On Approximating the Covering Radius and Finding Dense Lattice - - PowerPoint PPT Presentation
On Approximating the Covering Radius and Finding Dense Lattice - - PowerPoint PPT Presentation
On Approximating the Covering Radius and Finding Dense Lattice Subspaces Daniel Dadush Centrum Wiskunde & Informatica (CWI) ICERM April 2018 Outline 1. Integer Programming and the Kannan-Lovsz (KL) Conjecture. 2. 2 KL Conjecture
Outline
- 1. Integer Programming and
the Kannan-LovΓ‘sz (KL) Conjecture.
- 2. β2 KL Conjecture &
the Reverse Minkowski Conjecture.
- 3. Finding dense lattice subspaces.
Integer Programming (IP)
minππ¦ s.t. π΅π¦ β€ π π¦ π β€π π variables, π constraints Open Question: Is there a 2π(π) time algorithm? First result: 2π(π2) [Lenstra `83] Best known complexity: ππ(π) [Kannan `87]
πΏ π
π πΏ, β€π β smallest scaling π‘ such that every shift π‘πΏ + π’ contains an integer point.
β€2
πΏ
Main Dichotomy
Either covering radius π πΏ, β€π β€ 1.
πΏ
π πΏ, β€π β smallest scaling π‘ such that every shift π‘πΏ + π’ contains an integer point.
β€2
Main Dichotomy
Either covering radius π πΏ, β€π β€ 1.
πΏ
π πΏ, β€π β smallest scaling π‘ such that every shift π‘πΏ + π’ contains an integer point.
β€2
Main Dichotomy
Either covering radius π πΏ, β€π β€ 1.
π πΏ, β€π β smallest scaling π‘ such that every shift π‘πΏ + π’ contains an integer point.
β€2
Main Dichotomy
Either covering radius π πΏ, β€π β€ 1.
2 3πΏ
π πΏ, β€2 = 2
3
πΏ
β€2
Main Dichotomy
Either covering radius π πΏ, β€π β€ 1. Can find integer point in 2π(π) time [D. 12]
β€2
Main Dichotomy
There exists rank π β₯ 1 integer projection π β β€πΓπ such volπ ππΏ
1 π is small.
πΏ
Projection on π§-axis π = (0 1)
Or πΏ is βflatβ:
β€2
Main Dichotomy
There exists rank π β₯ 1 integer projection π β β€πΓπ such volπ ππΏ
1 π is small.
πΏ
Or πΏ is βflatβ:
Projection on π§-axis π = (0 1)
β€2
Main Dichotomy
There exists rank π β₯ 1 integer projection π β β€πΓπ such volπ ππΏ
1 π is small.
πΏ
Recurse on β volk PK subproblems [D. 12]
Or πΏ is βflatβ:
Duality Relation
1 β€ π πΏ, β€π min
πββ€πΓπ π π π =πβ₯1
volπ ππΏ
1 π β€ ?
βEasyβ side βHardβ side
Either covering radius π(πΏ, β€π) is small
- r πΏ is βflatβ.
Khinchine Flatness Theorem
1 β€ π πΏ, β€π min
πββ€1Γπ π π π =1
vol1(ππΏ) β€ ΰ·© O(π Ξ€
4 3)
πΏ
[Khinchine `48, Babai `86, Hastad `86, Lenstra-Lagarias-
Schnorr `87, Kannan-Lovasz `88, Banaszczyk `93-96, Banaszczyk-Litvak-Pajor-Szarek `99, Rudelson `00]
Kannan-LovΓ‘sz Flatness Theorem
1 β€ π πΏ, β€π min
πββ€πΓπ π π π =πβ₯1
volπ ππΏ
1 π β€ π
[Kannan `87, Kannan-LovΓ‘sz `88] πΏ
Kannan-LovΓ‘sz (KL) Conjecture
1 β€ π πΏ, β€π min
πββ€πΓπ π π π =πβ₯1
volπ ππΏ
1 π β€ π log π βΌ
πΏ
Faster Algorithm for IP?
1 β€ π πΏ, β€π min
πββ€πΓπ π π π =πβ₯1
volπ ππΏ
1 π β€ π log π
- D. `12: Assuming KL conjecture
+ π computable in (log π)π(π) time then there is log π π(π) time algorithm for IP.
β2 Kannan-LovΓ‘sz Conjecture
Does the conjecture hold for ellipsoids? β€π πΉ An ellipsoid is πΉ = ππΆ2
π
β2 Kannan-LovΓ‘sz Conjecture
Answer: YES* [Regev-S.Davidowitz 17] β€π πΉ
* up to polylogarithmic factors
β2 Kannan-LovΓ‘sz Conjecture
Can we compute the projection P? β€π πΉ
THIS TALK: YES, in 2π(π) time.
Outline
- 1. Integer Programming and
the Kannan-LovΓ‘sz (KL) Conjecture.
- 2. β2 KL Conjecture &
the Reverse Minkowski Conjecture.
- 3. Finding dense lattice subspaces.
ππΉ = πΆ2
π
Easier to think of Euclidean ball vs general lattice. β = πβ€π
β2 Kannan-LovΓ‘sz Conjecture
A lattice β β βπ is πΆβ€π for a basis πΆ = π1, β¦ , ππ . β(πΆ) denotes the lattice generated by πΆ. Note: a lattice has many equivalent bases.
π2 π2 π1 π2 π1
Lattices
β
A lattice β β βπ is πΆβ€π for a basis πΆ = π1, β¦ , ππ . β(πΆ) denotes the lattice generated by πΆ. The determinant of β is | det πΆ |.
Lattices
π1 π2 β
Lattices
π1 π2
A lattice β β βπ is πΆβ€π for a basis πΆ = π1, β¦ , ππ . β(πΆ) denotes the lattice generated by πΆ. The determinant of β is | det πΆ |. Equal to volume of any tiling set.
β
β2 Covering Radius
β
π
π β β π πΆ2
π, β
Distance of farthest point to the lattice β.
π²
Voronoi cell π² β all points closer to 0
Volumetric Lower Bounds
β
π
volπ πΆ2
ππ β
β₯ voln π² = det(β)
π²
Voronoi cell π² β all points closer to 0
Volumetric Lower Bounds
β
π
π(β) β₯ voln πΆ2
π β1 π det β 1 π
π²
Voronoi cell π² β all points closer to 0
Volumetric Lower Bounds
β
π
π(β) βΏ π det β
1 π
π²
Voronoi cell π² β all points closer to 0
Volumetric Lower Bounds
β
π
π β β₯ π ββπ ββπ projection onto π
π
πβπ
Volumetric Lower Bounds
β
π
π β β₯ π ββπ β³ π det ββπ
1 π
ββπ projection onto π dim(π) = π β₯ 1
π
πβπ
Volumetric Lower Bounds
β
π
π β β³ max
dim π =πβ₯1
π det ββπ
1 π
ββπ projection onto π dim(π) = π β₯ 1
π
πβπ
β2 Kannan-LovΓ‘sz Conjecture
Define π·πΏπ,2(π) to be smallest number such that π β β€ π·πΏπ,2(π) max
dim π =πβ₯1
π det ββπ
1 π
for all lattices of dimension at most π. π·πΏπ,2 π = Ξ©( log π) Lower bound for β with basis π1,
1 2 π2, β¦ , 1 π ππ.
KL Bounds
π β β€ π·πΏπ,2(π) max
dim π =πβ₯1
π det ββπ
1 π
Kannan-LovΓ‘sz `88: π
- D. Regev `16: logπ(1) π
Assuming Reverse Minkowski Conjecture. Regev, S.Davidowitz `17: log Ξ€
3 2 π
Reverse Minkowski Conjecture is proved!
Our Results
π dimensional lattice β β β(πΆ)
- 1. Can compute subspace π, dim π = π β₯ 1
π β β€ π(log2.5 π) π det ββπ
1 π
in 2π(π) time with high probability. Prior work: Kannan LovΓ‘sz `88: π in 2π(π) time.
- D. Micciancio `13: best subspace in ππ(π2) time.
Our Results
π dimensional lattice β β β(πΆ)
- 2. Can combine lower bounds over different
subspaces to certify π π up to the slicing constant ππ for βstableβ Voronoi cells*. π² * If voln π² = 1 can find hyperplane πΌ s.t. volnβ1 π² β© πΌ = Ξ©( 1
ππ)
πΌ
Our Results
π dimensional lattice β β β(πΆ)
- 2. Can combine lower bounds over different
subspaces to certify π π up to the slicing constant ππ for βstableβ Voronoi cells*. π² Slicing Conjecture: ππ = π(1) for all convex bodies! For βstableβ Voronoi cells: ππ = π(log π) [RS `17] πΌ
Notation
π β β sublattice of dimension π Convention: π = {0} then det π β 1. Normalized Determinant: nd π β det π
Ξ€ 1 π
Projected Sublattice: Ξ€ β π β β projected onto span π β₯
Lower Bounds for Chains
Theorem [D. 17]: For 0 = β0 β β1 β β― β βπ = β then π β 2 βΏ Οπ=1
π
dim( Ξ€ βπ βπβ1) nd Ξ€ β βπβ1 2 Only βmissing ingredientβ: Combined with techniques from [R.S. `17] easily get tightness within slicing constant ππ.
Lower Bounds for Chains
Theorem [D. 17]: For 0 = β0 β β1 β β― β βπ = β then π β 2 βΏ Οπ=1
π
dim( Ξ€ βπ βπβ1) nd Ξ€ β βπβ1 2 Proof Idea:
- 1. Establish SDP based lower bound: [D.R. `16]
π β 2 βΏ max Οπ rk ππ nd ππβ 2 s.t. Οπ ππ
βππ βΌ π½π
- 2. Build solution to above starting from any chain.
Lattice Density
β π
lim
π ββ
β β© π πΆ2
π
volπ(π πΆ2
π) =
1 det β
Lattice Density
β
lim
π ββ
β β© π πΆ2
π
volπ(π πΆ2
π) =
1 det β
π
Global density of lattice points per unit volume
Minkowskiβs First Theorem
1889
β
β β© π πΆ2
π β₯ 2βπ volπ(π πΆ2 π)
det β
π
Global density implies βlocal densityβ
Reverse Minkowski Theorem
Regev-S.Davidowitz `17: β lattice dimension π. If all sublattices of β have determinant at least 1 then: β has at most 2π(log2 π π 2) points at distance π . Almost tight: β€π has πΞ©(π) points at distance π for π βͺ π.
Outline
- 1. Integer Programming and
the Kannan-LovΓ‘sz (KL) Conjecture.
- 2. β2 KL Conjecture &
the Reverse Minkowski Conjecture.
- 3. Finding dense lattice subspaces.
Notation
π β β sublattice of dimension π Convention: π = {0} then det π β 1. Normalized Determinant: nd π β det π
Ξ€ 1 π
Projected Sublattice: Ξ€ β π β β projected onto span π β₯
Densest Subspace Problem
ndβ β β min
πββ πβ {0}
nd(π) π½-DSP: Given β find π β β, π β {0} such that nd π β€ π½ ndβ(β). Remark: dimension of π is not fixed! Key primitive for finding sparse lattice
- projections. Will focus on this problem.
Densest Subspace Problem
Theorem: Can solve π(log π)-DSP in 2π(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 β β© π§β₯
Canonical Polytope [Stuhler 76]
π dimensional lattice β
1 dim. π π-1 2 { π, log det π : sublattice π β β, dim π = π} (π, log det β) Log det (0,0) π¬(β)
Canonical Filtration [Stuhler 76]
π dimensional lattice β
1 dim. π π-1 2 Form Chain: 0 = β0 β β1 β β― β βπ = β (π, log det β) Log det (0,0) β1 β2 Vertices of π¬(β) β {0}
Canonical Filtration [Stuhler 76]
π dimensional lattice β
1 dim. π π-1 2 Form Chain: 0 = β0 β β1 β β― β βπ = β (π, log det β) Log det (0,0) β1 β2 Slope: ln nd( Ξ€ β2 β1) β {0}
Stable Lattice [Stuhler 76]
π dimensional lattice β is stable
1 dim. π π-1 2 If canonical filtration is trivial: 0 β β (π, log det β) Log det (0,0) β {0} I.e. no dense sublattices
Stable Lattice [Stuhler 76]
Example: β = β€π
1 dim. π π-1 2 β€π has trivial filtration: 0 β β€π Log det (0,0) π¬(β) {0}
Canonical Filtration [Stuhler 76]
- 1. Form Chain: 0 = β0 β β1 β β― β βπ = β .
- 2. Blocks
Ξ€ βπ βπβ1 are stable.
- 3. Slope increasing: nd
Ξ€ βπ βπβ1 < nd Ξ€ βπ+1 βπ . 1 π π-1 2 β1 β2 β {0}
Densest Subspace Problem
π dimensional lattice β
1 dim. π π-1 2 (π, log det β) Log det (0,0) β1 β2 Want sublattice with approx. minimum slope. β {0}
Densest Subspace Problem
High Level Approach: If β is not approximate minimizer: find π§ β 0, orthogonal to actual minimizer, and recurse on β β© π§β₯ Q: Where to find π§? A: The dual lattice ββ Q: How to find it in ββ? A: Discrete Gaussian sampling
π§ β ββ
A lattice β β βπ is πΆβ€π for a basis πΆ = π1, β¦ , ππ . The dual lattice is ββ = {π§ β span β : π§Tπ¦ β β€ βπ¦ β β} ββ is generated by πΆβT. Remark: β€π β = β€π
Dual Lattice
π§Tπ¦ = 0 π§Tπ¦ = 1 π§Tπ¦ = 2 π§Tπ¦ = 3 π§Tπ¦ = 4
β
π§ β ββ
A lattice β β βπ is πΆβ€π for a basis πΆ = π1, β¦ , ππ . The dual lattice is ββ = {π§ β span β : π§Tπ¦ β β€ βπ¦ β β} ββ is generated by πΆβT. det ββ = 1/det(β)
Dual Lattice
π§Tπ¦ = 0 π§Tπ¦ = 1 π§Tπ¦ = 2 π§Tπ¦ = 3 π§Tπ¦ = 4
β
Discrete Gaussian Distribution
Discrete Gaussian Distribution
ADRS `15: Can sample in 2π+π(π) time for any parameter.
Main Procedure
Repeat until β = {0} π‘ β nd β Update π β β if nd π > π‘ Sample π§ βΌ πΈββ,π/π‘ until π§ β 0 β β β β© π§β₯ Main Lemma: At any iteration, if β not π(log π) approximate minimizer, then β β© π§β₯ contains minimizer w.p. Ξ© 1 .
- Proc. finds apx minimizer with prob. 2βπ(π).
Proof of Main Lemma
wlog det β = det(ββ) = 1 β1 β β densest sublattice sample π§ βΌ πΈββ,π If nd β1 βͺ
1 log π
must show that π§ β 0 and π§ β₯ β1 w.p. Ξ©(1).
det β = det ββ = 1 β1 β β densest sublattice, nd β1 βͺ
1 log π
Sample π§ βΌ πΈββ,π Want:
- 1. Pr π§ = 0 β€ π
- 2. Pr π§ β ββ β© β1
β₯ β₯ 1 β π
det β = det ββ = 1 1. Pr
π§βΌπΈββ,π
π§ = 0 =
1 ππ(ββ)
β€
1 βββ© ππΆ2
π πβ Ξ€ π π2
(By Minkowski) β€
1 2ππβ Ξ€
π π2 = π(1)
ππ π΅ β Οπ¦βπ΅ πβ
Ξ€ π¦ π 2
det β = det ββ = 1 β1 β β densest sublattice, nd β1 βͺ Ξ€ 1 log π π β β1
β₯
2. Pr
π§βΌπΈββ,π
π§ β π =
ππ(βββ© π) ππ(ββ)
(ortho. is worst-case) β₯
ππ(βββ©π) ππ βββ©π ππ( Ξ€ ββ π)
=
1 ππ(ββ/π)
ππ π΅ β Οπ¦βπ΅ πβ
Ξ€ π¦ π 2
det β = det ββ = 1 β1 β β densest sublattice, nd β1 βͺ Ξ€ 1 log π π β β1
β₯
- 2. Need to show ππ
Ξ€ ββ π β€ 1 + π(1) Key: ndβ Ξ€ ββ π = Ξ€ 1 nd(β1) β« log π Reverse-Minkowski β ( Ξ€ ββ π) β© π πΆ2
π βͺ ππ π 2 , βπ β₯ 0
ππ π΅ β Οπ¦βπ΅ πβ
Ξ€ π¦ π 2
Canonical Polytope of ββ?
Assumption: det β = 1
1 dim. π π-1 2 (π, 0) Log det β1 β2 β {0}
det β = 1
Slope: ln nd(β1)
Canonical Polytope of ββ?
1 dim. π π-1 2 Log det ββ β© β1
β₯
ββ β© β2
β₯
(π, 0) ββ {0}
det ββ = 1 Map π β ββ β© πβ₯
Canonical Polytope of ββ?
1 dim. π π-1 2 Log det ββ β© β1
β₯
ββ β© β2
β₯
(π, 0) ββ {0}
π¬(ββ) is βreflectionβ of π¬(β)
Slope: βln nd(β1) Ξ€ ββ π
det ββ = 1
Conclusions
- 1. Algorithmic version of β2 Kannan-LovΓ‘sz
conjecture via discrete Gaussian sampling.
- 2. Lower bound certificates for covering radius that
are tight within π(1) under slicing conjecture.
Open Problem
- 1. Prove KL conjecture for general convex bodies.
- 2. Prove Slicing conjecture for Voronoi cells.