Practical Discrete Unit Disk Cover Using an Exact Line-Separable Algorithm

1. Practical Discrete Unit Disk Cover Using an Exact Line-Separable Algorithm Alejandro Salinger University of Waterloo Joint work with Francisco Claude, Gautam K. Das, Reza Dorrigiv, Stephane Durocher, Robert Fraser, Alejandro López-Ortiz, Bradford G. Nickerson Santiago, December 2010

2. Outline • Discrete Unit Disc Cover Problem • Line Separable Version • Simple Greedy Algorithm • Faster Dual Algorithm • Approximating the General Problem • Recent Advances

3. Discrete Unit Disk Cover (DUDC) • Given m unit disks D (facilities) and n points Q (clients) in the plane, the discrete unit disk cover problem is to select a minimum subset of the disks to cover the points.

4. Applications

5. Applications

6. Applications

7. Applications

8. About Discrete Unit Disk Cover • NP-Hard (Johnson, 1982) • Geometric version of SET-COVER • SET-COVER is not approximable within c log n • DUDC admits constant factor approximation • Related problems: • Minimum Geometric Disk Cover: disk centres are not restricted. • Discrete k-Centre: find set of k disks minimizing largest radius.

9. About Discrete Unit Disk Cover • Approximation algorithms for DUDC: • 108-approximate (Călinescu et al., 2004) • 72-approximate (Narayanappa & Voytechovsky, 2006) • 38-approximate (Carmi et al., 2007) • (1+ε)-approximate (Mustafa & Ray, 2009) • Uses local improvement approach • O(m65n) time in the worst case (3-approximation) • This paper: • 22-approximate, O(m2n4) algorithm.

10. Line-separated DUDC • Disk centres and points in Q are now separable by a line l, which we will call horizontal. • Formally, given sets of points and where is the set of unit disks centred at the points in P, find of minimum cardinality that covers Q. d2 d5 d1 d4 d3 p2 p5 p1 p3 p4 l q6 q9 q1 q3 q5 q7 q8 q2 q4

12. A Simple Greedy Algorithm • Simplification rules: • If a disk d1 covers no points from Q, we can remove it. • If a disk d1 is dominated by a disk d2, then we can remove d1 from the problem instance. • If a point q1 is only covered by a disk d1, then d1 must be part of the solution. by Rule 3 d2 d5 d1 d4 by Rule 3 d3 d3 d5 d2 Solution Set (D’) Discarded by Rule 2 p2 p5 p1 p3 by Rule 1 p4 l q6 q9 q1 q3 q5 q7 q8 q2 q4 d4 d1 by Rule 2

13. Greedy Step • Simplification rules are not always sufficient. • If no more simplification rules can be applied, then the leftmost disk is added to the solution set (disks are ordered by leftmost intersection with l). d2 d3 d1 p2 p1 p3 l q3 q1 q2

14. Sort disks from left to right While there are disks left Apply simplification rules Add leftmost disk to solution time (recall m facilities and n clients) The Greedy Algorithm

15. Faster Implementation 1 d2 2 2 d5 d1 d2 • Encode dominance in graph • Traverse nodes in order: • if , discard • if , is dominated by • if is not dominated by any disk to its right • time d1 d4 d3 1 1 5 p2 p5 d3 0 p1 p3 p4 2 1 q6 q9 q1 q3 q5 q7 3 4 4 q8 q2 q4 d5 d4

16. A faster algorithm • Dual setting • Goal: find of minimum cardinality such that each disk is stabbed by at least one point. p2 p5 p1 p3 p4 l q6 q9 q1 q3 q5 q7 q8 q2 q4 p1 p3 p5 Solution Set (P’)

17. A faster algorithm • From left to right, find maximal chain of disks with non-empty intersection • Add rightmost point to solution • Running time p2 p5 p1 p3 p4 l q6 q9 q1 q3 q5 q7 q8 q2 q4 Solution Set (P’) p3 p1 p5

18. Why is this optimal? • k: index such that and • s: rightmost point in • Any optimal solution requires 2 disks to cover • If that covers both and , then } not covered by • Another point must cover l d1 dk+1 da

19. Why is this optimal? (Case 1) • We can choose s instead of to cover (which also covers ) • may be covered by another point, possibly different from • Thus, the choice of s is correct s1 s2 l dk+1 d1 da

20. Why is this optimal? (Case 2) • covers and • However, since , there must be another disk not covered by • We can choose s to cover • may be covered by another point, possibly different from • Thus, the choice of s is correct s1 s2 l dk+1 d1 da

21. Why is this optimal? (Case 3) • We can choose s instead of to cover • may be covered by another point, possibly different from • Thus, the choice of s is correct s1 s2 l dk+1 d1 da

22. Back to DUDC • We have an exact algorithm for the line-separable case • The goal is to obtain an approximation algorithm for the general problem • We adapt the 38-approximation algorithm of Carmi et al. to obtain a 22-approximation to DUDC • Their algorithm uses a variant of the line-separable discrete unit disk cover

23. Minimum Assisted Cover (MAC) • All points Q on one side of the line l, and disks Dcentred both above and below l • All points in Q are covered by U • Want of minimum cardinality that covers Q • Carmi et al. provide a 4-approximation u2 u5 u1 u4 u3 p2 p5 p1 p3 p4 l q6 q9 q1 q3 q5 q7 q8 q2 q4 l1 l2

24. Minimum Assisted Cover (MAC) • Our LSDUDC algorithm plus greedy MAC gives a 2-approximation • Use our algorithm to obtain a set U’ • Use greedy MAC to obtain an improved solution A l

25. How good is A? • Separate A and OPT by l: OPTU, OPTL, AU, AL • Let ac(U’,OPTL) be the smallest subset of U’ that forms a cover when assisted by OPTL • A is the minimum size assisted cover based on U’ • |A| = |AU|+|AL|≤|ac(U’,OPTL)|+|OPTL| ac(U’,OPTL) U’ l OPTL

26. Approximation Ratio • Given a disk , there are 2 cases: • d is above the lower boundary of ac(U’,OPTL) • Otherwise, d contains one or more arc segments of the lower boundary of ac(U’,OPTL) d d Case 1 Case 2

27. Approximation Ratio Case 1: d is above the lower boundary of ac(U’,OPTL). • 2 disks from ac(U’,OPTL) can be used to cover d. • If β is the number of such disks in , then at most 2β disks are required to cover the same points in ac(U’,OPTL). l

28. Approximation Ratio Case 2: d contains one or more arc segments of the lower boundary of ac(U’,OPTL) • V: disks of arc segments covered by d • Let • Since W dominates V, there is at most one fully covered arc segment in V (otherwise U’ would not be optimal) • and must contain points not covered by d, so assign each to the left-most covering disk in OPT • At most two disks from V are associated with d vl vr l d

29. Approximation Ratio Therefore, each disk in OPTU has at most two associated disks in ac(U’,OPTL), thus • Line-Separable DUDC + Greedy Minimum Assisted Cover gives a 2-approximation for the Minimum Assisted Cover problem

30. Approximation of DUDC • Apply 2-approximation on each line in each direction. • Each disk can participate in 8 applications of the algorithm. • Carmi et al. give a 6-approximation for the single square problem. • Approximation factor: 2 x 8 + 1 x 6 = 22 • Worst case running time: O(m2n4) 3/2

31. Summary • We presented an exact algorithm for the case when clients and facilities are separated by a line • This allowed us to improve the approximation to the Minimum Assisted Cover problem • We improved the approximation ratio from 38 to 22 for the general Discrete Unit Disk Cover problem • O(m2n4) running time

32. Recent Advances • [Das et al., to appear in WALCOM 11] • The approximation ratio was improved to 18 • 6-approximation for within strip problem • 12-approximation for outside strip problem • Running time improved to • Ongoing work: improve approximation factor while keeping small running time