www.cs.technion.ac.il/~reuven ibm2006 1 lp rounding using fractional local ratio reuven bar-yehuda...

www.cs.technion.ac.il/~reuven IBM2006 1

LP RoundingLP Roundingusingusing

Fractional Local RatioFractional Local Ratio

Reuven Bar-Yehuda

www.cs.technion.ac.il/~reuven


General framework:General framework:Given a weight vector w.

Minimize [Maximize] w·x

Subject to: feasibility constraints F(x)

x is an r-approximation if F(x) and w·x rw·x*

[w·x rw·x* ]

An algorithm is an r-approximation if for any w, F

it returns an r-approximation


15

Min 5xBisli+8xTea+12xWater+10xBamba+20xShampoo+15xPopcorn+6xChocolate

+$4xWaterShampoo+ • • •

s.t. xShampoo + xWater + xWaterShampoo 1

5

812

20

6

10

$4

$1

$3

$1

$1

$2

$1$1


The generalized vertex cover problemThe generalized vertex cover problem

Minimize w·x

Subject to: xu + xv + xe 1 e={u,v} E

x {0,1}|V|+|E|


2-Approx 2-Approx GVC(G,w)GVC(G,w)

If E= return If e E w(e)=0 return {e}+GVC(G-e, w)

If v V w(v)=0 return {v}+GVC(G-E(v), w)

Let e={u,v} E s.t = min {w(u), w(v), w(e)}>0.

if x{u,v,e}w11(x) =

0 else

Notice:w1 x 2 w1 x for Good(x)

REC= GVC(G, VC(G, w2= w- w-ww11))

Induction hyp is: w2REC 2 w2x

so if Good(REC): w1REC 2 w1x we are done

If REC-e is a cover thenREC=REC-eIf REC-e is a cover thenREC=REC-e

Return RECReturn REC


““2 integral for the price of 1 fractional”: 2 integral for the price of 1 fractional”: The local ratio technique for roundingThe local ratio technique for rounding

Let x be the the fractional solution

Minimize w·x

Subject to: xu + xv + xe 1 e=(u,v) E

x [0,1]|V|+|E|

www.cs.technion.ac.il/~reuven IBM2006 7 ““d d integral for the price of integral for the price of ½(d+1) fractional”: fractional”: 2-2/(2-2/(ΔΔ+1)-Approx +1)-Approx GVC(G,w)GVC(G,w)If E= return If e E w(e)=0 return {e}+GVC(G-e, w)

If v V w(v)=0 return {v}+GVC(G-E(v)-v, w)

Let v V s.t xv is minimum and

Let =min(w(i) : i N[v]}

if i N[v]w11(i) =

0 else

Claim:w1 x rΔ w1 x for Good(x)

REC= GVC(G, VC(G, w2= w- w-ww11))

Induction hyp is: w2REC rΔ w2x

so if Good(REC): w1REC rΔ w1x we are done

If REC is not a minimal cover then make REC minimalIf REC is not a minimal cover then make REC minimal


Min xv

www.cs.technion.ac.il/~reuven IBM2006 8 ““d d integral for the price of integral for the price of ½(d+1) fractional”: fractional”: Claim: w1 x rΔ w1 x for Good(x)

Min xv

If Min xv ≥ ½

Then x(N[v]) ≥ ½(d+1)

Else x(N[v]) ≥ ½(d+1)

Thus w1 x ≥ ½(d+1)

But w1 x d

Hence: w1 x/ w1 x 2-2/(d+1)

2-2/(ΔΔ +1) = rΔ

www.cs.technion.ac.il/~reuven IBM2006 9 A Generalized Local-Ratio Schema for A Generalized Local-Ratio Schema for

M Minimizationinimization [ [MMaximization] problems:aximization] problems:Let x be any “fisible?” vector (e.g. an optimal solution)

Algorithm r-ApproxMin [Max](Set, w)

If Set = then return ;

If v Set w(v) = 0 then return {v} r-ApproxMin(Set-{v},w ) ;

[If v Set w(v) 0 then return r-ApproxMax(Set-{v},w ) ;]

Define “good” w1 ; i.e. Good(x): w1 x [] r w1 x

REC = r-ApproxMin [Max](Set, w2 ) ;

Induction hyp is: w2REC [] r w2x

so if Good(REC): w1REC [] r w1x we are done,

otherwise “fix it”; return REC’;


The maximum independent set problemThe maximum independent set problem

Maximize w·x

Subject to: xu + xv ≤ 1 e=(u,v) E

x {0,1}|V|


The maximum independent set problemThe maximum independent set problem “1 integral for the gain of “1 integral for the gain of rr fractional”: fractional”:

Let x be the the fractional solution

Maximize w·x

Subject to: xu + xv ≤ 1 e=(u,v) E

x [0,1]|V|

www.cs.technion.ac.il/~reuven IBM2006 12 Gain Gain 11 integral, lose integral, lose ½(d+1) fractional fractional

2/(2/(ΔΔ+1)-Approx +1)-Approx IS(G,w)IS(G,w)If v V w(v) 0 return IS(G-v, w)

If E= return V

Let v V s.t xv is maximum and

Let = w(v)

if i N[v]w11(i) =

0 else

Claim:w1 x ≥rΔ w1 x for Good(x)

REC= IS(G, (G, w2= w- w-ww11))

Induction hyp is: w2REC ≥ rΔ w2x

so if Good(REC): w1REC ≥ rΔ w1x we are done

If REC+v is an independent set then REC=REC+vIf REC+v is an independent set then REC=REC+v


Max xv

www.cs.technion.ac.il/~reuven IBM2006 13 Gain Gain 11 integral, lose integral, lose ½(d+1) fractional fractional Claim: w1 x ≥ rΔ w1 x for Good(x)

Max xv

If Max xv ≤ ½

Then x(N[v]) ≤ ½(d+1)

Else x (N[v]) ≤ ½(d+1)

Thus w1 x ≤ ½(d+1)

But w1 x ≥

Hens: w1 x/ w1 x ≥ 2/(d+1)

≥ 2/(ΔΔ +1) = rΔ

www.cs.technion.ac.il/~reuven IBM2006 14 Single Machine Scheduling :

Activity9Activity8Activity7Activity6Activity5Activity4Activity3Activity2Activity1 ????????????? time

Maximize s.t. For each instance I:

For each time t:

For each activity A:

I

IxIp )( }1,0{Ix

)()(:

1)(IetIsI

IxIw

1AI

Ix

Bar-Noy, Guha, Naor and Schieber STOC 99: 1/2 LP

Berman, DasGupta, STOC 00: 1/2

Bar-Noy at al, STOC 00 1/2


ÎÎ, and the weight decomposition:, and the weight decomposition:

• Let Î be the interval which ends first.

I in conflict with Î ,

• Define w1(I) = w2= w-w1

0 otherwise,

w1= w1= w1= w1= w1=

w1= w1=

w1= w1=

w1= 0

w1= 0

w1= 0w1= 0

w1= 0w1 = 0

w1= 0w1= 0

w1= 0 w1= 0

w1= 0

time

Activity9Activity8Activity7Activity6Activity5Activity4Activity3Activity2 Activity1


4-approximation for2 Dimentional Interval graphs


2t-approximation fort- Dimentional Interval graphs


2t-approximation for t-Interval Graphs

Maximize w·x

Subject to: vC xv ≤ 1 C Clique

x {0,1}|V|


2t-approximation for t- Interval Graphs

finding x

Maximize w·xSubject to: vC xv ≤ 1 C Interval Clique

x [0,1]|V|

e.g. x1+x4+x5 ≤ 1


2t-approximation for t- Interval Graphs

finding more relaxed x

Maximize w·xSubject to: vC xv ≤ t C t-Interval Clique

x [0,1]|V|

e.g. x1+x3+x4+x5 ≤ 3

www.cs.technion.ac.il/~reuven IBM2006 25 Gain Gain 11 integral, lose integral, lose 2t fractional fractional

1/(2t1/(2t)-Approx )-Approx IS(G,w)IS(G,w)If v V w(v) 0 return IS(G-v, w)

If E= return V

Let v V s.t x(N[v]) is minimum and

Let = w(v)

if i N[v]w11(i) =

0 else

Claim:w1 x ≥ rt w1 x for Good(x)


Induction hyp is: w2REC ≥ rt w2x

so if Good(REC): w1REC ≥ rt w1x we are done



Min x(N[v]) 2t

www.cs.technion.ac.il/~reuven IBM2006 26 Gain Gain 11 integral, lose integral, lose 2t fractional fractional Claim: w1 x ≥ rt w1 x for Good(x)

Min x(N[v])

We need to show that (next slide)

x(N[v]) ≤ 2t

Thus w1 x ≤ 2t

But w1 x ≥ 1

Hence: w1 x/ w1 x ≥ /(2t) = rt

www.cs.technion.ac.il/~reuven IBM2006 27 Claim: v u N[v] xu

≤ 2t

Define a directed graph G(V,E)

V = Set of t-splits

E = {ij : A right endpoind of i “hits” interval j}

Define xij = xi xj yi+ = ij xij and yi

- = j i xji

Thus yi+ t xi i yi = i yi

+ + i yi- 2t

i xi

Thus i yi 2t xi and therefore i i-j xj 2t


6t-apx for t-Interval Graphs with demands

finding x

Maximize w·xSubject to: vC dv xv ≤ 1 C Interval Clique

x [0,1]|V|

e.g. d1x1+d4x4+d5x5 ≤ 1


t-Interval Graphs with demands

6t = (fat)2t+(thin)4t (Assign zi=dixi )

R.Bar-Yehuda and D. Rawitz. ESA2005 and Discrete Optimization 2006.


2- Dimentional Interval graphs rectangles packing


MIS on axix-parallel rectangles: MIS on axix-parallel rectangles: • NP-Hard even on unit squares [Asano91]

• Divide and conquare O(logn)-apx [AKS98]

• PTAS where all heights are the same [AKS98]

• log(n)/ apx for any constant [BDMR01]

• 4c-apx where c=max #rects covering a point [LNO04]

• 12c-apx with demands [Rawitz06]


4c-apx

Liane Lewin-Eytan, Joseph (Seffi) Naor, and Ariel Orda1

Admission Control in Networks with Advance Reservations

Algorithmica (2004) 40: 293–304


4c-apx for rectangle packing

Types of intersections:

Stabbing:

Crossing:


4c-apx for rectangle packing

.

Result: 4c-apx

Algorithm:Partition the input into c crossing free setsApply 4-apx for each and pick the maximum.


4-approximation for MIS on axix-parallel rectanglesMIS on axix-parallel rectangles

finding x

Maximize w·x

Subject to: vC xv ≤ 1 C right upper corner Clique

x [0,1]|V|

e.g. x1+x3+x4 ≤ 1

2

1

3

5

4

www.cs.technion.ac.il/~reuven IBM2006 36 4-approximation for MIS on axix-parallel rectanglesMIS on axix-parallel rectangles

finding more relaxed x

Maximize w·x

Subject to: vC xv ≤ 2 C right segment Cliques

x [0,1]|V|

e.g. x1+x3+x4+x5 ≤ 2

2

1

3

5

4

www.cs.technion.ac.il/~reuven IBM2006 37 Gain Gain 11 integral, lose integral, lose 4 fractional fractional4-apx for crossing free recangles

If v V w(v) 0 return IS(G-v, w)

If E= return V

Let v V s.t x(N[v]) is minimum and

Let = w(v)

if i N[v]w11(i) =

0 else

Claim:w1 x ≥ ¼ w1 x for Good(x)


Induction hyp is: w2REC ≥ ¼ w2x

so if Good(REC): w1REC ≥ ¼ w1x we are done



Min x(N[v]) 4

0

0

0

0

0

www.cs.technion.ac.il/~reuven IBM2006 38 Claim: v u N[v] xu

≤ 4

Define a directed graph G(V,E)

V = Set of rectangles

E = {ij : Rectangle i “right-stubs” rectangle j}

Define xij = xi xj yi+ = ij xij and yi

- = j i xji

Thus yi+ 2*xi i yi = i yi

+ + i yi- 2*2

i xi

Thus i yi 4 xi and therefore i i-j xj 4


Max IS RECT with demand

• Admission Control with Advance Reservation in Simple Networks

Dror Rawitz 2006

Thin: Color with C colors Each factor 8

12c= fat 4c + thin 8c


Thank you !

www.cs.technion.ac.il/~reuven ibm2006 1 lp rounding using fractional local ratio reuven bar-yehuda...

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