0/1 Polytopes with Quadratic Chv atal Rank Thomas Rothvo and - - PowerPoint PPT Presentation

0/1 Polytopes with Quadratic Chv´ atal Rank

Thomas Rothvoß and Laura Sanit` a 3rd Cargese Workshop on Combinatorial Optimization

Gomory Chv´ atal Cuts

◮ Given: Polytope P ⊆ Rn

P

b b b b b b b b b b b b b b b b

Gomory Chv´ atal Cuts

◮ Given: Polytope P ⊆ Rn ◮ Want: Integral hull

PI := conv{P ∩ Zn} P PI

b b b b b b b b b b b b b b b b

Gomory Chv´ atal Cuts

◮ Given: Polytope P ⊆ Rn ◮ Want: Integral hull

PI := conv{P ∩ Zn}

◮ Idea: Let cx ≤ β valid

inequality for P (c ∈ Zn) P PI cx ≤ β

b b b b b b b b b b b b b b b b

Gomory Chv´ atal Cuts

◮ Given: Polytope P ⊆ Rn ◮ Want: Integral hull

PI := conv{P ∩ Zn}

◮ Idea: Let cx ≤ β valid

inequality for P (c ∈ Zn)

◮ The Gomory Chv´

atal cut cx ≤ ⌊β⌋ valid for PI P PI cx ≤ β cx ≤ ⌊β⌋

b b b b b b b b b b b b b b b b

Gomory Chv´ atal Cuts

◮ Given: Polytope P ⊆ Rn ◮ Want: Integral hull

PI := conv{P ∩ Zn}

◮ Idea: Let cx ≤ β valid

inequality for P (c ∈ Zn)

◮ The Gomory Chv´

atal cut cx ≤ ⌊β⌋ valid for PI P P ′ PI cx ≤ β cx ≤ ⌊β⌋

b b b b b b b b b b b b b b b b

◮ Gomory Chv´

atal closure P ′ =

• {all GC cuts for P} =
• c∈Zn

{x | cx ≤ ⌊max{cy | y ∈ P}⌋}

Gomory Chv´ atal Cuts

◮ Given: Polytope P ⊆ Rn ◮ Want: Integral hull

PI := conv{P ∩ Zn}

◮ Idea: Let cx ≤ β valid

inequality for P (c ∈ Zn)

◮ The Gomory Chv´

atal cut cx ≤ ⌊β⌋ valid for PI P P ′ PI cx ≤ β cx ≤ ⌊β⌋

b b b b b b b b b b b b b b b b

◮ Gomory Chv´

atal closure P ′ =

• {all GC cuts for P} =
• c∈Zn

{x | cx ≤ ⌊max{cy | y ∈ P}⌋}

◮ kth closure P (k) := P ′′′ . . .′ k times

Gomory Chv´ atal Cuts

◮ Given: Polytope P ⊆ Rn ◮ Want: Integral hull

PI := conv{P ∩ Zn}

◮ Idea: Let cx ≤ β valid

inequality for P (c ∈ Zn)

◮ The Gomory Chv´

atal cut cx ≤ ⌊β⌋ valid for PI P P ′ PI cx ≤ β cx ≤ ⌊β⌋

b b b b b b b b b b b b b b b b

◮ Gomory Chv´

atal closure P ′ =

• {all GC cuts for P} =
• c∈Zn

{x | cx ≤ ⌊max{cy | y ∈ P}⌋}

◮ kth closure P (k) := P ′′′ . . .′ k times ◮ Chv´

atal rank: P (rk(P)) = PI

What’s known

What’s known

◮ For each rational polyhedron P, rk(P) < ∞

What’s known

◮ For each rational polyhedron P, rk(P) < ∞ ◮ But for every k, there is P ⊆ R2 with rk(P) ≥ k

b b b b b b b b b b b b b b b b b b b b

(k, 1

2)

(0, 0) (0, 1) P

What’s known

◮ For each rational polyhedron P, rk(P) < ∞ ◮ But for every k, there is P ⊆ R2 with rk(P) ≥ k

b b b b b b b b b b b b b b b b b b b b

(k, 1

2)

(0, 0) (0, 1) P

◮ For the rest of the talk assume P ⊆ [0, 1]n

What’s known — if P ⊆ [0, 1]n

◮ rk(P) ≤ O(n3 log n)

[Bockmayer, Eisenbrand, Hartmann, Schulz ’98]

What’s known — if P ⊆ [0, 1]n

◮ rk(P) ≤ O(n3 log n)

[Bockmayer, Eisenbrand, Hartmann, Schulz ’98]

◮ rk(P) ≤

[Eisenbrand, Schulz ’99] O(n2 log n)

What’s known — if P ⊆ [0, 1]n

◮ rk(P) ≤ O(n3 log n)

[Bockmayer, Eisenbrand, Hartmann, Schulz ’98]

◮ rk(P) ≤

[Eisenbrand, Schulz ’99]

◮ For some P, rk(P) ≥ (1 + ε)n [Eisenbrand, Schulz ’99]

O(n2 log n)

What’s known — if P ⊆ [0, 1]n

◮ rk(P) ≤ O(n3 log n)

[Bockmayer, Eisenbrand, Hartmann, Schulz ’98]

◮ rk(P) ≤

[Eisenbrand, Schulz ’99]

◮ For some P, rk(P) ≥ (1 + ε)n [Eisenbrand, Schulz ’99] ◮ For some P, rk(P) ≥

[Pokutta, Stauffer ’11] O(n2 log n) 1.36n

What’s known — if P ⊆ [0, 1]n

◮ rk(P) ≤ O(n3 log n)

[Bockmayer, Eisenbrand, Hartmann, Schulz ’98]

◮ rk(P) ≤ O(n2 log n) [Eisenbrand, Schulz ’99] ◮ For some P, rk(P) ≥ (1 + ε)n [Eisenbrand, Schulz ’99] ◮ For some P, rk(P) ≥ 1.36n [Pokutta, Stauffer ’11]

What’s known — if P ⊆ [0, 1]n

◮ rk(P) ≤ O(n3 log n)

[Bockmayer, Eisenbrand, Hartmann, Schulz ’98]

◮ rk(P) ≤ O(n2 log n) [Eisenbrand, Schulz ’99] ◮ For some P, rk(P) ≥ (1 + ε)n [Eisenbrand, Schulz ’99] ◮ For some P, rk(P) ≥ 1.36n [Pokutta, Stauffer ’11]

Theorem (Sanit` a, R. ’12)

There exists a family of polytopes P ⊆ [0, 1]n with Chv´ atal rank rk(P) ≥ Ω(n2).

The polytope

cx = 1

2c1

The polytope

◮ Let c ∈ Zn ≥0 be a vector

P(c) := conv

• x ∈ {0, 1}n : cx ≤ c1

2

• Knapsack solutions

cx = 1

2c1

The polytope

◮ Let c ∈ Zn ≥0 be a vector

P(c) := conv

• x ∈ {0, 1}n : cx ≤ c1

2

• Knapsack solutions

cx = 1

2c1

( 1

2, . . . , 1 2)

The polytope

◮ Let c ∈ Zn ≥0 be a vector

P(c, ε) := conv x ∈ {0, 1}n : cx ≤ c1 2

• Knapsack solutions

∪ {x∗(ε)}

special vertex

• cx = 1

2c1

P PI ( 1

2, . . . , 1 2)

x∗(ε) = ( 1

2 + ε, . . . , 1 2 + ε)

Critical vectors

◮ Call ˜

c critical ⇔ ˜ c maximized at x∗ P x∗(ε)

Critical vectors

◮ Call ˜

c critical ⇔ ˜ c maximized at x∗ P x∗(ε) ˜ c

Critical vectors

◮ Call ˜

c critical ⇔ ˜ c maximized at x∗

◮ ˜

cx ≤ ⌊β⌋ cuts of x∗ = ⇒ ˜ c critical P x∗(ε) ˜ c

Critical vectors

◮ Call ˜

c critical ⇔ ˜ c maximized at x∗

◮ ˜

cx ≤ ⌊β⌋ cuts of x∗ = ⇒ ˜ c critical P

b b b b b b b b b b b b b b b b

b b

Critical vectors

◮ Call ˜

c critical ⇔ ˜ c maximized at x∗

◮ ˜

cx ≤ ⌊β⌋ cuts of x∗ = ⇒ ˜ c critical P

b b b b b b b b b b b b b b b b

b b

˜ c

Critical vectors

◮ Call ˜

c critical ⇔ ˜ c maximized at x∗

◮ ˜

cx ≤ ⌊β⌋ cuts of x∗ = ⇒ ˜ c critical P

b b b b b b b b b b b b b b b b

b b

˜ c ∼ ε

Overview

Overview critical vectors are long = ⇒ Ω(n2) rank

Overview critical vectors are good Simultaneous Diophantine Approximations to c critical vectors are long = ⇒ Ω(n2) rank

Overview random vector has no short, good SDA critical vectors are good Simultaneous Diophantine Approximations to c critical vectors are long = ⇒ Ω(n2) rank

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0

P (0)

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b ε1

P (1)

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b ε1 b ε2

P (2)

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b ε1 b ε2 b

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b ε1 b ε2 b b

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b ε1 b ε2 b b b εk

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b b b b b εk

x∗(εi) x∗(εi+1)

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b b b b b εk

˜ cx ≤ β

b

x∗(εi) x∗(εi+1)

◮ Let ˜

cx ≤ β be the GC cut “cutting furthest” in it. i

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b b b b b εk

˜ cx ≤ β

b

˜ cx ≤ ⌊β⌋

b

x∗(εi) x∗(εi+1)

◮ Let ˜

cx ≤ β be the GC cut “cutting furthest” in it. i

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b b b b b εk

˜ cx ≤ β

b

˜ cx ≤ ⌊β⌋

b

x∗(εi) x∗(εi+1)

◮ Let ˜

cx ≤ β be the GC cut “cutting furthest” in it. i ˜ cx∗(εi)−˜ cx∗(εi+1)

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b b b b b εk

˜ cx ≤ β

b

˜ cx ≤ ⌊β⌋

b

x∗(εi) x∗(εi+1)

◮ Let ˜

cx ≤ β be the GC cut “cutting furthest” in it. i 1 ≥ ˜ cx∗(εi)−˜ cx∗(εi+1)

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b b b b b εk

˜ cx ≤ β

b

˜ cx ≤ ⌊β⌋

b

x∗(εi) x∗(εi+1)

◮ Let ˜

cx ≤ β be the GC cut “cutting furthest” in it. i 1 ≥ ˜ cx∗(εi)−˜ cx∗(εi+1) = ˜ c1·(εi−εi+1)

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b b b b b εk

˜ cx ≤ β

b

˜ cx ≤ ⌊β⌋

b

x∗(εi) x∗(εi+1)

◮ Let ˜

cx ≤ β be the GC cut “cutting furthest” in it. i 1 ≥ ˜ cx∗(εi)−˜ cx∗(εi+1) = ˜ c1·(εi−εi+1) ≥ Ω( n εi )·(εi−εi+1)

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b b b b b εk

˜ cx ≤ β

b

˜ cx ≤ ⌊β⌋

b

x∗(εi) x∗(εi+1)

◮ Let ˜

cx ≤ β be the GC cut “cutting furthest” in it. i 1 ≥ ˜ cx∗(εi)−˜ cx∗(εi+1) = ˜ c1·(εi−εi+1) ≥ Ω( n εi )·(εi−εi+1)

◮ Then εi+1 εi

≥ 1 − Θ(1)

n

A lower bound strategy

Theorem

Assume: ∀ε ∈ [( 1

2)Θ(n), Θ(1)] :

˜ c critical = ⇒ ˜ c1 ≥ Ω( n

ε ).

Then rk(P) ≥ Ω(n2).

b ε0 b b b b b εk

˜ cx ≤ β

b

˜ cx ≤ ⌊β⌋

b

x∗(εi) x∗(εi+1)

◮ Let ˜

cx ≤ β be the GC cut “cutting furthest” in it. i 1 ≥ ˜ cx∗(εi)−˜ cx∗(εi+1) = ˜ c1·(εi−εi+1) ≥ Ω( n εi )·(εi−εi+1)

◮ Then εi+1 εi

≥ 1 − Θ(1)

n

= ⇒ k ≥ Ω(n2)

Overview critical vectors are long = ⇒ Ω(n2) rank

Overview critical vectors are good Simultaneous Diophantine Approximations to c critical vectors are long = ⇒ Ω(n2) rank

Simultaneous Diophantine Approximation

Lemma

Under magical assumptions ˜ c critical = ⇒ λ˜ c − c1 ≤ O(ε) · c1 (for some λ > 0)

◮ Intuition: ˜

c critical = ⇒ ˜ c well-approximates c P x∗(ε) c ˜ c

Simultaneous Diophantine Approximation

Lemma

Under magical assumptions ˜ c critical = ⇒ λ˜ c − c1 ≤ O(ε) · c1 (for some λ > 0)

◮ Intuition: ˜

c critical = ⇒ ˜ c well-approximates c P x∗(ε) c λ˜ c ˜ c

Simultaneous Diophantine Approximation

Lemma

Under magical assumptions ˜ c critical = ⇒ λ˜ c − c1 ≤ O(ε) · c1 (for some λ > 0)

◮ Intuition: ˜

c critical = ⇒ ˜ c well-approximates c P x∗(ε) c λ˜ c ˜ c

◮ Lemma follows from:

1 2 + ε

• ˜

c1

critical

≥ max{˜ cx | x ∈ PI} ≥ 1 2˜ c1 + Ω

• ˜

c − c λ

• 1

Simultaneous Diophantine Approximation

Lemma

Under magical assumptions ˜ c critical = ⇒ λ˜ c − c1 ≤ O(ε) · c1 (for some λ > 0)

◮ Intuition: ˜

c critical = ⇒ ˜ c well-approximates c P x∗(ε) c λ˜ c ˜ c

◮ Lemma follows from:

1 2 + ε

• ˜

c1

critical

≥ max{˜ cx | x ∈ PI} ≥ 1 2˜ c1 + Ω

• ˜

c − c λ

• 1
• to show:

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ˜

ci ci c1

1 2c1

c1 c2

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio).

˜ ci ci c1

1 2c1

c1 c2

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio).

˜ ci ci c1

1 2c1

c1 c2 ∈ J

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio).

˜ ci ci c1

1 2c1

c1 c2 ∈ J

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio).

˜ ci ci c1

1 2c1

c1 c2 ∈ J

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio).

˜ ci ci c1

1 2c1

c1 c2 ∈ J

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio).

˜ ci ci c1

1 2c1

c1 c2 ∈ J

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio).

˜ ci ci c1

1 2c1

c1 c2 ∈ J

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio).

˜ ci ci c1

1 2c1

c1 c2 ∈ J

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio). ◮ Pick λ > 0 s.t. i:˜ ci/ci>1/λ ci = c1 2

˜ ci ci c1

1 2c1

1 λ c1 c2 ∈ J

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio). ◮ Pick λ > 0 s.t. i:˜ ci/ci>1/λ ci = c1 2

˜ ci ci c1

1 2c1

1 λ c1 c2 ∈ J

• i∈J

˜ ci

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio). ◮ Pick λ > 0 s.t. i:˜ ci/ci>1/λ ci = c1 2

˜ ci ci c1

1 2c1

1 λ c1 c2 ∈ J

• i∈J

˜ ci = 1 2˜ c1+1 2

• i∈J

˜ ci −1 2

• i/

∈J

˜ ci

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio). ◮ Pick λ > 0 s.t. i:˜ ci/ci>1/λ ci = c1 2

˜ ci ci c1

1 2c1

1 λ c1 c2 ∈ J

• i∈J

˜ ci = 1 2˜ c1+1 2

• i∈J
• ˜

ci−ci λ

• −1

2

• i/

∈J

• ˜

ci−ci λ

Finding good knapsack solutions (the ideal case)

◮ We want: max{˜

cx | x ∈ PI} ≥ 1

2˜

c1 + Ω

• ˜

c − c

λ

• 1
• ◮ Sort ˜

c1 c1 > . . . > ˜ cn cn (i.e. according to profit cost ratio). ◮ Pick λ > 0 s.t. i:˜ ci/ci>1/λ ci = c1 2

˜ ci ci c1

1 2c1

1 λ c1 c2 ∈ J

• i∈J

˜ ci = 1 2˜ c1+1 2

• i∈J
• ˜

ci−ci λ

• −1

2

• i/

∈J

• ˜

ci−ci λ

• = 1

2˜ c1+1 2

• ˜

c− c λ

• 1

. . . now more realistic

˜ ci ci c1

1 2c1

c1 c2

. . . now more realistic

◮ Trick: Let c contain numbers 20, 21, 22, . . . , c∞ (3 copies)

˜ ci ci c1

1 2c1

c1 c2

◮ Knapsack capacity: 1 2c1

. . . now more realistic

◮ Trick: Let c contain numbers 20, 21, 22, . . . , c∞ (3 copies)

˜ ci ci c1

1 2c1

c1 c2

◮ Knapsack capacity: 1 2c1

. . . now more realistic

◮ Trick: Let c contain numbers 20, 21, 22, . . . , c∞ (3 copies)

˜ ci ci c1

1 2c1

c1 c2

◮ Knapsack capacity: 1 2c1

. . . now more realistic

◮ Trick: Let c contain numbers 20, 21, 22, . . . , c∞ (3 copies)

˜ ci ci c1

1 2c1

c1 c2

◮ Knapsack capacity: 1 2c1

. . . now more realistic

◮ Trick: Let c contain numbers 20, 21, 22, . . . , c∞ (3 copies)

˜ ci ci c1

1 2c1

c1 c2

◮ Knapsack capacity: 1 2c1

. . . now more realistic

◮ Trick: Let c contain numbers 20, 21, 22, . . . , c∞ (3 copies)

˜ ci ci c1

1 2c1

c1 c2

◮ Knapsack capacity: 1 2c1

. . . now more realistic

◮ Trick: Let c contain numbers 20, 21, 22, . . . , c∞ (3 copies)

˜ ci ci c1

1 2c1

c1 c2

◮ Knapsack capacity: 1 2c1

. . . now more realistic

◮ Trick: Let c contain numbers 20, 21, 22, . . . , c∞ (3 copies)

˜ ci ci c1

1 2c1

c1 c2

◮ Knapsack capacity: 1 2c1

. . . now more realistic

◮ Trick: Let c contain numbers 20, 21, 22, . . . , c∞ (3 copies)

˜ ci ci c1

1 2c1

1 λ c1 c2

◮ Knapsack capacity: 1 2c1

Overview critical vectors are good Simultaneous Diophantine Approximations to c critical vectors are long = ⇒ Ω(n2) rank

Overview random vector has no short, good SDA critical vectors are good Simultaneous Diophantine Approximations to c critical vectors are long = ⇒ Ω(n2) rank

Lemma

For D := 2n/8, pick ci ∈ {1, . . . , D} at random.

Lemma

For D := 2n/8, pick ci ∈ {1, . . . , D} at random. W.h.p. λ˜ c − c1 ≥ εc1 ∀ε ∀λ > 0 ∀˜ c1 ≤ o n ε

Lemma

For D := 2n/8, pick ci ∈ {1, . . . , D} at random. W.h.p. λ˜ c − c1 ≥ εc1 ∀ε ∀λ > 0 ∀˜ c1 ≤ o n ε

• ◮ Suffices: For fixed λ, ε,

(∗) Pr

• ∃˜

c∞ ≤ o 1 ε

• : λ˜

c − c∞ ≤ εD

• ≤ o(1)n

Lemma

For D := 2n/8, pick ci ∈ {1, . . . , D} at random. W.h.p. λ˜ c − c1 ≥ εc1 ∀ε ∀λ > 0 ∀˜ c1 ≤ o n ε

• ◮ Suffices: For fixed λ, ε,

(∗) Pr

• ∃˜

c∞ ≤ o 1 ε

• : λ˜

c − c∞ ≤ εD

• ≤ o(1)n

◮ Reason: Number of λ’s and ε’s is 2O(n); there must be a

Ω(n)-size subset of indices satisfying (∗)

Lemma

For D := 2n/8, pick ci ∈ {1, . . . , D} at random. W.h.p. λ˜ c − c1 ≥ εc1 ∀ε ∀λ > 0 ∀˜ c1 ≤ o n ε

• c1

c2 . . . . . . cn D

◮ Suffices: For fixed λ, ε,

(∗) Pr

• ∃˜

c∞ ≤ o 1 ε

• : λ˜

c − c∞ ≤ εD

• ≤ o(1)n

◮ Reason: Number of λ’s and ε’s is 2O(n); there must be a

Ω(n)-size subset of indices satisfying (∗)

Lemma

For D := 2n/8, pick ci ∈ {1, . . . , D} at random. W.h.p. λ˜ c − c1 ≥ εc1 ∀ε ∀λ > 0 ∀˜ c1 ≤ o n ε

• c1

c2 . . . . . . cn D λZ

• ( 1

ε) many ◮ Suffices: For fixed λ, ε,

(∗) Pr

• ∃˜

c∞ ≤ o 1 ε

• : λ˜

c − c∞ ≤ εD

• ≤ o(1)n

◮ Reason: Number of λ’s and ε’s is 2O(n); there must be a

Ω(n)-size subset of indices satisfying (∗)

Lemma

For D := 2n/8, pick ci ∈ {1, . . . , D} at random. W.h.p. λ˜ c − c1 ≥ εc1 ∀ε ∀λ > 0 ∀˜ c1 ≤ o n ε

• c1

c2 . . . . . . cn D λZ 2εD

◮ Suffices: For fixed λ, ε,

(∗) Pr

• ∃˜

c∞ ≤ o 1 ε

• : λ˜

c − c∞ ≤ εD

• ≤ o(1)n

◮ Reason: Number of λ’s and ε’s is 2O(n); there must be a

Ω(n)-size subset of indices satisfying (∗)

Lemma

For D := 2n/8, pick ci ∈ {1, . . . , D} at random. W.h.p. λ˜ c − c1 ≥ εc1 ∀ε ∀λ > 0 ∀˜ c1 ≤ o n ε

• c1

c2 . . . . . . cn D λZ 2εD

• (D)

◮ Suffices: For fixed λ, ε,

(∗) Pr

• ∃˜

c∞ ≤ o 1 ε

• : λ˜

c − c∞ ≤ εD

• ≤ o(1)n

◮ Reason: Number of λ’s and ε’s is 2O(n); there must be a

Ω(n)-size subset of indices satisfying (∗)

Lemma

For D := 2n/8, pick ci ∈ {1, . . . , D} at random. W.h.p. λ˜ c − c1 ≥ εc1 ∀ε ∀λ > 0 ∀˜ c1 ≤ o n ε

• c1

c2 . . . . . . cn D λZ 2εD

• (D)

◮ Suffices: For fixed λ, ε,

(∗) Pr

• ∃˜

c∞ ≤ o 1 ε

• : λ˜

c − c∞ ≤ εD

• ≤ o(1)n

◮ Reason: Number of λ’s and ε’s is 2O(n); there must be a

Ω(n)-size subset of indices satisfying (∗)

Overview random vector has no short, good SDA critical vectors are good Simultaneous Diophantine Approximations to c critical vectors are long = ⇒ Ω(n2) rank

The end Thanks for your attention

Where is the bottleneck for ω(n2) bound?

◮ Problem 1: Our proof technique does not extent! n 2 random numbers in {1, . . . , D} + n 2 “fill numbers”

cannot work for D ≫ 2n

◮ Problem 2: Set of normal vectors with ci ≥ 2Ω(n log n) is

extremely sparse! (2O(n2) potential normal vectors, but 2Ω(n2 log n) vectors with n log n bits per coefficient)

◮ Problem 3: For coefficients > 2ω(n), better SDAs exist!

For c ∈ [0, 1]n and N ∈ N. Find Q ∈ {1, . . . , N} s.t. minimize c − Zn

Q ∞.

◮ For Q := N, error ≤ 1

N

◮ Dirichlet’s Theorem: error ≤

1 Q·N 1/n