Optimal Backlog in the Plane

1. Optimal Backlog in the Plane joint work with Jukka Suomela Valentin Polishchuk

2. Sensor Field

3. Data Constantly Flows In Flow changes over time unpredictably Total Flow Rate = 1 over all sensors

4. Data Stored at Local Card • Needs to be gathered before card is overfull Cheaper to store data locally and gather than to transmit [Mathur, Desnoyers, Ganesan, Shenoy, Diao, ’06,’07] bandwidth is enough to report amount of data not to transmit

5. Maintenance Man • How large a memory card is needed? • What path to take?

6. Formally… • Set of “cups” • Adversary • pours water into cups • total rate 1 • Player • empties cups • moves with speed 1 • Player’s objective: min max amount of water (backlog) in any cup at any time • Adversary can guarantee Ω(D) ½ ½ D Online: Player doesn’t know how water will be distributed in the future

7. Our Result A strategy for the playerO(D) backlog Doesn’t depend on how water is poured cups positions number of cups • Adversary can guarantee Ω(D) ½ ½ D

8. Related Work • Geometric TSP • same rate at each cup, const over time • Data gathering [Falck,Floréen,Kaski,Kohonen,Orponen’04] • wireless communication • Our setting • physically visit sensors

9. Discrete Version • Cups at vertices of a graph • Timesteps • adversary pours total 1 of water • player moves along an edge • Regular nxn grid [Bender,Fekete,Kröller,Liberatore,Mitchell,P,S‘07] • O(n √ log log n)

10. Abstract Version • n cups • not on a graph nor in the plane • Timesteps • adversary pours total 1 of water • player empties any cup • Deamortization [Dietz,Sleator’87] • Empty-the-fullest: Hn

11. An Extension: (τ,k)-game • n cups • not on a graph nor in the plane • Timesteps • adversary pours total τ of water • player empties k fullest cups • After r steps • backlog in any cup is ≤ Hr •τ/ k • independent of n

12. Partition the Time τ0 = 10 D τ1 = 100 D …τi = 10i+1 D… τN = 10N+1 D • Into small intervals • And into larger intervals • And LARGER • … 0 time τ0 = 10 D τ1 = 10 τ0 τ2 = 10 τ1

13. Coroutine i (i = 0…N) τi • Invoked at times that are multiples of τi • Plays (τi,ki)-game with water after 10Lτi Empty ki fullest cups (and return back) ki= 25i * * * time ℓτi t =10Lτi + ℓτi 10Lτi Coroutines j>i are suspended Integer L , 1 ≤ ℓ≤ 10

14. For Example τ5 • Empty k5 = 255 cups, fullest with water after20τ5 * * * time 3τ5 20τ5 t =23τ5 Coroutines 6 (and higher) suspended

15. Global View τ0 = 10 D τ0 = 10 D τ1 = 100 D …τi = 10i+1 D… τN = 10N+1 D k0 = 1 k1 = 25 …ki = 25i… kN = 25N • Coroutine 0 (all other suspended) • Coroutine 1 (2 and higher suspended) • Coroutine 2 (3 and higher suspended) time τ1 = 10 τ0 τ2 = 10 τ1 Is there enough time to run as scheduled? Backlog?

16. Few’s Lemma τ0 = 10 D τ1 = 100 D …τi = 10i+1 D… τN = 10N+1 D k0 = 1 k1 = 25 …ki = 25i… kN = 25N • Any set of m2 pts in the plane • Tour of length 5mD (D – diameter of the set) • Coroutine i returns in 5i+1D time τi τi = 10i+1D > 1005i+1D + 1015iD + 1025i-1D + … + 10i51D

17. Backlog ℓN-1τN-1 ≤ H10τN/kN + H10τN-1/kN-1 ε + H10τN-1/kN-1 +…+ (τN,kN)-game for ℓN-1 rounds Coroutine N (τN-1,kN-1)-game for 10 rounds Coroutine N-1 τ0 = 10 D τ1 = 100 D …τi = 10i+1 D… τN = 10N+1 D k0 = 1 k1 = 25 …ki = 25i… kN = 25N Coroutine N-1 ℓN-1-1 rounds τi/ki = (2/5)i • 10D time O(D) ℓNτN ℓNτN + ℓN-1τN-1 + … + ℓ1τ1 + ℓ0τ0 + ε=tℓN , ℓN-1 , … , ℓ1 , 1≤ℓi ≤ 10, ε≤ τ0

18. Conclusion • O(D) backlog • In the plane • Continuous time • Doesn’t depend on • Adversary moves • Shape of the cup set • Number of cups

19. Open • Scalability with the number of players • 4 players: divide into 4 parts • diameter halves • backlog halves Better than ½ ?