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Approximation via Doubling

Approximation via Doubling. Marek Chrobak University of California, Riverside. Joint work with Claire Kenyon-Mathieu. Doubling method: (for a minimization problem) Choose d 1 < d 2 < d 3 … (typically powers of 2) For j = 1, 2, 3, … Assume that the optimum is ≤ d j

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Approximation via Doubling

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  1. Approximation via Doubling Marek Chrobak University of California, Riverside Joint work with Claire Kenyon-Mathieu 1

  2. Doubling method: (for a minimization problem) Choosed1 < d2 < d3 … (typically powers of 2) For j = 1, 2, 3, … Assume that the optimum is ≤ dj Use this bound to construct a solution of cost ≤ C·dj • Simple and effective (works for many problems, offline • and online) • Typically not best possible ratios 2

  3. Outline: Online bidding Cow-path Incremental medians (size approximation) Incremental medians (cost approximation) List scheduling on related machines Minimum latency tours Incremental clustering 3

  4. Outline: Online bidding Cow-path Incremental medians (size approximation) Incremental medians (cost approximation) List scheduling on related machines Minimum latency tours Incremental clustering 4

  5. 1 2 5 12 Online Bidding 5

  6. 1 2 5 12 Online Bidding 20 bags of gunpowder but… 6 bags could have been enough so ratio = 20/6 6

  7. Online Bidding Item for sale of value u(unknown to bidder) Buyer bids d1,d2,d3, … until some dj≥ u Cost: d1 + d2 + … + djOptimum = u Competitive ratio 7

  8. Deterministic Bidding - Upper Bound Doubling strategy: bid 1, 2, 4, … , 2i, … If 2j-1 <u ≤ 2j, the ratio is 8

  9. Online Bidding • Theorem: • The optimal competitive ratio for online bidding is: • 4 in the deterministic case • e 2.72 in the randomized case • Randomized e-ing strategy: choose uniformly random x  [0,1), and bid • e x, e x+1, e x+2 , e x+3 , … • [folklore] [Chrobak, Kenyon, Noga, Young, ‘06] 11

  10. Outline: Online bidding Cow-path Incremental medians (size approximation) Incremental medians (cost approximation) List scheduling on related machines Minimum latency tours Incremental clustering 12

  11. d1 u d2 d3 d4 0 Cow-Path 13

  12. For dj-1 <u ≤ dj+1 (j odd) dj+1 d2 d1 u d3 0 dj-1 dj 2  bidding ratio extra ratio 1 Analysis: So the ratio = 2  bidding ratio + 1 = 9 for dj= 2j 14

  13. Solution of (r-1)ln(r-1) = r 2e+1 Connection to online bidding does not work in randomized case -- why? • Theorem: • The optimal competitive ratio for the cow-path problem is • 9 in the deterministic case •  4.59 in the randomized case [Gal ‘80] [Baeza-Yates, Culberson, Rawlins ‘93] [Papadimitriou, Yannakakis ‘91] [Kao, Reif, Tate ‘94] … 16

  14. Outline: Online bidding Cow-path Incremental medians (size approximation) Incremental medians (cost approximation) List scheduling on related machines Minimum latency tours Incremental clustering 17

  15. The k-Median Problem X = set of facilities Y = set of customers X Y : metric space with distance function dxy For FX let cost(F) = y YdyF where dyF= minf Fdyf The k-Median Problem:Find a facility set F of size k for which cost(F) is minimized. optimal F = Qk (the k-median) 18

  16. customer facility (potential) 19

  17. k = 2 facilities cost = 17 3 4 1 1 1 3 2 2 20

  18. k = 4 facilities cost = 12 1 3 1 1 2 2 1 1 21

  19. Offline Case • k-Median is NP-hard • Offline approximations: given k, find F such that • |F | ≤ k and cost(F) ≤ C·optk • C-cost-approximation • Upper bound C = 3+ • [Arya, Garg, Khandekar, Munagala, Pandit ‘01] • C ≥ 1+2/e for polynomial algorithms • (unless P = NP)[Jain, Mahdian, Saberi ‘02] • cost(F) ≤ optk and |F| ≤ S·k • S-size-approximation • S = Ω(logn) for polynomial algorithms • (unless P = NP) 22

  20. Size-Competitive Incremental Medians • k not known, authorizations for additional facilities arrive over time • Algorithm produces a sequence of facility sets: F1F2 … Fn • An algorithm is S-size-competitive if • |Fk| ≤S·kand cost(Fk) ≤ optk • for all k. • Goal: small competitive ratio 23

  21. opt = 26 cost = 26 k = 1 2 5 2 4 3 3 5 2 24

  22. opt = 17 cost = 18 !!! k = 2 2 4 2 1 3 2 2 2 25

  23. opt = 17 cost = 15 k = 2 2 4 1 1 1 2 2 2 26

  24. … not a polynomial time algorithm … Size-Competitive Incremental Medians Algorithm: 1. choosed1 < d2 < d3 … 2. Compute Q1, Q2, … (optimal medians) 3. F1 = Qd(1) // d(j) = dj for k = 2, 3, … if k = di+1 Fk = Fk-1Qd(i+1) 27

  25. k k k k 1 2 3 4 5 … 11 12 13 … 19 20 21 … 31 32 … d1d2d3d4 Qd(1) Qd(3) Qd(2) Qd(4) Qk= optimal k-median 28

  26. k k k 1 2 3 4 5 … 11 12 13 … 19 20 21 … 31 32 … d1d2d3d4 Qd(1) Qd(3) Qd(2) Qd(4) Qk= optimal k-median 29

  27. k k k 1 2 3 4 5 … 11 12 13 … 19 20 21 … 31 32 … d1d2d3d4 Qd(1) Qd(3) Qd(2) Qd(4) Qk= optimal k-median 30

  28. k 1 2 3 4 5 … 11 12 13 … 19 20 21 … 31 32 … d1d2d3d4 Qd(1) Qd(3) Qd(2) Qd(4) Qk= optimal k-median 31

  29. Same as online bidding So we get ratio = 4 for dj = 2j Analysis: • At step k, for dj-1 <k ≤ dj • cost(Fk) ≤ cost(Qd(j)) = opt(dj)≤optk • |Fk| ≤ d1+d2+ … + dj • So the ratio is 32

  30. Theorem: • The optimal size-competitive ratio for incremental medians is: • 4 in the deterministic case • e ≈ 2.72 in the randomized case • (Lower bound: prove that online bidding reduces to incremental medians) • [Chrobak, Kenyon, Noga, Young, ‘06] 33

  31. Outline: Online bidding Cow-path Incremental medians (size approximation) Incremental medians (cost approximation) List scheduling on related machines Minimum latency tours Incremental clustering 34

  32. Cost-Competitive Incremental Medians • k not known, authorizations for additional facilities arrive over time • Algorithm produces a sequence of facility sets: F1F2 … Fn • An algorithm is C-cost-competitive if • |Fk|≤kand cost(Fk) ≤ C·optk • for all k. • Goal: small competitive ratio (in polynomial time, if possible …) 35

  33. 0 1 1 1 Example: Star with m arms, w farmers per cluster 36

  34. 0 cost = 2(m-1)w ≈ 2  opt cost 1 1 1 Example: Star with m arms, w farmers per cluster k = 1 So C 2 37

  35. 0 1 1 1 Example: Star with m arms, w farmers per cluster k = 1 2 3 4 … m cost = w opt cost = 0 So C ∞ 38

  36. use doubling to improve to 8 Cost-Competitive Incremental Medians • [Mettu, Plaxton ‘00]: • Lower bound of 2 • Upper bound C ≈ 30 (in polynomial time) 39

  37. Fk’ Fk” for k’ < k we want to show that Fk contains a cheap subset Fk’ Idea:construct sequence backwards, at each step extracting next set from previous one facilities customers Fk 40

  38. H |Q| = k’ < k Lemma: F, Q facility sets. |F| = k H = H(Q,F) = k’ facilities in F closest to the points in Q Then cost(H)  cost(F) + 2·cost(Q) 41

  39. H Q Proof:Choose fF : closest to x qQ : closest to x hH : closest to q (in F) customer x F f dxH≤ dxh ≤ dxq+ dqh ≤ dxq+ dqf ≤ dxq+ (dxf + dxq) =2dxq+ dxf =2dxQ+ dxF h q So cost(H) ≤ 2·cost(Q) + cost(F) 42

  40. Algorithm: 1. Choose d1 < d2 < d3 < … Wlog. optn = cost(X) = 1 2. Choose p(1) > … > p(m) = 1 s.t. cost(Qp(i)) = optp(i) = di (For simplicity assume they exist) 3. Construct sets Fk for k = n, p(1), p(2),… Fn X(all facilities) Fp(i+1)  H (Fp(i) , Qp(i+1) ) for i= 2,…,m 4.For p(i+1) < k < p(i) setFkFp(i+1) (So for these k we have |Fk| ≤ k) 5. Output F1, F2,…, Fn 43

  41. Fp(i-2) Qp(i-1) Fp(i-1) Fp(i) optimal Qp(i) Analysis: cost(Fp(i)) ≤ cost(Fp(i-1)) + 2·di ≤ cost(Fp(i-2)) + 2·di-1 + 2·di ≤ … ≤ 2 · (d1 + d2 + …. + di) 45

  42. This is 2  (bidding ratio) So we get ratio = 8 for dj = 2j • Suppose p(j) < k ≤ p(j-1) • Then • optk ≥ optp(j-1) = dj-1 • cost(Fk) = cost(Fp(j)) ≤ 2 · (d1 + d2 + …. + dj) 46

  43. Use (3+ )-approximate medians instead of optimal ones • Theorem: • Upper bounds for cost-competitive incremental • medians: • Deterministic • 8 • 24+  in polynomial time • Randomized • 2e • 6e +  ≈ 16.31 +  in polynomial time • [Lin, Nagarajan, Rajamaran, Williamson ‘06] • [Chrobak, Kenyon, Noga, Young ‘06] 47

  44. Current world records: • 16+, deterministic polynomial time • 4e +, randomized polynomial time • [Lin, Nagarajan, Rajamaran, Williamson ‘06] • Deterministic (not polynomial-time) • Lower bound of 2.0013 • Upper bound of 7.65 • [Chrobak, Hurand ‘07] 48

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