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Approximation via Doubling (Part II)

Approximation via Doubling (Part II). 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 (Part II)

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  1. Approximation via Doubling (Part II) 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. Online Bidding - Reminder 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 3

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

  5. 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] 5

  6. For dj-1 <u ≤ dj+1 (j odd) dj+1 d2 d1 u d3 0 dj-1 dj 2  bidding ratio extra ratio 1 Cow-Path Problem -- Reminder So the ratio = 2  bidding ratio + 1 = 9 for dj= 2j 6

  7. 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] … 7

  8. Outline: Online bidding Cow-path Incremental medians (size approximation) Incremental medians (cost approximation) List scheduling on related machines Minimum latency tours Incremental clustering 8

  9. Given a list of jobs (each with a specified processing time), assign them to processors to minimize makespan (max load) 7 4 2 5 1 6 time 3 jobs Online algorithm: assignment of a job does not depend on future jobs Goal: small competitive ratio processors List Scheduling 9

  10. makespan Greedy: Assign each job to the machine with the lightest load 7 7 4 2 5 1 6 4 3 6 5 jobs 2 1 3 processors 10

  11. 7 4 7 2 3 6 5 1 6 2 3 makespan 4 jobs 1 5 better schedule: processors 11

  12. y x m machines Analysis of Greedy: x = min load before placing last job y = length of last job • total load ≥ m·x, so optimum makespan ≥ x • optimum makespan ≥ y • so • greedy’s makespan = x+y • ≤ 2 ·optimum makespan 12

  13. List Scheduling • Greedy is (2-1/m)-competitive [Graham ’66] • Lower bound ≈1.88 • [Rudin III, Chandrasekaran’03] • Best known ratio ≈1.92 • [Albers ‘99] [Fleischer, Wahl ‘00] • Lots of work on randomized algorithms, • preemptive scheduling, … 13

  14. List Scheduling on Related Machines Related machines: machines may have different speeds 1 1 1  0.25  1  0.5 1 jobs processors 1 2 3 14

  15. 2L Algorithm2PACK(L):schedule each job on the slowest machine whose load will not exceed 2L L Hey, the opt makespan is at most L 6 3 5 4 7 1 1 4 5 2 6  0.25  1  0.5 3 2 7 jobs processors 1 2 3 15

  16. 2L L r r 1 2 … … m 1 2 … … m Lemma: If the little birdie is right (opt makespan ≤ L) then 2PACK will succeed. Proof: Suppose 2PACK fails on job h • h’s length on processor 1 ≤ L , so • so load of processor 1 > L • r = first processor with load ≤ L • (or m+1, if no such processor) • Claim: if opt executes k on • a machine in {r,r+1,…,m} then so • does 2PACK optimum 2PACK 16

  17. so k‘s length here ≤ L 2L Lemma: If the little birdie is right (opt makespan ≤ L) then 2PACK will succeed. so k fits on r k k suppose k executed here Proof: Suppose 2PACK fails on job h • h’s length on processor 1 ≤ L , so • so load of processor 1 > L L • r = first processor with load ≤ L • (or m+1, if no such processor) k • Claim: if opt executes k on • a machine in {r,r+1,…,m} then so • does 2PACK r r 1 2 … … m 1 2 … … m optimum 2PACK 17

  18. 2L L 1 2 … … m 1 2 … … m Lemma: If the little birdie is right (opt makespan ≤ L) then 2PACK will succeed. Proof: Suppose 2PACK fails on job h • h’s length on processor 1 ≤ L , so • so load of processor 1 > L • r = first processor with load ≤ L • (or m+1, if no such processor) • In other words: if 2PACK executes • k on a machine in {1,2,…,r-1} then so • does opt • So opt’s (speed-weighted) total load • on processors {1,2,…,r-1} is > (r-1)L r r • So some opt’s processor has load • > L -- contradiction optimum 2PACK 18

  19. Algorithm: 1. Choose d1 < d2 < d3 < … (makespan estimates) Let Bj= 2·( d1 + d2 + … + dj ) “bucket” j : time interval [Bj-1 , Bj ] 2. j = 0 while there are unassigned jobs apply 2PACK with L = dj in bucket j if2PACKfails on job k let j = j+1 and continue (starting with job k) 19

  20. … B1 Bj-1 Bj+1 B2 Bj processor 1 2 … m k bucket j k k’ 20

  21. 2  (bidding ratio) Analysis: • Suppose the optimal makespan is u • Choose j such that dj-1 < u≤ dj • Then 2PACK will succeed in j ’th bucket (L = dj) • so algorithm’s makespan ≤ 2·(d1+d2+ … + dj) • and We get ratio 8 for dj = 2j 21

  22. List Scheduling on Related Machines • Theorem: • There is an 8-competitive online algorithm for list scheduling on related machines (to minimize makespan). With randomization the ratio can be improved to 2e. • [Aspnes, Azar, Fiat, Plotkin, Waarts ‘06] • World records: • upper bound ≈ 5.828 (4.311 randomized) • lower bound ≈2.438 (2 randomized) • [Berman, Charikar, Karpinski ‘97] • [Epstein, Sgall ‘00] 22

  23. Outline: Online bidding Cow-path Incremental medians (size approximation) Incremental medians (cost approximation) List scheduling on related machines Minimum latency tours Incremental clustering 23

  24. Minimum Latency Tour X = metric space P = v1v2…vh : path in X Latency of vi on P latencyP(vi) = d(v1,v2) + … + d(vi-1,vi) (Total) latency of P = i latencyP(vi) Minimum Latency Tour Problem: Given X, find a tour (path visiting all vertices) of minimum total latency Goal: polynomial-time approximation algorithm 24

  25. E 15 A D 2 8 F B 11 4 C Total latency = 2 + 4 + 8 + 11 + 15 = 40 25

  26. E A D 2-tour F 4-tour B C Minimum k-Tour Problem: find a shortest k-tour (a path that starts and ends at v1 and visits ≥ k different vertices) 26

  27. Algorithm: 1. Choose d1 < d2 < d3 < … 2. For each k compute the optimal k-tour Tk 3. Choose p(1) < … < p(m) = n s.t. length(Tp(i)) = di (For simplicity assume they exist) 4. Output Q = Tp(1) Tp(2) …Tp(m) (concatenation) Denote Q = q1q2…qn (qi = first point on Q different from q1, q2,…,qi-1) 27

  28. Tp(1) v1 Tp(2) Tp(3) Q 28

  29. s2 T S s1 s3 sk Lemma: S = s1s2…sn : tour with optimum latency. Then latencyS(sk) ≥ (1/2)·length(Tk) Proof: T is a k-tour, so 2·latencyS(sk) = length(T) ≥ length(Tk) 29

  30. 2  (bidding ratio) Analysis: • For p(j-1)<k ≤ p(j) • latencyS(sk) ≥ (1/2)·length(Tk) • ≥ (1/2)·length(Tp(j-1)) = dj-1/2 • qk will be visited in Tp(j) (or earlier), so • latencyQ(qk) ≤ d1+d2+ … + dj We get ratio 8 for dj = 2j … if we can compute k-tours efficiently !!! 30

  31. 7 4 6 12 11 9 10 3 7 5 8 2 5 u 7 x y v t If X is a weighted tree, optimal k-tours can be computed in polynomial time… • Dynamic programming: • W.l.o.g. assume X is a rooted binary tree • optk(u) = minimum of • 2x+optk-1(v), 2y+optk-1(t) and • minj {2x+optj(v)+2y+optk-1-j(t) } Theorem: There is a polynomial-time 8-approximation algorithm for maximum latency tours on weighted trees [Blum, Chalasani, Coppersmith, Pulleyblank, Raghavan, Sudan ‘94] 31

  32. Tp(j) v1 Can we do better? Choose a random direction (clockwise or counter-clockwise) and traverse each Tp(j) in this direction … u Expected latency of u = d1+d2+ …+ dj-1 + dj/2 We get ratio 6 for dj = 2j 32

  33. Can we do even better? Instead of dj = 2j choose dj = cj+x, where c is the constant from the Cow Path problem and x is random in [0,1) • We don’t really really randomization: • choose better direction (clockwise or counter-clockwise) • There are only O(n) x’s that matter, so try them all Theorem: There is a polynomial-time 3.591 -approximation algorithm for maximum latency tours on weighted trees [Goemans, Kleinberg ‘98] Can be extended to arbitrary spaces, with ratio 3.591 [Chauduri, Godfrey, Rao, Talwar ‘03] 33

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

  35. k-Clustering • X = metric space • For C  X , • diameter(C) = maximum distance between points in C • k-Clustering Problem: Given k, partition X into k disjoint clusters C1,…,Ck to minimize the maximum diameter(Cj) • Offline: • approximable with ratio 2 • [Gonzales ‘85] [Hochbaum, Shmoys ‘85] • lower bound of 2 for polynomial algorithms (unless P = NP) • [Feder, Greene ‘88] [Bern, Eppstein ‘96] 35

  36. E A H D F G B C k=3 3-Clustering with maximum diameter 5 36

  37. E A H D F G B C k=3 3-Clustering with maximum diameter 3 37

  38. different model than incremental medians !!! Incremental k-Clustering • Problem: Maintain k-clustering when • points in X arrive online • allowed operations: • add point to a cluster • merge clusters • create a new singleton cluster • Goal: online competitive algorithm (polynomial-time) 38

  39. A D G C k=3 diameter = 0 39

  40. E A D G C k=3 diameter = 2 40

  41. E A H D G C k=3 diameter = 3 41

  42. E A H D G C k=3 diameter = 3 42

  43. Ci oi radius • Notation and terminology: • Algorithm’s clusters C1,C2,…,Ck’ with k’ ≤ k • in each Ci fix a centeroi • radius of Ci = max distance between xCi and oi • diameter of Ci ≤ 2 · (radius of Ci) 43

  44. Procedure CleanUp(z). Goal: merge some clusters C1,C2,…,Ck’ so that afterwards all inter-center distances are > z Find a maximal set J of clusters with all inter-center distances > z 2. for each cluster Ca J choose Cb J with d(oa,ob) ≤ z merge Cainto Cb (with center ob) 44

  45. z Lemma: If the max radius before CleanUp is h then after CleanUp it is ≤ h+z. Proof: follows from the ∆ inequality 45

  46. z h z Lemma: If the max radius before CleanUp is h then after CleanUp it is ≤ h+z. Proof: follows from the ∆ inequality v 46

  47. Invariant: • inter-center distance > dj checkpoint j Algorithm: 1. Choose d1 < d2 < d3 < … 2. Initially C1,C2,…,Ck are singletons (first k points) (Assume min distance between these points is > 1) 3. j 1 4. repeat when a new point x arrives ifd(x,oi) ≤ dj for some i add x to Ci else if k’ < k k’k’+1; Ck’  {x} else create a temporary cluster Ck+1 {x} while there k+1 clusters jj+1 do CleanUp with z = dj (merge clusters) 47

  48. Example:k = 4 48

  49. dj+1 dj+2 Example:k = 4 49

  50. 2  (bidding ratio) Analysis: • At checkpoint j: • Before clean-up • k+1 clusters with inter-center distances > dj-1 • so opt diameter > dj-1 • After clean-up • max radius ≤ d1 + d2 + … + dj • so max diameter ≤ 2·(d1 + d2+ … + dj) We get ratio 8 for dj = 2j 50

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