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Daniel Dadush Centrum Wiskunde & Informatica (CWI) Aussois 2019
On Approximating the Covering Radius and Finding Dense Lattice Subspaces Daniel Dadush Centrum Wiskunde & Informatica (CWI) Aussois 2019
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Lattices A lattice ββ β π is π΅ β€ π for a basis π΅= π 1 ,β¦, π π .
β(π΅) denotes the lattice generated by π΅. Note: a lattice has many equivalent bases. π 1 π 2 β
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Lattices A lattice ββ β π is π΅ β€ π for a basis π΅= π 1 ,β¦, π π .
β(π΅) denotes the lattice generated by π΅. The determinant of β is | det π΅ | . π 1 π 2 β
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Lattices A lattice ββ β π is π΅ β€ π for a basis π΅= π 1 ,β¦, π π .
β(π΅) denotes the lattice generated by π΅. The determinant of β is | det π΅ | . Equal to volume of any tiling set. π 1 π 2 β
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Notation πββ sublattice of dimension π Normalized Determinant:
nd π β det π 1 π Projected Sublattice: β π β Ξ span π β₯ (β)
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β 2 Covering Radius π β βπ π΅ 2 π ,β
π β βπ π΅ 2 π ,β Distance of farthest point to the lattice β. π π± β Voronoi cell π±β all points closest to 0
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Distance of farthest point to the lattice β.
β 2 Covering Radius π β βπ π΅ 2 π ,β Distance of farthest point to the lattice β. π π± β Question: Does the covering radius admit a βgoodβ dual characterization?
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Volumetric Lower Bounds
vol π π΅ 2 π π β β₯ vol π π± = det (β) π π± β
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Volumetric Lower Bounds
π(β) β₯ vol π π΅ 2 π β 1 π det β 1 π π π± β
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Volumetric Lower Bounds
π β βΏ π det β 1 π β dim β nd(β) π π± β
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Volumetric Lower Bounds
π β β₯π β/π βΏ dim β π nd(β/π) β π β/π π
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β 2 Kannan-LovΓ‘sz Constant
Define πΆ πΎπΏ,2 (π) to be smallest number such that π β β€ πΆ πΎπΏ,2 (π) max πββ dim β π nd(β/π) for all lattices of dimension at most π. πΆ πΎπΏ,2 π =Ξ©( log π ) : Basis π 1 , π 2 ,β¦, 1 π π π . KannanβLovΓ‘sz `88: πΆ πΎπΏ,2 π =π( π )
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β 2 Kannan-LovΓ‘sz Conjecture
Define πΆ πΎπΏ,2 (π) to be smallest number such that π β β€ πΆ πΎπΏ,2 (π) max πββ dim β π nd(β/π) for all lattices of dimension at most π. πΆ πΎπΏ,2 π =Ξ©( log π ) : Basis π 1 , π 2 ,β¦, 1 π π π . KannanβLovΓ‘sz `88: πΆ πΎπΏ,2 π =π( log π )
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β 2 KL Conjecture Resolution
D. Regev 16: Reduction to Reverse Minkowski Conj. If all sublattices of β have determinant at least 1 then βπ>0 ββ©π π΅ 2 π β€expβ‘(polylog π π 2 ). Regev-S. Davidowitz 17: Reverse Minkowski proved. πΆ πΎπΏ,2 π =π( log π ) Questions: 1. Can we compute KL projections? Is there a better characterization?
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Canonical Polygon [Stuhler 76]
π dimensional lattice β π«(β) Log det (π, log det β ) (0,0) dim. 1 2 π-1 π { π, log det π :sublattice πββ, dim π =π }
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Canonical Filtration [Stuhler 76]
π dimensional lattice β β Log det Vertices of π«(β) β 2 (π, log det β ) (0,0) β 1 {0} dim. 1 2 π-1 π Form Chain: 0 = β 0 β β 1 ββ―β β π =β
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Canonical Filtration [Stuhler 76]
π dimensional lattice β Slope: ln nd( β 2 β 1 ) β Log det β 2 (π, log det β ) (0,0) β 1 {0} dim. 1 2 π-1 π Form Chain: 0 = β 0 β β 1 ββ―β β π =β
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Stable Lattice [Stuhler 76]
π dimensional lattice β is stable I.e. no dense sublattices β Log det (π, log det β ) (0,0) {0} dim. 1 2 π-1 π If canonical filtration is trivial: 0 ββ
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β 2 KL for stable lattices
πΆ π := max β π β dim β , where β ranges over all stable lattices of det 1 and dimension β€π. Conjecture [Shapira-Weiss 16]: πΆ π =π(1) ( β€ π is worst-case) Regev-S. Davidowitz 17: πΆ π =π( log π )
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Canonical filtration and β 2 KL
Regev-S. Davidowitz 17: For the canonical filtration 0 = β 0 β β 1 ββ―β β π =β we have that π β 2 β€ πΆ π 2 πβ π dim ( β π β πβ1 ) nd β π β πβ1 2 β€ log π πΆ π 2 max iβ[π] dim ( β β πβ1 ) nd β β πβ1 2
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Above NP certificate tight up to πΆ π !
Main Results Theorem [D. 18]: 1. Can compute π( log π ) ``approximateββ canonical filtration in 2 π(π) time. 2. For any chain 0 = β 0 β β 1 ββ―β β π =β π β 2 βΏ πβ π dim ( β π β πβ1 ) nd β β πβ1 2 Above NP certificate tight up to πΆ π !
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Densest Sublattice Problem
π β β min πββ πβ {0} nd(π) πΌ-DSP: Given β find πββ, πβ {0} such that nd π β€πΌ π(β). Remark: dimension of π is not fixed! Key primitive for finding sparse lattice projections.
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Densest Sublattice Problem
π dimensional lattice β Want sublattice with slope β€ log πΌ + log nd ( β 1 ) β Log det β 2 (π, log det β ) (0,0) β 1 {0} dim. 1 2 π-1 π
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Densest Sublattice Problem
Theorem: Can solve π( log π )-DSP in 2 π(π) time with high probability. High Level Approach: If β is not approximate minimizer: find π¦β 0, orthogonal to actual minimizer, and recurse on ββ© π¦ β₯
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Densest Sublattice Problem
High Level Approach: If β is not approximate minimizer: find π¦β 0, orthogonal to actual minimizer, and recurse on ββ© π¦ β₯ Q: Where to find π¦? A: The dual lattice β β Q: How to find it in β β ? A: Discrete Gaussian sampling
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Discrete Gaussian Distribution
6:30-8:30 Aharonov and Regev use the periodic Gaussian function. You can see how oracle access to this can be helpful.
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Discrete Gaussian Distribution
Aggarwal-D.-Regev-S. Davidowitz: `15: Can sample in 2 π+π(π) time for any parameter. 6:30-8:30 Aharonov and Regev use the periodic Gaussian function. You can see how oracle access to this can be helpful.
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Conclusions Open Problem Algorithmic version of β 2 KL conjecture.
Lower bound certificates for covering radius that are conjecturally tight within π(1). Open Problem Prove that β 2 KL for stable lattices is π(1). Prove KL conjecture for general convex bodies.
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Kannan-LovΓ‘sz (KL) Conjecture
π= πΎ If πΎβ© β€ π =β
then β πβ[π], πβ β€ πΓπ , ππππ π =π such that vol ππΎ 1 π β€π( log π ). [Kannan `87, Kannan-LovΓ‘sz `88]
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