AT&T Labs 1

The Sensor Cover Problem

Alon Efrat (U. Arizona)

Adam Buchsbaum (AT&T Labs)

Shaili Jain (Harvard)

Suresh Venkatasubramanian (AT&T Labs)

Ke Yi (AT&T Labs)

AT&T Labs 2

Sensors

Low-power devices with sensing capabilities Sensors monitoring regions

We would like to cover a given region for as long as possible

Since sensors have limited life, we need to schedule sensor uptime to maximize coverage time

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Preemptive and Non-preemptive Preemptive: Sensors can be turned on and

off multiple times, and the total duration is constrained by battery

Non-preemptive: Each sensor can be turned on only once, and then it runs till death

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Problem Definition

A sensor s is defined by a region R(s) U and a duration d(s).

A schedule for a sensor is an assignment of a start time t(s), so that the sensor is active for all t(s) t t(s) + d(s).

A point x U is covered at time t if there is some active sensor s at time t such that x R(s).

Problem: Given n sensors defined by (Ri , di), i [n], schedule all sensors so that U is covered for all 0 t T, and T is maximized.

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Prior Work

Previous work assumes that U is an arbitrary finite set, Ri’s are arbitrary subsets of U, and di = 1 (Set Cover Packing) (log n)-approximation [Feige et al. 02] Various heuristics in the networks community

Does geometry help?

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1-D: Restricted Strip Covering Each sensor covers an

interval Problem: Slide rectangles

up and down to cover a rectangular area of maximum height

Load L = 5

3

5

232

54

52

43

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Restricted Strip Covering: Related Results Packing instead of covering: The dynamic sto

rage allocation problem (2+)-approx. [Buchsbaum et al. ’04]

Further allow rectangles to move both vertically and horizontally: Strip packing (1+)-approx. [Kenyon and Remila ’00]

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Our Results

Sensor shape Uniform duration Variable duration

Intervals Exact in P NP-hard

O(log log log n)-approx.

(2+)-approx. for non-nested intervals

Disks NP-hard

O(log(ndmax/L))-approx.

O(log(Lmax/L))-approx. when congruent

NP-hard

O(log n)-approx.

Arbitrary sets (log n)-approx.[Feige et al. ’02]

(log n)-approx.

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An Easy Start: Uniform Duration A greedy algorithm can achieve optimal

solution of duration L

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Variable Durations: NP-hard

O(log n)2004

O(log log n)2005

O(log log log n)2006

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Roadmap to “Triple-Log”

Dynamic programming for small L Find the optimal schedule in O(nLO(L)) time Poly time for L = O(log n / log log n) (1+)-approx. for L < c1dmin log n / log log n

The “grouping” technique for large L (1+)-approx. for L > c2dmax log n

Combine the two techniques to get an O(log log log n)-approx. for L in between

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Dynamic Programming

Check if T is achievable Try all T=1,2,…,L

At any x-coordinate i, consider all possible schedules of sensors covering i.

Mark a state “valid” if there is previous compatible valid state

i=1 i=2

…

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Dynamic Programming Analysis A crude analysis yields

an nO(n) bound Observation 1: at most

2T sensors in any state Observation 2: can skip

x-coordinates where there are >5T sensors # states per step: total running time is O(nLO

(L)).

i=1 i=2

…

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Dynamic Programming Analysis An exact solution for L = O(log n / log log n) (1+)-approx. for L < c1dmin log n / log log n

(by truncating) Replace di by

New L’ < c1/ ∙ log n / log log n Run the dynamic program

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Roadmap to “Triple-Log”

Dynamic programming for small L Solves the problem in O(nLO(L)) time (1+)-approx. for L < c1dmin log n / log log n

The “grouping” technique for large L (1+)-approx. for L > c2 / 5 ∙ dmax log n

Combine the two techniques to get an O(log log log n)-approx. for L in between

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Basic idea: group short sensors into longer sensors, then apply the greedy algorithm

Need to bound the amount wasted

The Grouping Algorithm

wasted

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Grouping Sensors Stabbed by a Common Anchor

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Grouping Sensors Stabbed by a Common Anchor

Lnew(x) = Lold(x) / (1+) - 4D/

D: height of new sensors

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Partitioning

A -grouping is a partition of sensors into a number of subsets such that: Sensors in a subset are stabbed by a common an

chor For any x-coordinate i, the set of sensors live at i a

re drawn from no more than subsets Lemma: We can group a set of unit-duration s

ensors into sensor with duration D such that

Lnew(x) > Lold(x) / (1+) - O(D / )

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Partitioning for a Set of Intervals = (log n) for a general set of intervals = 2 for non-nested intervals

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The Algorithm

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The Algorithm

By repeatedly applying the lemma to make all sensors have equal duration

Theorem: For sufficiently small positive , there is a poly-time algorithm that gives a schedule of duration at least

L/(1 + ) - O(dmax / 4).

Corollary: There is a constant c2, such that the algorithm gives a schedule of duration at least L/(1 + ) when L > c2dmax / 5.

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Roadmap to “Triple-Log”

Dynamic programming for small L Solves the problem in O(nLO(L)) time (1+)-approx. for L < c1dmin log n / log log n

The “grouping” technique for large L (1+)-approx. for L > c2dmax / 5

( = O(log n))

Combine the two techniques to get an O(log log log n)-approx. for L in between

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Putting Pieces Together

Consider the case when = O(1). Fix parameter = 5/(c2). Partition sensors into blue set of those

with duration >L (large) and red set of the rest (small). For each x-coordinate, one of the sets has load at least L/2. Use

each set to cover those x-coordinates where it has the load. Dynamic programming for the blue set gives solution of duration L

/(2(1 + )). Grouping for the red set gives same result. Putting solutions together gives a schedule of duration L/(2 + 2).

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The General Case

Set h = L log log n / log n; fix parameter l . Partition sensors:

R0 gets large sensors, of duration > h. Ri for i = 1, …, l gets middle sensors, of duration [h/2i, h/2i-1); tru

ncate to h/2i . R +1 gets small sensors, of duration < h/2l = L log log n/(2l log n).

For each x-coordinate, one of the sets has load at least L/(l + 2). Use each set to cover those x-coordinate where it has the load.

Dynamic programming for R0 gives solution of durationL/((1 + )(l + 2)).

Greedy on the middle Ri ’s gives solution of duration L/(2(l + 2)). Grouping for Rl +1 with = 1 gives solution of duration

Choose l= O(log log log n)

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Conclusion and Open Problems 1D:

For non-nested intervals, (2 + )-approx. General case: O(log log log n)-approx. We conjec

ture that constant approx is possible. 2D:

O(log(ndmax / L))-approx. Not much better than the O(log n)-approx. for the

arbitrary-set version Can we improve?

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More Open Problems

Preemptive Schedules 1D: Can be solved exactly using max-flow

Load = Duration of optimal preemptive schedule 2D: NP-hard

No approximation known

Sensor networks connectivity

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Thank you!