New Conjectures in the Geometry of Numbers Daniel Dadush Centrum - - PowerPoint PPT Presentation

New Conjectures in the Geometry of Numbers

Daniel Dadush

Centrum Wiskunde & Informatica (CWI)

Oded Regev

New York University

1. A Reverse Minkowski Inequality & its conjectured Strengthening.

• 2. Strong Reverse Minkowski implies

the Kannan & Lovász conjecture for ℓ2.

• 3. From Decomposing Integer Programs to

the general Kannan & Lovász conjecture.

Talk Outline

The standard integer lattice ℤ𝑜.

Lattices

ℤ𝑜

𝑓1 𝑓2

A lattice ℒ ⊆ ℝ𝑜 is 𝐶ℤ𝑜 for a basis 𝐶 = 𝑐1, … , 𝑐𝑜 . ℒ(𝐶) denotes the lattice generated by 𝐶. Note: a lattice has many equivalent bases.

𝑐2 𝑐2 𝑐1 𝑐2 𝑐1

Lattices

ℒ

A lattice ℒ ⊆ ℝ𝑜 is 𝐶ℤ𝑜 for a basis 𝐶 = 𝑐1, … , 𝑐𝑜 . ℒ(𝐶) denotes the lattice generated by 𝐶. The determinant of ℒ is | det 𝐶 |.

𝑐1 𝑐2

Lattices

ℒ

Lattices

ℒ 𝑐1 𝑐2

A lattice ℒ ⊆ ℝ𝑜 is 𝐶ℤ𝑜 for a basis 𝐶 = 𝑐1, … , 𝑐𝑜 . ℒ(𝐶) denotes the lattice generated by 𝐶. The determinant of ℒ is | det 𝐶 |. Equal to volume of any tiling set.

A lattice ℒ ⊆ ℝ𝑜 is 𝐶ℤ𝑜 for a basis 𝐶 = 𝑐1, … , 𝑐𝑜 . The dual lattice is ℒ∗ = {𝑧 ∈ span ℒ : 𝑧T𝑦 ∈ ℤ ∀𝑦 ∈ ℒ} ℒ∗ is generated by 𝐶−T. The identity det ℒ∗ = Τ 1 det(ℒ) holds.

Lattices

ℒ

𝑧 ∈ ℒ∗

A lattice ℒ ⊆ ℝ𝑜 is 𝐶ℤ𝑜 for a basis 𝐶 = 𝑐1, … , 𝑐𝑜 . The dual lattice is ℒ∗ = {𝑧 ∈ span ℒ : 𝑧T𝑦 ∈ ℤ ∀𝑦 ∈ ℒ} ℒ∗ is generated by 𝐶−T.

Lattices

ℒ

𝑧T𝑦 = 0 𝑧T𝑦 = 1 𝑧T𝑦 = 2 𝑧T𝑦 = 3 𝑧T𝑦 = 4

Reversing Minkowski’s Convex Body Theorem

Minkowski’s Convex Body Theorem

Theorem [Minkowski]: For an 𝑜-dimensional symmetric convex body 𝐿 and lattice ℒ, we have that 𝐿 ∩ ℒ ≥ ⌈ Τ 2−𝑜vol𝑜 𝐿 det ℒ ⌉ ℒ 𝐿

Minkowski’s Convex Body Theorem

Theorem [Minkowski]: For an 𝑜-dimensional symmetric convex body 𝐿 and lattice ℒ, we have that 𝐿 ∩ ℒ ≥ ⌈ Τ 2−𝑜vol𝑜 𝐿 det ℒ ⌉ Question: Is the above lower bound also “close” to being an upper bound?

Minkowski’s Convex Body Theorem

Theorem [Minkowski]: For an 𝑜-dimensional symmetric convex body 𝐿 and lattice ℒ, we have that 𝐿 ∩ ℒ ≥ ⌈ Τ 2−𝑜vol𝑜 𝐿 det ℒ ⌉ ℒ 𝐿 Clearly NO!

Minkowski’s Convex Body Theorem

Theorem [Minkowski]: For an 𝑜-dimensional symmetric convex body 𝐿 and lattice ℒ, we have that 𝐿 ∩ ℒ ≥ ⌈ Τ 2−𝑜vol𝑜 𝐿 det ℒ ⌉ ℒ 𝐿 Can we strengthen the lower bound? Points all lie in a lattice subspace.

Reverse Minkowski Inequality

For a symmetric convex body 𝐿 and lattice ℒ, let MB 𝐿, ℒ = max

𝑒≥0

max

𝑋 𝑚𝑏𝑢. 𝑡𝑣𝑐. dim 𝑋 =𝑒

2−𝑒 vold(𝐿 ∩ 𝑋) det(ℒ ∩ 𝑋) ℒ 𝐿 𝑋 𝑋 is a lattice subspace of ℒ if dim 𝑋 ∩ ℒ = dim(𝑋).

Reverse Minkowski Inequality

For a symmetric convex body 𝐿 and lattice ℒ, let MB 𝐿, ℒ = max

𝑒≥0

max

𝑋 𝑚𝑏𝑢. 𝑡𝑣𝑐. dim 𝑋 =𝑒

2−𝑒 vold(𝐿 ∩ 𝑋) det(ℒ ∩ 𝑋) ℒ 𝐿 𝑋 Is this bound any better?

Weak Reverse Minkowski

For a symmetric convex body 𝐿 and lattice ℒ, let MB 𝐿, ℒ = max

𝑒≥0

max

𝑋 𝑚𝑏𝑢. 𝑡𝑣𝑐. dim 𝑋 =𝑒

2−𝑒 vold(𝐿 ∩ 𝑋) det(ℒ ∩ 𝑋) Let 𝐶2

𝑜 denote the unit Euclidean ball.

Theorem [D.,Regev `16]: For an 𝑜-dimensional lattice ℒ 𝐶2

𝑜 ∩ 𝑀 ≤ MB(6 𝑜𝐶2 𝑜, ℒ).

Furthermore, for any symmetric convex body 𝐿 𝐿 ∩ 𝑀 ≤ MB(6𝑜𝐿, ℒ).

Strong Reverse Minkowski

For a symmetric convex body 𝐿 and lattice ℒ, let MB 𝐿, ℒ = max

𝑒≥0

max

𝑋 𝑚𝑏𝑢. 𝑡𝑣𝑐. dim 𝑋 =𝑒

2−𝑒 vold(𝐿 ∩ 𝑋) det(ℒ ∩ 𝑋) Let 𝐶2

𝑜 denote the unit Euclidean ball.

Conjecture [D.,Regev `16]: For an 𝑜-dimensional lattice ℒ 𝐶2

𝑜 ∩ 𝑀 ≤ MB(𝑃( log 𝑜)𝐶2 𝑜, ℒ).

Furthermore, for any symmetric convex body 𝐿 𝐿 ∩ 𝑀 ≤ MB(𝑃(log 𝑜)𝐿, ℒ). Tight example: 𝐿 = 𝐶1

𝑜 and ℒ = ℤ𝑜.

𝑧1

• 𝑧1

𝜇1𝐿

Symmetric convex body 𝐿 and lattice ℒ in ℝ𝑜. 𝜇𝑗 𝐿, ℒ = inf 𝑡 ≥ 0: dim ℒ ∩ 𝑡𝐿 ≥ 𝑗 , 𝑗 𝜗[𝑜]

Successive Minima

𝑧2 𝜇2𝐿

• 𝑧2

𝑧2 𝜇2𝐿

• 𝑧2

𝑧1

• 𝑧1

𝜇1𝐿

Symmetric convex body 𝐿 and lattice ℒ in ℝ𝑜. Π𝑗=1

𝑜 𝜇𝑗 𝐿, ℒ ≤ 2𝑜 det ℒ

vol𝑜 𝐿 ≤ 𝑜! Π𝑗=1

𝑜 𝜇𝑗(𝐿, ℒ)

Minkowski’s Second Theorem

Theorem [Henk `02]: 𝐿 ∩ ℒ ≤ 2𝑜−1 ς𝑗=1

𝑜 ⌊1 + 2 𝜇𝑗 𝐿,ℒ ⌋

Lattice points bounds via Minima

ℒ 𝐿

Must show 𝐿 ∩ ℒ ≤ max

𝑒≥0

max

𝑋 𝑚𝑏𝑢. 𝑡𝑣𝑐. dim 𝑋 =𝑒

2−𝑒 vold(6𝑜𝐿∩𝑋)

det(ℒ∩𝑋)

If max

𝑗∈[𝑜] 𝜇𝑗 𝐿, ℒ > 1, then 𝐿 ∩ ℒ is lower dimensional and

we can induct. If max

𝑗∈[𝑜] 𝜇𝑗 𝐿, ℒ ≤ 1, then we have that

𝐿 ∩ ℒ ≤ 2𝑜−1 ς𝑗=1

𝑜

1 +

2 𝜇𝑗 𝐿,ℒ

(by Henk) ≤ 2𝑜−1 ς𝑗=1

𝑜 3 𝜇𝑗 𝐿,ℒ ≤ 6𝑜 ς𝑗=1 𝑜 1 𝜇𝑗 𝐿,ℒ

≤ 𝑜! 3𝑜 Τ voln(K) det ℒ (by Minkowski) ≤ Τ voln 3nK det ℒ ≤ MB(6𝑜𝐿, ℒ)

Proof of Weak Minkowski

The Kannan & Lovász conjecture for ℓ2

ℓ2 covering radius

ℒ Let 𝜈 𝐶2

𝑜, ℒ = max 𝑢∈ℝ𝑜 min 𝑧∈ℒ 𝑢 − 𝑧 2 , i.e. the

farthest distance from the lattice.

𝜈

Main question: How to get good lower bounds on the covering radius?

Lower bounds for covering radius

Lemma: 𝜈 𝐶2

𝑜, ℒ ≥ 1 2𝜇1(𝐶2

𝑜,ℒ∗)

𝑧 ∈ ℒ∗

ℒ

𝑧T𝑦 = 1 𝑧T𝑦 = 2 𝑧T𝑦 = 3 𝑧T𝑦 = 4

𝑢

Lower bounds for covering radius

Lemma: 𝜈 𝐶2

𝑜, ℒ ≥ det ℒ

1 𝑜

voln 𝐶2

𝑜 1 𝑜

= Ω( 𝑜) det ℒ 1/𝑜 ℒ

𝜈

Since 𝒲 ⊆ 𝜈𝐶2

𝑜,

voln 𝒲 = det(ℒ) ≤ voln 𝜈B2

𝑜 = 𝜈𝑜voln 𝐶2 𝑜

𝒲

Lower bounds for covering radius

Lemma: Let 𝜌𝑋 denote the orthogonal projection

• nto a 𝑒-dimensional subspace 𝑋 (*). Then

𝜈 𝐶2

𝑜, ℒ ≥ det 𝜌𝑋 ℒ

1 𝑒

vold 𝐶2

𝑒 1 𝑒

=

1 det ℒ∗∩ 𝑋

1 𝑒vold 𝐶2 𝑒 1 𝑒

= Ω 1

𝑒 det ℒ∗∩𝑋

1 𝑒

. Pf: For the first inequality, since orthogonal projections shrink distances we get 𝜈 𝐶2

𝑜, ℒ ≥ 𝜈 𝜌𝑋 𝐶2 𝑜 , 𝜌𝑋 ℒ

. The second equality follows from the identity 𝜌𝑋 ℒ ∗ = ℒ∗ ∩ 𝑋.

Kannan & Lovász for ℓ2

Theorem [Kannan-Lovasz `88]: Ω(1) ≤ 𝜈 𝐶2

𝑜, ℒ

min

𝑚𝑏𝑢.subspace W 1≤dim 𝑋 =𝑒≤𝑜 det ℒ∗∩𝑋

1 𝑒

𝑒

≤ 𝑃( 𝑜)

Kannan & Lovász for ℓ2

Conjecture [Kannan-Lovász `88]: Ω 1 ≤ 𝜈 𝐶2

𝑜, ℒ

min

𝑚𝑏𝑢.subspace W 1≤dim 𝑋 =𝑒≤𝑜 det ℒ∗∩𝑋 1/𝑒 𝑒

= O( log 𝑜) Remark: implies that there are very good NP-certificates for showing that the covering radius is large.

Reverse Minkowski vs Kannan & Lovász

Theorem [D. Regev `16]: If for any 𝑜-dimensional lattice ℒ 𝐶2

𝑜 ∩ ℒ ≤ MB(𝑔 𝑜 𝐶2 𝑜, ℒ)

for a non-decreasing function 𝑔(𝑜), then the Kannan & Lovász conjecture for ℓ2 holds with bound 𝑃 log 𝑜 𝑔 𝑜 .

MB 𝐶2

𝑜, ℒ = max 𝑒≥0

max

𝑋 𝑚𝑏𝑢. 𝑡𝑣𝑐. dim 𝑋 =𝑒

2−𝑒 vold(𝐶2

𝑜 ∩ 𝑋)

det(ℒ ∩ 𝑋)

Main Approach

1. Use convex programming relaxation for 𝜈 𝐶2

𝑜, ℒ 2:

𝜈 𝐶2

𝑜, ℒ 2 ≤ 𝑃 1

min trace 𝐵 s.t. σ𝑧∈ℒ∗∖{0} 𝑓−𝑧T𝐵𝑧 ≤ 1, A psd.

• 2. Use Reverse Minkowski to formulate an approximate

dual to the above program.

• 3. Round / massage optimal dual solution to get the

subspace 𝑋.

A Sufficient Conjecture

𝜈 𝐶2

𝑜, ℒ 2 ≤ 𝑃 1

min trace 𝐵 s.t. σ𝑧∈ℒ∗∖{0} 𝑓−𝑧T𝐵𝑧 ≤ 1, A psd. To formulate the required dual for the above program, we can rely on the following weaker conjecture: Conjecture [D. Regev `16]: Ω 1 ≤ max

𝑠>0

log 𝑠𝐶2

𝑜 ∩ ℒ

𝑠 min

𝑚𝑏𝑢.𝑡𝑣𝑐. 𝑋 1≤dim 𝑋 =𝑒

det ℒ ∩ 𝑋

1 𝑒

≤ 𝑃 log 𝑜

The Relaxed Program

min trace 𝐵 s.t. σ𝑧∈ℒ∗∖{0} 𝑓−𝑧T𝐵𝑧 ≤ 1, A psd. We relax the above using the following weaker constraints det𝑋 𝐵 ≥

1 det ℒ∗∩ 𝑋 2 ∀ 𝑚𝑏𝑢. 𝑡𝑣𝑐𝑡𝑞𝑏𝑑𝑓 𝑋

where det𝑋(𝐵) ≔ det(𝑃𝑋

𝑈 𝐵𝑃𝑋), where 𝑃𝑋 is any

• rthonormal basis of 𝑋.

This relaxation makes the value of the program drop by at most an 𝑔 𝑜 2 factor (the Reverse Minkowski bound).

The Dual Program

min trace 𝐵 s.t. det𝑋 𝐵 ≥

1 det ℒ∗∩ 𝑋 2 ∀ 𝑚𝑏𝑢. 𝑡𝑣𝑐𝑡𝑞𝑏𝑑𝑓 𝑋, A psd.

The strong dual for the above program is max

k≥0 σ𝑗=1 𝑙

𝑒𝑗

det𝑋𝑗 𝑌𝑗

Τ 1 𝑒𝑗

det ℒ∗∩𝑋𝑗

Τ 2 𝑒𝑗

s.t. σ𝑗=1

𝑙

𝑌𝑗 ≼ 𝐽𝑜 Here we range over all finite 𝑙, where for 𝑗 ∈ [𝑙], 𝑋

𝑗 is a lattice subspace of ℒ∗, dim 𝑋 𝑗 = 𝑒𝑗 and

𝑌𝑗 is a PSD matrix with image 𝑋

𝑗.

The Dual Program

max

k≥0 σ𝑗=1 𝑙

𝑒𝑗

det𝑋𝑗 𝑌𝑗

Τ 1 𝑒𝑗

det ℒ∗∩𝑋𝑗

Τ 2 𝑒𝑗

s.t. σ𝑗=1

𝑙

𝑌𝑗 ≼ 𝐽𝑜 Theorem [D., Regev `16]: above value is at most 𝑃 log2 𝑜 max

𝑚𝑏𝑢.subspace W 1≤dim 𝑋 =𝑒≤𝑜 𝑒 det ℒ∗∩𝑋𝑗

Τ 2 𝑒

Corresponds to setting 𝑙 = 1 and 𝑌1 = projection on 𝑋

1.

Main idea: disentangle the subspaces via “uncrossing”.

The Dual Program

max

k≥0 σ𝑗=1 𝑙

𝑒𝑗

det𝑋𝑗 𝑌𝑗

Τ 1 𝑒𝑗

det ℒ∗∩𝑋𝑗

Τ 2 𝑒𝑗

s.t. σ𝑗=1

𝑙

𝑌𝑗 ≼ 𝐽𝑜 Theorem [D., Regev `16]: above value is at most 𝑃 log2 𝑜 max

𝑚𝑏𝑢.subspace W 1≤dim 𝑋 =𝑒≤𝑜 𝑒 det ℒ∗∩𝑋𝑗

Τ 2 𝑒

Implies ℓ2 Kannan & Lovász bound of 𝑃 log 𝑜 𝑔 𝑜 . (note above programs approximate 𝜈 𝐶2

𝑜, ℒ 2)

Integer Programming and the general Kannan & Lovász conjecture

Anatomy of an IP Algorithm

ℤ𝑜 𝐿

Problem: Find point in 𝐿 ∩ ℤ𝑜 or decide that 𝐿 is integer free.

Anatomy of an IP Algorithm

ℤ𝑜 𝐿

Problem: Find point in 𝐿 ∩ ℤ𝑜 or decide that 𝐿 is integer free.

𝑧

𝑑

Anatomy of an IP Algorithm

ℤ𝑜

Main dichotomy:

• 1. Either 𝐿 contains a “deep” integer point:

find 𝑧 by “rounding” from center 𝑑 of 𝐿.

𝐿

𝑧

Anatomy of an IP Algorithm

ℤ𝑜

Main dichotomy:

• 1. Either 𝐿 contains a “deep” integer point:

find 𝑧 by “rounding” from center 𝑑 of 𝐿.

2.

𝐿 is “flat”: decompose feasible region into lower dimensional subproblems and recurse.

𝐿

𝑦2=1 𝑦2=2 subproblems

IP Algorithms

Time Space Authors 2𝑃(𝑜3) poly(𝑜) Lenstra 83 2𝑃(𝑜)𝑜2.5𝑜 poly(𝑜) Kannan 87 2𝑃(𝑜)𝑜2𝑜 2𝑜 Hildebrand Köppe 10 2𝑃(𝑜)𝑜𝑜 2𝑜

• D. Peikert Vempala 11,
• D. 12

𝑜𝑃(𝑜) barrier: don’t know how to create less than 𝑜𝑃(1) subproblems per dimension…

IP Algorithms

Time Space Authors 2𝑃(𝑜3) poly(𝑜) Lenstra 83 2𝑃(𝑜)𝑜2.5𝑜 poly(𝑜) Kannan 87 2𝑃(𝑜)𝑜2𝑜 2𝑜 Hildebrand Köppe 10 2𝑃(𝑜)𝑜𝑜 2𝑜

• D. Peikert Vempala 11,
• D. 12

Conjecture Kannan & Lovász `88: 𝑃(log 𝑜) subproblems per dimension should be possible!

General Covering Radius & Deep Points

The covering radius of 𝐿 with respect to ℤ𝑜 is 𝜈 𝐿, ℤ𝑜 = min {𝑡 ≥ 0: ∀𝑢 ∈ ℝ𝑜, (𝑡𝐿 + 𝑢) ∩ ℤ𝑜 ≠ ∅}

• r, smallest scaling 𝑡 such that every shift 𝑡𝐿 + 𝑢

contains an integer point.

ℤ2

𝐿

General Covering Radius & Deep Points

The covering radius of 𝐿 with respect to ℤ𝑜 is 𝜈 𝐿, ℤ𝑜 = min {𝑡 ≥ 0: ∀𝑢 ∈ ℝ𝑜, (𝑡𝐿 + 𝑢) ∩ ℤ𝑜 ≠ ∅}

• r, smallest scaling 𝑡 such that every shift 𝑡𝐿 + 𝑢

contains an integer point.

ℤ2

𝐿+ 𝑢

𝜈 𝐿, ℤ2 > 1

The covering radius of 𝐿 with respect to ℤ𝑜 is 𝜈 𝐿, ℤ𝑜 = min {𝑡 ≥ 0: ∀𝑢 ∈ ℝ𝑜, (𝑡𝐿 + 𝑢) ∩ ℤ𝑜 ≠ ∅}

General Covering Radius & Deep Points

ℤ2

2𝐿

𝜈 𝐿, ℤ2 = 2

General Covering Radius & Deep Points

Integral shifts of 2𝐿 “cover” ℝ2, i.e. 2𝐿 + ℤ2 = ℝ2.

2𝐿

ℤ2

𝑑

If covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, then any shift of 𝐿 contains a “deep” integer point.

General Covering Radius & Deep Points

ℤ2

𝐿 + t1

𝑑

If covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, then any shift of 𝐿 contains a “deep” integer point.

General Covering Radius & Deep Points

ℤ2

𝐿 + t1

𝑑

If covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, then any shift of 𝐿 contains a “deep” integer point.

General Covering Radius & Deep Points

ℤ2

𝐿 + t2

𝑑

If covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, then any shift of 𝐿 contains a “deep” integer point.

General Covering Radius & Deep Points

ℤ2

𝐿 + t3

𝑑

Thus, if 𝜈 𝐿, ℤ𝑜 ≤ 1/2 finding an integer point in 𝐿 is “easy”.

General Covering Radius & Deep Points

ℤ2

𝐿

Thus, if 𝜈 𝐿, ℤ𝑜 ≤ 1/2 finding an integer point in 𝐿 is “easy”. If not, we should decompose 𝐿 into lower dimensional pieces…

General Covering Radius & Deep Points

ℤ2

𝐿

𝐿

Lower Bounds on the Covering Radius

𝑧 𝑧T𝑦 = .9 𝑧T𝑦 = .1

Lemma: 1 ≤ min

𝑧∈ℤ𝑜∖ 0 width𝐿 𝑧

𝜈 𝐿, ℤ𝑜

To cover space, 𝐿 must have width at least 1. .8

𝐿 𝐿 𝐿

Theorem: For 𝐿 ⊆ ℝ𝑜 convex, either

• 1. covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, or
• 2. ∃ y ∈ ℤ𝑜 such that 𝐿 has width ෨

𝑃(𝑜 Τ

4 3) w.r.t. 𝑧:

Khinchine’s Flatness Theorem

ℤ2

𝐿 𝐿

Theorem: For 𝐿 ⊆ ℝ𝑜 convex, either

• 1. covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, or
• 2. ∃ y ∈ ℤ𝑜 ∖ {0} such that 𝐿 has width ෨

𝑃(𝑜 Τ

4 3) w.r.t. 𝑧:

Khinchine’s Flatness Theorem

ℤ2

𝐿

Theorem: For 𝐿 ⊆ ℝ𝑜 convex, either

• 1. covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, or
• 2. ∃ y ∈ ℤ𝑜 ∖ 0 such that 𝐿 has width ෨

𝑃(𝑜 Τ

4 3) w.r.t. 𝑧:

Khinchine’s Flatness Theorem

ℤ2 𝑧 𝑧T𝑦 = 2.3 𝑧T𝑦 = .8 1.5

𝐿

Theorem: For 𝐿 ⊆ ℝ𝑜 convex, either

• 1. covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, or
• 2. ∃ y ∈ ℤ𝑜 ∖ 0 such that 𝐿 has width ෨

𝑃(𝑜 Τ

4 3) w.r.t. 𝑧.

Khinchine’s Flatness Theorem

ℤ2 𝑧 𝑧T𝑦 = 0 𝑧T𝑦 = 2 𝑧T𝑦 = 1 𝑧T𝑦 = 3

Khinchine’s Flatness Theorem

Theorem: 1 ≤ min

𝑧∈ℤ𝑜∖ 0 width𝐿 𝑧

𝜈 𝐿, ℤ𝑜 ≤ ෨ 𝑃(𝑜 Τ

4 3)

To cover space, suffices to scale 𝐿 to have width somewhere between 1 and ෨ 𝑃(𝑜 Τ

4 3).

[Khinchine `48, Babai `86, Hastad `86, Lenstra-Lagarias- Schnorr `87, Kannan-Lovasz `88, Banaszczyk `93-96, Banaszczyk-Litvak-Pajor-Szarek `99, Rudelson `00]

Bounds improvements for general bodies: Khinchine `48: (𝑜 + 1)! Babai `86: 2𝑃(𝑜) Lenstra-Lagarias-Schnorr `87: 𝑜5/2 Kannan-Lovasz `88: 𝑜2 Banaszczyk et al `99: 𝑜3/2 Rudelson `00: 𝑜4/3logc 𝑜

Khinchine’s Flatness Theorem

Bound conjectured to be 𝑃(𝑜) (best possible). Best known bounds: Ellipsoids: Θ(𝑜)

[Banaszczyk `93]

Centrally Symmetric: 𝑃(𝑜 log 𝑜) [Banaszczyk `96] General: ෨ 𝑃(𝑜 Τ

4 3)

[Banaszczyk `96, Rudelson `00]

ℤ𝑜 𝐿 = {𝑦: σ𝑗 𝑦𝑗 ≤ 𝑜, 𝑦 ≥ 0} 𝐿

Khinchine’s Flatness Theorem

Recursion only uses integrality of “one variable” at a time. Why?

How can we improve decompositions?

𝐿 ℤ2 𝑧 𝑧T𝑦 = 0 𝑧T𝑦 = 2 𝑧T𝑦 = 1 𝑧T𝑦 = 3

Reason: It’s very efficient to enumerate integer points in an interval.

2.3 .8

When else is it efficient to enumerate integer points? Theorem [D. Peikert Vempala `11, D. `12]: For 𝐿 ⊆ ℝ𝑒 convex, can enumerate 𝐿 ∩ ℤ𝑒 in time 2𝑃(𝑒) max

𝑢∈ℝ𝑒 |𝐿 + 𝑢| .

How can we improve decompositions?

𝐿

When else is it efficient to enumerate integer points? Theorem [D. Peikert Vempala `11, D. `12]: For 𝐿 ⊆ ℝ𝑒 convex, can enumerate 𝐿 ∩ ℤ𝑒 in time 2𝑃(𝑒) max

𝑢∈ℝ𝑒 |𝐿 + 𝑢| .

How can we improve decompositions?

𝐿+𝑢

Generalized Decomposition Strategy for 𝐿: Choose 𝑒 ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜 rank 𝑒 such that

How can we improve decompositions?

𝑄(𝐿)

max

𝑢∈ℝ𝑒 𝑄 𝐿 + 𝑢 1/𝑒 is “small”

Generalized Decomposition Strategy for 𝐿: Choose 𝑒 ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜 rank 𝑒 such that

How can we improve decompositions?

𝑄(𝐿)

max

𝑢∈ℝ𝑒 𝑄 𝐿 + 𝑢 1/𝑒 is “small”

Enumerate 𝑄 𝐿 ∩ ℤ𝑒 and recurse on subproblems.

subproblems

How should we choose 𝑒 ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜? Hardest bodies to decompose satisfy 𝜈 𝐿, ℤ𝑜 = 1/2, i.e. as “fat” as possible without containing deep points.

How can we improve decompositions?

𝐿

How should we choose 𝑒 ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜? If 𝜈 𝐿, ℤ𝑜 = 1/2 then also 𝜈 𝑄 𝐿 , ℤ𝑒 ≤ 1/2, so integer projections of 𝐿 are only “fatter”.

How can we improve decompositions?

𝐿 𝑄(𝐿)

How should we choose 𝑒 ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜? If 𝜈 𝑄 𝐿 , ℤ𝑒 ≤ 1/2, then max

𝑢∈ℝ𝑒 𝑄 𝐿 + 𝑢 1/𝑒 = 𝑃 1 vold P K 1/𝑒.

How can we improve decompositions?

Good choice of 𝑒 ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜 should approximately minimize min

𝑒≥1

min

𝑄∈ℤ𝑒×𝑜 𝑠𝑏𝑜𝑙 𝑒

vold P K

1/𝑒.

Lemma: If 𝐿 + ℤ𝑜 = ℝ𝑜, then voln 𝐿 ≥ 1.

𝐿

ℤ2

Volumetric bounds on the Covering Radius

Lemma: If 𝐿 + ℤ𝑜 = ℝ𝑜, then voln 𝐿 ≥ 1.

Pf: Let 𝐷 = 0,1 𝑜 be the cube. voln 𝐿 = σ𝑧∈ℤ𝑜 voln 𝐿 ∩ 𝐷 + 𝑧 = σ𝑧∈ℤ𝑜 vol𝑜( 𝐿 − 𝑧 ∩ 𝐷) ≥ voln 𝐿 + ℤ𝑜 ∩ 𝐷 = vol 𝐷 = 1. 𝐿

ℤ2

Volumetric bounds on the Covering Radius

𝐷

Lemma: If 𝐿 + ℤ𝑜 = ℝ𝑜, then voln 𝐿 ≥ 1. Lemma: Let 𝑄 ∈ ℤ𝑙×𝑜 have rank 𝑙. If 𝐿 + ℤ𝑜 = ℝ𝑜, then volk 𝑄𝐿 ≥ 1. Pf: ℝ𝑙 = 𝑄 𝐿 + ℤ𝑜 = 𝑄 𝐿 + 𝑄 ℤ𝑜 ⊆ 𝑄 𝐿 + ℤ𝑙.

Volumetric bounds on the Covering Radius

Lemma: Let 𝑄 ∈ ℤ𝑒×𝑜 have rank 𝑒. If 𝐿 + ℤ𝑜 = ℝ𝑜, then vold 𝑄𝐿 ≥ 1. Lemma: For a convex body 𝐿 ⊆ ℝ𝑜 𝜈 𝐿, ℤ𝑜 ≥ max

𝑒∈[𝑜] max P∈ℤ𝑒×𝑜 rank 𝑒 1 vol𝑒 𝑄𝐿 1/𝑒

Pf: Must scale 𝐿 to have volume at least 1 in each integer projection.

Volumetric bounds on the Covering Radius

Kannan Lovász Conjecture

Theorem [Kannan-Lovász `88]: For 𝐿 ⊆ ℝ𝑜 convex, either

• 1. Covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, or
• 2. ∃ d ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜 rank 𝑒 such that vold 𝑄𝐿 ≤ 2𝑜 𝑒.

In particular, 1 ≤ 𝜈 𝐿, ℤ𝑜 min

𝑒∈ 𝑜

min

P∈ℤ𝑒×𝑜 rank 𝑒

vol𝑒 𝑄𝐿

1 𝑒 ≤ 2𝑜

Proof is constructive. Enables a 2𝑃(𝑜)𝑜𝑜 time IP algorithm [D. `12].

𝐿 ℤ2

Kannan Lovász Conjecture

Theorem [Kannan-Lovász `88]: For 𝐿 ⊆ ℝ𝑜 convex, either

• 1. Covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, or
• 2. ∃ d ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜 rank 𝑒 such that vold 𝑄𝐿 ≤ 2𝑜 𝑒.

𝑓𝑦: 𝑒 = 2, 𝑄 = 1 1

𝐿 ℤ2

Kannan Lovász Conjecture

subproblems 𝑓𝑦: 𝑒 = 2, 𝑄 = 1 1

Theorem [Kannan-Lovász `88]: For 𝐿 ⊆ ℝ𝑜 convex, either

• 1. Covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, or
• 2. ∃ d ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜 rank 𝑒 such that vold 𝑄𝐿 ≤ 2𝑜 𝑒.

𝐿 ℤ2

Kannan Lovász Conjecture

𝑓𝑦: 𝑒 = 1, 𝑄 = 0 1 subproblems Same as standard flatness.

Theorem [Kannan-Lovász `88]: For 𝐿 ⊆ ℝ𝑜 convex, either

• 1. Covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, or
• 2. ∃ d ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜 rank 𝑒 such that vold 𝑄𝐿 ≤ 2𝑜 𝑒.

Kannan Lovász Conjecture

Conjecture [Kannan-Lovász `88]: For 𝐿 ⊆ ℝ𝑜 convex, either

• 1. Covering radius 𝜈 𝐿, ℤ𝑜 ≤ 1/2, or
• 2. ∃ d ≥ 1, 𝑄 ∈ ℤ𝑒×𝑜 rank 𝑒 such that

vold 𝑄𝐿 ≤ 𝑃 log 𝑜 𝑒. In particular, 1 ≤ 𝜈 𝐿, ℤ𝑜 min

𝑒∈ 𝑜

min

P∈ℤ𝑒×𝑜 rank 𝑒

vol𝑒 𝑄𝐿

1 𝑒 ≤ 𝑃(log 𝑜)

An efficient enough construction would imply a 2𝑃 𝑜 log 𝑜 𝑃(𝑜) time IP algorithm!

Conclusions

1. New natural conjectured converse for Minkowski’s convex body theorem. 2. Showed that it implies an important special case of a conjecture of Kannan & Lovász.

Open Problems

1. Prove or disprove any of the conjectures!

• 2. Find new applications.

THANK YOU!