A Logarithmic Additive Integrality Gap for Bin Packing Rebecca - PowerPoint PPT Presentation

Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction x 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each y direction x 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each y direction x 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each y direction x 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction y x 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction x y 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction y x 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction x y 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction x y 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction x (2) If hit hyperplane, stay on it y 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction x (2) If hit hyperplane, stay on it y 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction x (2) If hit hyperplane, stay on it y 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction x (2) If hit hyperplane, stay on it y 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction x (2) If hit hyperplane, stay on it y 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each direction y x (2) If hit hyperplane, stay on it 0 0 1

b Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each y direction x (2) If hit hyperplane, stay on it 0 0 1

Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion with std. deviation 3 in each b y direction x (2) If hit hyperplane, stay on it 0 0 1

Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion v i with std. deviation 3 in each λ i b y direction x (2) If hit hyperplane, stay on it ◮ Analysis: ◮ Pr[hit � v i , y − x � = λ i ] ≤ e − Ω( λ 2 i ) 0 0 1

Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion v i with std. deviation 3 in each λ i b y direction x (2) If hit hyperplane, stay on it ◮ Analysis: ◮ Pr[hit � v i , y − x � = λ i ] ≤ e − Ω( λ 2 i ) E [# constraints hit] ≤ m ◮ 0 16 0 1

Constructive Partial Coloring Lemma Lemma [Lovett-Meka ’12] Given x ∈ [0 , 1] m , unit vectors v i , parameters λ i ≥ 0 s.t. i / 16 ≤ m 16 . Then one can find y ∈ [0 , 1] m with i e − λ 2 � ◮ y j ∈ { 0 , 1 } for at least half of the indices j ◮ | � v i , y − x � | ≤ λ i · � v i � 2 ∀ i ◮ Algorithm: 1 (1) Perform Brownian motion v i with std. deviation 3 in each λ i b y direction x (2) If hit hyperplane, stay on it ◮ Analysis: ◮ Pr[hit � v i , y − x � = λ i ] ≤ e − Ω( λ 2 i ) E [# constraints hit] ≤ m ◮ 0 16 0 1

The algorithm (1) Compute a fractional LP solution x (2) FOR log n iterations DO (3) Rebuild incidence matrix (4) Apply Lovett-Meka rounding x → x ′

The algorithm (1) Compute a fractional LP solution x (2) FOR log n iterations DO (3) Rebuild incidence matrix Effect: Rows of A get small � · � 2 -norm (4) Apply Lovett-Meka rounding x → x ′

The algorithm (1) Compute a fractional LP solution x (2) FOR log n iterations DO (3) Rebuild incidence matrix Effect: Rows of A get small � · � 2 -norm (4) Apply Lovett-Meka rounding x → x ′ Effect: ◮ | frac( x ′ ) | ≤ 1 2 · | frac( x ) |

The algorithm (1) Compute a fractional LP solution x (2) FOR log n iterations DO (3) Rebuild incidence matrix Effect: Rows of A get small � · � 2 -norm (4) Apply Lovett-Meka rounding x → x ′ Effect: ◮ | frac( x ′ ) | ≤ 1 2 · | frac( x ) | ◮ cost( x ′ ) ≤ cost( x ) + O (1)

Assigning items to patterns b 1 = 2 b 2 = 1 b 3 = 7 Items: Bins: x p 1 = 1 x p 2 = 1 x p 3 = 1 2 2 2

Assigning items to patterns b 1 = 2 b 2 = 1 b 3 = 7 Items: 1 1 1 1 2 2 2 2 Bins: x p 1 = 1 x p 2 = 1 x p 3 = 1 2 2 2

Assigning items to patterns b 1 = 2 b 2 = 1 b 3 = 7 Items: 1 1 1 1 1 2 2 2 1 2 2 1 1 2 1 1 2 2 2 2 Bins: x p 1 = 1 x p 2 = 1 x p 3 = 1 2 2 2

Assigning items to patterns b 1 = 2 b 2 = 1 b 3 = 7 Items: y C 1 = 1 y C 2 = 1 y C 3 = 2 Containers: Bins: x p 1 = 1 x p 2 = 1 x p 3 = 1 2 2 2

Assigning items to patterns b 1 = 2 b 2 = 1 b 3 = 7 Items: 1 1 2 2 1 1 1 y C 1 = 1 y C 2 = 1 y C 3 = 2 Containers: 1 2 1 1 1 2 2 1 1 1 2 2 2 2 Bins: x p 1 = 1 x p 2 = 1 x p 3 = 1 2 2 2

Assigning items to patterns b 1 = 2 b 2 = 1 b 3 = 7 Items: 1 1 2 2 1 1 1 y C 1 = 1 y C 2 = 1 y C 3 = 2 Containers: 1 2 1 1 1 2 2 1 1 1 2 2 2 2 Bins: x p 1 = 1 x p 2 = 1 x p 3 = 1 2 2 2 ◮ deficiency = total size of non-assignable items+containers

Container building ◮ Consider fract. solution x at beginning of iteration 2 A

Container building ◮ Consider fract. solution x at beginning of iteration supp(x) 2 A

Container building ◮ Consider fract. solution x at beginning of iteration supp(x) 2 A . . . pattern p

Container building ◮ Consider fract. solution x at beginning of iteration supp(x) 2 A container C . . . pattern p

Container building ◮ Consider fract. solution x at beginning of iteration supp(x) Lemma Can reassign containers so that for container C of size s C ∈ [ 1 k , 2 k ]: 2 ◮ Each row has ∗ � A C � 1 ≥ k 1 / 2 A container C ◮ All entries ≤ k 1 / 4 Deficiency increase is O (1). . . ∗ modulo technicalities . pattern p

Container building (2) container: patterns: For each size class [ 1 k , 2 k ] (starting with smallest items) do:

Container building (2) container: patterns: For each size class [ 1 k , 2 k ] (starting with smallest items) do: → cost O ( k 1 / 2 (1) Group containers so that � A C � 1 ≥ k 1 / 2 k )

Container building (2) k 1 / 4 container: patterns: k 1 / 4 For each size class [ 1 k , 2 k ] (starting with smallest items) do: → cost O ( k 1 / 2 (1) Group containers so that � A C � 1 ≥ k 1 / 2 k )

Container building (2) container: patterns: For each size class [ 1 k , 2 k ] (starting with smallest items) do: → cost O ( k 1 / 2 (1) Group containers so that � A C � 1 ≥ k 1 / 2 k ) (2) Replace any multiples of k 1 / 4 copies of same container in same pattern by super-container

Container building (2) container: patterns: For each size class [ 1 k , 2 k ] (starting with smallest items) do: → cost O ( k 1 / 2 (1) Group containers so that � A C � 1 ≥ k 1 / 2 k ) (2) Replace any multiples of k 1 / 4 copies of same container in same pattern by super-container ◮ Deficiency increase: O ( k 1 / 2 · 1 k 1 / 2 ) for class 1 1 k ) = Θ( k .

Container building (2) container: patterns: For each size class [ 1 k , 2 k ] (starting with smallest items) do: → cost O ( k 1 / 2 (1) Group containers so that � A C � 1 ≥ k 1 / 2 k ) (2) Replace any multiples of k 1 / 4 copies of same container in same pattern by super-container ◮ Deficiency increase: O ( k 1 / 2 · 1 k 1 / 2 ) for class 1 1 k ) = Θ( k . ◮ Over all k : � k Θ( k − 1 / 2 ) = O (1)

Applying the Partial Coloring Lemma . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 A 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . .

Applying the Partial Coloring Lemma . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 A 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . ◮ Given: x . Find: y with | ( � j ≤ i A j )( x − y ) | small

Applying the Partial Coloring Lemma . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . ◮ Given: x . Find: y with | ( � j ≤ i A j )( x − y ) | small ◮ Suppose 1 k ≤ s i ≤ 2 k

Applying the Partial Coloring Lemma . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 I 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . ◮ Given: x . Find: y with | ( � j ≤ i A j )( x − y ) | small ◮ Suppose 1 k ≤ s i ≤ 2 k ◮ For interval I ⊆ [ n ]:

Applying the Partial Coloring Lemma . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 I 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 + . . . . . . . . . . . . . . . . . . . . . . . . v I = ( 1 0 2 1 1 1 1 1 ) ◮ Given: x . Find: y with | ( � j ≤ i A j )( x − y ) | small ◮ Suppose 1 k ≤ s i ≤ 2 k ◮ For interval I ⊆ [ n ]: v I := � i ∈ I A i

Applying the Partial Coloring Lemma 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 � v I � 1 = ck 17 / 16 I 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . ◮ Given: x . Find: y with | ( � j ≤ i A j )( x − y ) | small ◮ Suppose 1 k ≤ s i ≤ 2 k ◮ For interval I ⊆ [ n ]: v I := � i ∈ I A i

Applying the Partial Coloring Lemma 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 � v I � 1 = 2 − 1 ck 17 / 16 I 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ◮ Given: x . Find: y with | ( � j ≤ i A j )( x − y ) | small ◮ Suppose 1 k ≤ s i ≤ 2 k ◮ For interval I ⊆ [ n ]: v I := � i ∈ I A i

Applying the Partial Coloring Lemma 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 � v I � 1 = 2 − 2 ck 17 / 16 I 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ◮ Given: x . Find: y with | ( � j ≤ i A j )( x − y ) | small ◮ Suppose 1 k ≤ s i ≤ 2 k ◮ For interval I ⊆ [ n ]: v I := � i ∈ I A i

Applying the Partial Coloring Lemma 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 � v I � 1 = 2 − 3 ck 17 / 16 I 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ◮ Given: x . Find: y with | ( � j ≤ i A j )( x − y ) | small ◮ Suppose 1 k ≤ s i ≤ 2 k ◮ For interval I ⊆ [ n ]: v I := � i ∈ I A i

Applying the Partial Coloring Lemma 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 � v I � 1 = 2 − 3 ck 17 / 16 I 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ◮ Given: x . Find: y with | ( � j ≤ i A j )( x − y ) | small ◮ Suppose 1 k ≤ s i ≤ 2 k ◮ For interval I ⊆ [ n ]: v I := � i ∈ I A i , λ I := level( I )

Applying the Partial Coloring Lemma 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m columns ◮ Run Partial coloring with v I := � i ∈ I A i and λ I := level( I )

Applying the Partial Coloring Lemma 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m columns ◮ Run Partial coloring with v I := � i ∈ I A i and λ I := level( I ) � e − λ 2 I / 16 I

Applying the Partial Coloring Lemma sum ≤ k 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m columns ◮ Run Partial coloring with v I := � i ∈ I A i and λ I := level( I ) k · m I / 16 ≤ � e − λ 2 � ck 17 / 16 2 − ℓ · e − ℓ 2 / 16 I ℓ ≥ 0

Applying the Partial Coloring Lemma sum ≤ k 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m columns ◮ Run Partial coloring with v I := � i ∈ I A i and λ I := level( I ) k · m m I / 16 ≤ ck 17 / 16 2 − ℓ · e − ℓ 2 / 16 ≤ � e − λ 2 � 100 · k 1 / 16 I ℓ ≥ 0

Applying the Partial Coloring Lemma 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 i 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m columns ◮ Bound error for item i : � | ( A j )( x − y ) | ≤ j ≤ i

Applying the Partial Coloring Lemma 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 i 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m columns ◮ Bound error for item i : � | ( A j )( x − y ) | ≤ j ≤ i

Applying the Partial Coloring Lemma 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m columns ◮ Bound error for item i : � � | ( A j )( x − y ) | ≤ ℓ · � v I on level ℓ � 2 j ≤ i ℓ ≥ 0

Applying the Partial Coloring Lemma 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 � v I � 1 = Θ( k 17 / 16 ) I 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m columns ◮ Bound error for item i : � � | ( A j )( x − y ) | ≤ ℓ · � v I on level ℓ � 2 ≤ O (1) · � v I � 2 j ≤ i ℓ ≥ 0

Applying the Partial Coloring Lemma 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 � v I � 1 = Θ( k 17 / 16 ) I 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m columns ◮ Bound error for item i : � � | ( A j )( x − y ) | ≤ ℓ · � v I on level ℓ � 2 ≤ O (1) · � v I � 2 j ≤ i ℓ ≥ 0 � k 1 / 4 H¨ older ≤ � v I � 1 · k 1 / 2

Applying the Partial Coloring Lemma 3 2 1 0 ← level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 � v I � 1 = Θ( k 17 / 16 ) I 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m columns ◮ Bound error for item i : � � | ( A j )( x − y ) | ≤ ℓ · � v I on level ℓ � 2 ≤ O (1) · � v I � 2 j ≤ i ℓ ≥ 0 � k 1 / 4 H¨ older k 1 / 2 ≤ O ( k 15 / 16 ) ≤ � v I � 1 ·

The algorithm (1) Compute a fractional LP solution x (2) FOR log n iterations DO (3) Rebuild incidence matrix Effect: Rows of A get small � · � 2 -norm (4) Apply Lovett-Meka rounding x → x ′ Effect: | frac( x ′ ) | ≤ 1 2 · | frac( x ) |

The algorithm (1) Compute a fractional LP solution x (2) FOR log n iterations DO (3) Rebuild incidence matrix Effect: Rows of A get small � · � 2 -norm Deficiency increase: O (1) (4) Apply Lovett-Meka rounding x → x ′ Effect: | frac( x ′ ) | ≤ 1 2 · | frac( x ) |

The algorithm (1) Compute a fractional LP solution x (2) FOR log n iterations DO (3) Rebuild incidence matrix Effect: Rows of A get small � · � 2 -norm Deficiency increase: O (1) (4) Apply Lovett-Meka rounding x → x ′ Effect: | frac( x ′ ) | ≤ 1 2 · | frac( x ) | k O ( k 15 / 16 ) · 1 Deficiency increase: � k = O (1)