Optimizing Sensor Movement for Barrier Coverage in Linear Domain

1. Algorithms on Minimizing the Maximum Sensor Movement forBarrier Coverage of a Linear Domain Danny Z. Chen1, Yan Gu2, Jian Li2, and Haitao Wang1 1University of Notre Dame 2Tsinghua University SWAT 2012

2. Motivation • A sensor has a sensing range r, detecting objects within the range r a sensor

3. Motivation • A region can be protected by sensors • Full coverage of region

4. Motivation • Barrier coverage: only covering its border • The border is the barrier intruder

5. Mobile sensors

6. Our problem: the segment barrier • A barrier B, represented by a line segment, and n sensors on the line containing B • Each sensor has a covering interval • Move the sensors such that • Barrier coverage: the barrier is covered by all sensors • Min-max: the maximum sensor movement is minimized

7. Previous work and our result • The uniform case: covering intervals of all sensors have the same size • O(n2) time, Czyzowicz et al. 09’ • Our result: O(nlog n) time • The non-uniform case: different ranges • Open problem: polynomial solvable? • Our result: O(n2 logn) time (a recent result) • O(n2 logn loglogn) time in the paper

8. Related work • Min-Sum • Uniform: O(n2) time, Czyzowicz et al. 10’ • Non-uniform: NP-hard, Czyzowicz et al. 10’ • Min-Number: minimize the number of moving sensors • Uniform: O(n3) time, Mehrandish et al. 11’ • Non-uniform: NP-hard, Mehrandish et al. 11’

9. The decision versions • Given a value d, determine whether it is possible to cover the barrier such that the maximum sensor movement is at most d • The uniform case • O(n) time, Czyzowicz et al. 09’ • The non-uniform case • Our result: O(nlog n) time • Optimization versions: the original problems

10. The uniform case • Why is it easier? • The following order preserving property holds • The order of the sensors in an optimal solution is the same as that in the input 4 1 2 3

11. Our algorithm for the uniform case • d* : the maximum sensor movement in an optimal solution • Determine a set D of candidate values, such that • d* is in D and |D|=O(n2) • Determine D implicitly • First attempt: arrange D in a sorted list, and then do binary search using the decision algorithm • Difficulty: It is unclear how to arrange D in a sorted list • Our approach • arrange D in O(n) sorted lists • a technique: binary search on sorted lists

12. The non-uniform case • The difficulty • The order preserving property does not hold!!! • The key • Determine the order of sensors in an optimal solution • The decision version • The optimization version • Parameterize the decision algorithm • Similar to parametric search • but no parallel scheme is involved • the algorithm is practical

13. The decision algorithm • Problem: given a value d, determine whether d* ≤ d • A greedy approach • try to cover the barrier from left to right as much as possible • Initially, move each sensor to the right for distance d • Later, we only consider moving sensors to the left for distance at most 2d

14. The decision algorithm • Use a vertical line L sweeping from left to right on B • Each step determines a sensor for covering B from the current position of L to the right as much as possible • The rule: • S1 : the set of intervals intersecting L • If S1 is not empty, • take the interval in S1 with rightmost right endpoint sweeping line L

15. The decision algorithm • The rule (S1 is empty): • Consider the line L’ with distance 2d right of L • S2: intervals with left endpoints between L and L’ • If S2 is empty, d* ≤ d is not true • Else, take the interval in S2 with the leftmostright endpoint L’ sweeping line L 2d

16. The decision algorithm • The rule (S1 is empty): • Consider the line L’ with distance 2d right of L • S2: intervals with left endpoints between L and L’ • If S2 is empty, d* ≤ d is not true • Else, take the interval in S2 with the leftmostright endpoint L’ sweeping line L

17. The decision algorithm • The rule (S1 is empty): • Consider the line L’ with distance 2d right of L • S2: intervals with left endpoints between L and L’ • If S2 is empty, d* ≤ d is not true • Else, take the interval in S2 with the leftmostright endpoint L’ sweeping line L

18. The algorithm implementation • Running time: O(nlog n) • sweeping • With O(n log n) time preprocessing (sorting) • for each value d, • determine whether d* ≤ d in O(n) time (a recent result) • O(n loglog n) time in the paper

19. The optimization algorithm • Key: find an order for sensors • Every step determines a sensor • by simulating the decision algorithm for d = d* • The i-th step • Input: an interval (xi-1, yi-1) • d* is in (xi-1, yi-1) • Goal: determine the sensor which is the one in the i-th step of the decision algorithm when d=d* • Output: d* or a sub-interval (xi , yi) of (xi-1, yi-1) • d* is in (xi , yi)

20. The first step i=1, (xi-1, yi-1)=(-∞,+∞) • Determine the set S1 • Move all sensors rightwards, find a set of distance values d1 d2 d3 … such that • for any two consecutive diand di+1, when the moving distance is d with di< d < di+1, the set S1is the same • Find index i such that di< d* ≤ di+1 • Determine whether d* = di+1 d1 d2 d3

21. The first step (cont.) • If d* = di+1, output d* • If d* ≠ di+1, (di,di+1) and S1 are available • If S1 is not empty, • determine the sensor • output (di, di+1) as (xi , yi) and proceed on the next step • If S1 is empty, • determine the set S2 similarly

22. Time analysis • Each step takes O(n log n) time • Dominated by the binary search • Overall running time: O(n2 log n)

23. Conclusion • The uniform case • An improved algorithm: O(n logn) time • Previous work: O(n2) • The non-uniform case • First-known polynomial-time algorithm: O(n2 logn) time • Open problem answered!

24. Thank you