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

Discrepancy theory

◮ Set system with m sets, n elements ◮ Coloring x ∈ {−1, +1}n

i S

b b

−1

b

+1 −1

Discrepancy theory

◮ Set system with m sets, n elements ◮ Coloring x ∈ {−1, +1}n ◮ A =

  1 1 1 1 1 1   i S

b b

−1

b

+1 −1

Discrepancy theory

◮ Set system with m sets, n elements ◮ Coloring x ∈ {−1, +1}n ◮ A =

  1 1 1 1 1 1  

◮ disc(A) =

min

x∈{−1,1}n Ax∞

i S

b b

−1

b

+1 −1

Discrepancy theory

◮ Set system with m sets, n elements ◮ Coloring x ∈ {−1, +1}n ◮ A =

  1 1 1 1 1 1  

◮ disc(A) =

min

x∈{−1,1}n Ax∞

i S

b b

−1

b

+1 −1 Known results:

◮ n sets, n elements: disc(A) = O(√n) [Spencer ’85]

Discrepancy theory

◮ Set system with m sets, n elements ◮ Coloring x ∈ {−1, +1}n ◮ A =

  1 1 1 1 1 1  

◮ disc(A) =

min

x∈{−1,1}n Ax∞

i S

b b

−1

b

+1 −1 Known results:

◮ n sets, n elements: disc(A) = O(√n) [Spencer ’85] ◮ Every element in ≤ t sets: disc(A) < 2t [Beck & Fiala ’81]

Discrepancy theory

◮ Set system with m sets, n elements ◮ Coloring x ∈ {−1, +1}n ◮ A =

  1 1 1 1 1 1  

◮ disc(A) =

min

x∈{−1,1}n Ax∞

i S

b b

−1

b

+1 −1 Known results:

◮ n sets, n elements: disc(A) = O(√n) [Spencer ’85] ◮ Every element in ≤ t sets: disc(A) < 2t [Beck & Fiala ’81] ◮ Beck-Fiala Conjecture: disc(A) ≤ O(

√ t)

Discrepancy theory

◮ Set system with m sets, n elements ◮ Coloring x ∈ {−1, +1}n ◮ A =

  1 1 1 1 1 1  

◮ disc(A) =

min

x∈{−1,1}n Ax∞

i S

b b

−1

b

+1 −1 Known results:

◮ n sets, n elements: disc(A) = O(√n) [Spencer ’85] ◮ Every element in ≤ t sets: disc(A) < 2t [Beck & Fiala ’81] ◮ Beck-Fiala Conjecture: disc(A) ≤ O(

√ t)

Theorem (H., Rothvoss ’18)

Suppose n ≥ ˜ Θ(m2), entries of A ∈ {0, 1}m×n chosen indep. with prob. p. Then with high probability disc(A) ≤ 1.

Fourier Analysis

◮ Suppose X ∈ Zm a random variable, θ ∈ Rm. ◮ We define

ˆ X(θ) = E[e2πiX,θ]

Fourier Analysis

◮ Suppose X ∈ Zm a random variable, θ ∈ Rm. ◮ We define

ˆ X(θ) = E[e2πiX,θ] =

• λ∈Zm

Pr[X = λ]e2πiX,θ

Fourier Analysis

◮ Suppose X ∈ Zm a random variable, θ ∈ Rm. ◮ We define

ˆ X(θ) = E[e2πiX,θ] =

• λ∈Zm

Pr[X = λ]e2πiX,θ

◮ Fourier inversion formula:

For λ ∈ Zm we have Pr[X = λ] =

• θ∈[− 1

2, 1 2)m

ˆ X(θ)e−2πiλ,θdθ

Fourier Analysis

◮ Suppose X ∈ Zm a random variable, θ ∈ Rm. ◮ We define

ˆ X(θ) = E[e2πiX,θ] =

• λ∈Zm

Pr[X = λ]e2πiX,θ

◮ Fourier inversion formula:

For λ = 0 we have Pr[X = 0] =

• θ∈[− 1

2, 1 2)m

ˆ X(θ)dθ

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 ∈ Zm. We have

Pr[X = 0] =

• θ∈[− 1

2 , 1 2 )m

ˆ X(θ)dθ

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 ∈ Zm. We have

Pr[X = 0] =

• θ∈[− 1

2 , 1 2 )m

ˆ X(θ)dθ =

• θ∈[− 1

2 , 1 2 )m

ˆ D(θ) ˆ R(θ)dθ

The Fourier Coefficients

ˆ D(θ) = E

• e2πiD,θ

The Fourier Coefficients

ˆ D(θ) = E

x∼{−1,1}n

• e2πiAx,θ

The Fourier Coefficients

ˆ D(θ) = E

x∼{−1,1}n

• e2πiAx,θ

=

n

• j=1

E

xj[e2πixjAj,θ]

where Aj ∈ {0, 1}m the jth column of A.

The Fourier Coefficients

ˆ D(θ) = E

x∼{−1,1}n

• e2πiAx,θ

=

n

• j=1

cos

• 2πAj, θ
• where Aj ∈ {0, 1}m the jth column of A.

The Fourier Coefficients

ˆ D(θ) = E

x∼{−1,1}n

• e2πiAx,θ

=

n

• j=1

cos

• 2πAj, θ
• where Aj ∈ {0, 1}m the jth column of A.

b b b b b b b b b

θ ∈ [−1

2, 1 2]m

The Fourier Coefficients

ˆ D(θ) = E

x∼{−1,1}n

• e2πiAx,θ

=

n

• j=1

cos

• 2πAj, θ
• where Aj ∈ {0, 1}m the jth column of A.

◮ If |Aj, θ| ≤ 1 4 for all j, then ˆ

D(θ) ≥ 0.

b b b b b b b b b

θ ∈ [−1

2, 1 2]m

The Fourier Coefficients

ˆ D(θ) = E

x∼{−1,1}n

• e2πiAx,θ

=

n

• j=1

cos

• 2πAj, θ
• where Aj ∈ {0, 1}m the jth column of A.

◮ If |Aj, θ| ≤ 1 4 for all j, then ˆ

D(θ) ≥ 0.

b b b b b b b b b

θ ∈ [−1

2, 1 2]m

B2(0, O(

• 1

t))

The Fourier Coefficients

ˆ D(θ) = E

x∼{−1,1}n

• e2πiAx,θ

=

n

• j=1

cos

• 2πAj, θ
• where Aj ∈ {0, 1}m the jth column of A.

◮ If |Aj, θ| ≤ 1 4 for all j, then ˆ

D(θ) ≥ 0.

◮ If Aj, θ ≈ half-integral for all j, then | ˆ

D(θ)| large.

b b b b b b b b b

θ ∈ [−1

2, 1 2]m

The Fourier Coefficients

ˆ D(θ) = E

x∼{−1,1}n

• e2πiAx,θ

=

n

• j=1

cos

• 2πAj, θ
• where Aj ∈ {0, 1}m the jth column of A.

◮ If |Aj, θ| ≤ 1 4 for all j, then ˆ

D(θ) ≥ 0.

◮ If Aj, θ ≈ half-integral for all j, then | ˆ

D(θ)| large.

b b b b b b b b b

θ ∈ [−1

2, 1 2]m

The Fourier Coefficients

ˆ D(θ) = E

x∼{−1,1}n

• e2πiAx,θ

=

n

• j=1

cos

• 2πAj, θ
• where Aj ∈ {0, 1}m the jth column of A.

◮ If |Aj, θ| ≤ 1 4 for all j, then ˆ

D(θ) ≥ 0.

◮ If Aj, θ ≈ half-integral for all j, then | ˆ

D(θ)| large.

◮ If Aj, θ far from half-integral for many j, then | ˆ

D(θ)| small

b b b b b b b b b

θ ∈ [−1

2, 1 2]m

Analyzing ˆ D(θ)

b b b b b b b b b

θ ∈ [−1

2, 1 2]m

Analyzing ˆ D(θ)

b b b b b b b b b

θ ∈ [−1

2, 1 2]m ◮ With high probability,

• θ2≤1/

√ t

ˆ D(θ)dθ > n−Θ(m)

Analyzing ˆ D(θ)

b b b b b b b b b

θ ∈ [−1

2, 1 2]m ◮ With high probability,

• θ2≤1/

√ t

ˆ D(θ)dθ > n−Θ(m)

◮ With high probability,

• d2(θ, 1

2 Zm)>1/

√ t

| ˆ D(θ)| < e−Θ(n/m)

Defining R

Given A, define R ∈ Zm by Ri = Ai1 even ±1 Ai1 odd (chosen uniformly)

Defining R

Given A, define R ∈ Zm by Ri = Ai1 even ±1 Ai1 odd (chosen uniformly) Recall D = Ax where x ∈ {−1, 1}n. = ⇒ X = D + R ∈ 2Zm.

Defining R

Given A, define R ∈ Zm by Ri = Ai1 even ±1 Ai1 odd (chosen uniformly) Recall D = Ax where x ∈ {−1, 1}n. = ⇒ X = D + R ∈ 2Zm. Pr[X = 0] = 2m

• θ∈[− 1

4 , 1 4 )m

ˆ X(θ)dθ

Defining R

Given A, define R ∈ Zm by Ri = Ai1 even ±1 Ai1 odd (chosen uniformly) Recall D = Ax where x ∈ {−1, 1}n. = ⇒ X = D + R ∈ 2Zm. Pr[X = 0] = 2m

• θ∈[− 1

4 , 1 4 )m

ˆ X(θ)dθ

b b b b b b b b b

θ ∈ [−1

2, 1 2]m

Analyzing ˆ R(θ)

We can compute ˆ R(θ) = E[e2πiR,θ] =

• Ai1 odd

E[e2πiRiθi] =

• Ai1 odd

cos(2πθi)

Analyzing ˆ R(θ)

We can compute ˆ R(θ) = E[e2πiR,θ] =

• Ai1 odd

E[e2πiRiθi] =

• Ai1 odd

cos(2πθi) and therefore

◮ For θ ∈ [−1 4, 1 4]m, 0 ≤ ˆ

R(θ) ≤ 1

Analyzing ˆ R(θ)

We can compute ˆ R(θ) = E[e2πiR,θ] =

• Ai1 odd

E[e2πiRiθi] =

• Ai1 odd

cos(2πθi) and therefore

◮ For θ ∈ [−1 4, 1 4]m, 0 ≤ ˆ

R(θ) ≤ 1

◮ For θ2 ≤ 1 √n, ˆ

R(θ) ≥ 1

2.

Analyzing ˆ R(θ)

We can compute ˆ R(θ) = E[e2πiR,θ] =

• Ai1 odd

E[e2πiRiθi] =

• Ai1 odd

cos(2πθi) and therefore

◮ For θ ∈ [−1 4, 1 4]m, 0 ≤ ˆ

R(θ) ≤ 1

◮ For θ2 ≤ 1 √n, ˆ

R(θ) ≥ 1

2.

So with high probability (over the choice of A) we have

• θ2≤1/

√ t

ˆ D(θ)dθ > n−Θ(m)

• d2(θ, 1

2 Zm)>1/

√ t

| ˆ D(θ)| < e−Θ(n/m)

Analyzing ˆ R(θ)

We can compute ˆ R(θ) = E[e2πiR,θ] =

• Ai1 odd

E[e2πiRiθi] =

• Ai1 odd

cos(2πθi) and therefore

◮ For θ ∈ [−1 4, 1 4]m, 0 ≤ ˆ

R(θ) ≤ 1

◮ For θ2 ≤ 1 √n, ˆ

R(θ) ≥ 1

2.

So with high probability (over the choice of A) we have

• θ2≤1/

√ t

ˆ D(θ) · ˆ R(θ)dθ > n−Θ(m)

• d2(θ, 1

2Zm)>1/

√ t

| ˆ D(θ) · ˆ R(θ)| < e−Θ(n/m)

Summary

θ ∈ [−1

4, 1 4]m

With high probability (over the choice of A), we have

• θ2≤1/

√ t

ˆ D(θ) · ˆ R(θ)dθ > n−Θ(m)

• θ2>1/

√ t

| ˆ D(θ) · ˆ R(θ)|dθ < e−Θ(n/m)

Summary

θ ∈ [−1

4, 1 4]m

With high probability (over the choice of A), we have

• θ2≤1/

√ t

ˆ D(θ) · ˆ R(θ)dθ > n−Θ(m)

• θ2>1/

√ t

| ˆ D(θ) · ˆ R(θ)|dθ < e−Θ(n/m) Then for n ≥ Θ(m2 log n), we compute Pr[D+R = 0] = 2m

• θ∈[− 1

4 , 1 4 )

ˆ D(θ) ˆ R(θ)dθ

Summary

θ ∈ [−1

4, 1 4]m

With high probability (over the choice of A), we have

• θ2≤1/

√ t

ˆ D(θ) · ˆ R(θ)dθ > n−Θ(m)

• θ2>1/

√ t

| ˆ D(θ) · ˆ R(θ)|dθ < e−Θ(n/m) Then for n ≥ Θ(m2 log n), we compute Pr[D+R = 0] = 2m

• θ∈[− 1

4 , 1 4 )

ˆ D(θ) ˆ R(θ)dθ > n−Θ(m)−e−Θ(n/m) > 0

Summary

θ ∈ [−1

4, 1 4]m

With high probability (over the choice of A), we have

• θ2≤1/

√ t

ˆ D(θ) · ˆ R(θ)dθ > n−Θ(m)

• θ2>1/

√ t

| ˆ D(θ) · ˆ R(θ)|dθ < e−Θ(n/m) Then for n ≥ Θ(m2 log n), we compute Pr[D+R = 0] = 2m

• θ∈[− 1

4 , 1 4 )

ˆ D(θ) ˆ R(θ)dθ > n−Θ(m)−e−Θ(n/m) > 0 = ⇒ disc(A) ≤ 1.

Bound for θ2 < 1/16 √ t

Recall that ˆ D(θ) = n

j=1 cos(2πAj, θ). ◮ t = pm is the expected column sum. ◮ t ≥ log n =

⇒ whp Aj2 ≤ √ 2t for all j. With high probability |Aj, θ| ≤ 1

8 for all j,

Bound for θ2 < 1/16 √ t

Recall that ˆ D(θ) = n

j=1 cos(2πAj, θ). ◮ t = pm is the expected column sum. ◮ t ≥ log n =

⇒ whp Aj2 ≤ √ 2t for all j. With high probability |Aj, θ| ≤ 1

8 for all j, and so

ˆ D(θ) = exp(−2π2θT AATθ ± O(nt2)θ4

2)

Bound for θ2 < 1/16 √ t

Recall that ˆ D(θ) = n

j=1 cos(2πAj, θ). ◮ t = pm is the expected column sum. ◮ t ≥ log n =

⇒ whp Aj2 ≤ √ 2t for all j. With high probability |Aj, θ| ≤ 1

8 for all j, and so

ˆ D(θ) = exp(−2π2θT AATθ ± O(nt2)θ4

2)

≥ exp(−2π2mnθ2

2)

Bound for θ2 < 1/16 √ t

Recall that ˆ D(θ) = n

j=1 cos(2πAj, θ). ◮ t = pm is the expected column sum. ◮ t ≥ log n =

⇒ whp Aj2 ≤ √ 2t for all j. With high probability |Aj, θ| ≤ 1

8 for all j, and so

ˆ D(θ) = exp(−2π2θT AATθ ± O(nt2)θ4

2)

≥ exp(−2π2mnθ2

2)

= ˆ Y (θ) where Y a gaussian with expectation 0, cov matrix mn · I.

Bound for θ2 < 1/16 √ t

Recall that ˆ D(θ) = n

j=1 cos(2πAj, θ). ◮ t = pm is the expected column sum. ◮ t ≥ log n =

⇒ whp Aj2 ≤ √ 2t for all j. With high probability |Aj, θ| ≤ 1

8 for all j, and so

ˆ D(θ) = exp(−2π2θT AATθ ± O(nt2)θ4

2)

≥ exp(−2π2mnθ2

2)

= ˆ Y (θ) where Y a gaussian with expectation 0, cov matrix mn · I.

• θ2≤

1 16 √ t

ˆ D(θ)dθ ≥

• θ2≤ 1

√n

ˆ Y (θ)dθ ≥ n−Θ(m)

Bound for θ2 > 1/16 √ t

Lemma

For c > 0 some constant, θ∞ ≤ 1

4,

E

Aj[| cos(2πAj, θ)|] ≤ 1 − min{1

4pθ2

2, c}

Bound for θ2 > 1/16 √ t

Lemma

For c > 0 some constant, θ∞ ≤ 1

4, θ2 ≥ 1 16 √ t

E

Aj[| cos(2πAj, θ)|] ≤ 1 − min{1

4pθ2

2, c} ≤ 1 − 1

m

Bound for θ2 > 1/16 √ t

Lemma

For c > 0 some constant, θ∞ ≤ 1

4, θ2 ≥ 1 16 √ t

E

Aj[| cos(2πAj, θ)|] ≤ 1 − min{1

4pθ2

2, c} ≤ 1 − 1

m Write φk(θ) = k

j=1 | cos(2πAj, θ)|.

For θ2 ≥

1 16 √ t, we have E[φk(θ)|φk−1(θ)] ≤ (1 − 1 m)φk−1(θ).

Bound for θ2 > 1/16 √ t

Lemma

For c > 0 some constant, θ∞ ≤ 1

4, θ2 ≥ 1 16 √ t

E

Aj[| cos(2πAj, θ)|] ≤ 1 − min{1

4pθ2

2, c} ≤ 1 − 1

m Write φk(θ) = k

j=1 | cos(2πAj, θ)|.

For θ2 ≥

1 16 √ t, we have E[φk(θ)|φk−1(θ)] ≤ (1 − 1 m)φk−1(θ). ◮ Let Φk =

• θ2≥

1 16 √ t φk(θ)dθ.

Bound for θ2 > 1/16 √ t

Lemma

For c > 0 some constant, θ∞ ≤ 1

4, θ2 ≥ 1 16 √ t

E

Aj[| cos(2πAj, θ)|] ≤ 1 − min{1

4pθ2

2, c} ≤ 1 − 1

m Write φk(θ) = k

j=1 | cos(2πAj, θ)|.

For θ2 ≥

1 16 √ t, we have E[φk(θ)|φk−1(θ)] ≤ (1 − 1 m)φk−1(θ). ◮ Let Φk =

• θ2≥

1 16 √ t φk(θ)dθ.

◮ E[Φk | Φk−1] ≤ (1 − 1 m) · Φk−1

Bound for θ2 > 1/16 √ t

Lemma

For c > 0 some constant, θ∞ ≤ 1

4, θ2 ≥ 1 16 √ t

E

Aj[| cos(2πAj, θ)|] ≤ 1 − min{1

4pθ2

2, c} ≤ 1 − 1

m Write φk(θ) = k

j=1 | cos(2πAj, θ)|.

For θ2 ≥

1 16 √ t, we have E[φk(θ)|φk−1(θ)] ≤ (1 − 1 m)φk−1(θ). ◮ Let Φk =

• θ2≥

1 16 √ t φk(θ)dθ.

◮ E[Φk | Φk−1] ≤ (1 − 1 m) · Φk−1 ◮ Can use martingale concentration to get that

Φn ≤ e−Θ(n/m) with probability at least 1 − e−Θ(n/m).

Bound for θ2 > 1/16 √ t

Lemma

For c > 0 some constant, θ∞ ≤ 1

4, θ2 ≥ 1 16 √ t

E

Aj[| cos(2πAj, θ)|] ≤ 1 − min{1

4pθ2

2, c} ≤ 1 − 1

m Write φk(θ) = k

j=1 | cos(2πAj, θ)|.

For θ2 ≥

1 16 √ t, we have E[φk(θ)|φk−1(θ)] ≤ (1 − 1 m)φk−1(θ). ◮ Let Φk =

• θ2≥

1 16 √ t φk(θ)dθ.

◮ E[Φk | Φk−1] ≤ (1 − 1 m) · Φk−1 ◮ Can use martingale concentration to get that

Φn ≤ e−Θ(n/m) with probability at least 1 − e−Θ(n/m).

◮ But Φn =

• θ2≥

1 16 √ t | ˆ

D(θ)|dθ is exactly the quantity we wanted to bound.

Summary

θ ∈ [−1

4, 1 4]m

With high probability (over the choice of A), we have

• θ2≤1/

√ t

ˆ D(θ) · ˆ R(θ)dθ > n−Θ(m)

• θ2>1/

√ t

| ˆ D(θ) · ˆ R(θ)|dθ < e−Θ(n/m) Then for n ≥ Θ(m2 log n), we compute Pr[D+R = 0] = 2m

• θ∈[− 1

4 , 1 4 )

ˆ D(θ) ˆ R(θ)dθ > n−Θ(m)−e−Θ(n/m) > 0 = ⇒ disc(A) ≤ 1.

Open Questions

(1) Can we come up with an algorithm to actually find good colorings?

Open Questions

(1) Can we come up with an algorithm to actually find good colorings? (2) What about random set systems for n < m2?

Open Questions

(1) Can we come up with an algorithm to actually find good colorings? (2) What about random set systems for n < m2? (3) Can we use similar techniques to prove Spencer’s theorem (or Beck-Fiala conjecture?)

Open Questions

(1) Can we come up with an algorithm to actually find good colorings? (2) What about random set systems for n < m2? (3) Can we use similar techniques to prove Spencer’s theorem (or Beck-Fiala conjecture?)

◮ Would need to get stronger decay from R

Open Questions

(1) Can we come up with an algorithm to actually find good colorings? (2) What about random set systems for n < m2? (3) Can we use similar techniques to prove Spencer’s theorem (or Beck-Fiala conjecture?)

◮ Would need to get stronger decay from R ◮ Candidate R: For i = 1, ..., m Ri chosen as a sum of ∆

• ind. {−1, 0, 1} with Pr[Ri = 1] = Pr[Ri = −1] = 1

4.

Thanks!