Approximation via Doubling

1 Wroclaw University, Sept 18, 2007

Approximation via Doubling

Marek Chrobak

University of California, Riverside

Joint work with Claire Kenyon-Mathieu


Doubling method:

(for a minimization problem)

Choose d1 < 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


Outline:

1. Online bidding2. Cow-path3. Incremental medians (size approximation)4. Incremental medians (cost approximation)5. List scheduling on related machines6. Minimum latency tours7. Incremental clustering


Outline:



Online Bidding

1

2

5

12


Online Bidding

1

2

5

12

20 bags of gunpowder

but… 6 bags could have been enough

so ratio = 20/6


Online Bidding

Item for sale of value u (unknown to bidder)

Buyer bids d1,d2,d3, … until some dj ≥ u

Cost: d1 + d2 + … + dj Optimum = u

Competitive ratio

€

maxu, jd1 +d2 + ...+ d j

u: d j−1 < u ≤ d j

⎧ ⎨ ⎩

⎫ ⎬ ⎭

≅ max j

d1 +d2 + ...+ d jd j−1

⎧ ⎨ ⎩

⎫ ⎬ ⎭


Deterministic Bidding - Upper Bound

If 2j-1 < u ≤ 2j, the ratio is

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

€

21 +22 + ...+ 2 j

2 j−1≤

2 j+1

2 j−1= 4


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]


Outline:



d1 d2d3 d40 u

Cow-Path


Analysis:

d1 d2d3dj+10 udj-1dj

For dj-1 < u ≤ dj+1 (j odd)

2 bidding ratio extra ratio 1

So the ratio = 2 bidding ratio + 1 = 9 for dj = 2j


Theorem:The optimal competitive ratio for the cow-path problem is


• 4.59 in the randomized case

Solution of (r-1)ln(r-1) = r 2e+1

Connection to online bidding does not work in randomized case -- why?

[Gal ‘80] [Baeza-Yates, Culberson, Rawlins ‘93]

[Papadimitriou, Yannakakis ‘91] [Kao, Reif, Tate ‘94] …


Outline:



The k-Median Problem

X = set of facilitiesY = set of customers X Y : metric space with distance function dxy

For F X let cost(F) = y Y dyF

where dyF = minf F dyf

The k-Median Problem: Find a facility set F of size k for which cost(F) is minimized.

optimal F = Qk (the k-median)


customer

facility (potential)


k = 2 facilities

3

1

2 2

3

1

4

1

cost = 17


k = 4 facilities

1

2 21

1

1

1

3

cost = 12


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)


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·k and cost(Fk) ≤ optk for all k.

Goal: small competitive ratio


k = 1

2

4

53

2

25

3

cost = 26opt = 26


k = 2

2

1

22

2

24

3

cost = 18 !!! opt = 17


k = 2

2

1

22

2

14

1

cost = 15 opt = 17


Size-Competitive Incremental Medians

Algorithm:

1. choose d1 < 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)

… not a polynomial time algorithm …


Qd(1)

Qk = optimal k-median

Qd(4)

1 2 3 4 5 … 11 12 13 … 19 20 21 … 31 32 … d1 d2 d3 d4

k k k k

Qd(2)

Qd(3)


Qd(2)

1 2 3 4 5 … 11 12 13 … 19 20 21 … 31 32 … d1 d2 d3 d4

kkk

Qd(1)

Qd(3)

Qd(4)



Qd(3)

Qd(2)

1 2 3 4 5 … 11 12 13 … 19 20 21 … 31 32 … d1 d2 d3 d4

k

Qd(1)

k k

Qd(4)



Qd(4)

1 2 3 4 5 … 11 12 13 … 19 20 21 … 31 32 … d1 d2 d3 d4

k

Qd(3)

Qd(2)

Qd(1)



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

€

maxk, jd1 +d2 + ...+ d j

k: d j−1 < k ≤ d j

⎧ ⎨ ⎩

⎫ ⎬ ⎭

Same as online bidding

So we get ratio = 4 for dj = 2j


Theorem:

The optimal size-competitive ratio for incremental medians is:


• e ≈ 2.72 in the randomized case

(Lower bound: prove that online bidding reduces to incremental medians)

[Chrobak, Kenyon, Noga, Young, ‘06]


Outline:



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|≤ k and cost(Fk) ≤ C·optk for all k.

Goal: small competitive ratio (in polynomial time, if possible …)


Example: Star with m arms, w farmers per cluster

1

0

1

1



1

0

1

1

k = 1

cost = 2(m-1)w ≈ 2 opt cost

So C 2



1

0

1

1

k = 1

cost = w

opt cost = 0

2 3 4 … m

So C ∞


Cost-Competitive Incremental Medians

[Mettu, Plaxton ‘00]:

• Lower bound of 2

• Upper bound C ≈ 30 (in polynomial time)

use doubling to improve to 8


Idea: construct sequence backwards, at each step extracting next set from previous one

for k’ < k we want to show that Fk contains a cheap subset Fk’

customers

facil

ities

Fk

Fk’

Fk”


Lemma: F, Q facility sets.

|F| = k

H |Q| = k’ < k

H = H(Q,F) = k’ facilities in F closest to the points in Q

Then

cost(H) cost(F) + 2·cost(Q)


Proof: Choose

H Q

F

customer x

f

h q

f F : closest to xq Q : closest to xh H : closest to q (in F)

dxH ≤ dxh

≤ dxq + dqh

≤ dxq + dqf

≤ dxq + (dxf + dxq) = 2dxq + dxf

= 2dxQ + dxFSo

cost(H) ≤ 2·cost(Q) + cost(F)


Algorithm:

1. Choose d1 < d2 < d3 < … Wlog. optn = cost(X) = 12. 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) set Fk Fp(i+1)

(So for these k we have |Fk| ≤ k)

5. Output F1, F2,…, Fn


Fp(i-2)

cost(Fp(i)) ≤ cost(Fp(i-1)) + 2·di

≤ cost(Fp(i-2)) + 2·di-1 + 2·di

≤ …≤ 2 · (d1 + d2 + …. + di)

Analysis:

Fp(i-1)

Fp(i)

Qp(i-1)

optimal Qp(i)


Suppose p(j) < k ≤ p(j-1)

Then

• optk ≥ optp(j-1) = dj-1

• cost(Fk) = cost(Fp(j)) ≤ 2 · (d1 + d2 + …. + dj)

€

ratio ≤ 2 ⋅d1 +d2 + ...+ d j

d j−1

This is 2 (bidding ratio)

So we get ratio = 8 for dj = 2j


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]

Use (3+ )-approximate medians instead of optimal ones


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]

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Approximation via Doubling