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IBM LP Rounding using Fractional Local Ratio Reuven Bar-Yehuda

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IBM 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 r w·x* [w·x r w·x* ] An algorithm is an r-approximation if for any w, F it returns an r-approximation

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IBM Min 5x Bisli +8x Tea +12x Water +10x Bamba +20x Shampoo +15x Popcorn +6x Chocolate +$4x WaterShampoo + s.t. x Shampoo + x Water + x WaterShampoo $4 $1 $3 $1 $2 $1

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IBM The generalized vertex cover problem Minimize w·x Subject to:x u + x v + x e 1 e={u,v} E x {0,1} |V|+|E|

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IBM Approx 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} 1 w 1 (x) = 0 else Notice:w 1 x 2 w 1 x for Good(x) VC(G, w-w 1 ) REC= GVC(G, w 2 = w-w 1 ) Induction hyp is: w 2 REC 2 w 2 x so if Good(REC): w 1 REC 2 w 1 x we are done If REC-e is a cover thenREC=REC-e If REC-e is a cover thenREC=REC-e Return REC

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IBM integral for the price of 1 fractional: The local ratio technique for rounding Let x be the the fractional solution Minimize w·x Subject to:x u + x v + x e 1 e=(u,v) E x [0,1] |V|+|E|

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IBM d integral for the price of fractional: 2-2/(Δ+1)-Approx GVC(G,w)d integral for the price of ½(d+1) fractional: 2-2/(Δ+1)-Approx 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 x v is minimum and Let =min(w(i) : i N[v]} if i N[v] 1 w 1 (i) = 0 else Claim:w 1 x r Δ w 1 x for Good(x) VC(G, w-w 1 ) REC= GVC(G, w 2 = w-w 1 ) Induction hyp is: w 2 REC r Δ w 2 x so if Good(REC): w 1 REC r Δ w 1 x we are done If REC is not a minimal cover then make REC minimal If REC is not a minimal cover then make REC minimal Return REC Min x v

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IBM d integral for the price of fractional:d integral for the price of ½(d+1) fractional: Claim: w 1 x r Δ w 1 x for Good(x) Min x v If Min x v ½ Then x (N[v]) ½(d+1) Else x (N[v]) ½(d+1) Thus w 1 x ½(d+1) But w 1 x d Hence : w 1 x/ w 1 x 2-2/(d+1) Δ 2-2/( Δ +1) = r Δ

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IBM A Generalized Local-Ratio Schema for M inimization [ M 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 w 1 ; i.e. Good(x): w 1 x [ ] r w 1 x REC = r-ApproxMin [Max](Set, w 2 ) ; Induction hyp is: w 2 REC [ ] r w 2 x so if Good(REC): w 1 REC [ ] r w 1 x we are done, otherwise fix it; return REC;

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IBM The maximum independent set problem Maximize w·x Subject to:x u + x v 1 e=(u,v) E x {0,1} |V|

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IBM The maximum independent set problem 1 integral for the gain of r fractional: Let x be the the fractional solution Maximize w·x Subject to:x u + x v 1 e=(u,v) E x [0,1] |V|

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IBM Gain 1 integral, lose fractional 2/(Δ+1)-Approx IS(G,w) Gain 1 integral, lose ½(d+1) fractional 2/(Δ+1)-Approx IS(G,w) If v V w(v) 0 return IS(G-v, w) If E= return V Let v V s.t x v is maximum and Let = w(v) if i N[v] 1 w 1 (i) = 0 else Claim:w 1 x r Δ w 1 x for Good(x) (G, w-w 1 ) REC= IS(G, w 2 = w-w 1 ) Induction hyp is: w 2 REC r Δ w 2 x so if Good(REC): w 1 REC r Δ w 1 x we are done If REC+v is an independent set then REC=REC+v If REC+v is an independent set then REC=REC+v Return REC Max x v

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IBM Gain 1 integral, lose fractional Gain 1 integral, lose ½(d+1) fractional Claim: w 1 x r Δ w 1 x for Good(x) Max x v If Max x v ½ Then x (N[v]) ½(d+1) Else x (N[v]) ½(d+1) Thus w 1 x ½(d+1) But w 1 x Hens : w 1 x/ w 1 x 2/(d+1) Δ 2/( Δ +1) = r Δ

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IBM Single Machine Scheduling : Activity9 Activity8 Activity7 Activity6 Activity5 Activity4 Activity3 Activity2 Activity1 ????????????? time Maximize s.t. For each instance I: For each time t: For each activity A: 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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IBM Î, and the weight decomposition: Let Î be the interval which ends first. I in conflict with Î, Define w 1 (I) = w 2 = w-w 1 0 otherwise, w 1 = w 1 = 0 time Activity9 Activity8 Activity7 Activity6 Activity5 Activity4 Activity3 Activity2 Activity1

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IBM approximation for 2 Dimentional Interval graphs

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IBM approximation for 2 Dimentional Interval graphs

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IBM approximation for 2 Dimentional Interval graphs

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IBM approximation for 2 Dimentional Interval graphs

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IBM approximation for 2 Dimentional Interval graphs

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IBM t-approximation for t- Dimentional Interval graphs

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IBM t-approximation for t-Interval Graphs Maximize w·x Subject to: v C x v 1 C Clique x {0,1} |V|

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IBM t-approximation for t- Interval Graphs finding x Maximize w·x Subject to: v C x v 1 C Interval Clique x [0,1] |V| e.g. x 1 +x 4 +x 5 1

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IBM t-approximation for t- Interval Graphs finding more relaxed x Maximize w·x Subject to: v C x v t C t-Interval Clique x [0,1] |V| e.g. x 1 +x 3 +x 4 +x 5 3

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IBM Gain 1 integral, lose fractional 1/(2t)-Approx IS(G,w) Gain 1 integral, lose 2 t fractional 1/(2t)-Approx 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] 1 w 1 (i) = 0 else Claim:w 1 x r t w 1 x for Good(x) (G, w-w 1 ) REC= IS(G, w 2 = w-w 1 ) Induction hyp is: w 2 REC r t w 2 x so if Good(REC): w 1 REC r t w 1 x we are done If REC+v is an independent set then REC=REC+v If REC+v is an independent set then REC=REC+v Return REC Min x (N[v]) 2t

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IBM Gain 1 integral, lose fractional Gain 1 integral, lose 2t fractional Claim: w 1 x r t w 1 x for Good(x) Min x (N[v]) We need to show that (next slide) x (N[v]) 2t Thus w 1 x 2t But w 1 x 1 Hence : w 1 x/ w 1 x /(2t ) = r t

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IBM Claim: v u N[v] x u 2t Define a directed graph G(V,E) V = Set of t-splits E = {i j : A right endpoind of i hits interval j} Define x ij = x i x j y i + = i j x ij and y i - = j i x ji Thus y i + t x i i y i = i y i + + i y i - 2t i x i Thus i y i 2t x i and therefore i i-j x j 2t

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IBM t-apx for t-Interval Graphs with demands finding x Maximize w·x Subject to: v C d v x v 1 C Interval Clique x [0,1] |V| e.g. d 1 x 1 +d 4 x 4 +d 5 x 5 1

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IBM t-Interval Graphs with demands 6t = ( fat ) 2t +( thin ) 4t (Assign z i =d i x i ) R.Bar-Yehuda and D. Rawitz. ESA2005 and Discrete Optimization 2006.

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IBM Dimentional Interval graphs rectangles packing

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IBM 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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IBM c-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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IBM c-apx for rectangle packing Types of intersections: Stabbing: Crossing:

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IBM c-apx for rectangle packing. Result: 4c -apx Algorithm: Partition the input into c crossing free sets Apply 4 -apx for each and pick the maximum.

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IBM MIS on axix-parallel rectangles 4-approximation for MIS on axix-parallel rectangles finding x Maximize w·x Subject to: v C x v 1 C right upper corner Clique x [0,1] |V| e.g. x 1 +x 3 +x

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IBM MIS on axix-parallel rectangles 4-approximation for MIS on axix-parallel rectangles finding more relaxed x Maximize w·x Subject to: v C x v 2 C right segment Cliques x [0,1] |V| e.g. x 1 +x 3 +x 4 +x

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IBM Gain 1 integral, lose fractional Gain 1 integral, lose 4 fractional 4-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] 1 w 1 (i) = 0 else Claim:w 1 x ¼ w 1 x for Good(x) (G, w-w 1 ) REC= IS(G, w 2 = w-w 1 ) Induction hyp is: w 2 REC ¼ w 2 x so if Good(REC): w 1 REC ¼ w 1 x we are done If REC+v is an independent set then REC=REC+v If REC+v is an independent set then REC=REC+v Return REC Min x (N[v])

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IBM Claim: v u N[v] x u 4 Define a directed graph G(V,E) V = Set of rectangles E = {i j : Rectangle i right-stubs rectangle j} Define x ij = x i x j y i + = i j x ij and y i - = j i x ji Thus y i + 2*x i i y i = i y i + + i y i - 2*2 i x i Thus i y i 4 x i and therefore i i-j x j 4

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IBM 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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IBM Thank you !

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