A Distributed O(1)-Approximation Algorithm for the Uniform Facility Location Problem

Abstract

We investigate a metric facility location problem in a distributed setting. In this problem, we assume that each point is a client as well as a potential location for a facility and that the opening costs for the facilities and the demands of the clients are uniform. The goal is to open a subset of the input points as facilities such that the accumulated cost for the whole point set, consisting of the opening costs for the facilities and the connection costs for the clients, is minimized.

We present a randomized distributed algorithm that computes in expectation an ${\mathcal {O}}(1)$-approximate solution to the metric facility location problem described above. Our algorithm works in a synchronous message passing model, where each point is an autonomous computational entity that has its own local memory and that communicates with the other entities by message passing. We assume that each entity knows the distance to all the other entities, but does not know any of the other pairwise distances. Our algorithm uses three rounds of all-to-all communication with message sizes bounded to $\mathcal{O}(\log(n))$ bits, where n is the number of input points.

We extend our distributed algorithm to constant powers of metric spaces. For a metric exponent ℓ≥1, we obtain a randomized ${\mathcal {O}}(1)$-approximation algorithm that uses three rounds of all-to-all communication with message sizes bounded to $\mathcal{O}(\log(n))$ bits.

Appendix: Proof of Lemma 2

In the proof, we use a modified version of the Mettu-Plaxton algorithm [14]. The original Mettu-Plaxton algorithm computes for each point x _i ∈X a special value, which the authors call the radius of x _i . Then, it goes through all the points in X in non-decreasing order of their radii and opens a facility at a point x _i ∈X if x _i has no open facility in the ball whose center is x _i and whose radius is twice the radius of x _i . Our modified version works exactly as the Mettu-Plaxton algorithm except that, in the first step, it computes, for each point x _i ∈X, the radius r _i that satisfies (1), instead of the original radius proposed by Mettu and Plaxton [14]. A pseudocode listing of this algorithm is given by Algorithm A.1.

Algorithm A.1

Modified-Mettu-Plaxton(X)

We will first show that this modified Mettu-Plaxton algorithm yields a constant-factor approximation. Based on this result, we will then prove that the sum of the exponentiated radii approximates the optimal cost FacLoc^∗(X,f,ℓ) within a constant factor.

Let F _MP be the set of open facilities computed by the modified Mettu-Plaxton algorithm. In the following, we will show that $\operatorname {FacLoc}(X,F_{\textsc{MP}},f,\ell) \le 3^{\ell}\cdot\allowbreak \mathrm {FacLoc}^{*}(X,f,\ell)$. The argumentation is basically the same as in [14]. Only a few minor adaptations to our scenario have been made.

Proposition 1

For any point x _i ∈X, there exists an open facility x _j ∈F _MP such that r _j ≤r _i and D(x _i ,x _j )≤2⋅r _i .

Proof

If there is no such open facility x _j with r _j ≤r _i in $\mathcal{B}(x_{i}, 2 \cdot r_{i})$, then we open a facility at x _i and x _i belongs to F _MP . □

Proposition 2

Let x _i and x _j be distinct open facilities in F _MP . Then, we have D(x _i ,x _j )>2⋅max{r _i ,r _j }.

Proof

Without loss of generality, we assume that r _j ≤r _i and x _j is previous to x _i in the sorted sequence processed by the algorithm. It follows that $x_{j} \notin \mathcal{B}(x_{i},2 \cdot r_{i})$. Otherwise, the point x _i would not be an open facility. Thus, we have

$$\mathrm {D}(x_i,x_j) > 2 \cdot r_i \ge 2 \cdot r_j . $$

 □

For any point x _j ∈X and an arbitrary set of open facilities F′⊆X, let

$$\operatorname {charge}\bigl(x_j,F'\bigr) := \mathrm {D}\bigl(x_j,F'\bigr)^\ell + \sum_{x_i \in F'} \max\bigl\{ 0,r_i^\ell - \mathrm {D}(x_i,x_j)^\ell\bigr\} . $$

Proposition 3

For an arbitrary set of open facilities F′⊆X, we have

$$\sum_{x_j \in X} \operatorname {charge}\bigl(x_j,F' \bigr) = \operatorname {FacLoc}\bigl(X,F',f,\ell\bigr) . $$

Proof

Due to the definition of $\operatorname {charge}(\cdot,\cdot)$ and Equation (1), we get

 □

Proposition 4

Let x _j ∈X be any point, let F′⊆X be an arbitrary set of open facilities, and let x _i ∈F′ be any open facility. If we have D(x _j ,x _i )=D(x _j ,F′), then $\operatorname {charge}(x_{j},F') \ge \max\{ r_{i}^{\ell},\mathrm {D}(x_{j},x_{i})^{\ell}\}$.

Proof

If $x_{j} \notin \mathcal{B}(x_{i},r_{i})$, then

Otherwise, we have

 □

Proposition 5

Let x _j ∈X be any point, and let x _i be any open facility in F _MP . If $x_{j} \in \mathcal{B}(x_{i},r_{i})$, then $\operatorname {charge}(x_{j},F_{\textsc{MP}}) \le r_{i}^{\ell}$.

Proof

By Proposition 2, there is no open point x _m ∈F _MP such that we have i≠m and $x_{j} \in \mathcal{B}(x_{m},r_{m})$. Since D(x _j ,F _MP )≤D(x _j ,x _i ), we obtain

 □

Proposition 6

Let x _j ∈ be any point, and let x _i be any open facility in F _MP . If $x_{j} \notin \mathcal{B}(x_{i},r_{i})$, then we have $\operatorname {charge}(x_{j},F_{\textsc{MP}}) \le \mathrm {D}(x_{j},x_{i})^{\ell}$.

Proof

The correctness of the assertion follows immediately, unless there is an open facility x _m ∈F _MP such that $x_{j} \in \mathcal{B}(x_{m},r_{m})$. If such an open facility x _m exists, then Propositions 2 and 5 imply D(x _i ,x _m )>2⋅max{r _i ,r _m } and $\operatorname {charge}(x_{j},F_{\textsc{MP}}) \le r_{m}^{\ell}$. Furthermore, by triangle inequality, we obtain

which proves $\operatorname {charge}(x_{j},F_{\textsc{MP}}) \le r_{m}^{\ell}\le \mathrm {D}(x_{j},x_{i})^{\ell}$. □

Proposition 7

For any point x _j ∈X and an arbitrary set of open facilities F′⊆X, we have $\operatorname {charge}(x_{j},F_{\textsc{MP}}) \le 3^{\ell}\cdot \operatorname {charge}(x_{j},F')$.

Proof

Let x _i be some open facility in F′ such that we have D(x _j ,x _i )=D(x _j ,F′). By Proposition 1, there exists a facility x _m ∈F _MP such that we have r _m ≤r _i and D(x _i ,x _m )≤2⋅r _i .

If $x_{j} \in \mathcal{B}(x_{m},r_{m})$, then we get $\operatorname {charge}(x_{j},F_{\textsc{MP}}) \le r_{m}^{\ell}$ by Proposition 5. Since Proposition 4 implies $\operatorname {charge}(x_{j},F') \ge r_{i}^{\ell}$, we can conclude

This proves the assertion in case that we have $x_{j} \in \mathcal{B}(x_{m},r_{m})$.

If $x_{j} \notin \mathcal{B}(x_{m},r_{m})$, then $\operatorname {charge}(x_{j},F_{\textsc{MP}}) \le \mathrm {D}(x_{j},x_{m})^{\ell}$ by Proposition 6. Thus, by triangle inequality, we get

Now, the assertion follows by Proposition 4. □

Lemma 11

$\operatorname {FacLoc}(X,F_{\textsc{MP}},f,\ell) \le 3^{\ell}\cdot \mathrm {FacLoc}^{*}(X,f,\ell)$

Proof

The correctness of the assertion follows from Propositions 3 and 7. □

Based on the results above, we can prove Lemma 2. We first prove the lower bound and then the upper bound. The argumentation is basically the same as in [2]. Only a few minor adaptations to our scenario have been made.

Lower bound

Let F _MP be the set of open facilities computed by the modified Mettu-Plaxton algorithm. Then, it follows from Proposition 1 that

$$ 2^{\ell}\cdot\sum_{x_i \in X} r_i^{\ell} \ge \sum_{x_i \in X} \mathrm {D}(x_i,F_\textsc{MP})^\ell. $$

(6)

Next, we show that we also have

$$ 2^{\ell}\cdot\sum_{x_i \in X} r_i^{\ell} \ge f \cdot \vert F_\textsc{MP} \vert . $$

(7)

Due to Proposition 2, each point x _i ∈X is contained in at most one ball $\mathcal{B}(x_{j},r_{j})$ for some open facility x _j ∈F _MP . Furthermore, for any point $x_{m} \in \mathcal{B}(x_{j},r_{j})$, we must have r _j ≤2⋅r _m . Let us assume that this assumption is false and r _m <r _j /2. Then, we have that x _m is previous to x _j in the ordered sequence processed by the algorithm. Furthermore, the algorithm either opens a facility at x _m or there is an open facility in the ball $\mathcal{B}(x_{m},2\cdot r_{m})$. Since

$$x_m \in \mathcal{B}(x_m,2\cdot r_m) \subseteq \mathcal{B}(x_m,r_j) \subseteq \mathcal{B} \bigl(x_m,r_j + \mathrm {D}(x_j,x_m) \bigr) \subseteq \mathcal{B}(x_j,2\cdot r_j) , $$

the modified Mettu-Plaxton algorithm would not open a facility at x _j , which is a contradiction. Hence, we obtain

which proves inequality (7). Due to inequalities (6) and (7), we get

Upper bound

Due to Lemma 11, we know that

$$\operatorname {FacLoc}(X,F_\textsc{MP},f,\ell) \le 3^\ell \cdot \mathrm {FacLoc}^*(X,f,\ell) . $$

Thus, to prove the upper bound, it remains to show that

$$\sum_{x_i \in X} r_i^{\ell} \,\le\, 2^{\ell} \cdot \operatorname {FacLoc}(X,F_\textsc{MP},f,\ell) . $$

Due to Proposition 3, we have

where δ(j) denotes the index of the facility in F _MP that is closest to x _j . Thus, if we can show that

$$ 2^{\ell} \cdot \biggl( \sum_{x_i \in F_\textsc{MP}} r_i^\ell + \sum_{x_j \in X \backslash F_\textsc{MP}} \max\bigl\{ r_{\delta(j)}^\ell,\mathrm {D}(x_j,x_{\delta(j)})^\ell \bigr\} \biggr) \ge \sum_{x_i \in X} r_i^\ell , $$

(8)

then we are done. It is sufficient to prove

$$ r_j^{\ell} \,\le\, 2^{\ell-1} \cdot \bigl( \mathrm {D}(x_j, x_{\delta (j)})^{\ell} + r_{\delta (j)}^{\ell} \bigr) $$

(9)

because this implies $\max\{ r_{\delta(j)}^{\ell},\mathrm {D}(x_{j},x_{\delta(j)})^{\ell}\} \ge r_{j}^{\ell}/2^{\ell}$ and Inequality (8) follows. We prove the correctness of Inequality (9) by contradiction. Hence, we assume that

$$r_j^{\ell} > 2^{\ell -1} \cdot \bigl( \mathrm {D}(x_j, x_{\delta (j)})^{\ell} + r_{\delta (j)}^{\ell}\bigr) . $$

We can easily prove by induction that 2^ℓ−1⋅(a ^ℓ+b ^ℓ)≥(a+b)^ℓ for any a,b≥0. Thus, we obtain

$$r_j^{\ell} > \bigl( \mathrm {D}(x_j, x_{\delta (j)}) + r_{\delta (j)}\bigr) ^{\ell} , $$

which, in turn, would imply $\mathcal{B}(x_{\delta (j)}, r_{\delta (j)}) \subseteq \mathcal{B}(x_{j}, r_{j})$. Furthermore, by applying triangle inequality and 2^ℓ−1⋅(a ^ℓ+b ^ℓ)≥(a+b)^ℓ for an a,b≥0, we get

as upper bound on the exponentiated distance between x _j and any point $x_{m} \in \mathcal{B}(x_{\delta (j)}, r_{\delta (j)})$. Now, we obtain

which is a contradiction because the definition of r _j requires

$$\sum_{x_m \in X \cap \mathcal{B}(x_j,r_j)} r_j^{\ell} - \mathrm {D}(x_j,x_m)^{\ell} = f . $$

It follows that inequality (9) is true, which was the only thing left to prove the assertion of the lemma.