approximation via doubling (part ii)
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
1 Wroclaw University, Sept 18, 2007
Approximation via Doubling(Part II)
Marek Chrobak
University of California, Riverside
Joint work with Claire Kenyon-Mathieu
2 Wroclaw University, Sept 18, 2007
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
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Online Bidding - Reminder
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
⎧ ⎨ ⎩
⎫ ⎬ ⎭
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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
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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]
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Cow-Path Problem -- Reminder
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
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Theorem:The optimal competitive ratio for the cow-path problem is
• 9 in the deterministic case
• 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] …
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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
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List Scheduling
123
456
7
jobs
Given a list of jobs (each with a specified processing time), assign them to processors to minimize makespan (max load)
Online algorithm: assignment of a job does not depend on future jobs
Goal: small competitive ratio processors
tim
e
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123
456
7
processors
jobs
123
4 56
7
make
span
Greedy: Assign each job to the machine with the lightest load
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123
456
7
jobs
processors
1
2
3
45
67
make
span
better schedule:
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Analysis of Greedy:
x = min load before placing last joby = length of last job
• so
greedy’s makespan = x+y ≤ 2 ·optimum makespan
x
y
m machines
• total load ≥ m·x, so optimum makespan ≥ x• optimum makespan ≥ y
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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, …
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List Scheduling on Related Machines
processors 1 2 3
Related machines: machines may have different speeds
0.25 0.5 1
1
jobs
1
11
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0.25 0.5 1
1
2
3
4
5
6
7
jobs
1
2
3
4
5
6
7
Algorithm 2PACK(L): schedule each job on the slowest machine whose load will not exceed 2L
L
2L
processors 1 2 3
Hey, theopt makespan is
at most L
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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)
1 2 … … m 1 2 … … m
L
2L
• Claim: if opt executes k on a machine in {r,r+1,…,m} then so does 2PACK
optimum 2PACK
r r
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1 2 … … m 1 2 … … m
L
2L
k
so k‘s length here ≤ L
so k fits on r
k
r r
optimum 2PACK
k
suppose kexecuted here
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
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1 2 … … m 1 2 … … m
L
2L
r r
optimum 2PACK
• So opt’s (speed-weighted) total load on processors {1,2,…,r-1} is > (r-1)L
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 some opt’s processor has load > L -- contradiction
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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 jif 2PACK fails on job k let j = j+1 and continue (starting with job
k)
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k
bucket j
1
2
m
…
processor B1 B2 Bj-1 BjBj+1…
k
k’
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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
€
ratio ≤2 ⋅(d1 +d2 + ...+ d j )
d j−1
We get ratio 8 for dj = 2j
2 (bidding ratio)
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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]
List Scheduling on Related Machines
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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
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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
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Total latency = 2 + 4 + 8 + 11 + 15 = 40
A15E
2
D
11
C4
B
8 F
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AE
D
C
B
F
Minimum k-Tour Problem: find a shortest k-tour (a path that starts and ends at v1 and visits ≥ k different vertices)
2-tour
4-tour
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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)
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v1
Tp(1)
Tp(2)
Tp(3)
Q
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Lemma:S = s1s2…sn : tour with optimum latency.Then
latencyS(sk) ≥ (1/2)·length(Tk)
Proof:
s1
s2
s3
sk
ST T is a k-tour, so
2·latencyS(sk) = length(T)
≥ length(Tk)
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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
€
ratio ≤d1 +d2 + ...+ d j
d j−1 /2= 2 ⋅
d1 +d2 + ...+ d jd j−1
We get ratio 8 for dj = 2j
€
ratio ≤latency(Q)
latency(S)≤ maxk
latencyQ (qk )
latencyS (sk )
… if we can compute k-tours efficiently !!!
2 (bidding ratio)
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If X is a weighted tree, optimal k-tours can be computed in polynomial time… 10
4
5
1112
7
6
82
39
7
75
Theorem:There is a polynomial-time 8-approximation algorithm for maximum latency tours on weighted trees[Blum, Chalasani, Coppersmith, Pulleyblank, Raghavan, Sudan ‘94]
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) }
u
v t
x y
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Choose a random direction (clockwise or counter-clockwise) and traverse each Tp(j) in this direction …
v1Tp(j)
u
Expected latency of u = d1+d2+ …+ dj-1 + dj/2
€
ratio ≤d1 +d2 + ...+ d j−1 + d j /2
d j−1 /2= 2 ⋅
d1 +d2 + ...+ d j−1 + d j /2
d j−1
We get ratio 6 for dj = 2j
Can we do better?
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Can be extended to arbitrary spaces, with ratio 3.591[Chauduri, Godfrey, Rao, Talwar ‘03]
Theorem:There is a polynomial-time 3.591 -approximation algorithm for maximum latency tours on weighted trees [Goemans, Kleinberg ‘98]
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
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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
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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]
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E
D
C
B
F
A
G
H
3-Clustering with maximum diameter 5k=3
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E
D
C
B
F
A
G
H
3-Clustering with maximum diameter 3k=3
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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)
different model than incremental medians !!!
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D
C
G
diameter = 0
A
k=3
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D
C
G
A
E
diameter = 2k=3
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D
C
G
A
E
H
diameter = 3k=3
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D
C
G
A
E
H
diameter = 3k=3
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Notation and terminology:
• Algorithm’s clusters C1,C2,…,Ck’ with k’ ≤ k• in each Ci fix a center oi
• radius of Ci = max distance between x Ci and oi • diameter of Ci ≤ 2 · (radius of Ci)
Ci
oiradius
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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 Jchoose Cb J with d(oa,ob) ≤ zmerge Ca into Cb (with center ob)
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Lemma: If the max radius before CleanUp is h then after CleanUp it is ≤ h+z.
Proof: follows from the ∆ inequality
z
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Lemma: If the max radius before CleanUp is h then after CleanUp it is ≤ h+z.
Proof: follows from the ∆ inequality
z
v
z
h
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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 14. repeat when a new point x arrives
if d(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)checkpoint j
Invariant:• inter-center distance > dj
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Example: k = 4
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Example: k = 4
dj+1
dj+2
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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)
€
ratio ≤2 ⋅(d1 +d2 + ...+ d j )
d j−1
We get ratio 8 for dj = 2j
2 (bidding ratio)
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Theorem: There is an 8-competitive online algorithm for incremental clustering (ratio 2e with randomization).[Charikar, Chekuri, Feder, Motwani ‘06]
Other results:
• upper bound 6 (4.33 randomized) (not polynomial-time)
• lower bound 2.414 (2 randomized)
Incremental Clustering
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1. Online bidding2. Cow-path3. Incremental medians (size approximation)4. Incremental medians (cost approximation)5. List scheduling on related machines6. Minimum latency tours7. Incremental clustering8. Other scheduling problems:
• List scheduling, related machines with preemption
• Scheduling with min-sum criteria• Non-clairvoyant scheduling
9. Hierarchical clustering10. Load balancing11. Online algorithms for set cover (combined
with primal-dual)….
Doubling Method Applications:
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Thank you !
Questions?