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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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slide1
Approximation via Doubling

Marek Chrobak

University of California, Riverside

Joint work with Claire Kenyon-Mathieu

1

slide2
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

slide3
Outline:

Online bidding

Cow-path

Incremental medians (size approximation)

Incremental medians (cost approximation)

List scheduling on related machines

Minimum latency tours

Incremental clustering

3

slide4
Outline:

Online bidding

Cow-path

Incremental medians (size approximation)

Incremental medians (cost approximation)

List scheduling on related machines

Minimum latency tours

Incremental clustering

4

slide5
1

2

5

12

Online Bidding

5

slide6
1

2

5

12

Online Bidding

20 bags of gunpowder

but… 6 bags could have been enough

so ratio = 20/6

6

slide7
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

slide8
Deterministic Bidding - Upper Bound

Doubling strategy: bid 1, 2, 4, … , 2i, …

If 2j-1

8

slide9
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

slide10
Outline:

Online bidding

Cow-path

Incremental medians (size approximation)

Incremental medians (cost approximation)

List scheduling on related machines

Minimum latency tours

Incremental clustering

12

slide11
d1

u

d2

d3

d4

0

Cow-Path

13

slide13
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

slide14
Outline:

Online bidding

Cow-path

Incremental medians (size approximation)

Incremental medians (cost approximation)

List scheduling on related machines

Minimum latency tours

Incremental clustering

17

slide15
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

slide16
customer

facility (potential)

19

slide17
k = 2 facilities

cost = 17

3

4

1

1

1

3

2

2

20

slide18
k = 4 facilities

cost = 12

1

3

1

1

2

2

1

1

21

slide19
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

slide20
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

slide21
opt = 26

cost = 26

k = 1

2

5

2

4

3

3

5

2

24

slide22
opt = 17

cost = 18 !!!

k = 2

2

4

2

1

3

2

2

2

25

slide23
opt = 17

cost = 15

k = 2

2

4

1

1

1

2

2

2

26

slide24
… 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

slide25
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

slide26
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

slide27
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

slide28
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

slide29
Same as online bidding

So we get ratio = 4 for dj = 2j

Analysis:

  • At step k, for dj-1
    • cost(Fk) ≤ cost(Qd(j)) = opt(dj)≤optk
    • |Fk| ≤ d1+d2+ … + dj
  • So the ratio is

32

slide30
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

slide31
Outline:

Online bidding

Cow-path

Incremental medians (size approximation)

Incremental medians (cost approximation)

List scheduling on related machines

Minimum latency tours

Incremental clustering

34

slide32
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

slide33
0

1

1

1

Example: Star with m arms, w farmers per cluster

36

slide34
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

slide35
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

slide36
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

slide37
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

slide38
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

slide39
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

slide40
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

slide41
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

slide42
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

slide43
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

slide44
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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