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Approximation via Doubling

This paper explores approximation algorithms via doubling methods for optimization problems, emphasizing online bidding and incremental median scenarios. The presentation includes a detailed examination of algorithmic strategies, including deterministic and randomized approaches, competitive ratios, and problem-solving efficiency. Key topics covered are online bidding strategies, cow-path problems, size and cost approximations, and incremental clustering techniques. The findings illustrate optimal performance ratios in both deterministic and randomized cases, showcasing the applicability and effectiveness of the doubling method across various algorithms.

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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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