An approximation algorithm for the Euclidean incremental median problem

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1 Discrete Optimization 22 (2016) Contents lists available at ScienceDirect Discrete Optimization An approximation algorithm for the Euclidean incremental median problem Vladimir Shenmaier Sobolev Institute of Mathematics, Novosibirsk, Russia a r t i c l e i n f o a b s t r a c t Article history: Received 16 July 2015 Received in revised form 25 March 2016 Accepted 25 August 2016 Available online 17 September 2016 Keywords: Incremental medians Euclidean medians Hierarchical clustering Approximation algorithm Online algorithm In the incremental version of the k-median problem, we find a sequence of facility sets F 1 F 2 F n, where each F k contains at most k facilities. This sequence is said to be δ-competitive if the cost of each F k is at most δ times the optimum cost of k facilities. The best deterministic (randomized) algorithm available for the metric space has a competitive ratio of 8 (7.656). The best one for the one-dimensional problem finds a competitive sequence. We give a competitive solution for the high-dimensional Euclidean space Elsevier B.V. All rights reserved. 1. Introduction In this paper, we study the incremental version of the well-known k-median problem. The feature of this version is that the total number k of facilities to be placed is unknown. Instead, we must place facilities one at a time so that at each step the facilities already placed constitute an optimal or approximate solution for the median problem of the corresponding cardinality. An instance of the problem is specified by a finite set of customers C, a finite set of facilities F, the distance dist(u, f) 0 defined for each customer u and each facility f, and the weight w(u) 0 defined for each customer u. The cost of an arbitrary set of facilities X is the weighted sum cost(x) = u C w(u) dist(u, X), where the distance dist(u, X) from the customer u to the set X is defined as the distance from u to the nearest facility in X. This research is supported by RFBR (projects , and ). address: shenmaier@mail.ru / 2016 Elsevier B.V. All rights reserved.

2 V. Shenmaier / Discrete Optimization 22 (2016) In the classical k-median problem, we are given a positive integer k, the objective is to find a k-element facility set of the minimum cost. In the incremental version of the problem, we find an incremental sequence of facility sets F 1 F 2 F n, where each F k contains at most k facilities, n is the cardinality of F. This sequence is called δ-competitive if each F k is a δ-approximate k-median: cost(f k ) δ cost(x) for any k-element facility set X. We say that an algorithm for the incremental problem is δ-competitive, or has a competitive ratio of δ, if it produces a δ-competitive solution F 1, F 2,..., F n for any instance of the problem. Related work The first constant-factor solution for the metric incremental median problem was proposed by Mettu and Plaxton [1]. They obtained a linear-time algorithm with competitive ratio Chrobak et al. [2] (and, independently, Lin et al. [3]) proved that each instance has an 8-competitive incremental medians sequence. This produces an algorithm with competitive ratio 8c, where c is the approximation ratio of available k-median problem solutions. Chrobak and Hurand [4] proved, via a probabilistic argument, the existence of an improved solution with competitive ratio at most which produces a randomized algorithm with competitive ratio 7.656c. They also showed that in the general metric case, no competitive ratio better than 2.01 is possible. Note that, for c = 1, the above algorithms become non-polynomial since they use optimal k-medians. Lin et al. [3] gave a polynomial-time algorithm with competitive ratio 16. For the one-dimensional case, a competitive solution was proposed in the work [5]. Our result We present a 7.076c-competitive algorithm for the Euclidean problem. In this case, both customers and facilities lie in the space R d and the distance function is induced by Euclidean metric: dist(u, f) = u f 2 = d (u(i) f(i)) 2. If the dimension of space is fixed, we can use PTAS for Euclidean k-medians [6]. In this case, since c = 1 + ε, our algorithm gives a ( ε)-competitive solution for the incremental median problem in polynomial time. 2. Algorithm description The basic idea of the algorithm is similar to that of the algorithms from [2,5]. We firstly find c-approximate solutions F 1, F 2,..., F n of the usual k-median problem for k = 1, 2,..., n. Next, the algorithm determines the inextensible sequence of indices i 1, i 2,..., i t in which i 1 = 1 and, for k = 2,..., t, the index i k is minimum such that i=1 cost(f i k ) < cost(f i k 1 )/α, where α is a parameter of the algorithm, α = Note that, when the previous algorithms are used, the analogous constant is 2 for the 8-competitive algorithm of [2] and for the one-dimensional algorithm of [5]. Then, we construct a partial solution for the incremental median problem corresponding to the indices i k. To wit, we find an increasing sequence of sets F i1 F i2 F it constructed recursively in the

3 314 V. Shenmaier / Discrete Optimization 22 (2016) Fig. 1. Determining the set F ik. reverse order: we determine F it firstly, then we determine F it 1, and so on. We put F it equal to Fi t and, for k = t 1,..., 1, we define F ik as follows. For each facility f Fi k, we consider the set C(f) of customers for which f is the nearest facility in Fi k (if the nearest is not unique, break the tie arbitrarily). We pick the facility f F ik+1 best from the viewpoint of minimizing the weighted sum of the distances to the customers of C(f). The facilities chosen in this way for all f Fi k constitute the set F ik (see Fig. 1). Henceforth, we denote this construction of F ik by Γ : F ik = Γ (F i k, F ik+1 ) = {f f F i k }. Note that in the algorithm of [2], during the construction of F ik, the replacement for a facility f Fi k is determined as the facility in F ik+1 nearest to f. The one-dimensional algorithm of [5] determines the facilities in F ik+1 nearest to f both from the left and the right and then picks the best of them. Clearly, the number of elements in F ik is at most i k and F ik F ik+1. Therefore, a partial solution for the incremental median problem is constructed. This solution is extended arbitrarily to a complete one by adding to F ik for k = 1, 2,..., t either points of F ik+1 \ F ik if k < t or points of F \ F it if k = t. The running time of the algorithm is determined by that of finding the initial sequence F1, F2,..., Fn of c-approximate k-medians in Euclidean space. Particularly, if we use PTAS from [6], the value of c becomes 1 + ε and the running time of the algorithm becomes O(2 O((log(1/ε)/ε)d 1) n 2 log d+6 n). 3. Competitive ratio analysis Theorem 1. The algorithm described above has a competitive ratio of 4ζβc 7.076c, where ζ = 3 and β = α/2 = Proof. Let us introduce the following concept: a configuration is a triple consisting of an arbitrary weighted set of customers C and two facility sets F and F. The justification of the upper bound for the competitive ratio of solutions obtained by the algorithm rests on the following Lemma 1. Given an arbitrary configuration C, F, F, we have cost(γ (F, F )) ζ cost(f ) + β cost(f ). (1) We prove this lemma in the next section.

4 V. Shenmaier / Discrete Optimization 22 (2016) By Lemma 1, the cost of F ik for k = 1, 2,..., t is at most the sum of the series ζ cost(f i k ) + βζ cost(f i k+1 ) + β 2 ζ cost(f i k+2 ) + + β t ζ cost(f i t ). Taking the choice of the indices i 1, i 2,..., i t into account, we have cost(f ik ) ζ cost(fi k )(1 + β/α + (β/α) (β/α) t ) cost(fi ζα k ) α β. Therefore, we obtain an upper bound for the competitive ratio of the partial solution F i1, F i2,..., F it. Let us estimate the cost of an arbitrary set F s for s = 1, 2,..., n and s {i 1, i 2,..., i t }. Take the maximum index i k such that i k < s. Then cost(f s ) cost(f i k )/α by the construction of the sequence i 1, i 2,..., i t. Since the objective function does not increase as the number of facilities placed increases, this implies cost(f s ) cost(f ik ) cost(fi ζα k ) α β cost(f s ) ζα2 α β. Recall that α = 2β (this is the optimal value of the parameter α) and that the solutions F1, F2,..., Fn are c-approximate. It follows that the competitive ratio of the algorithm is at most 4ζβc < 7.076c. The proof of Theorem 1 is complete. 4. Proof of Lemma 1 To prove Lemma 1 we extend the technique of simplifications used in [5]. Suppose that, for some configuration C, F, F, we have the converse: cost(γ (F, F )) > ζ cost(f ) + β cost(f ). (2) We will arrive at a contradiction in two stages: the construction of a substantially simpler configuration in which (2) or a similar inequality holds (we refer to such configurations as bad); the derivation of a contradiction using the simplicity of this configuration. The first stage consists in nine simplifications of a configuration described below. Simplification 1 By the inequality (2), there is at least one facility f F such that the contribution of the customers of C(f ) to the left side of (2) is greater than to the right one. Consequently, if we put F = {f } and C = C(f ), the inequality still holds. Thus, we have a bad configuration with a one-point set F. Definition 1. The cost of an arbitrary point x R d is the value cost(x) = cost({x}) = u C w(u) u x. The geometric median of C is the point x R d where the function cost(x) reaches the minimum. Move the facility f to the geometric median of C. The right side of (2) does not increase, the left one remains unchanged, so (2) still holds. Without loss of generality, we will assume that f lies at the origin: f = 0.

5 316 V. Shenmaier / Discrete Optimization 22 (2016) Simplification 2 Definition 2. Given a constant l 0, the isosurface of level l for the function cost(x) is the set of points in R d where this function takes the value l. Since the function cost(x) is convex (as a sum of convex ones), its isosurfaces represent boundaries of convex bodies in the space R d. Let G be the isosurface of level l = cost(γ (F, F )) and [G] be the convex body bounded by this surface. By (2), we have l > cost(f ), so the facility f lies strictly inside [G]. On the other hand, since the operator Γ replaces f with the best facility from F, all the elements of F lie either outside the body [G] or on the surface G. Further, points that lie strictly inside (outside) [G] will be called internal (external). Perform the following three actions: (a) add all the points of G to the facility set F ; (b) move each external customer to the surface G along the beam connecting its initial position with f = 0; (c) remove all the external facilities from the set F. Proposition 1. The obtained configuration still satisfies (2). Proof. During the action (a), none of the new facilities gets inside [G]. But [G] is bounded by the isosurface of level cost(γ (F, F )), so the left side of (2) is unchanged. On the other hand, the right one does not increase. It follows that, as before, (2) holds. During the action (b), by the triangle inequality, the decrease of the distance from any external customer u to any facility is at most the decrease of the distance from u to f. So the left side of (2) decreases at most as the first term of the right one does. The second term, β cost(f ), does not increase since the distance from F to u becomes zero. Thus, the left side of (2) decreases at most as the right one does, therefore, (2) still holds. After the actions (a) and (b) all the external facilities do not serve any customers, so the action (c) will change neither right nor left side of (2). Proposition 1 is proved. As a result, we have a bad configuration in which F coincides with G and all the customers belong to [G]. The surface G is not necessarily an isosurface for the updated function cost(x), but the convexity of [G] will be enough. Simplification 3 Split each internal customer u into two ones u 1 and u 2, where u 1 is at the intersection point of the beam 0u with the surface G and u 2 is at the origin: u 2 = 0. The weight of u 1 is defined as w 1 = wr/r 1, where w = w(u), r = u, and r 1 = u 1. The weight of u 2 is defined as w 2 = w w 1 = w(1 r/r 1 ). Proposition 2. The obtained configuration still satisfies (2). Proof. Note that w = w 1 + w 2 and the barycenter of u 1 and u 2 coincides with u: (w 1 u 1 + w 2 u 2 )/(w 1 + w 2 ) = w 1 u 1 /w = (w u / u 1 )u 1 /w = u. Estimate the changes of both sides of (2). The left one, the cost of the best facility in F, does not decrease since, by the norm axioms, the distance from any facility f to the barycenter of two customers is at most the weighted mean of the distances from f to them: w u f w 1 u 1 f + w 2 u 2 f. The first term of the right side, ζ cost(f ), is unchanged

6 V. Shenmaier / Discrete Optimization 22 (2016) Fig. 2. Splitting an internal customer u. since the weighted sum of the distances from the point f = 0 to the new customers is equal to the weighted distance from f to the original one: w 1 u 1 + w 2 u 2 = w 1 r 1 = wr. It remains to show that the second term of the right side of (2), β cost(f ), does not increase. Denote by B(O, r) and S(O, r) the ball and the sphere of radius r centered at point O R d : B(O, r) = {v R d v O r}, S(O, r) = {v R d v O = r}. Let x 2 = dist(0, G) and H be the convex hull of the set B(0, x 2 ) {u 1 }. Denote by x the distance from u to the boundary of H (see Fig. 2). Then x 2 = u 2 p 2, where p 2 is some orthogonal projection of the point u 2 onto the conical part of H, and x = u p, where p is the orthogonal projection of the point u onto the segment u 1 p 2. The contribution of the new customers u 1, u 2 to cost(f ) is w 1 dist(u 1, G) + w 2 dist(u 2, G) = w 2 x 2 since u 1 lies on G. The contribution of the original one is w dist(u, G). Since the body [G] is convex and it contains the ball B(0, x 2 ) and the point u 1, it contains the whole set H as well. Therefore, dist(u, G) x. On the other hand, since the triangles u 1 u 2 p 2 and u 1 up are similar, x/x 2 = (r 1 r)/r 1 = 1 r/r 1. Then w dist(u, G) wx = w(1 r/r 1 )x 2 = w 2 x 2, so the contribution of the new customers to cost(f ) is at most that of the original one. Consequently, the second term of the right side of (2) does not increase. Proposition 2 is proved. Finally, merge customers which get to the same point. As a result, we obtain a bad configuration in which all the customers belong to G {0}. Simplification 4 Let a be the point of G closest to the origin. Scale the coordinates of all points so that a = 1. Set the coordinate axes so that the first one passes through the points a and 0 in the direction opposite to a. Then a = ( 1, 0,..., 0). Consider the ball B defined by the inequality ζ x x a. Its radius is equal to ζ/(ζ 2 1) = 3/2, the center is at the point (1/(ζ 2 1), 0,..., 0) = (0.5, 0,..., 0) (see Fig. 3). Note that the set B G is not empty. Indeed, otherwise, for any customer u G, we have ζ u > u a. Therefore, cost(a) < ζ cost(0) + w(0) = ζ cost(f ) + cost(f ) which contradicts (2). Denote by b the point of the set B G closest to the origin. Since a, b F, we have min{cost(a), cost(b)} > ζ cost(0) + β cost(f ). (3) Our future goal is to get a contradiction with this inequality. So we will further mean that a configuration is bad if it satisfies (3).

7 318 V. Shenmaier / Discrete Optimization 22 (2016) Fig. 3. The ball B and the body T. Consider the body T = B(0, 1) (B B(0, r)), where r = b (see Fig. 3). By the choice of the points a and b, the interior of this body contains neither facilities of F nor any customers, except for 0. Move both facilities and customers from the surface G to the boundary of T along the beams connecting them with the origin. By the triangle inequality, the decrease of the distance from any customer u to any facility is at most the decrease of the distance from u to f. So the left side of (3) decreases at most as the first term of the right one does. The second term, β cost(f ), is unchanged since, as before, all the customers, except for 0, are at points with facilities and, at the same time, dist(0, F ) remains the same. Thus, (3) still holds. Finally, remove each customer u located on the boundary of the ball B. The contribution of u to cost(0) is w(u) u. At the same time, its contribution to cost(a) is w(u) u a = w(u)ζ u and that to cost(b) is w(u) u b w(u) 3 w(u)ζ u. Therefore, the decrease of the left side of (3) is at most that of the right one. Thus, we obtain a bad configuration in which all the customers, except for 0, lie on the two spherical surfaces S(0, 1) \ B and S(0, r) B, where 1 r 1/(ζ 1) Simplification 5 If the point b does not lie on the first coordinate axis, set the other axes so that the second one gets into the plane a0b and the point b gets into its lower half-plane. Then, turn each customer around the first coordinate axis so that it gets into the plane of the first two axes, into its upper closed half-plane (see Fig. 4). In other words, if a customer u lies at the point with coordinates (u(1), u(2),..., u(d)), replace it with one at the point (u(1), u 2 u(1) 2, 0,..., 0). Proposition 3. The obtained configuration still satisfies (3). Proof. The distances from u to the facilities 0 and a are unchanged. On the other hand, that to the facility b does not decrease since u gets to the most distant point from b among those that have the same norm and the value of the first coordinate as u. Consequently, the left side of (3) does not decrease, the right one remains unchanged. Thus, we obtain a two-dimensional bad configuration.

8 V. Shenmaier / Discrete Optimization 22 (2016) Fig. 4. Reduction to a two-dimensional case. Simplification 6 Definition 3. Suppose that customers u 1 and u 2 lie on a circle centered at a point O R 2. A circular barycenter of u 1 and u 2 is a point u on this circle such that u 1 Ou = w(u 2 ) w(u 1 ) + w(u 2 ) u 1Ou 2, where the angles u 1 Ou and u 1 Ou 2 are measured in one direction (either clockwise or counterclockwise) and u 1 Ou 2 [0, 2π]. The circular barycenter u is said to be natural if in the chosen direction u 1 Ou 2 [0, π]. This is said to be exterior to a point f R 2 if u and f lie in the different sectors of the plane between the beams Ou 1, Ou 2 or on these beams. Note that in the current configuration, all the customers, except for 0, lie on the arcs S(0, 1) \ B and S(0, r) B, in the upper closed half-plane. At the same time, both facilities a and b lie in the lower closed one. Therefore, the natural circular barycenter of each pair of customers from the same arc is exterior both to a and b. Lemma 2. Let a point f and customers u 1, u 2 lie on the circle S(0, 1) and u be the circular barycenter of u 1, u 2 exterior to f. Then the distance from f to u is at least the weighted mean of the distances from f to u 1 and u 2 : (w(u 1 ) + w(u 2 )) f u w(u 1 ) f u 1 + w(u 2 ) f u 2. Proof. Let ϕ = f0u, ϕ 1 = f0u 1, and ϕ 2 = f0u 2. Then, by the conditions, ϕ = (w(u 1 )ϕ 1 + w(u 2 )ϕ 2 )/(w(u 1 ) + w(u 2 )). Using the cosine theorem, we can express the distance from f to u through the angle ϕ: f u = 2 1 cos ϕ. The derivative of this expression over ϕ is sin ϕ/( 2 1 cos ϕ). Consequently, the second derivative f u is cos ϕ sin 2 ϕ 2(1 cos ϕ) 1/2 2 2(1 cos ϕ). 3/2 Show that f u 0 when cos ϕ < 1. It needs to prove that cos ϕ sin 2 ϕ/(2(1 cos ϕ)). But this is equivalent to 2 cos ϕ 1 + cos ϕ which holds for any ϕ. Thus, the function f u = f u(ϕ) is concave over the variable ϕ on the interval (0, 2π), therefore, f u(ϕ) (w(u 1 ) f u(ϕ 1 ) + w(u 2 ) f u(ϕ 2 ) )/(w(u 1 ) + w(u 2 )). Lemma 2 is proved.

9 320 V. Shenmaier / Discrete Optimization 22 (2016) Lemma 3. Let a point f lie on the horizontal segment [0.5, 2], customers u 1, u 2 lie on the circle S(0, 1) so that f0u 1, f0u 2 [π/3, 2π π/3], and u be the circular barycenter of u 1, u 2 exterior to f. Then the distance from f to u is at least the weighted mean of the distances from f to u 1 and u 2. Proof. By the cosine theorem, f u = 1 + x 2 2x cos ϕ = c 1 c2 cos ϕ, where ϕ = f0u, x = f, c 1 = 2x, and c 2 = (1 + x 2 )/(2x). The derivative of this expression over ϕ is c 1 sin ϕ/(2 c 2 cos ϕ). Consequently, the second derivative f u is c 1 cos ϕ 2 (c 2 cos ϕ) sin 2 ϕ. 1/2 2(c 2 cos ϕ) 3/2 Note that cos ϕ 0.5 by the choice of the points u 1, u 2 and c 2 1, so the radicand in the denominators of both terms is positive. Show that f u 0. It needs to prove that 2 cos ϕ(c 2 cos ϕ) sin 2 ϕ which is equivalent to 2c 2 cos ϕ cos 2 ϕ 1 0. (4) If cos ϕ = 0, the inequality is obvious. For cos ϕ > 0, this holds if and only if c 2 (cos 2 ϕ + 1)/(2 cos ϕ). Since c 2 = (x 2 + 1)/(2x), this is true for cos ϕ x 1/ cos ϕ. But cos ϕ 0.5, so the inequality (4) holds for any x [0.5, 2]. For cos ϕ < 0, the inequality holds if and only if c 2 (cos 2 ϕ + 1)/(2 cos ϕ). Since c 2 = (x 2 + 1)/(2x), this is true if either x 1/ cos ϕ or x cos ϕ. But x 0.5 and cos ϕ 0.5, so the second holds. Thus, in all of these cases, (4) is true, so the second derivative of the function f u is non-positive for cos ϕ 0.5. Therefore, this function is concave over the variable ϕ on the interval (π/3, 2π π/3) which implies the required inequality. Lemma 3 is proved. Based on the proved lemmas, let us replace each pair of customers u 1, u 2 that lie on the same arc, S(0, 1) \ B or S(0, r) B, with the new customer of the weight w(u 1 ) + w(u 2 ) that lies at the natural circular barycenter of u 1 and u 2. Perform this operation until all customers located on the same arc collapse to a single point. Proposition 4. The obtained configuration still satisfies (3). Proof. Since a circular barycenter lies on the same circle as original customers, the right side of (3) remains unchanged. Let us estimate the change of the left one. Suppose that f is one of the facilities a and b. Case 1 : f lies on the same circle as the customers u 1 and u 2. In this case, the weighted distance from f to the new customer is at least the weighted sum of the distances to the original ones by Lemma 2. This means that cost(f) does not decrease. Case 2 : f lies on one circle, the customers u 1, u 2 lie on the other, say S(0, R), where R {1, r}. By properties of the ball B (see Fig. 3) and the mutual arrangement of customers and facilities in the current configuration, the angles f0u 1 and f0u 2 belong to [π/3, π + π/3] [π/3, 2π π/3]. On the other hand, f belongs to [R/r, Rr] [0.5R, 2R]. Then the weighted distance from f to the new customer is at least the weighted sum of the distances to the original ones by Lemma 3. Therefore, cost(f) does not decrease. Thus, the left side of (3) does not decrease. Proposition 4 is proved. So we obtain a bad configuration with only three customers: 0, u, and v, where u lies on the arc S(0, 1)\B and v lies on the arc S(0, r) B.

10 V. Shenmaier / Discrete Optimization 22 (2016) Fig. 5. Moving the facility b and the customer u. Simplification 7 Let us modify the current configuration so that the facility b gets to the lowest point P of the arc S(0, r) B. For this purpose, turn the customer u and the facility b around the origin by the same angle min{ b0p, u0q} clockwise, where Q is the upper intersection point of the circle S(0, 1) with the ball B (see Fig. 5). Since the norms of the points b, u and the angle between them remain the same, b u is unchanged. On the other hand, both a u and b v do not decrease. Finally, a v is unchanged since a and v stay in place. Thus, the left side of (3) does not decrease. At the same time, the right one is unchanged. Then, (3) still holds and either facility b gets to the point P or the customer u gets to the point Q. In the second case, move the facility b to the point P. By properties of the ball B (see Fig. 3), Q is its upper point. Therefore, u b does not decrease since P is the point of S(0, r) B most distant from Q. Thus, in both cases, the facility b gets to the point P without decreasing the costs of a and b. Simplification 8 Denote by P b the highest point of the arc S(0, r) B and let P a be the rightmost point of the arc S(0, 1) A, where A is the ball defined by the inequality ζ x x b (see Fig. 6). The radius of this ball is r times the radius of B, the center is at the point b/(ζ 2 1). Lemma 4. The point P a lies not to the right of the point Q on the circle S(0, 1). Proof. The angle a0p a is equal to the sum of the angles Θ r and Ω r, where Θ r is the angle between the vectors b and P a, while Ω r is the angle between the vectors a and P b (see Fig. 6). Determine the angle Θ r. Let x = cos Θ r and y = sin Θ r. Then, since the point P a lies at the intersection of the circle S(0, 1) with the boundary of the ball A, we have b P a 2 = (r + x) 2 + y 2 = ζ 2. Since x 2 + y 2 = 1, this is equivalent to r 2 + 2xr + 1 = ζ 2. Consequently, x = (ζ 2 1 r 2 )/(2r), so Θ r = arccos((2 r 2 )/(2r)). Determine the angle Ω r. Let x = cos Ω r and y = sin Ω r. Then, since the point P b lies at the intersection of the circle S(0, r) with the boundary of the ball B, we have a P b 2 = (1 + rx) 2 + (ry) 2 = ζ 2 r 2. Since x 2 + y 2 = 1, this is equivalent to 1 + 2rx + r 2 = ζ 2 r 2. Consequently, x = ((ζ 2 1)r 2 1)/(2r), so Ω r = arccos((2r 2 1)/(2r)). Since a0p a = Ω r + Θ r and a0q = 2π/3, to prove the lemma it is sufficient to show that cos(ω r + Θ r ) 0.5. But cos(ω r + Θ r ) = cos Ω r cos Θ r sin Ω r sin Θ r = gh 1 g 2 1 h 2, where

11 322 V. Shenmaier / Discrete Optimization 22 (2016) Fig. 6. The points P a, P b and the angles Θ r, Ω r. g = cos Ω r = (2r 2 1)/(2r) and h = cos Θ r = (2 r 2 )/(2r). So the required inequality can be written as (gh + 0.5) 2 (1 g 2 )(1 h 2 ) which is equivalent to gh + g 2 + h 2 3/4 0. If we substitute the expressions for g and h, this inequality becomes the following: (2r 2 1)(2 r 2 ) + (2r 2 1) 2 + (2 r 2 ) 2 3r 2 = 3r 4 6r But this holds for any values of r. Lemma 4 is proved. Lemma 5. Let the customer u be to the left of the point P a on the circle S(0, 1) and the customer v be to the right of the point P b on the circle S(0, r). Then the derivative of the function b v over the angle ϕ(v) = b0v is at least the derivative of the function b u over the angle ϕ(u) = b0u: b v ϕ(v) b u ϕ(u). Proof. Using the cosine theorem, we can write explicitly the expressions for b v and b u : b v = r 2 1 cos ϕ(v), b u = 1 + r 2 2r cos ϕ(u). Note that ϕ(v) ϕ(p b ) = 2Ω r, where ϕ(p b ) = b0p b (see Fig. 6). On the other hand, ϕ(u) ϕ(p a ) = π Θ r, where ϕ(p a ) = b0p a. By properties of chords of circles, the angle between the chord vb and the upper half of the tangent to the circle S(0, r) at the point v decreases with increasing the length of this chord and, consequently, with increasing the angle ϕ(v) (see Fig. 7). Therefore, the minimum of the derivative of the function b v for ϕ(v) ϕ(p b ) is reached at the point ϕ(p b ). Let b be the lower intersection point of the line ub with the circle S(0, 1) (see Fig. 7). Note that the angle between the chord ub and the left half of the tangent to the circle S(0, 1) at the point u decreases with increasing the length of this chord if ϕ(u) π and with decreasing its length if ϕ(u) > π. Therefore, in both cases, this angle decreases with increasing the angle ϕ(u). It follows that the maximum of the derivative of the function b u for ϕ(u) ϕ(p a ) is reached at the point ϕ(p a ). Thus, it is sufficient to prove the required inequality for ϕ(u) = ϕ(p a ), ϕ(v) = ϕ(p b ). The derivative b v of the function b v at the point ϕ(v) = ϕ(p b ) is equal to r sin ϕ(p b )/( 2 1 cos ϕ(p b )). The derivative b u of the function b u at the point ϕ(u) = ϕ(p a )

12 V. Shenmaier / Discrete Optimization 22 (2016) Fig. 7. A visualization of b u and b v. is equal to r sin ϕ(p a )/ 1 + r 2 2r cos ϕ(p a ). Therefore, b v b u if and only if sin ϕ(p b ) 1 + r 2 2r cos ϕ(p a ) sin ϕ(p a ) 2 1 cos ϕ(p b ). Since sin ϕ(p b ) and sin ϕ(p a ) are non-negative, this is equivalent to (1 + cos ϕ(p b ))(1 + r 2 2r cos ϕ(p a )) 2 sin 2 ϕ(p a ). But cos ϕ(p b ) = cos(2ω r ) = 2 cos 2 Ω r 1 = (2r 2 1) 2 /(2r 2 ) 1 and cos ϕ(p a ) = cos(π Θ r ) = cos Θ r = (r 2 2)/(2r). Therefore, this inequality becomes the following: (2r 2 1) 2 2r r 2 2r(r2 2) 2 1 (r2 2) 2 2r 4r 2 = 8r2 r 4 4 2r 2. After collecting like terms this is simplified to 3(2r 2 1) 2 8r 2 r 4 4 which can be written as 13t 2 20t + 7 0, where t = r 2. The roots of this quadratic form are 7/13 and 1, so the inequality holds for any r 1. Lemma 5 is proved. Lemma 6. Let the customer u be to the left of the point P a on the circle S(0, 1) and the customer v be to the right of the point P b on the circle S(0, r). Then the derivative of the function a u over the angle ψ(u) = a0u is at least the derivative of the function a v over the angle ψ(v) = a0v: a u ψ(u) a v ψ(v). Proof. Using the cosine theorem, we can write explicitly the expressions for a u and a v : a u = 2 1 cos ψ(u), a v = 1 + r 2 2r cos ψ(v). Note that ψ(u) ψ(p a ) = Ω r + Θ r, where ψ(p a ) = a0p a (see Fig. 6). On the other hand, ψ(v) ψ(p b ) = π Ω r, where ψ(p b ) = a0p b, and ψ(v) π since all the customers lie in the upper closed half-plane. By properties of chords of circles, the angle between the chord ua and the right half of the tangent to the circle S(0, 1) at the point u decreases with increasing the length of this chord and, consequently, with increasing the angle ψ(u) (see Fig. 8). Therefore, the minimum of the derivative of the function a u for ψ(u) ψ(p a ) is reached at the point ψ(p a ).

13 324 V. Shenmaier / Discrete Optimization 22 (2016) Fig. 8. A visualization of a u and a v. Let a be the left intersection point of the line va with the circle S(0, r) (see Fig. 8). Note that, if ψ(v) increases, the distance from 0 to the chord va decreases, therefore, the length of this chord increases. So the angle between the chord va and the lower half of the tangent to the circle S(0, r) at the point v, which decreases with increasing the length of this chord, decreases with increasing the angle ψ(v). Consequently, the maximum of the derivative of the function a v for ψ(v) ψ(p b ) is reached at the point ψ(p b ). Thus, it is enough to prove the required inequality for ψ(u) = ψ(p a ), ψ(v) = ψ(p b ). The derivative a u of the function a u at the point ψ(u) = ψ(p a ) is equal to sin ψ(p a ) 1 + cos 2 1 cos ψ(pa ) = ψ(pa ) 1 + cos Ωr cos Θ r sin Ω r sin Θ = r 2 2 = (2r2 1)(2 r 2 ) 2 4r 2 1 (2r2 1) 2 4r 2 1 (2 r2 ) 2 4r 2. The derivative a v of the function a v at the point ψ(v) = ψ(p b ) is equal to 1 1 r sin ψ(p b ) + r2 2r cos ψ(p b ) = r sin Ω r + r2 + 2r cos Ω r = r 1 (2r2 1) 2 4r r 2 + 2r(2r2 1) = 1 (2r2 1) r 4r 2 Therefore, a u a v if and only if 1 + (2r2 1)(2 r 2 ) 4r 2 1 (2r2 1) 2 4r 2 1 (2 r2 ) 2 4r (2r2 1) 2 3 4r 2 which is equivalent to 1 (2r2 1) 2 4r 2 1 (2 r2 ) 2 4r (2r2 1) 4r 2 2 r (2r2 1) = (2r2 1)(4 + r 2 ) 3 4r 2 = 2r4 + 11r r 2. In other words, 12r 2 1 (2r2 1) 2 4r 2 1 (2 r2 ) 2 4r 2 = 3 8r 2 4r 4 1 8r 2 r 4 4 2r r 2 4.

14 V. Shenmaier / Discrete Optimization 22 (2016) Note that the right side of this inequality and both radicands in the left one are non-negative for the allowable values of r: 1 r 1/(ζ 1). Get rid of the roots by raising both sides in the square: 9(8r 2 4r 4 1)(8r 2 r 4 4) = 9(4r 8 40r r 4 40r 2 + 4) (2r r 2 4) 2 = 4r r r 4 88r After collecting like terms this inequality becomes the following: 32r 8 404r r 4 272r The polynomial in this inequality is equal to (t 1)(32t 3 372t t 20), where t = r 2. The first of these multipliers is non-negative for t 1, the second one, on the contrary, is negative at least on the segment 1 t 10. Therefore, the inequality holds for any r from the interval [1, 1/(ζ 1)]. Lemma 6 is proved. Proposition 5. If the customer u lies on the circle S(0, 1) so that a u ζ and the customer v lies on the circle S(0, r), then w(v) > 0 and v is to the right of the point P b. Proof. If w(v) = 0, we have cost(a) = w(u) a u + w(0) ζ cost(0) + cost(f ) which contradicts (3). On the other hand, if v is not to the right of P b, we have a v ζ v. It follows that which contradicts (3) again. cost(a) = w(u) a u + w(v) a v + w(0) w(u)ζ + w(v)ζ v + w(0) = ζ cost(0) + cost(f ) Based on the proved statements, let us get rid of the customer u. If the weight of u is zero, we can remove it without changing the costs of any facilities. Consider the case when w(u) > 0. Since u lies on the arc S(0, 1) \ B, we have a u ζ. Therefore, w(v) > 0 and v is to the right of P b by Proposition 5. Suppose that u is to the left of the point P a. Then we can perform the following actions. Split the customer u into two ones u 1 and u 2 that lie at the same point as u and put w(u 1 ) = w, w(u 2 ) = w(u) w, where w = min{w(u), w(v)}. Similarly, split the customer v into two ones v 1 and v 2 that lie at the same point as v and put w(v 1 ) = w, w(v 2 ) = w(v) w. Then, move the customers u 1 and v 1 along the circles S(0, 1) and S(0, r) respectively toward each other by the same small angle ϕ. By Lemmas 5 and 6, the costs of both facilities a and b do not decrease. Next, replace u 1, u 2 with the customer u of the weight w(u 1 ) + w(u 2 ) = w(u) that lies at their natural circular barycenter on the arc S(0, 1) and replace v 1, v 2 with the customer v of the weight w(v 1 ) + w(v 2 ) = w(v) that lies at their natural circular barycenter on the arc S(0, r). Then the costs of a and b still do not decrease by Lemmas 2 and 3. But the new customer u lies to the right of the original one u by the angle ϕ w w(u) and the customer v lies to the left of v by the angle ϕ w w(v). It follows that, since w > 0, we can move the customers u and v along their circles toward each other without decreasing the costs of a and b until either u gets to the point P a or v gets to the point P b. By Lemma 4, the customer u is not to the right of the point Q. Then a u ζ which implies that, by Proposition 5, the customer v cannot get to the point P b. Therefore, u gets to the point P a or its initial position is not to the left of P a. In both cases, we have a configuration where u lies on the closed segment of S(0, 1) between the points P a and Q. By the definition of these points, it follows that b u ζ u = ζ and a u ζ. So we can remove the customer u since the decrease of the left side of (3) will be at most that of the right one. Thus, we get a bad configuration with only two customers: 0 and v, where v is on the arc S(0, r) B. Proposition 6. In the obtained configuration, r β. Proof. Suppose that r < β. Then, since b v 3 and v 1, we have cost(b) = b v w(v) + rw(0) < 3 w(v) + βw(0) ζ cost(0) + β cost(f ) which contradicts (3).

15 326 V. Shenmaier / Discrete Optimization 22 (2016) Simplification 9 Normalize the weights of the customers so that their sum becomes equal to one. Then we have w(0) = 1 w(v). Write out explicitly the members of (3): cost(a) = 1 w(v) + a v w(v), cost(b) = r(1 w(v)) + b v w(v), cost(0) = rw(v), cost(f ) = 1 w(v). Since all of them are linear over the variable w(v), the difference between the left and right sides of (3), the expression min{cost(a), cost(b)} ζ cost(0) β cost(f ), (5) reaches the maximum if either w(v) {0, 1} or w(v) is such that cost(a) = cost(b). But the first is impossible: if only one of the customers v and 0 has a non-zero weight, the cost of its service by the closest facility from {a, b} is at most max{ 3 cost(0), cost(f )} which contradicts (3). Thus, (5) reaches the maximum at the value of w(v) such that cost(a) = cost(b): w(v) = r 1 a v b v + r 1. As a result, we obtain a bad configuration with only two degrees of freedom: the radius r and the angle ϕ which determines the position of the customer v on the circle S(0, r). To be specific, the angle ϕ will be measured counterclockwise from the first coordinate axis. Since cost(a) = cost(b), the inequality (3) becomes the following: 1 w(v) + a v w(v) ζrw(v) β(1 w(v)) > 0. If we substitute the expression for w(v) obtained above and collect like terms, this can be written as (r 1)( a v 1 ζr + β) a v b v + r β > 0. (6) Since r β > 1 and a v 3 b v, the denominator in (6) is positive. It follows that this inequality holds if and only if which is equivalent to the inequality where (r 1)( a v 1 ζr + β) > (β 1)( a v b v + r 1) L(r, ϕ) > 0, (7) L(r, ϕ) = ζ(r r 2 ) + (r β) a v + (β 1) b v, a v = 1 + r 2 + 2r cos ϕ, b v = r 2 1 cos(ω r + ϕ), Ω r = arccos((2r 2 1)/(2r)), β r 1/(ζ 1), 0 ϕ Ω r.

16 V. Shenmaier / Discrete Optimization 22 (2016) Verification of the inequality (7) It remains to analyze the function L(r, ϕ) on the compact set M of all the possible values of its variables: M = {(r, ϕ) r [β, 1/(ζ 1)], ϕ [0, Ω r ]}. Finding the maximum of L(r, ϕ) analytically seems to be impossible, but we can estimate it using a computer. To justify computer calculations find upper bounds for both derivatives of L. Lemma 7. If (r, ϕ) M, then L(r,ϕ) r Proof. The derivative over the variable r is equal to < and L(r,ϕ) ϕ < ζ(1 2r) + a v + (r β) a v + (β 1) b v, where a v and b v are the derivatives of a v and b v over r. Note that a v [ 3, r+1] and a v = (r +cos ϕ)/ a v (r +1)/ 3 1/( 3 1). On the other hand, b v 0 since the angle Ω r decreases with increasing r. So the derivative of L is at most 2 3 r r (r β)/( 3 1) which reaches the maximum at the minimum possible value of r: r = β. Therefore, it is at most (1 2 3)β < The derivative over the variable ϕ is (r β)r sin ϕ 1 + r2 + 2r cos ϕ + (β 1)r sin(ω r + ϕ) 2 1 cos(ωr + ϕ). Since r β, the first term is at most 0. The second one is β 1 2 r 1 + cos(ω r + ϕ) β 1 2 r 1 + cos Ω r = β 1 2 r 1 + 2r2 1 2r which reaches the maximum at the maximum possible value of r: r = 1/( 3 1). So this term is at most (β 1)/( 3 1) < Thus, to prove the non-positivity of L on the set M it is sufficient to establish that, for some small h > 0, this function is not greater than h( ) = 0.246h at the nodes of the two-dimensional grid For h = 2 22, we calculate that G(h) = {(β + ih, jh) i, j are integer} M. max{l(r, ϕ) (r, ϕ) G(h)} < 0.322h. It follows that L(r, ϕ) < 0 on the whole set M which contradicts (7). The proof of Lemma 1 is complete. Comment. The chosen value of the parameter β = is about the minimum for which (7) fails. References [1] R. Mettu, G. Plaxton, The online median problem, SIAM J. Comput. 32 (3) (2003) [2] M. Chrobak, C. Kenyon, J. Noga, N. Young, Incremental medians via online bribery, Algorithmica 50 (4) (2008) [3] G. Lin, C. Nagarajan, R. Rajamaran, D.P. Williamson, A general approach for incremental approximation and hierarchical clustering, SIAM J. Comput. 39 (8) (2010) [4] M. Chrobak, M. Hurand, Better bounds for incremental medians, Theoret. Comput. Sci. 412 (7) (2011) [5] V.V. Shenmaier, An approximate solution algorithm for the one-dimensional online median problem, J. Appl. Ind. Math. 2 (3) (2008) [6] S.G. Kolliopoulos, S. Rao, A nearly linear-time approximation scheme for the Euclidean k-median problem, SIAM J. Comput. 37 (3) (2007)