Speeding up Dynamic Programming in the Line-Constrained k-median

Abstract

In the planar k-median problem we are given a set of demand points and want to open up to k facilities as to minimize the sum of the transportation costs from each demand point to its nearest facility. In the line-constrained version the medians are required to lie on a given line. We present a new dynamic programming formulation for this problem, based on constructing a weighted DAG over a set of median candidates. We prove that, for any convex distance metric and any line, this DAG satisfies the concave Monge property. This allows us to construct efficient algorithms in $L_\infty$ and $L_1$ and any line, while the previously known solution (Wang and Zhang, ISAAC 2014) works only for vertical lines. We also provide an asymptotically optimal $\mathcal {O}(n)$ solution for the case of $k=1$.

1 Introduction

The planar k-median problem is a variation of the well-known facility location problem. For a given set P of demand points, we want to find a set Q of k facilities, such that the sum of all transportation costs from a demand point to its closest facility is minimized. Each $p\in P$ is associated with its own (positive) cost per unit of distance to assigned facility, denoted w(p). Formally, we want to minimize:

$$ S(P) = \sum _{p \in P} \min _{q \in Q} \, w(p) \cdot d(p, q) $$

Because the problem is NP-hard for many metrics [7], we further restrict it by introducing a line-constraint on the set Q. We require that all facilities should belong to a specified facility line $\chi$ defined by an equation $ax + by = c$, where $a,b,c \in \mathbb {R} \>$ and $\>a \cdot b \ne 0$. Such a constraint is natural when all facilities are by design placed along a path that can be locally treated as linear, e.g., pipeline, railroad, highway, country border, river, longitude or latitude.

For $k=1$ we obtain the line-constrained 1-median problem. Despite the additional restriction, the complexity of this simplest variant strongly depends on the metric. For a point $p \in \mathbb {R}^2$, let x(p) and y(p) denote its x- and y-coordinate. The most natural metric is the Euclidean distance, where ${L_2}(p,q) = \sqrt{(x(p) - x(q))^2 + (y(p) - y(q))^2}$. It is known that even for 5 points, it is not possible to construct the 1-median with a ruler and compass. It can also be proven that the general, line-constrained and 3-dimension versions of the k-median problem are not solvable over the field of rationals [2]. Hence it is natural to consider also other distance functions, for example:

• Chebyshev distance ${L_\infty }(p,q) = \max \{ |x(p) - x(q) |, |y(p) - y(q) |\}$,

• Manhattan distance ${L_1}(p, q) = |x(p) - x(q) |+ |y(p) - y(q) |$,

• squared Euclidean distance ${L_{2}^{2}}(p,q) = (x(p) - x(q))^2 + (y(p) - y(q))^2.$ ¹

All these distances functions have been recently considered by Wang and Zhang [10] in the context of line-constrained k-median problem. They designed efficient algorithms based on a reduction to the minimum weight k-link path problem. However, their $L_1$ and $L_\infty$ solutions work only in the special case of a horizontal facility line.

We provide a different dynamic programming formulation of the problem that works for any facility line $\chi$ in $L_1$ and $L_\infty$. The new formulation can also be seen as a minimum weight k-link path in a DAG, where the weights are Monge. However, looking up the weight of an edge in this DAG is more expensive. We show how to implement edge lookups in $\mathcal {O}(\log n)$ after $\mathcal {O}(n\log n)$ time and space preprocessing which then allows us to apply the SMAWK algorithm [1] or, if $k=\varOmega (\log n)$, the algorithm of Schieber [9] to obtain the following complexities.

Metric	Facility line	Time complexity
Wang and Zhang [10]
$L_1$	horizontal	$\min \{\mathcal {O}(nk), n2^{\mathcal {O}(\sqrt{\log k \log \log n})}\log n\}$
$L_\infty$	horizontal	$\min \{\mathcal {O}(nk \log n), n2^{\mathcal {O}(\sqrt{\log k \log \log n})}\log ^2 n\}$
Our results
$L_1$	general	$\min \{\mathcal {O}(nk \log n), n2^{\mathcal {O}(\sqrt{\log k \log \log n})}\log n\}$
$L_\infty$	general	$\min \{\mathcal {O}(nk \log n), n2^{\mathcal {O}(\sqrt{\log k \log \log n})}\log n\}$

In $L_\infty$, our general solution is faster than the one given by Wang and Zhang for the special case of horizontal facility line. We also provide a specialized procedure solving the problem for $k=1$ in linear time.

2 Preliminaries

A basic tool for speeding up dynamic programming is the so-called Monge property. It can often be used to improve the time complexity by an order of magnitude, especially in geometric problems.

Definition 1

A weight function w is concave Monge if, for all $a < b$ and $c < d$, $w(a,c) + w(b,d) \le w(b,c) + w(a,d)$.

Dynamic programming can often be visualized as finding the row minima in a $n\times n$ matrix. Naively, this takes $\mathcal {O}(n^2)$ time. However, Aggarwal et al. [1] showed how to decrease the time complexity to $\mathcal {O}(n)$ if the matrix has the so-called total monotonicity property, which is often established through the Monge property. Their method is usually referred to as the SMAWK algorithm. There is a deep connection between SMAWK and other methods for speeding up dynamic programming, such as the Knuth-Yao inequality used for building optimal binary search trees, as observed by Bein et al. [3].

Let D be a DAG on n nodes $0,1,\ldots ,n-1$ with a concave Monge weight function w(i, j) defined for $0 \le i< j < n$ that corresponds to the weight of the edge $\left\langle i,j \right\rangle$. A minimum diameter path in D is a path from 0 to $n-1$ with the minimum weight. Galil and Park showed how to find such path in optimal $\mathcal {O}(n)$ time using the SMAWK algorithm [5]. A minimum weight k-link path is a minimum weight path from 0 to $n-1$ consisting of exactly k edges (links).

Lemma 2

Minimum weight k-link path can be found in $\mathcal {O}(nk)$ and, for $k=\varOmega (\log n)$, $n2^{\mathcal {O}(\sqrt{\log k \log \log n})}$ time.

Proof

To obtain $\mathcal {O}(nk)$ time complexity, we iteratively compute minimum weight 1-link, 2-link, ..., $(k-1)$-link and finally k-link paths from 0 to every other node. This can be seen as k layers of dynamic programming, each requiring only $\mathcal {O}(n)$ time thanks to the SMAWK algorithm. Alternatively, $n2^{\mathcal {O}(\sqrt{\log k \log \log n})}$ time algorithm for $k=\varOmega (\log n)$ was given by Schieber [9].    $\square$

The weights of the edges in our DAG will be computed on-the-fly with orthogonal queries. We will use the following tool: preprocess a given set of n weighted points in a plane for computing the sum of the weights of all points in a given query range $[x,+\infty ] \times [y,+\infty ]$. We call this problem orthogonal range sum. The following is well-known.

Lemma 3

There exists a data structure for the orthogonal range sum problem that can be built in $\mathcal {O}(n \log n)$ time and answers any query in $\mathcal {O}(\log n)$ time.

Proof

We convert the points into a sequence by sorting them according to their x-coordinates (without losing the generality, these coordinates are all distinct) and writing down the corresponding y-coordinates. The y-coordinates are further normalized by replacing with the ranks on a sorted list of all y-coordinates (again, we assume that they are all distinct). Hence we obtain a sequence of length n over an alphabet [n], where each character has its associated weight. We build a wavelet tree [6] of this sequence in $\mathcal {O}(n\log n)$ time. Each node of the wavelet tree is augmented with an array of partial sums of the prefixes of its subsequence. Given an orthogonal query, we first normalize it by looking at the sorted list of all x- and y-coordinates. Then we traverse the wavelet tree starting from the root and accumulate appropriate partial sums. The details can be found in [8].    $\square$

3 Normalizing Problem Instances

$L_1$ and $L_\infty$ metrics are equivalent, which can be seen by rotating the plane by $45^{\circ }$. Hence from now on we will work in $L_1$ metric. This simplification was not possible in the previous approach [10], since it required the facility line to be horizontal, which is no longer true after rotation.

We further modify the problem instance so that the line $\chi$ is expressed in a slope intercept form $y=ax$, where $a \in [0, 1]$, and all coordinates of points in P are non-negative. This is always possible by reflecting along the horizontal axis, then along the line $y=x$, and finally translating. Such transformations do not modify the distances in $L_1$, so computing the k-median solution Q for the transformed instance gives us the answer for the original instance. Because any solution Q can be transformed so that the x-coordinates of all facilities are distinct without increasing the cost, we will consider only such solutions and identify each facility with its x-coordinate.

4 Computing 1-median

Let $D (p, x)$ be the weighted distance between $p\in P$ and $(x, a\cdot x)\in \chi$:

$$ D (p, x) = w(p) \cdot d(p, (x, a\cdot x)) $$

Whenever we say that $p\in P$ is closer to coordinate $x_i$ than $x_j$, we mean that $D(p,x_i)<D(p,x_j)$. For a set of points $A\subseteq P$, $D (A, x)$ is the sum of weighted distances:

$$ D (A, x) = \sum _{p \in A} w(p) \cdot d(p, (x, a\cdot x)) $$

The 1-median is simply $\min \limits _{x \in \mathbb {R}} D (P, x)$.

A function $f : \mathbb {R} \rightarrow \mathbb {R}$ is convex if the line segment between any two points on its graph lies above or on the graph. Such functions have the following properties:

• 1. $f(x) = |x - y|$ is convex for any y.

• 2. if f(x) is convex, then $g(x)= c \cdot f(x)$ is convex for any positive c.

• 3. if f(x) and g(x) are convex, then $h(x)=f(x) + g(x)$ is also convex.

Lemma 4

For any point p, $D (p, x)$ is convex. For any set of points P, $D (P, x)$ is also convex.

Proof

Consider any point $p\in P$. From the definition:

$$ D (p, x) = w(p) \cdot L_1(p, (x, a\cdot x)) = w(p) \cdot \left( | x(p) - x | + |y(p) - a \cdot x | \right) . $$

This is a sum of absolute values functions multiplied by the (positive) weight of p. Hence by the properties of convex functions $D (p, x)$ is convex. Then $D (P, x)$ is also convex since it is a sum of convex functions over $p \in P$.    $\square$

Since $D (p, x)$ is convex, any of its local minima is a global minimum. Similarly to the function $f(x) = |x|$, it is only semi-differentiable. Its derivative $D'(p, x)$ is a staircase nondecreasing function, undefined for at most two values $x=x_1$ and $x=x_2$. We call $x_1$ and $x_2$ the median candidates and for convenience assume that $D'(p,x)$ is equal to its right derivative there. When $a=0$ or $p\in \chi$, $D'(p, x)$ has exactly one median candidate $x_1=x(p)$, that is the minimum. Otherwise, there are two median candidates $x_1=x(p)$ and $x_2=\frac{y(p)}{a}$. For $a \in (0,1)$, $x_1$ is the only minimum, whereas for $a=1$ every value in range $[x_1, x_2]$ is a minimum. Because the derivative of a sum of functions is the sum of their derivatives, $D'(P, x)$ can only change at a median candidate of some $p \in P$. This means that a minimum of D(p, x) corresponds to one of at most 2n median candidates of P. In other words, there exists a solution $(x,y)\in \chi$, such that $x=x(p)$ or $y=y(p)$ for some $p\in P$. From now on, we use $\mathcal {M}(P)$ to denote the set of median candidates of P. $\mathcal {M}(P)$ can be computed in $\mathcal {O}(n)$ time by simply iterating over $p\in P$ and adding $x=x(p)$ and $x=\frac{y(p)}{a}$ to the result (note that this might give us a multiset, i.e., some median candidates might be included multiple times).

Theorem 5

We can solve line-constrained 1-median problem in $\mathcal {O}(n)$ time.

Proof

Because $D'(p,x)$ is nondecreasing, we can binary search for the largest x such that $D'(p,x) \le 0$. Then we return x as the solution. In every step of the binary search we use the median selection algorithm [4] to narrow down the current search range $X = (x_{\text {left}}, x_{\text {right}})$. At the beginning of every step:

• 1. M is a multiset of all median candidates of P that are in X.

• 2. S contains all points from P with at least one median candidate in M.

• 3. $r = D'(P \setminus S, x)$ for some $x \in X$.

We select the median $x_m$ of M and compute $D'(P,x_m)$. If $D'(p,x_m) > 0$, we continue the search in $(x_{\text {left}},x_m)$, and otherwise in $(x_m,x_{\text {right}})$, updating S and M accordingly. Eventually $x_{\text {left}}=x_{\text {right}}$ and we return $x_{\text {left}}$.

The key observation is that when a point p is removed from S, it does no longer have a median candidate within X and its $D'(p,x)$ remains constant in all further computations. This means that $D'(P \setminus S, x)$ is constant for all $x\in X$ and r can be updated after removing every point p from S in $\mathcal {O}(1)$ time. $x_m$ can be found in $\mathcal {O}(|M|)$ time. Calculating $D'(P,x_m)= r + D'(S, x_m)$ then takes $\mathcal {O}(1+|S|)$ time. For a point p to be in S, one of its median candidates must belong to M, so $|S| \le |M|$. Hence the complexity of a single iteration is $\mathcal {O}(|M|)$. After each iteration the size of M decreases by a factor of two, so the running time is described by $T(n)=\mathcal {O}(n)+T(n/2)$, which solves to $\mathcal {O}(n)$.    $\square$

Theorem 6

We can calculate D(P, x) for every $x \in \mathcal {M}(P)$ in $\mathcal {O}(n \log n)$ time.

Proof

The elements of $\mathcal {M}(P)$ can be sorted in $\mathcal {O}(n\log n)$ time, and we can assume that every point generates exactly two median candidates. Let $\mathcal {M}(P)=\{x_1,x_2,\ldots ,x_{2n}\}$, where $x_i \le x_{i+1}$ for all $i=1,2,\ldots ,2n-1$. Recall that $D'(P,x)=D'(P,x_{i})$ for any $x\in (x_i,x_{i+1})$. We compute $D(p,x_1)$ together with $D'(P,x_1)$ in $\mathcal {O}(n)$ time. Then all other $D(p,x_i)$ are computed sequentially for $i=2,3,\ldots ,2n$ in $\mathcal {O}(1)$ time each using the formula:

$$\begin{aligned} D(P,x_{i})&= D(P, x_{i-1}) + D'(P, x_{i-1}) \cdot (x_{i} - x_{i-1}) \\ D'(P, x_{i})&= D'(P, x_{i-1}) + 2 \cdot w(p) \cdot \sigma \end{aligned}$$

where $x_i$ is generated by the point p, $\sigma = 1$ if $x_i=x(p)$ and $\sigma =a$ otherwise.    $\square$

5 Computing k-median

Consider now any optimal solution Q of the k-median problem for the given set of weighted points P. For any facility $q \in Q$, let $P_q$ be the set of points of P assigned to q. By interchanging the order of the summation, Q should minimize

$$ \sum _{q \in Q} \sum _{p \in P_q} w(p) \cdot d(p, q). $$

Hence q must be an optimal solution of the 1-median problem for $P_q$. Since replacing q will not increase the sum of distances of points in $P\setminus P_q$, q can be chosen to be a median candidate of $P_q$. We deduce that there exists an optimal solution $Q'$ such that

For $k \ge \min (n, |\mathcal {M}(P)|)$, every $p\in P$ can be assigned to its closest possible facility. Such an assignment can be easily computed in $\mathcal {O}(n)$ time. If we are required to return exactly k medians, then we add enough additional points to $\mathcal {M}(P)$. From now on, we assume that $k < \min (n, |\mathcal {M}(P)|)$. Thus there exists an optimal k-median solution, where all facilities are 1-median candidates of P.

Fig. 1.

Path in the DAG ending at the candidate $x_i$. Dashed lines represent current assignment of points from P to the closest chosen facility.

By arranging all median candidates in a sequence according to their x-coordinates, we can view choosing k facilities as selecting a $(k+1)$-link path in a DAG between two artificial elements infinitely to the left and to the right of the sequence, called source and sink, respectively.

Imagine that we traverse the sequence from left to right while deciding if we should open a new facility at the current median candidate, see Fig. 1. Initially, all points are assigned to the artificial facility source and the cost of the current solution S is set to $+\infty$. If we decide to open a new facility at the current median candidate $x_j$, for every $p\in P$ we check if $x_j$ is closer to p than the facility p is currently assigned to. If so, we reassign p to $x_j$, see Fig. 2.

Fig. 2.

We follow the edge $\left\langle i, j \right\rangle$. All (black) points now assigned to $x_j$ were previously assigned to $x_i$, see Fig. 1.

We claim that $p\in P$ can be closer to $x_j$ than the facility p is currently assigned to only if the currently assigned facility is the most recently chosen facility $x_i$, that is, the current solution does not contain any facilities between $x_i$ and $x_j$. Assuming that the claim holds, we define the weight of an edge $\left\langle \textsf {source},i \right\rangle$ to be $D(P,x_i)$, and the weight of an internal edge $\left\langle i, j \right\rangle$ to be total decrease of the cost after giving each point $p\in P$ the possibility to switch from $x_i$ to $x_j$. Finally, the weight of an edge $\left\langle j,\textsf {sink} \right\rangle$ is 0. Then selecting k medians corresponds to selecting an $(k+1)$-link from $\textsf {source}$ to $\textsf {sink}$ in the DAG. However, we need to show the claim. To this end we consider the following properties of convex functions:

Proposition 7

For any convex function f and $a< b < c$:

• (a) If $f(c) < f(b)$ then $f(b) < f(a)$.

• (b) If $f(c) < f(a)$ then $f(b) < f(a)$.

Proof

Assume otherwise for any of the two implications. This means that $f(a) \le f(b) > f(c)$ and the segment AC where $A=(a, f(a))$ and $C=(c, f(c))$ lies below f(b), contradicting the assumption that the function f is convex.    $\square$

Now we can prove the claim. Consider a point $p\in P$ such that its currently assigned facility is $x_i$ and, for some $k>i$, facility $x_k$ was not selected as a better option. Then, for any $j>k$, facility $x_j$ cannot be a better option either, because $x_i< x_k < x_j$ so by Proposition 7(a) $D(p,x_i) \le D(p,x_k)$ implies $D(p,x_j) \ge D(p,x_k)$. This means that if $x_i$ was the most recently opened facility and $x_j$ is the current median candidate, opening a new facility at $x_j$ changes the total cost by

$$ \sum _{p \in P} \min ( D (p, x_j) - D (p, x_i), 0). $$

Definition 8

Let $x_1, x_2,\ldots , x_{n-1}, x_{2n}$ be the sorted sequence of median candidates of P. We define its median DAG over nodes $0,1,\ldots ,2n,2n+1$ with edge weight function w(i, j) as follows:

The total cost of any k-median solution is equal to the sum of weights on its corresponding path of length $k+1$ between 0 and $2n+1$, so finding k-median reduces to finding the minimum weight $(k+1)$-link path in the median DAG.

Because a sum of Monge functions is also Monge, to prove that w(i, j) is Monge we argue that $w_p(i,j)$ is Monge, where $w(i,j) = \sum \limits _{p \in P} w_p(i,j)$ and:

Proposition 9

For any convex function f, if $a<b<c$ then:

$$ \min (f(c) - f(a), 0) \le \min (f(c) - f(b), 0). $$

Proof

If $f(c) \ge f(b)$ then the right side of the equation is equal to 0 and left side is non-positive. If $f(c) < f(b)$ then by Proposition 7(a) also $f(b) < f(a)$, so

$$ \min (f(c)-f(a),0) \le f(c)-f(a) < f(c)-f(b) = \min (f(c)-f(b),0) $$

so the claim holds.    $\square$

Proposition 10

For any convex function f, if $a<b<c$ then

$$ f(b) + \min (f(c) - f(a), 0) \le f(c) + \min (f(b) - f(a), 0). $$

Proof

If $f(b) \ge f(a)$ then by Proposition 7(a) also $f(c) \ge f(b)$. Hence also $f(c)\ge f(a)$ and

$$ f(b) + \min (f(c) - f(a), 0) = f(b) \le f(c) = f(c) + \min (f(b)-f(a), 0) $$

so the property holds. Otherwise, $f(b)<f(a)$ and the property becomes

$$ f(b)+\min (f(c) - f(a), 0) \le f(c) + f(b) - f(a) $$

which is always true due to $\min (f(c)-f(a),0)\le f(c)-f(a)$.    $\square$

Proposition 11

For any convex function f, if $a<b<c<d$ then

$$ \min (f(c) - f(a), 0) + \min (f(d) - f(b), 0) \le \min (f(d) - f(a), 0) + \min (f(c) - f(b), 0). $$

Proof

If $f(d) \ge f(a)$, then

$$ \min (f(d) - f(b), 0) \le 0 = \min (f(d) - f(a), 0). $$

Combined with Proposition 9 applied to $a<b<c$ we obtain the claim. Otherwise, $f(d) < f(a)$ and by Proposition 7(b) applied to $a<c<d$ also $f(c) < f(a)$, so the property becomes

$$ f(c) + \min (f(d) - f(b), 0) \le f(d) + \min (f(c) - f(b), 0) $$

which holds by Proposition 10 applied to $b< c < d$.    $\square$

Theorem 12

For any point p, $w_p(i,j)$ is concave Monge.

Proof

Consider any $s,t,u,v \in [0, 2n+1]$ such that $s<t<u<v$. We need to prove that for any $p\in P$:

$$ w_p(s,u) + w_p(t,v) \le w_p(s,v) + w_p(t, u). $$

Case 1. $s=0$ and $v=2n+1$

Straightforward, since $w_p(s,v)=\infty$ and all other edges have finite weights.

Case 2. $s > 0$ and $v = 2n+1$

$$\begin{aligned} w_p(s,u) + w_p(t,v)&= w_p(s,u) + 0 \\&= \min ( D (p, u) - D (p, s), 0) \\&\mathop {\le }\limits ^{9} \min ( D (p, u) - D (p, t), 0) \\&= 0 + w_p(t,u) \\&= w_p(s,v) + w_p(t,u) \end{aligned}$$

Case 3. $s = 0$ and $v < 2n+1$

$$\begin{aligned} w_p(s,u) + w_p(t,v)&= D (p, u) + \min ( D (p, v) - D (p, t), 0) \\&\mathop {\le }\limits ^{10} D (p, v) + \min ( D (p, u) - D (p, t), 0) \\&= w_p(s,v) + w_p(t,u) \end{aligned}$$

Case 4. $s> 0$ and $v < 2n+1$

$$\begin{aligned} w_p(s,u) + w_p(t,v)&= \min ( D (p, u) - D (p, s), 0) + \min ( D (p, v) - D (p, t), 0) \\&\mathop {\le }\limits ^{11} \min ( D (p, v) - D (p, s), 0) + \min ( D (p, u) - D (p, t), 0) \\&= w_p(s,v) + w_p(t,u) \end{aligned}$$

So in all cases $w_p(s,u)+w_p(t,v)\le w_p(s,v)+w_p(t,u)$ and hence $w_p(i,j)$ is concave Monge.    $\square$

In order to apply the known algorithms for finding minimum weight k-link path in the k-median problem, we need to answer queries for w(i, j).

Lemma 13

After $\mathcal {O}(n\log n)$ time and space preprocessing, we can answer queries for w(i, j) in $\mathcal {O}(\log n)$ time per query.

Proof

All edges from the source can be computed in $\mathcal {O}(n \log n)$ time via Theorem 6. All edges to sink have zero weight. It remains to show how to calculate the weight of an internal edge $\left\langle i, j \right\rangle$. Consider the set of points $p\in P$ that are closer to $x_j$ than to $x_i$:

$$ V(i,j) = \left\{ (x,y) \in P : |x - x_i| + |y - a\cdot x_i| > |x - x_j| + |y - a\cdot x_j| \right\} $$

By definition, $w(i,j) = D(V(i,j),x_j)-D(V(i,j),x_i)$. We describe how to compute $D(V(i,j),x_i)$. $D(V(i,j), x_j)$ can be computed using the formula:

$$\begin{aligned} D(V(i,j), x_j) = D(P, x_j) - D(P \setminus V(i,j), x_j) \end{aligned}$$

where $D(P, x_j)$ is the already preprocessed weight of the edge $\left\langle \textsf {source}, j \right\rangle$, and $D(P \setminus V(i,j), x_j)$ can be calculated by rotating the plane by $180^{\circ }$ and using the same method as the one described below.

First we argue that if $(x,y)\in V(i,j)$ then $x > x_i$. Otherwise

$$ |y-a\cdot x_i| - |y-a\cdot x_j| > x_j - x_i \ge a\cdot x_j - a\cdot x_i \ge 0 $$

and we obtain a contradiction in each of the three cases:

• 1. $y < a\cdot x_i$ then the inequality becomes $a\cdot x_i - a\cdot x_j > 0$ but $x_i < x_j$.

• 2. $y \in [a\cdot x_i,a\cdot x_j)$ then the inequality becomes $2y > 2a\cdot x_j$ but $y < a\cdot x_j$.

• 3. $y > a\cdot x_j$ then the inequality becomes $a\cdot x_j - a\cdot x_i > a\cdot x_j - a\cdot x_i$.

We partition V(i, j) into $V_1(i,j)$ and $V_2(i,j)$ with a horizontal line $y=a\cdot x_i$:

$$\begin{aligned} V_1(i,j)&= V(i,j) \cap \{(x,y): y \ge a\cdot x_i \} \\ V_2(i,j)&= V(i,j) \cap \{(x,y): y < a\cdot x_i \}. \end{aligned}$$

The median candidate $(x_i,a\cdot x_i)$ is on the left and bottom of all points in $V_1(i,j)$ and on the left and top of all points in $V_2(i,j)$. Consider the minimum area rectangle enclosing P with sides parallel to the coordinate axes, and enumerate its corners clockwise starting from the top left as $c_1,c_2,c_3,c_4$. In $L_1$ metric, one of the shortest routes from any point in $V_1(i,j)$ to the bottom left corner point $c_4$ goes via $x_i$, see Fig. 3. Therefore our desired sum of distances to $x_i$ can be described in respect to $c_4$ as:

$$\begin{aligned} D (V_1(i,j), x_i)&= \sum _{p \in V_1(i,j)} w(p) \cdot d(p, (x_i, a\cdot x_i)) \\&= \left( \sum _{p \in V_1(i,j)} w(p) \cdot d(p, c_4)\right) - \left( d(c_4, (x_i, a\cdot x_i)) \cdot \sum _{p \in V_1(i,j)} w(p)\right) . \end{aligned}$$

Similarly, one of the shortest routes from any point in $V_2(i,j)$ to $c_1$ goes via $x_i$:

$$\begin{aligned} D (V_2(i,j), x_i) =\left( \sum _{p \in V_2(i,j)} w(p) \cdot d(p, c_1)\right) - \left( d(c_1, (x_i, a\cdot x_i)) \cdot \sum _{p \in V_2(i,j)} w(p)\right) . \end{aligned}$$

Fig. 3.

Shortest route in $L_1$ from $p_2$ to $c_1$ and from $p_1$ to $c_4$ passing through the median candidate $x_i$.

Fig. 4.

$V_1$ represented as the gray $V_3$ minus the striped $V_4$.

The distances $d(c_1,(x_i,a\cdot x_i))$ and $d(c_4,(x_i,a\cdot x_i))$ can be computed in $\mathcal {O}(1)$ time. The expressions $\sum _{p \in V_2(i,j)} w(p) \cdot d(p, c_1)$ and $\sum _{p \in V_2(i,j)} w(p)$ can be evaluated in $\mathcal {O}(\log n)$ with orthogonal queries. To calculate $\sum _{p \in V_1(i,j)} w(p) \cdot d(p, c_4)$ and $\sum _{p \in V_1(i,j)} w(p)$, we represent $V_1(i,j)$ as $V_3(i,j)\setminus V_4(i,j)$, see Fig. 4 where $\delta _x = x_j - x_i$, $\delta _y = a(x_j - x_i)$ and

$$\begin{aligned} V_3(i,j)&= \left\{ (x,y) \in P : y> ax_i \wedge \left( x+ y> \frac{(\delta _x + \delta _y)}{2} \right) \right\} \\ V_4(i,j)&= \left\{ (x,y) \in P : x \le x_i + \delta _x - \delta _y \wedge \left( x+ y > \frac{(\delta _x + \delta _y)}{2} \right) \right\} . \end{aligned}$$

Now each of $V_2(i,j)$, $V_3(i,j)$ and $V_4(i,j)$ is defined by an intersection of two half-planes. By transforming every point $p\in P$ into $(x(p)+y(p),y(p))$ for $V_3(i,j)$ and into $(x(p)+y(p),x(p))$ for $V_4(i,j)$, we can assume that the lines defining the half-planes are parallel to the coordinate axes. Hence each sum can be calculated with orthogonal queries in $\mathcal {O}(\log n)$ time and $\mathcal {O}(n\log n)$ time and space preprocessing by Lemma 3.    $\square$

We reduced the line-constrained k-median problem in $L_1$ to the minimum k-link path problem. The weight of any edge can be retrieved in $\mathcal {O}(\log n)$ time by decomposing it into a constant number of orthogonal queries. By plugging in an appropriate algorithm for the minimum k-link path problem, we obtain the final theorem.

Theorem 14

We can solve the line-constrained k-median problem in $L_1$ and $L_\infty$ using $\mathcal {O}(kn\log n)$ time or, if $k=\varOmega (\log n)$, $n2^{\mathcal {O}(\sqrt{\log k \log \log n})}\log n$ time.