Loading in 5 sec....

Approximation via DoublingPowerPoint Presentation

Approximation via Doubling

- 76 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Approximation via Doubling' - nevada-mcfarland

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Marek Chrobak

University of California, Riverside

Joint work with Claire Kenyon-Mathieu

1

(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

Online bidding

Cow-path

Incremental medians (size approximation)

Incremental medians (cost approximation)

List scheduling on related machines

Minimum latency tours

Incremental clustering

3

Online bidding

Cow-path

Incremental medians (size approximation)

Incremental medians (cost approximation)

List scheduling on related machines

Minimum latency tours

Incremental clustering

4

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

Deterministic Bidding - Upper Bound

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

If 2j-1 <u ≤ 2j, the ratio is

8

- 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

Online bidding

Cow-path

Incremental medians (size approximation)

Incremental medians (cost approximation)

List scheduling on related machines

Minimum latency tours

Incremental clustering

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

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

Online bidding

Cow-path

Incremental medians (size approximation)

Incremental medians (cost approximation)

List scheduling on related machines

Minimum latency tours

Incremental clustering

17

X = set of facilities

Y = set of customers

X Y : metric space with distance function dxy

For FX 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

- 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

Size-Competitive Incremental Medians

- k not known, authorizations for additional facilities arrive over time
- Algorithm produces a sequence of facility sets: F1F2 … Fn
- An algorithm is S-size-competitive if
- |Fk| ≤S·kand cost(Fk) ≤ optk
- for all k.
- Goal: small competitive ratio

23

… 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-1Qd(i+1)

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

28

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

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

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

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

- 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

Online bidding

Cow-path

Incremental medians (size approximation)

Incremental medians (cost approximation)

List scheduling on related machines

Minimum latency tours

Incremental clustering

34

Cost-Competitive Incremental Medians

- k not known, authorizations for additional facilities arrive over time
- Algorithm produces a sequence of facility sets: F1F2 … 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

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

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

Cost-Competitive Incremental Medians

- [Mettu, Plaxton ‘00]:
- Lower bound of 2
- Upper bound C ≈ 30 (in polynomial time)

39

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

|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

Q

Proof:Choose

fF : closest to x

qQ : closest to x

hH : 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

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) setFkFp(i+1)

(So for these k we have |Fk| ≤ k)

5. Output F1, F2,…, Fn

43

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

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

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

- 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

Download Presentation

Connecting to Server..