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Ryan O’Donnell (CMU, IAS) joint work with Ankur Moitra (MIT)

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Binary optimization, linear objective F ⊆ {0,1} n, “feasible solutions” max

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Binary optimization, linear objectives F ⊆ {0,1} n, “feasible solutions” “max”

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Pareto optima def: x ∈ F is Pareto optimal if not dominated on every objective by some y ∈ F. Obj 1 Obj 2 x y

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Pareto optima Obj 1 Obj 2 x y def: x ∈ F is Pareto optimal if not dominated on every objective by some y ∈ F.

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Obj 1 Obj 2

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thm: If d+1 linear objectives are “semi-random” (in the Smoothed Analysis sense) then for any F ⊆ {0,1} n, E[# Pareto optima in F ] ≤ O(n 2d ).

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Application: 0-1 Knapsack [Nemhauser−Ullmann69]: For i=1…n, compute Pareto optimal 〈 weight,value 〉 pairs achievable by items 1…i. [Beier−Vöcking03]: With 2 semi-rand objs, E[# PO’s] ≤ O(n 4 ). ⇒ O(n 5 )-time alg for 0-1 Knapsack in Smoothed Analysis model!

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Model [BV03] F ⊆ {0,1} n arbitrary d+1 linear objs x = Obj(x) ϕ ≥ 1 a parameter Each is a r.v. on [0,1] with pdf bdd by ϕ. All independent.

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Model [BV03] Each is a r.v. on [0,1] with pdf bdd by ϕ. All independent. 1/ ϕ ϕ

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Prior work

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[BV03]: d = 1, E[# PO’s] ≤ O( ϕ n 4 ) [B04]: Same for arbit. F. Conj: ≤ O(n f(d) ) [BRV07]: d = 1, E[# PO’s] ≤ O( ϕ n 2 ) [RT09]: E[# PO’s] ≤ roughly [MO11]: E[# PO’s] ≤ 2(4 ϕ d) d(d+1)/2 n 2d Remark: All bounds here hold even assuming Obj 0 adversarial, nonlinear, arbitrary. for n ≥ exp(exp(O(d 2 log d)) ( F = {0,1} n ) [BV03]: Ω(n 2 ) when Obj 0 adversarial, ϕ = 1 [BR11]: Ω( ϕ n 2 ), d = 1; Ω( ϕ n) d-logd roughly, d > 1

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Warmup: Alternate proof (sketch) for [BRV07]’s d = 1 bound

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F ⊆ {0,1} n Obj 0 (x) arbitrary Obj 1 (x) ϕ -semi-randomE[# PO’s] ≤ O( ϕ n 2 ) ? Obj Obj 1

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.8.4 Obj Obj 0 (x) arbitrary Obj 1 (x) ϕ -semi-random Obj 1 F ⊆ {0,1} n E[# PO’s] ≤ O( ϕ n 2 ) ?

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Goal: For each -strip S, E[# PO’s in S] ≈ Pr[ ∃ a PO in S] ≤ O(n) ϕ Obj Obj 1 Union-bound over (n/) strips ⇒ E[# PO’s] ≤ O( ϕ n 2 ) S

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Obj Obj 1 Given x, Pr[ Vx ∈ S ] ? S

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Obj Obj 1 Given x, Pr[ Vx ∈ S ] ? S Take any j s.t. x j =

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Obj Obj 1 Given x, Pr[ Vx ∈ S ] ? S Take any j s.t. x j = = Pr[ V j ∈ certain width- interval] ≤ ϕ.

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Obj Given x, Pr[ Vx ∈ S ] ? S Take any j s.t. x j = 1. = Pr[ V j ∈ certain width- interval] ≤ ϕ. Boundedness Lemma: Pr[ Vx ∈ S] ≤ O(), even just using the randomness of V j.

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Obj Obj 1 S event T x,S = “x is PO and Vx ∈ S”

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Obj Obj 1 S event T x,S = “x is PO and Vx ∈ S” Fantasy: Now a unique x s.t. T x,S may yet occur

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more complicated event: T x,j,S = “x is PO and x j = 1 and Vx ∈ S and first PO to x’s left, call it y, has y j = 0” Uniqueness Lemma: Draw all V i ’s except V j. Then ∃ at most 1 x s.t. T x,j,S may still occur.

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T x,j,S = “x is PO and x j = 1 and Vx ∈ S and first PO to x’s left, call it y, has y j = 0” Obj Obj 1 S

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Remainder of the d=1 proof sketch

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T x,j,S = “x is PO and x j = 1 and Vx ∈ S and …” Uniqueness Lemma: Draw all V i ’s except V j. Then ∃ at most 1 x s.t. T x,j,S may still occur. Bddness Lemma: For that x, Pr[Vx ∈ S] ≤ ϕ. Union-bound over all j, S: E[ # PO x s.t. first PO to x’s left, y, has y j ≠ 1 = x j for some j ] ≤ n(n/) ϕ = n 2 ϕ. For each PO x, ∃ j s.t. y j ≠ c = x j. Maybe c = 0… Union-bound over c ∈ {0,1}… + another trick.

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Our result: Larger d

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Obj 0 arbitrary; Obj 1, Obj 2 ϕ -semi-random d=2: Obj 0 Obj 1 Obj 2

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Obj Obj 1 Obj 2 V =

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Obj Obj 2 S Obj 1 V =

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more complicated event: T x,j,S = “x is PO and x j = 1 and Vx ∈ S and first PO to x’s left, call it y, has y j = 0”

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more complicated event: T x,j,c,S = “x is PO and x j = c and Vx − ∈ S and first PO to x’s left, call it y, has y j ≠ c”

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more complicated event: T x,J,C, S = “x is PO and…” still J: list of d coordinates C: d×d matrix of bits w/ certain pattern S : d-dimensional strip, plus “nearby” d′-dim. strips for all d′ < d

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more complicated event: J: list of d coordinates C: d×d matrix of bits w/ certain pattern S : d-dimensional strip, plus “nearby” d’-dim. strips for all d’ < d B : “blueprint” [AMR09] + = T x, B = “x is PO and conditions involving V, x, B.” still

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Given B, entries of V are partitioned into two sets, V[few B ] and V[most B ]. V = T x, B = “x is PO and conditions involving V, x, B.”

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Boundedness Lemma: For any V[most B ] outcome, Pr [condits involving V, x, B ] ≤ ϕ d(d+1)/2 d(d+1)/2. Uniqueness Lemma: Draw V[most B ]. Then ∃ at most 1 x s.t. T x, B may still occur. T x, B = “x is PO and conditions involving V, x, B.” V[few B ] Counting Lemma: O(d/) d(d+1)/2 n 2d possible B. Union bound ⇒ E[# PO’s] ≤ O(d ϕ ) d(d+1)/2 n 2d. Given B, entries of V are partitioned into two sets, V[few B ] and V[most B ].

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T x,j,c,S = “x is PO and x j = c and Vx − ∈ S and first PO to x’s left, call it y, has y j ≠ c” J: list of d coordinates C: d×d matrix of bits w/ certain pattern S : d-dimensional strip, plus “nearby” d′-dim. strips for all d′ < d B : “blueprint” + = T x, B = “x is PO and conditions involving V, x, B.”

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T x, B = “x is PO and SKETCH (x,V) = B ” where SKETCH () is a certain deterministic algorithm.

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SKETCH(x,V):

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Bddness Lemma: For any V[most B ] outcome, Pr [ SKETCH (x,V) = B ] ≤ ϕ d(d+1)/2 d(d+1)/2. Uniqueness Lemma: Draw V[most B ]. Then ∃ at most 1 x s.t. T x, B may still occur. V[few B ] T x, B = “x is PO and SKETCH (x,V) = B ”

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Uniqueness Lemma: Draw V[most B ]. Then ∃ at most 1 x s.t. T x, B may still occur. T x, B = “x is PO and SKETCH (x,V) = B ” Equivalent Lemma: Suppose x, V are such that x is PO. Assume SKETCH (x,V) = B. Then RECONSTRUCT ( B,V[most B ]) = x.

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RECONSTRUCT( B,V[most]):

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Bddness Lemma: For any V[most B ] outcome, Pr [ SKETCH (x,V) = B ] ≤ ϕ d(d+1)/2 d(d+1)/2. V[few B ] Uniqueness Lemma: Suppose x, V are such that x is PO. Assume SKETCH (x,V) = B. Then RECONSTRUCT ( B,V[most B ]) = x. Counting Lemma: O(d/) d(d+1)/2 n 2d possible B.

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A sketch of SKETCH (x,V) for d = 2

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SKETCH (x,V): 1.Let F 0 = F 2.Let D 1 = { z ∈ F 0 : V 1 z > V 1 x and V 2 z > V 2 x } 3.Let y 1 = argmax{ Obj 0 (y) : y ∈ D 1 } 4.Let j 1 = least index s.t. 5.Let F 1 = { z ∈ F 0 : Obj 0 (z) > Obj 0 (y 1 ) and } 6.Let D 2 = { z ∈ F 1 : V 2 z > V 2 x } 7.Let y 2 = argmax{ Obj 1 (y) : y ∈ D 2 } 8.Let j 2 = least index s.t. 9.Let F 2 = { z ∈ F 1 : Obj 1 (z) > Obj 1 (y 1 ) and } // if Vx is PO then it must be argmax{ Obj 2 (z) : z ∈ F 2 } 10. Output: J = (j 1, j 2 ), C = S = ( strip of Obj 1,2 (x) − diag(V j 1,j 2 C), strip of Obj 2 (x) − ) SKETCH (x,V):

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Close gap of n d vs. n 2d. Lower bounds when all objs semirandom? F ⊆ {0,1,2,…,k} n ? Bounds on Var[ # POs ] ? More Smoothed Analysis via “blueprints”? Future directions?

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