Optimization problems INSTANCE FEASIBLE SOLUTIONS COST.

Optimization problems

INSTANCE

FEASIBLE SOLUTIONS

COST

Vertex Cover problem

INSTANCE graph G

FEASIBLE SOLUTIONS SV, such that (eE) Se

COST c(S) = |S|

Set Cover problem

INSTANCE family of sets A1,...,An

FEASIBLE SOLUTIONS S[n], such that Ai COST c(S) = |S|

iS

Set Cover problem


FEASIBLE SOLUTIONS S[n], such that Ai

COST c(S) = |S|

INSTANCE graph G


COST c(S) = |S|


iS

Set Cover problem


FEASIBLE SOLUTIONS S[n], such that Ai

COST c(S) = |S|

INSTANCE graph G


COST c(S) = |S|


iS

= E

Ai E is the set of edges adjacent to i V

Optimization problems

INSTANCE

FEASIBLE SOLUTIONS

COST

OPTIMAL SOLUTION OPT= min c(T)T FEASIBLE SOLUTIONS

approximation algorithm

INSTANCE T

such that

c(T) OPT

Last Class:

2-approximation algorithm for Vertex-Cover

2-approximation algorithm for Metric TSP

1.5-approximation algorithm for Metric TSP

This Class:

(1+)-approximation algorithm for Knapsack

O(log n)-approximation algorithm for Set Cover

KnapsackINSTANCE: value vi, weight wi, for i {1,...,n} weight limit W

FEASIBLE SOLUTION: collection of items S {1,...,n} with total weight W

COST (MAXIMIZE): sum of the values of items in S

Knapsack INSTANCE: value vi, weight wi, for i {1,...,n} weight limit W



We had:

pseudo-polynomial algorithm, time = O(Wn)

pseudo-polynomial algorithm, time = O(Vn), where V = v1 + ... +vn

Knapsack INSTANCE: value vi, weight wi, for i {1,...,n} weight limit W



pseudo-polynomial algorithm, time = O(Vn), where V = v1 + ... +vn

GOAL

convert

into an approximation algorithm

IDEA = rounding

Knapsack

wlog all wi WM = maximum of vi

vi vi’ := n vi / (M)

OPT’ n2/

S = optimal solution in originalS’ = optimal solution in modified

Will show: optimal solution in modified is an approximately optimal solution in original

Knapsack vi vi’ := n vi / (M)


(n/(M)) vi vi’ vi’ ( nvi / (M) -1 ) i S’ i S’ i S i S


Knapsack vi vi’ := n vi / (M)


(n/(M)) vi vi’ vi’ ( nvi / (M) -1 ) i S’ i S’ i S i S


vi ( vi - ) i S’ i S

n/(M) 1 OPT – MOPT(1– )

Running time? pseudo-polynomial algorithm, time = O(V’n), where V’ = v’1 + ... +v’n

M = maximum of vi


Running time? pseudo-polynomial algorithm, time = O(V’n), where V’ = v’1 + ... +v’n

M = maximum of vi


v’i n/

V’ n

running time = O(n3/)

FPTASFully polynomial-time approximation scheme

(1+)-approximation algorithm running in time poly(INPUT,1/)

We have an algorithm for the Knapsack problem,which outputs a solution with

value (1-) OPT

and runs in time O(n3/)

Weighted set cover problem

INSTANCE: A1,...,Am , weights w1,...,wm

FEASIBLE SOLUTION: collection S of the Ai covering OBJECTIVE (minimize): the cost of the collection

(in the unweighted version we have wi =1)


Greedy algorithm:

pick Ai with minimal wi / |Ai| remove elements in Ai from repeat

Negative example (last class)

approximation ratio = (log n)


Greedy algorithm:


Theorem: O(log n) approximation algorithm.

Weighted set cover problemGreedy algorithm:



Ai

everybody pays wi / |Ai|

when Ai picked, cost of the solution increases by wi


Let B be a set of weight w. How muchdid the guys in B pay?

B

pick me!

cost=w

/

B

pick me!

cost=w

/

B




B

pick me!

cost=w

/

B

pick me!

cost=w

/

B

sorry Ai was cheaper

sorry Ai was cheaper

Ai

paid less than w/B




B


continue, size of B went down by 1

B

pick me!

cost=w

/(B-1)

pick me!

cost=w

/(B-1)




B

sorry Aj was cheaper

sorry Aj was cheaper

Aj

paid less than w/(B-1)

pick me!

cost=w

/(B-1)

pick me!

cost=w

/(B-1)




B


continue, size of B went down by 1

B

pick me!

cost=w

/(B-2)

pick me!

cost=w

/(B-2)


B

vertices in order they are covered by greedy

paidw/B

paidw/(B-1)

paidw/(B-2) paidw

paidw/2


B

TOTAL PAID w (1/B + 1/(B-1) + ... +1/2 + 1) = w O(ln B) = w O(ln n)

paidw/B

paidw/(B-1)

paidw/(B-2) paidw

paidw/2

Weighted set cover problemINSTANCE: A1,...,Am , weights w1,...,wm

FEASIBLE SOLUTION: collection S of the Ai covering OBJECTIVE (minimize): the cost of the collection

Greedy algorithm:



Clustering

n points in Rm

d(i,j) = distance between points i,j

partition the points into k clusters of small diameter

diam(C) = max d(i,j)i,jC

Clustering k = 3

Clustering k = 2

k-Clustering

INSTANCE n points in Rm

FEASIBLE SOLUTION partition of [n] into C1,...,Ck

COST max diam(Ci)

i[k]

diam(C) = max d(i,j)i,jC

k-Clustering

GREEDY ALGORITHM

pick s1 [n] for i from 2 to k do pick si the farthest point from s

1,...,s

i-1

Ci = {x [n] whose closest center is si}

k-Clustering

GREEDY ALGORITHM


1,...,s

i-1


s1

k-Clustering

GREEDY ALGORITHM


1,...,s

i-1


s1

s2

k-Clustering

GREEDY ALGORITHM


1,...,s

i-1


s1

s2

s3

k-Clustering

GREEDY ALGORITHM


1,...,s

i-1


Theorem: GREEDY ALGORITHM IS A 2-APPROXIMATION ALGORITHM

Theorem: GREEDY ALGORITHM IS A 2-APPROXIMATION ALGORITHM

s1

s2

sk

sk+1

d(si,sj) d(sk+1,{s1,...,sk}) = r

OPT r cost of greedy 2r

Optimization problems INSTANCE FEASIBLE SOLUTIONS COST.

Documents

Transcript of Optimization problems INSTANCE FEASIBLE SOLUTIONS COST.