An Improved Approximation for k-median, and Positive Correlation in Budgeted Optimization

Transcription

1 An Improved Approximation for k-median, and Positive Correlation in Budgeted Optimization Jaroslaw Byrka 1 Thomas Pensyl 2 Bartosz Rybicki 1 Aravind Srinivasan 2 Khoa Trinh 2 1 University of Wroclaw, Poland 2 University of Maryland, USA ISMP 2015

2 k-median Input: set of facilities F set of clients C symmetric distance metric d on C F an integer k > 0 Goal: pick k facilities which minimize total connection cost of clients. Example: k = 2, open facilities A, C and pay cost = 7. A B C

3 Uncapcitated Facility Location Input: set of facilities F set of clients C symmetric distance metric d on C F each facility has an opening cost Goal: pick some facilities which minimize total opening cost and connection cost. Example: open facilities A, C and pay cost = A B C

4 Bi-point solution as middle step Bi-point solution: a feasible solution which is a convex combination of two integral solutions. Step I: Construct a bi-point solution. Use some special UFL algorithms to get a cheap bi-point solution. Step II: Round bi-point solution to integral one.

5 Previous Work Year% Authors% Bi.point% Construc3on% Bi.point% Rounding% 99# Charikar#et.#al.# (LP#Rounding)# 6.667# 99# Jain#&#Vazirani# 3#+#ε# 2# 6#+#ε# 02# Jain,Mahdian#&#Saberi# 2#+#ε# 2# 4#+#ε# 01# Arya#et.#al.# (Local#Search)# 3#+#ε# k.median% Approxima3on% 12# Li#&#Svensson# 2#+#ε# 1.366#+#ε# 2.732#+#ε# 15% Our%work% 2#+#ε# 1.337%+%ε% 2.674%+%ε%

6 Bi-point Rounding Again, bi-point solution is convex combination of two integral solutions, F 1 and F 2. a F 1 + b F 2 = k, where (a, b > 0, a + b = 1) F 2 F 1

7 Stars Form stars by attaching each facility in F 1 to closest facility in F 1. F 2 F 1

8 Li-Svensson s rounding algorithm Key property: if some leaf is closed then its root must be open. Pr[i 1 is open] a for all i 1 F 1 Pr[i 2 is open] b for all i 2 F 2 approximation factor = #open facilities = a F 1 + b F 2 + O(1) = k + O(1). F 2 F 1

9 Distance bounds i 2 d 2 d 1 j d 3 i 1 i 3 There is an open facility in {i 1, i 2, i 3 } d 3 d 2 + d(i 2, i 3 ) d 1 + 2d 2 The total (expected) connection cost is a (non-linear) function of cost (F 1 ), cost (F 2 ), a, b which can be related to OPT.

10 Tight Instance In tight case, almost all stars have exactly 1 leaf. BUT still many clients near 2-leafed stars.

11 Tight Instance Why not close some 1-leafed stars to open more leaves of the 2-leafed stars?

12 Long 1-leafed stars i 2 i 3 i 4 i 5 Consider any 1-leafed star (i 2, i 3 ) and let i 4 be the closest leaf of a 2-leafed star to i 3. Then the star (i 2, i 3 ) is long if for some constant g > 0. d(i 2, i 3 ) gd(i 3, i 4 ),

13 Long 1-leafed stars can be closed apple d d 1 d 1 + d 2 2 d apple d1+d2 2 g apple d1+d2 g i 1 j i 2 i 3 i 4 i 4 i 5

14 Bi-point Rounding Improved Bi-point rounding algorithm: Consists of 8 different rounding strategies. Analysis uses a non-linear factor revealing program. Approximation ratio:

15 Dependent Rounding Given a vector P = (p 1,..., p n ) with 0 p i 1, Dependent Rounding algorithm rounds P into a 0-1 vector X such that the sum is preserved: n i=1 X i = n i=1 p i, the marginals are preserved: Pr[X i = 1] = p i, negative correlation between X i s: e.g. Pr[X i = 1 X j = 1] p i p j Pr[Xi = 0 X j = 0] (1 p i )(1 p j )

16 A Typical Example of Negative Correlation in Facility Location Problems Assume that we have a client j and a set S of facilities that are close to j. Each facility i S has an opening variable y i. Applying dependent rounding to opening variables in S gives [ ] Pr[all facilities in S are closed] = Pr {i is closed} i S i S(1 Pr[i is open]) exp ( i S ) y i.

17 Positive Correlation i 2 d 2 d 1 j d 3 i 1 i 3 For each root i F 1, let X i be the indicator whether i or its leaves is opened.

18 Positive Correlation i 2 d 2 d 1 j d 3 i 1 i 3 For each root i F 1, let X i be the indicator whether i or its leaves is opened. We want to upper-bound E [ cost (j)]: E [ cost (j)] Pr[X i3 = 0]d 2 + Pr[X i3 = 1 X i1 = 1]d 1 + Pr[X i3 = 1 X i1 = 0]d 3

19 Positive Correlation i 2 d 2 d 1 j d 3 i 1 i 3 For each root i F 1, let X i be the indicator whether i or its leaves is opened. We want to upper-bound E [ cost (j)]: E [ cost (j)] Pr[X i3 = 0]d 2 + Pr[X i3 = 1 X i1 = 1]d 1 + Pr[X i3 = 1 X i1 = 0]d 3 Pr[X i3 = 1 X i1 = 1] Pr[X i3 = 1] Pr[X i1 = 1] = a 2, by negative correlation

20 Positive Correlation i 2 d 2 d 1 j d 3 i 1 i 3 For each root i F 1, let X i be the indicator whether i or its leaves is opened. We want to upper-bound E [ cost (j)]: E [ cost (j)] Pr[X i3 = 0]d 2 + Pr[X i3 = 1 X i1 = 1]d 1 + Pr[X i3 = 1 X i1 = 0]d 3 Pr[X i3 = 1 X i1 = 1] Pr[X i3 = 1] Pr[X i1 = 1] = a 2, by negative correlation How about Pr[Xi3 = 1 X i1 = 0]?

21 Near Independence Independence: Near-Independence: Pr[A B C] = p A p B p C (1 ɛ)p A p B p C Pr[A B C] (1 + ɛ)p A p B p C Can dependent rounding achieve this property? Answer: Yes, but with limits...

22 Extreme Examples Observation 1: if p 1 = p 2 =... = p n = 1/2, we have Pr[X 1 = 1 X 2 = 0... X n = 0] = 0. On the other hand, Pr[X 1 = 1] n i=2 Pr[X i = 0] = 2 n > 0. Get around this by considering a small subset of events.

23 Extreme Examples Observation 1: if p 1 = p 2 =... = p n = 1/2, we have Pr[X 1 = 1 X 2 = 0... X n = 0] = 0. On the other hand, Pr[X 1 = 1] n i=2 Pr[X i = 0] = 2 n > 0. Get around this by considering a small subset of events. Observation 2: if p 1 = p 2 =... = p n = 1/n. Then, Pr[X 1 = 1 X 2 = 1] = 0. On the other hand, Pr[X 1 = 1] Pr[X 2 = 1] = 1/n 2 > 0. Get around this by by requiring that the values p i are not too close to 0 or 1.

24 Modified Dependent Rounding Modification: randomly permute the variables before applying dependent rounding.

25 Modified Dependent Rounding Modification: randomly permute the variables before applying dependent rounding. Our results: if there is some α > 0 that α p i 1 α for all i, then for any t events of the form X i = 0 or X i = 1, their joint probability lies within ( )) 1 (1 t2 nα 2 and ( 1 + ( 1 + t ) ) t 1 nα 2, of the value it would if these events were independent.

26 Modified Dependent Rounding Modification: randomly permute the variables before applying dependent rounding. Our results: if there is some α > 0 that α p i 1 α for all i, then for any t events of the form X i = 0 or X i = 1, their joint probability lies within ( )) 1 (1 t2 nα 2 and ( 1 + ( 1 + t ) ) t 1 nα 2, of the value it would if these events were independent. in particular, these factors are (1 + o(1)) and (1 o(1)) if t = o(α n).

27 Modified Dependent Rounding Application to k-median: reduce #extra facilities from O(1/ɛ) to O(log(1/ɛ)) reduce run-time from n O(1/ɛ2) to n O(1/ɛ log(1/ɛ)).

28 Open questions Closing the gap for k-median? Our algorithm: ( ɛ)-apx Best known lower bound: 1 + 2/e Other applications of dependent rounding with near-independence property?

29 Thank you!