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
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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
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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
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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|
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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
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““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
Return RECReturn REC
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’;
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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|
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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
Return RECReturn REC
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
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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
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4-approximation for2 Dimentional Interval graphs
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4-approximation for2 Dimentional Interval graphs
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4-approximation for2 Dimentional Interval graphs
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4-approximation for2 Dimentional Interval graphs
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4-approximation for2 Dimentional Interval graphs
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2t-approximation fort- Dimentional Interval graphs
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2t-approximation for t-Interval Graphs
Maximize w·x
Subject to: vC xv ≤ 1 C Clique
x {0,1}|V|
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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
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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)
REC= IS(G, (G, w2= w- w-ww11))
Induction hyp is: w2REC ≥ rt w2x
so if Good(REC): w1REC ≥ rt 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
Return RECReturn REC
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
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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
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t-Interval Graphs with demands
6t = (fat)2t+(thin)4t (Assign zi=dixi )
R.Bar-Yehuda and D. Rawitz. ESA2005 and Discrete Optimization 2006.
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2- Dimentional Interval graphs rectangles packing
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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]
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4c-apx
Liane Lewin-Eytan, Joseph (Seffi) Naor, and Ariel Orda1
Admission Control in Networks with Advance Reservations
Algorithmica (2004) 40: 293–304
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4c-apx for rectangle packing
Types of intersections:
Stabbing:
Crossing:
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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.
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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)
REC= IS(G, (G, w2= w- w-ww11))
Induction hyp is: w2REC ≥ ¼ w2x
so if Good(REC): w1REC ≥ ¼ 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
Return RECReturn REC
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
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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
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Thank you !