlo $(1+\epsilon)$-approximation algorithm outputs with a positive probability a solution without approximation guarantees. In contrast, we show that metric $1$-median has a randomized algorithm that always outputs a $(2+\epsilon)$-approximate solution in expected $O(n/\epsilon^{2})$ time for all constants $\epsilon>0$. So, excluding the known deterministic algorithms (which are Las Vegas only in the degenerate sense), this paper gives the first Las Vegas approximation algorithm for metric $1$-median with an expected sublinear running time. Note that deterministic sublinear-time algorithms for metric $1$-median can be $4$-approximate but not $(4-\epsilon)$-approximate for any constant $\epsilon>0$ [1, 4]. So our approximation ratio of $2+\epsilon$ beats that of any deterministic sublinear-time algorithm. Inheriting Indyk’s algorithm, our algorithm outputs a $(1+\epsilon)$-approximate $1$-median in $O(n/\epsilon^{2})$ time with probability $\Omega(1)$ for all constants $\epsilon>0$.

Below is our high-level and inaccurate sketch of proof, where $\epsilon$, $\delta>0$ are small constants:

1. (i)

Run Indyk’s algorithm to find a probably $(1+\epsilon/10^{10})$-approximate $1$-median, $z$. Then let $r=\sum_{x\in[n]}\,d(z,x)/n$ be the average distance from $z$ to all points. 2. (ii)

For all $R>0$, denote by $B(z,R)$ the open ball with center $z$ and radius $R$. Use the triangle inequality (with details omitted here) to show $z$ to be a solution no worse than the points in $[n]\setminus B(z,8r)$, i.e.,

[TABLE] 3. (iii)

Take a uniformly random bijection $\pi\colon[\,|B(z,\delta nr)|\,]\to B(z,\delta nr)$. Then observe that

[TABLE]

where the first (resp., second) inequality follows from the injectivity of $\pi$ (resp., the triangle inequality). 4. (iv)

Assume $B(z,\delta nr)=[n]$ for simplicity. So by inequalities (1)–(3), if the following inequality holds, then it serves as a witness that $z$ is $(2+\epsilon)$-approximate:

[TABLE]

333Assuming $B(z,\delta nr)=[n]$, inequalities (2)–(4) imply $\sum_{x\in[n]}\,d(z,x)\leq(2+\epsilon)\cdot\sum_{x\in[n]}\,d(y,x)$ for all $y\in B(z,8r)$. Furthermore, $\sum_{x\in[n]}\,d(z,x)\leq\sum_{x\in[n]}\,d(y,x)$ for all $y\in[n]\setminus B(z,8r)$ by inequality (1).

To guarantee outputting a $(2+\epsilon)$-approximate $1$-median, output $z$ only when inequality (4) holds. Restart from item (i) whenever inequality (4) is false.

More details of item (iv) follow: For a $1$-median $z^{\prime}$ of $B(z,\delta nr)$, it will be easy to show

[TABLE]

444Though not directly stated in later sections, this is a consequence of Lemmas 7 and 12 in Sec. 4.

When $z$ in item (i) is indeed $(1+\epsilon/10^{10})$-approximate,

[TABLE]

Assuming $B(z,\delta nr)=[n]$, inequalities (5)–(6) make inequality (4) hold with high probability as long as $\sum_{i=1}^{\lfloor|B(z,\delta nr)|/2\rfloor}\,d(\pi(2i-1),\pi(2i))$ is highly concentrated around its expectation. The need for such concentration is why we restrict the radius of the codomain of $\pi$ to be $\delta nr$ in item (iii)—Large distances ruin concentration bounds. To accommodate for the points in $[n]\setminus B(z,\delta nr)$, our witness for the approximation ratio of $z$ actually differs slightly from inequality (4), unlike in item (iv).555Our witness for the approximation ratio of $z$ is as in line 6 of Las Vegas median in Fig. 1.

2 Definitions and preliminaries

For a metric space $([n],d)$, $x\in[n]$ and $R>0$, define

[TABLE]

to be the open ball with center $x$ and radius $R$. For brevity,

[TABLE]

The pairs in $B^{2}(x,R)$ are ordered.

An algorithm $A$ with oracle access to $d\colon[n]\times[n]\to[\,0,\infty\,)$ is denoted by $A^{d}$ and may query $d$ on any $(x,y)\in[n]\times[n]$ for $d(x,y)$. In this paper, all Landau symbols (such as $O(\cdot)$, $o(\cdot)$, $\Theta(\cdot)$ and $\Omega(\cdot)$) are w.r.t. $n$. The following result is due to Indyk.

Fact 1** ([8, 9]).**

For all $\epsilon>0$, metric $1$-median has a Monte Carlo $(1+\epsilon)$-approximation $O(n/\epsilon^{2})$-time algorithm with a failure probability of at most $1/e$.

Henceforth, denote Indyk’s algorithm in Fact 1 by Indyk median. It is given $n\in\mathbb{Z}^{+}$, $\epsilon>0$ and oracle access to a metric $d\colon[n]\times[n]\to[\,0,\infty\,)$. By convention, denote the expected value and the variance of a random variable $X$ by $\mathop{\mathrm{E}}[\,X\,]$ and $\mathop{\mathrm{var}}(X)$, respectively.

Chebyshev’s inequality** ([12]).**

Let $X$ be a random variable with a finite expected value and a finite nonzero variance. Then for all $k\geq 1$,

[TABLE]

3 Algorithm and approximation ratio

Throughout this paper, take any small constant $\epsilon>0$, e.g., $\epsilon=10^{-100}$. By line 1 of Las Vegas median in Fig. 1, $\delta>0$ is likewise a small constant. The following lemma implies that $z$ in line 3 of Las Vegas median is a solution (to metric $1$-median) no worse than those in $[n]\setminus B(z,8r)$, where $r$ is as in line 4.

Lemma 2**.**

In each iteration of the while loop of Las Vegas median,

[TABLE]

Proof.

For each $y\in[n]\setminus B(z,8r)$,

[TABLE]

where the first inequality follows from the triangle inequality, the second follows from $y\notin B(z,8r)$ and the last equality follows from line 4 of Las Vegas median. ∎

Lemma 3**.**

When line 7 of Las Vegas median is run,

[TABLE]

Proof.

Pick any $y\in B(z,8r)$. We have

[TABLE]

where the first and the second inequalities follow from the injectivity of $\pi$ in line 5 of Las Vegas median and the triangle inequality, respectively.666Note that $\pi(1)$, $\pi(2)$, $\ldots$, $\pi(2\,\lfloor|B(z,\delta nr)|/2\rfloor)$ are distinct elements of $B(z,\delta nr)$. Furthermore,

[TABLE]

where the first and the second inequalities follow from the triangle inequality and $y\in B(z,8r)$, respectively. Summing up inequalities (3)–(8),

[TABLE]

This and lines 6–7 of Las Vegas median imply

[TABLE]

when line 7 is run. Finally, $nr=\sum_{x\in[n]}\,d(z,x)$ by line 4. ∎

Lemmas 2–3 and line 1 of Las Vegas median yield the following.

Lemma 4**.**

When line 7 of Las Vegas median is run,

[TABLE]

i.e., $z$ is a $(2+\epsilon)$-approximate $1$-median.

By Lemma 4, Las Vegas median outputs a $(2+\epsilon)$-approximate $1$-median at termination.

4 Probability of termination in any iteration

This section analyzes the probability of running line 7 in any particular iteration of the while loop of Las Vegas median. The following lemma uses an easy averaging argument.

Lemma 5**.**

[TABLE]

and, therefore,

[TABLE]

Proof.

Clearly,

[TABLE]

Then use line 4 of Las Vegas median. ∎

Henceforth, assume $n\geq 1/\delta+4$ without loss of generality; otherwise, find a $1$-median by brute force. So $|B(z,\delta nr)|\geq 4$ by Lemma 5. Define

[TABLE]

to be the average distance in $B(z,\delta nr)$.

Lemma 6**.**

$r^{\prime}\leq 2r$.

Proof.

By equation (9) and the triangle inequality,

[TABLE]

Obviously, the average distance from $z$ to the points in $B(z,\delta nr)$ is at most that from $z$ to all points, i.e.,

[TABLE]

Inequalities (4)–(11) and line 4 of Las Vegas median complete the proof. ∎

To analyze the probability that the condition in line 6 of Las Vegas median holds, we shall derive a concentration bound for

[TABLE]

whose expected value and variance are examined in the next four lemmas.

Lemma 7**.**

With expectations taken over $\pi$,

[TABLE]

Proof.

For each $i\in[\,\lfloor|B(z,\delta nr)|/2\rfloor\,]$, $\{\pi(2i-1),\pi(2i)\}$ is a uniformly random size-$2$ subset of $B(z,\delta nr)$ by line 5 of Las Vegas median. Therefore,

[TABLE]

where the second (resp., last) equality follows from the identity of indiscernibles (resp., equation (9) and Lemma 5). Finally, use equations (4)–(14), the linearity of expectation and Lemma 5. ∎

Clearly,

[TABLE]

where the last equality follows from the linearity of expectation and the separation of pairs $(i,j)$ according to whether $i=j$.

Lemma 8**.**

With expectations taken over $\pi$,

[TABLE]

Proof.

Pick any distinct $i$, $j\in[\,\lfloor|B(z,\delta nr)|/2\rfloor\,]$. By line 5 of Las Vegas median,

[TABLE]

is a uniformly random size-$4$ subset of $B(z,\delta nr)$. So

[TABLE]

Clearly,

[TABLE]

In summary,

[TABLE]

Together with Lemma 5 and equation (9), this completes the proof. ∎

Lemma 9**.**

With expectations taken over $\pi$,

[TABLE]

Proof.

By line 5 of Las Vegas median, $\{\pi(2i-1),\pi(2i)\}$ is a uniformly random size-$2$ subset of $B(z,\delta nr)$ for each $i\in[\,\lfloor|B(z,\delta nr)|/2\rfloor\,]$. Therefore,

[TABLE]

For all $u$, $v\in B(z,\delta nr)$,

[TABLE]

where the first inequality follows from the triangle inequality.

By equations (9) and (4)–(19), the left-hand side of inequality (17) cannot exceed the optimal value of the following problem, called max square sum:

Find $d_{u,v}\in\mathbb{R}$ for all $u$, $v\in B(z,\delta nr)$ to maximize

[TABLE]

subject to

[TABLE]

Above, constraint (21) (resp., (22)) mimics equation (9) (resp., inequality (19) and the non-negativeness of distances). Appendix A bounds the optimal value of max square sum from above by

[TABLE]

This evaluates to be at most $(1+o(1))(\delta n^{2}rr^{\prime}+2\delta^{2}nr^{2})$ by Lemma 5. ∎

Recall that the variance of any random variable $X$ equals $\mathop{\mathrm{E}}[X^{2}]-(\mathop{\mathrm{E}}[X])^{2}$.

Lemma 10**.**

With variances taken over $\pi$,

[TABLE]

Proof.

By equations (4)–(16) and Lemmas 8–9,

[TABLE]

This and Lemma 7 imply

[TABLE]

Finally, invoke Lemma 6. ∎

Lemma 11**.**

For all $k>1$,

[TABLE]

where the probability is taken over $\pi$.

Proof.

Use Chebyshev’s inequality and Lemmas 7 and 10. ∎

Let $z^{\prime}\in B(z,\delta nr)$ be a $1$-median of $B(z,\delta nr)$, i.e.,

[TABLE]

breaking ties arbitrarily. So by the averaging argument,

[TABLE]

Lemma 12**.**

[TABLE]

Proof.

We have

[TABLE]

Clearly, $|B\left(z,\delta nr\right)|\leq n$. ∎

Lemma 13**.**

For all sufficiently large $n$,

[TABLE]

Proof.

We have

[TABLE]

where the first inequality (resp., the first equality) follows from the triangle inequality (resp., line 4 of Las Vegas median). By Lemmas 6 and 12,

[TABLE]

By inequalities (4)–(25) and Lemma 5, $d(z^{\prime},z)\leq(3+o(1))r$.777In fact, this is stronger than the lemma to be proved. ∎

Lemma 14**.**

For all sufficiently large $n$,

[TABLE]

Proof.

By the triangle inequality,

[TABLE]

Now sum up the above with the inequality in Lemma 12. ∎

Lemma 15**.**

For all sufficiently large $n$ and with probability greater than $1/2$,

[TABLE]

where the probability is taken over $\pi$ and the internal coin tosses of Indyk median in line 3 of Las Vegas median.

Proof.

By Lemma 11 with $k=5$,

[TABLE]

with probability at least $1-1/25$. By Fact 1 and line 3 of Las Vegas median,

[TABLE]

with probability at least $1-1/e$. Now by the union bound, inequalities (27)–(29) hold simultaneously with probability at least $1-1/25-1/e>1/2$. It remains to derive inequality (26) from inequalities (27)–(29) for all sufficiently large $n$.

Line 4 of Las Vegas median, inequalities (28)–(29) and Lemma 14 give

[TABLE]

This and inequality (27) imply

[TABLE]

888To see this, rewrite inequality (27) as

$nr^{\prime}<2\left(1\pm o(1)\right)\left[\left(\sum_{i=1}^{\lfloor|B(z,\delta nr)|/2\rfloor}\,d\left(\pi\left(2i-1\right),\pi\left(2i\right)\right)\right)+5\sqrt{2\left(1+o(1)\right)\delta}\,nr\right].$

Clearly, $16r/\delta\leq 0.01\cdot\sqrt{\delta}\,nr$ for all sufficiently large $n$. So inequality (31) implies, for all sufficiently large $n$ and after laborious calculations,

[TABLE]

This implies inequality (26) for all sufficiently large $n$ (note that $\epsilon/10^{10}<\sqrt{\delta}$ by line 1 of Las Vegas Median).999Divide both sides by $(2+2\epsilon/10^{10})(1+o(1))$ so that the coefficient before $\sum_{i=1}^{\lfloor|B(z,\delta nr)|/2\rfloor}\,d(\pi(2i-1),\pi(2i))$ becomes $1$ in the right-hand side. Then verify the left-hand side (which is now $(nr-(1+\epsilon/10^{10})11\sqrt{2(1+o(1))\delta}\,nr)/((2+2\epsilon/10^{10})(1+o(1)))$) to be at least $(1-100\sqrt{\delta})nr/2$ for all sufficiently large $n$. ∎

Lemma 15 and lines 6–7 of Las Vegas median show the probability of termination in any iteration to be $\Omega(1)$. Because the proof of Lemma 15 implies that inequalities (26)–(29) hold simultaneously with probability $\Omega(1)$ in any iteration of Las Vegas median, it happens with probability $\Omega(1)$ that in the first iteration, $z$ is returned in line 7 (because of inequality (26)) and is $(1+\epsilon/10^{10})$-approximate (because of inequality (28)). So Las Vegas median outputs a $(1+\epsilon/10^{10})$-approximate $1$-median with probability $\Omega(1)$ in the first iteration. In summary, we have the following.

Lemma 16**.**

The first iteration of the while loop of Las Vegas median outputs a $(1+\epsilon)$-approximate $1$-median with probability $\Omega(1)$.

5 Putting things together

We now show that metric $1$-median has a Las Vegas $(2+\epsilon)$-approximation algorithm with an expected $O(n/\epsilon^{2})$ running time for all constants $\epsilon>0$. Our algorithm also outputs a $(1+\epsilon)$-approximate $1$-median in time $O(n/\epsilon^{2})$ with probability $\Omega(1)$.

Theorem 17**.**

For each constant $\epsilon>0$, metric $1$-median has a randomized algorithm that (1) always outputs a $(2+\epsilon)$-approximate solution in an expected $O(n/\epsilon^{2})$ time and that (2) outputs a $(1+\epsilon)$-approximate solution in time $O(n/\epsilon^{2})$ with probability $\Omega(1)$.

Proof.

By Lemma 4, Las Vegas median outputs a $(2+\epsilon)$-approximate $1$-median at termination. To prevent Las Vegas median from running forever, find a $1$-median by brute force (which obviously takes $O(n^{2})$ time) after $n^{2}$ steps of computation.

By Fact 1, line 3 of Las Vegas median takes $O(n/\epsilon^{2})$ time. Line 5 takes time $O(|B(z,\delta nr)|)=O(n)$ by the Knuth shuffle. Clearly, the other lines also take $O(n)$ time. Consequently, each iteration of the while loop of Las Vegas median takes $O(n/\epsilon^{2})$ time. By Lemma 15 and lines 6–7, Las Vegas median runs for at most $1/\Omega(1)=O(1)$ iterations in expectation. So its expected running time is $O(1)\cdot O(n/\epsilon^{2})=O(n/\epsilon^{2})$.

Having shown each iteration of Las Vegas median to take $O(n/\epsilon^{2})$ time, establish condition (2) of the theorem with Lemma 16. ∎

By Fact 1, Indyk median satisfies condition (2) in Theorem 17. But it does not satisfy condition (1).

We briefly justify the optimality of the ratio of $2+\epsilon$ in Theorem 17. Let $A$ be a randomized algorithm that always outputs a $(2-\epsilon)$-approximate $1$-median. Furthermore, denote by $p\in[n]$ (resp., $Q\subseteq[n]\times[n]$) the output (resp., the set of queries as unordered pairs) of $A^{d_{1}}(n)$, where $d_{1}$ is the discrete metric (i.e., $d_{1}(x,y)=1$ and $d_{1}(x,x)=0$ for all distinct $x$, $y\in[n]$). Without loss of generality, assume $(p,y)\in Q$ for all $y\in[n]\setminus\{p\}$ by adding dummy queries. So $A$ knows that

[TABLE]

Furthermore, assume that $A$ never queries for the distance from a point to itself.

In the sequel, consider the case that $|Q|<\epsilon\cdot(n-1)^{2}/4$. By the averaging argument, there exists a point $\hat{p}\in[n]\setminus\{p\}$ involved in at most $2\cdot|Q|/(n-1)$ queries in $Q$. Clearly, $A$ cannot exclude the possibility that $d_{1}(\hat{p},y)=1/2$ for all $y\in[n]\setminus\{\hat{p}\}$ satisfying $(\hat{p},y)\notin Q$. In summary, $A$ cannot rule out the case that

[TABLE]

Equations (32)–(33) contradict the guarantee that $p$ is $(2-\epsilon)$-approximate. In summary, any randomized algorithm that always outputs a $(2-\epsilon)$-approximate $1$-median must always make at least $\epsilon\cdot(n-1)^{2}/4=\Omega(\epsilon n^{2})$ queries given oracle access to the discrete metric.

Appendix A Analyzing max square sum

Max square sum has an optimal solution, denoted $\{\tilde{d}_{u,v}\in\mathbb{R}\}_{u,v\in B(z,\delta nr)}$, because its feasible solutions (i.e., those satisfying constraints (21)–(22)) form a closed and bounded subset of $\mathbb{R}^{(|B(z,\delta nr)|^{2})}$. (Recall from elementary mathematical analysis that a continuous real-valued function on a closed and bounded subset of $\mathbb{R}^{k}$ has a maximum value, where $k<\infty$.) Note that $\{\tilde{d}_{u,v}\in\mathbb{R}\}_{u,v\in B(z,\delta nr)}$ must be feasible to max square sum. Below is a consequence of constraint (21).

Lemma A.1**.**

[TABLE]

Proof.

Clearly,

[TABLE]

Furthermore, the left-hand side of inequality (34) is an integer. ∎

Lemma A.2**.**

[TABLE]

Proof.

Assume otherwise. Then

[TABLE]

So by constraint (22) (and the feasibility of $\{\tilde{d}_{u,v}\}_{u,v\in B(z,\delta nr)}$ to max square sum),

[TABLE]

Consequently, there exist distinct $(x,y)$, $(x^{\prime},y^{\prime})\in B^{2}(z,\delta nr)$ satisfying

[TABLE]

By symmetry, assume $\tilde{d}_{x,y}\geq\tilde{d}_{x^{\prime},y^{\prime}}$. By inequality (35), there exists a small real number $\beta>0$ such that increasing $\tilde{d}_{x,y}$ by $\beta$ and simultaneously decreasing $\tilde{d}_{x^{\prime},y^{\prime}}$ by $\beta$ will preserve constraints (21)–(22). I.e., the solution $\{\hat{d}_{u,v}\in\mathbb{R}\}_{u,v\in B(z,\delta nr)}$ defined below is feasible to max square sum:

[TABLE]

Clearly, objective (20) w.r.t. $\{\hat{d}_{u,v}\}_{u,v\in B(z,\delta nr)}$ exceeds that w.r.t. $\{\tilde{d}_{u,v}\}_{u,v\in B(z,\delta nr)}$ by

[TABLE]

where the inequality holds because $\tilde{d}_{x,y}\geq\tilde{d}_{x^{\prime},y^{\prime}}$ and $\beta>0$.

In summary, $\{\hat{d}_{u,v}\}_{u,v\in B(z,\delta nr)}$ is a feasible solution achieving a greater objective (20) than the optimal solution $\{\tilde{d}_{u,v}\}_{u,v\in B(z,\delta nr)}$ does, a contradiction. ∎

We now bound the optimal value of max square sum.

Theorem A.3**.**

The optimal value of max square sum is at most

[TABLE]

Proof.

W.r.t. the optimal (and thus feasible) solution $\{\tilde{d}_{u,v}\}_{u,v\in B(z,\epsilon nr)}$, objective (20) equals

[TABLE]

where $\chi[P]=1$ if $P$ is true and $\chi[P]=0$ otherwise, for any predicate $P$. Now invoke Lemma A.2. ∎