Discrepancy of Random Set Systems Rebecca Hoberg and Thomas Rothvo - PowerPoint PPT Presentation

Discrepancy of Random Set Systems Rebecca Hoberg and Thomas Rothvoß

Discrepancy theory ◮ Set system with m sets, n elements i S

b b b Discrepancy theory − 1 ◮ Set system with m sets, n elements ◮ Coloring x ∈ {− 1 , +1 } n − 1 +1 i S

b b b Discrepancy theory − 1 ◮ Set system with m sets, n elements ◮ Coloring x ∈ {− 1 , +1 } n − 1 +1   1 1 0 i S ◮ A = 1 0 1   0 1 1

b b b Discrepancy theory − 1 ◮ Set system with m sets, n elements ◮ Coloring x ∈ {− 1 , +1 } n − 1 +1   1 1 0 i S ◮ A = 1 0 1   0 1 1 ◮ disc( A ) = x ∈{− 1 , 1 } n � Ax � ∞ min

b b b Discrepancy theory − 1 ◮ Set system with m sets, n elements ◮ Coloring x ∈ {− 1 , +1 } n − 1 +1   1 1 0 i S ◮ A = 1 0 1   0 1 1 ◮ disc( A ) = x ∈{− 1 , 1 } n � Ax � ∞ min Known results: ◮ n sets, n elements: disc( A ) = O ( √ n ) [Spencer ’85]

b b b Discrepancy theory − 1 ◮ Set system with m sets, n elements ◮ Coloring x ∈ {− 1 , +1 } n − 1 +1   1 1 0 i S ◮ A = 1 0 1   0 1 1 ◮ disc( A ) = x ∈{− 1 , 1 } n � Ax � ∞ min Known results: ◮ n sets, n elements: disc( A ) = O ( √ n ) [Spencer ’85] ◮ Every element in ≤ t sets: disc( A ) < 2 t [Beck & Fiala ’81]

b b b Discrepancy theory − 1 ◮ Set system with m sets, n elements ◮ Coloring x ∈ {− 1 , +1 } n − 1 +1   1 1 0 i S ◮ A = 1 0 1   0 1 1 ◮ disc( A ) = x ∈{− 1 , 1 } n � Ax � ∞ min Known results: ◮ n sets, n elements: disc( A ) = O ( √ n ) [Spencer ’85] ◮ Every element in ≤ t sets: disc( A ) < 2 t [Beck & Fiala ’81] √ ◮ Beck-Fiala Conjecture: disc( A ) ≤ O ( t )

b b b Discrepancy theory − 1 ◮ Set system with m sets, n elements ◮ Coloring x ∈ {− 1 , +1 } n − 1 +1   1 1 0 i S ◮ A = 1 0 1   0 1 1 ◮ disc( A ) = x ∈{− 1 , 1 } n � Ax � ∞ min Known results: ◮ n sets, n elements: disc( A ) = O ( √ n ) [Spencer ’85] ◮ Every element in ≤ t sets: disc( A ) < 2 t [Beck & Fiala ’81] √ ◮ Beck-Fiala Conjecture: disc( A ) ≤ O ( t ) Theorem (H., Rothvoss ’18) Suppose n ≥ ˜ Θ( m 2 ) , entries of A ∈ { 0 , 1 } m × n chosen indep. with prob. p . Then with high probability disc ( A ) ≤ 1 .

Fourier Analysis ◮ Suppose X ∈ Z m a random variable, θ ∈ R m . ◮ We define ˆ X ( θ ) = E [ e 2 πi � X,θ � ]

Fourier Analysis ◮ Suppose X ∈ Z m a random variable, θ ∈ R m . ◮ We define ˆ � X ( θ ) = E [ e 2 πi � X,θ � ] = Pr[ X = λ ] e 2 πi � X,θ � λ ∈ Z m

Fourier Analysis ◮ Suppose X ∈ Z m a random variable, θ ∈ R m . ◮ We define ˆ � X ( θ ) = E [ e 2 πi � X,θ � ] = Pr[ X = λ ] e 2 πi � X,θ � λ ∈ Z m For λ ∈ Z m we have ◮ Fourier inversion formula: � ˆ X ( θ ) e − 2 πi � λ,θ � dθ Pr[ X = λ ] = θ ∈ [ − 1 2 , 1 2 ) m

Fourier Analysis ◮ Suppose X ∈ Z m a random variable, θ ∈ R m . ◮ We define ˆ � X ( θ ) = E [ e 2 πi � X,θ � ] = Pr[ X = λ ] e 2 πi � X,θ � λ ∈ Z m ◮ Fourier inversion formula: For λ = 0 we have � ˆ Pr[ X = 0] = X ( θ ) dθ θ ∈ [ − 1 2 , 1 2 ) m

Fourier Analysis for Discrepancy Theory For fixed A ∈ { 0 , 1 } m × n : ◮ Choose x ∼ {− 1 , 1 } n uniformly, and let D = Ax .

Fourier Analysis for Discrepancy Theory For fixed A ∈ { 0 , 1 } m × n : ◮ Choose x ∼ {− 1 , 1 } n uniformly, and let D = Ax . ◮ Suffices to show Pr[ D + R = 0] > 0 where � R � ∞ ≤ ∆ chosen at random

Fourier Analysis for Discrepancy Theory For fixed A ∈ { 0 , 1 } m × n : ◮ Choose x ∼ {− 1 , 1 } n uniformly, and let D = Ax . ◮ Suffices to show Pr[ D + R = 0] > 0 where � R � ∞ ≤ ∆ chosen at random ◮ X = D + R ∈ Z m . We have � ˆ Pr[ X = 0] = X ( θ ) dθ θ ∈ [ − 1 2 , 1 2 ) m

Fourier Analysis for Discrepancy Theory For fixed A ∈ { 0 , 1 } m × n : ◮ Choose x ∼ {− 1 , 1 } n uniformly, and let D = Ax . ◮ Suffices to show Pr[ D + R = 0] > 0 where � R � ∞ ≤ ∆ chosen at random ◮ X = D + R ∈ Z m . We have � ˆ Pr[ X = 0] = X ( θ ) dθ θ ∈ [ − 1 2 , 1 2 ) m � D ( θ ) ˆ ˆ = R ( θ ) dθ θ ∈ [ − 1 2 , 1 2 ) m

The Fourier Coefficients ˆ e 2 πi � D,θ � � � D ( θ ) = E

The Fourier Coefficients ˆ e 2 πi � Ax,θ � � � D ( θ ) = E x ∼{− 1 , 1 } n

The Fourier Coefficients n ˆ � x j [ e 2 πix j � A j ,θ � ] e 2 πi � Ax,θ � � � D ( θ ) = = E E x ∼{− 1 , 1 } n j =1 where A j ∈ { 0 , 1 } m the j th column of A .

The Fourier Coefficients n ˆ � e 2 πi � Ax,θ � � 2 π � A j , θ � � � � D ( θ ) = = cos E x ∼{− 1 , 1 } n j =1 where A j ∈ { 0 , 1 } m the j th column of A .

b b b b b b b b b The Fourier Coefficients n ˆ � e 2 πi � Ax,θ � � 2 π � A j , θ � � � � D ( θ ) = = cos E x ∼{− 1 , 1 } n j =1 where A j ∈ { 0 , 1 } m the j th column of A . θ ∈ [ − 1 2 , 1 2 ] m 0

b b b b b b b b b The Fourier Coefficients n ˆ � e 2 πi � Ax,θ � � 2 π � A j , θ � � � � D ( θ ) = = cos E x ∼{− 1 , 1 } n j =1 where A j ∈ { 0 , 1 } m the j th column of A . 4 for all j , then ˆ ◮ If |� A j , θ �| ≤ 1 D ( θ ) ≥ 0. θ ∈ [ − 1 2 , 1 2 ] m 0

b b b b b b b b b The Fourier Coefficients n ˆ � e 2 πi � Ax,θ � � 2 π � A j , θ � � � � D ( θ ) = = cos E x ∼{− 1 , 1 } n j =1 where A j ∈ { 0 , 1 } m the j th column of A . 4 for all j , then ˆ ◮ If |� A j , θ �| ≤ 1 D ( θ ) ≥ 0. θ ∈ [ − 1 2 , 1 2 ] m 0 � B 2 ( 0 , O ( 1 t ))

b b b b b b b b b The Fourier Coefficients n ˆ � e 2 πi � Ax,θ � � 2 π � A j , θ � � � � D ( θ ) = = cos E x ∼{− 1 , 1 } n j =1 where A j ∈ { 0 , 1 } m the j th column of A . 4 for all j , then ˆ ◮ If |� A j , θ �| ≤ 1 D ( θ ) ≥ 0. ◮ If � A j , θ � ≈ half-integral for all j , then | ˆ D ( θ ) | large. θ ∈ [ − 1 2 , 1 2 ] m 0

b b b b b b b b b The Fourier Coefficients n ˆ � e 2 πi � Ax,θ � � 2 π � A j , θ � � � � D ( θ ) = = cos E x ∼{− 1 , 1 } n j =1 where A j ∈ { 0 , 1 } m the j th column of A . 4 for all j , then ˆ ◮ If |� A j , θ �| ≤ 1 D ( θ ) ≥ 0. ◮ If � A j , θ � ≈ half-integral for all j , then | ˆ D ( θ ) | large. θ ∈ [ − 1 2 , 1 2 ] m 0

b b b b b b b b b The Fourier Coefficients n ˆ � e 2 πi � Ax,θ � � 2 π � A j , θ � � � � D ( θ ) = = cos E x ∼{− 1 , 1 } n j =1 where A j ∈ { 0 , 1 } m the j th column of A . 4 for all j , then ˆ ◮ If |� A j , θ �| ≤ 1 D ( θ ) ≥ 0. ◮ If � A j , θ � ≈ half-integral for all j , then | ˆ D ( θ ) | large. ◮ If � A j , θ � far from half-integral for many j , then | ˆ D ( θ ) | small θ ∈ [ − 1 2 , 1 2 ] m 0

b b b b b b b b b Analyzing ˆ D ( θ ) θ ∈ [ − 1 2 , 1 2 ] m 0

b b b b b b b b b Analyzing ˆ D ( θ ) θ ∈ [ − 1 2 , 1 2 ] m 0 ◮ With high probability, � ˆ D ( θ ) dθ > n − Θ( m ) √ � θ � 2 ≤ 1 / t

b b b b b b b b b Analyzing ˆ D ( θ ) θ ∈ [ − 1 2 , 1 2 ] m 0 ◮ With high probability, � ˆ D ( θ ) dθ > n − Θ( m ) √ � θ � 2 ≤ 1 / t ◮ With high probability, � | ˆ D ( θ ) | < e − Θ( n/m ) √ d 2 ( θ, 1 2 Z m ) > 1 / t

Defining R Given A , define R ∈ Z m by � 0 � A i � 1 even R i = ± 1 � A i � 1 odd (chosen uniformly)

Defining R Given A , define R ∈ Z m by � 0 � A i � 1 even R i = ± 1 � A i � 1 odd (chosen uniformly) Recall D = Ax where x ∈ {− 1 , 1 } n . ⇒ X = D + R ∈ 2 Z m . =

Defining R Given A , define R ∈ Z m by � 0 � A i � 1 even R i = ± 1 � A i � 1 odd (chosen uniformly) Recall D = Ax where x ∈ {− 1 , 1 } n . ⇒ X = D + R ∈ 2 Z m . = � ˆ Pr[ X = 0] = 2 m X ( θ ) dθ θ ∈ [ − 1 4 , 1 4 ) m

b b b b b b b b b Defining R Given A , define R ∈ Z m by � 0 � A i � 1 even R i = ± 1 � A i � 1 odd (chosen uniformly) Recall D = Ax where x ∈ {− 1 , 1 } n . ⇒ X = D + R ∈ 2 Z m . = � ˆ Pr[ X = 0] = 2 m X ( θ ) dθ θ ∈ [ − 1 4 , 1 4 ) m θ ∈ [ − 1 2 , 1 2 ] m 0

Analyzing ˆ R ( θ ) We can compute ˆ � � R ( θ ) = E [ e 2 πi � R,θ � ] = E [ e 2 πiR i θ i ] = cos(2 πθ i ) � A i � 1 odd � A i � 1 odd

Analyzing ˆ R ( θ ) We can compute ˆ � � R ( θ ) = E [ e 2 πi � R,θ � ] = E [ e 2 πiR i θ i ] = cos(2 πθ i ) � A i � 1 odd � A i � 1 odd and therefore 4 ] m , 0 ≤ ˆ ◮ For θ ∈ [ − 1 4 , 1 R ( θ ) ≤ 1

Analyzing ˆ R ( θ ) We can compute ˆ � � R ( θ ) = E [ e 2 πi � R,θ � ] = E [ e 2 πiR i θ i ] = cos(2 πθ i ) � A i � 1 odd � A i � 1 odd and therefore 4 ] m , 0 ≤ ˆ ◮ For θ ∈ [ − 1 4 , 1 R ( θ ) ≤ 1 √ n , ˆ 1 R ( θ ) ≥ 1 ◮ For � θ � 2 ≤ 2 .