New Conjectures in the Geometry of Numbers

Transcription

1 New Conjectures in the Geometry of Numbers Daniel Dadush Centrum Wiskunde & Informatica (CWI) Oded Regev New York University

2 Talk Outline 1. A Reverse Minkowski Inequality & its conjectured Strengthening. 2. Strong Reverse Minkowski implies the Kannan & Lovász conjecture for l From Decomposing Integer Programs to the general Kannan & Lovász conjecture.

3 Lattices The standard integer lattice Z n. e 2 e 1 Z n

4 Lattices A lattice L R n is BZ n for a basis B = b 1,, b n. L(B) denotes the lattice generated by B. Note: a lattice has many equivalent bases. b 1 b 1 b 2 b 2 b 2 L

5 Lattices A lattice L R n is BZ n for a basis B = b 1,, b n. L(B) denotes the lattice generated by B. The determinant of L is det B. b 1 b 2 L

6 Lattices A lattice L R n is BZ n for a basis B = b 1,, b n. L(B) denotes the lattice generated by B. The determinant of L is det B. Equal to volume of any tiling set. b 1 b 2 L

7 Lattices A lattice L R n is BZ n for a basis B = b 1,, b n. The dual lattice is L = {y span L : y T x Z x L} L is generated by B T. The identity det L = Τ 1 det(l) holds. L

8 Lattices A lattice L R n is BZ n for a basis B = b 1,, b n. The dual lattice is L = {y span L : y T x Z x L} L is generated by B T. y L L y T x = 0 y T x = 1 y T x = 2 y T x = 3 y T x = 4

9 Reversing Minkowski s Convex Body Theorem

10 Minkowski s Convex Body Theorem Theorem [Minkowski]: For an n-dimensional symmetric convex body K and lattice L, we have that K L 2 n vol n K Τdet L K 0 L

11 Minkowski s Convex Body Theorem Theorem [Minkowski]: For an n-dimensional symmetric convex body K and lattice L, we have that K L 2 n vol n K Τdet L Question: Is the above lower bound also close to being an upper bound?

12 Minkowski s Convex Body Theorem Theorem [Minkowski]: For an n-dimensional symmetric convex body K and lattice L, we have that K L 2 n vol n K Τdet L 0 K L Clearly NO!

13 Minkowski s Convex Body Theorem Theorem [Minkowski]: For an n-dimensional symmetric convex body K and lattice L, we have that K L 2 n vol n K Τdet L Points all lie in a lattice subspace. 0 K Can we strengthen the lower bound? L

14 Reverse Minkowski Inequality For a symmetric convex body K and lattice L, let MB K, L = max 2 d vol d(k W) d 0 det(l W) max W lat. sub. dim W =d W 0 K L W is a lattice subspace of L if dim W L = dim(w).

15 Reverse Minkowski Inequality For a symmetric convex body K and lattice L, let MB K, L = max 2 d vol d(k W) d 0 det(l W) max W lat. sub. dim W =d W 0 K L Is this bound any better?

16 Weak Reverse Minkowski For a symmetric convex body K and lattice L, let MB K, L = max 2 d vol d(k W) d 0 det(l W) max W lat. sub. dim W =d Let B 2 n denote the unit Euclidean ball. Theorem [D.,Regev `16]: For an n-dimensional lattice L B 2 n L MB(6 nb 2 n, L). Furthermore, for any symmetric convex body K K L MB(6nK, L).

17 Strong Reverse Minkowski For a symmetric convex body K and lattice L, let MB K, L = max 2 d vol d(k W) d 0 det(l W) max W lat. sub. dim W =d Let B 2 n denote the unit Euclidean ball. Conjecture [D.,Regev `16]: For an n-dimensional lattice L B 2 n L MB(O( log n)b 2 n, L). Furthermore, for any symmetric convex body K K L MB(O(log n)k, L). Tight example: K = B 1 n and L = Z n.

18 Successive Minima Symmetric convex body K and lattice L in R n. λ i K, L = inf s 0: dim L sk i, i ε[n] y 2 -y 1 λ 1 K 0 y 1 λ2k -y 2

19 Minkowski s Second Theorem Symmetric convex body K and lattice L in R n. Π n det L i=1 λ i K, L 2n vol n K n! Π i=1 n λ i (K, L) y 2 -y 1 λ 1 K 0 y 1 λ2k -y 2

20 Lattice points bounds via Minima Theorem [Henk `02]: K L 2 n 1 ς n i= λ i K,L K 0 L

21 Proof of Weak Minkowski Must show K L max d 0 max W lat. sub. dim W =d 2 d vol d(6nk W) det(l W) If max i [n] λ i K, L > 1, then K L is lower dimensional and we can induct. If max i [n] λ i K, L 1, then we have that K L 2 n 1 ςn i= λ i K,L (by Henk) 2 n 1 ςn 3 i=1 λ i K,L 6n ςn 1 i=1 λ i K,L n! 3 n vol n (K) Τdet L (by Minkowski) vol n 3nK Τdet L MB(6nK, L)

22 The Kannan & Lovász conjecture for l 2

23 l 2 covering radius Let μ B n 2, L = max min t y t Rn 2, i.e. the y L farthest distance from the lattice. μ Main question: How to get good lower bounds on the covering radius? L

24 Lower bounds for covering radius Lemma: μ B 2 n, L 1 2λ 1 (B 2 n,l ) t y L L y T x = 1 y T x = 2 y T x = 3 y T x = 4

25 Lower bounds for covering radius Lemma: μ B 2 n, L det L 1 n vol n B 2 n 1 n = Ω( n) det L 1/n V μ L Since V μb 2 n, vol n V = det(l) vol n μb 2 n = μ n vol n B 2 n

26 Lower bounds for covering radius Lemma: Let π W denote the orthogonal projection onto a d-dimensional subspace W (*). Then μ B 2 n, L det π W L = vol d B 2 d det L W 1 d 1 d 1 1 dvol d B 2 d 1 d = Ω 1 d det L W Pf: For the first inequality, since orthogonal projections shrink distances we get μ B 2 n, L μ π W B 2 n, π W L. The second equality follows from the identity π W L = L W. 1. d

27 Kannan & Lovász for l 2 Theorem [Kannan-Lovasz `88]: Ω(1) μ B 2 n, L min lat.subspace W 1 dim W =d n det L W d 1 d O( n)

28 Kannan & Lovász for l 2 Conjecture [Kannan-Lovász `88]: Ω 1 μ B 2 n, L min lat.subspace W 1 dim W =d n det L W 1/d d = O( log n) Remark: implies that there are very good NP-certificates for showing that the covering radius is large.

29 Reverse Minkowski vs Kannan & Lovász MB B 2 n, L = max d 0 Theorem [D. Regev `16]: If for any n-dimensional lattice L B 2 n L MB(f n B 2 n, L) for a non-decreasing function f(n), then the Kannan & Lovász conjecture for l 2 holds with bound O log n f n. max 2 d vol d(b n 2 W) W lat. sub. det(l W) dim W =d

30 Main Approach 1. Use convex programming relaxation for μ B 2 n, L 2 : μ B 2 n, L 2 O 1 min trace A s.t. σ y L {0} e yt Ay 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 W.

31 A Sufficient Conjecture μ B 2 n, L 2 O 1 min trace A s.t. σ y L {0} e yt Ay 1, A psd. To formulate the required dual for the above program, we can rely on the following weaker conjecture: Conjecture [D. Regev `16]: log rb n 2 L Ω 1 max r>0 r O log n min lat.sub. W 1 dim W =d det L W 1 d

32 The Relaxed Program mintrace A s.t. σ y L {0} e yt Ay 1, A psd. We relax the above using the following weaker constraints det W A 1 det L W 2 lat. subspace W where det W (A) det(o W T AO W ), where O W is any orthonormal basis of W. This relaxation makes the value of the program drop by at most an f n 2 factor (the Reverse Minkowski bound).

33 The Dual Program min trace A s.t. det W A 1 det L W 2 lat. subspace W, A psd. The strong dual for the above program is max σ k k 0 i=1 s.t. σ i=1 det Wi X i d i det L W i k X i I n Τ 1 d i Τ 2 d i Here we range over all finite k, where for i [k], W i is a lattice subspace of L, dim W i = d i and X i is a PSD matrix with image W i.

34 The Dual Program max σ k k 0 i=1 s.t. σ i=1 det Wi X i d i det L W i k X i I n Τ 1 d i Τ 2 d i Theorem [D., Regev `16]: above value is at most O log 2 n max lat.subspace W 1 dim W =d n d det L W i Τ 2 d Corresponds to setting k = 1 and X 1 = projection on W 1. Main idea: disentangle the subspaces via uncrossing.

35 The Dual Program max σ k k 0 i=1 s.t. σ i=1 det Wi X i d i det L W i k X i I n Τ 1 d i Τ 2 d i Theorem [D., Regev `16]: above value is at most O log 2 n max lat.subspace W 1 dim W =d n d det L W i Τ 2 d Implies l 2 Kannan & Lovász bound of O log n f n. (note above programs approximate μ B 2 n, L 2 )

36 Integer Programming and the general Kannan & Lovász conjecture

37 Anatomy of an IP Algorithm K Z n Problem: Find point in K Z n or decide that K is integer free.

38 Anatomy of an IP Algorithm y K Z n Problem: Find point in K Z n or decide that K is integer free.

39 Anatomy of an IP Algorithm y c K Z n Main dichotomy: 1. Either K contains a deep integer point: find y by rounding from center c of K.

40 Anatomy of an IP Algorithm x 2 =2 subproblems x 2 =1 K Z n Main dichotomy: 1. Either K contains a deep integer point: find y by rounding from center c of K. 2. K is flat : decompose feasible region into lower dimensional subproblems and recurse.

41 IP Algorithms Time Space Authors 2 O(n3 ) poly(n) Lenstra 83 2 O(n) n 2.5n poly(n) Kannan 87 2 O(n) n 2n 2 n Hildebrand Köppe 10 2 O(n) n n 2 n D. Peikert Vempala 11, D. 12 n O(n) barrier: don t know how to create less than n O(1) subproblems per dimension

42 IP Algorithms Time Space Authors 2 O(n3 ) poly(n) Lenstra 83 2 O(n) n 2.5n poly(n) Kannan 87 2 O(n) n 2n 2 n Hildebrand Köppe 10 2 O(n) n n 2 n D. Peikert Vempala 11, D. 12 Conjecture Kannan & Lovász `88: O(log n) subproblems per dimension should be possible!

43 General Covering Radius & Deep Points The covering radius of K with respect to Z n is μ K, Z n = min {s 0: t R n, (sk + t) Z n } or, smallest scaling s such that every shift sk + t contains an integer point. K Z 2

44 General Covering Radius & Deep Points The covering radius of K with respect to Z n is μ K, Z n = min {s 0: t R n, (sk + t) Z n } or, smallest scaling s such that every shift sk + t contains an integer point. μ K, Z 2 > 1 K+ t Z 2

45 General Covering Radius & Deep Points The covering radius of K with respect to Z n is μ K, Z n = min {s 0: t R n, (sk + t) Z n } μ K, Z 2 = 2 2K Z 2

46 General Covering Radius & Deep Points Integral shifts of 2K cover R 2, i.e. 2K + Z 2 = R 2. 2K Z 2

47 General Covering Radius & Deep Points If covering radius μ K, Z n 1/2, then any shift of K contains a deep integer point. K + t 1 c Z 2

48 General Covering Radius & Deep Points If covering radius μ K, Z n 1/2, then any shift of K contains a deep integer point. K + t 1 c Z 2

49 General Covering Radius & Deep Points If covering radius μ K, Z n 1/2, then any shift of K contains a deep integer point. c K + t 2 Z 2

50 General Covering Radius & Deep Points If covering radius μ K, Z n 1/2, then any shift of K contains a deep integer point. K + t 3 c Z 2

51 General Covering Radius & Deep Points Thus, if μ K, Z n 1/2 finding an integer point in K is easy. K c Z 2

52 General Covering Radius & Deep Points Thus, if μ K, Z n 1/2 finding an integer point in K is easy. If not, we should decompose K into lower dimensional pieces K Z 2

53 Lower Bounds on the Covering Radius Lemma: 1 min y Z n 0 width K y μ K, Z n.8 y T x =.9 y T x =.1 K y To cover space, K must have width at least 1.

54 Khinchine s Flatness Theorem Theorem: For K R n convex, either 1. covering radius μ K, Z n 1/2, or 2. y Z n such that K has width O(n Τ 4 3 ) w.r.t. y: K K K Z 2

55 Khinchine s Flatness Theorem Theorem: For K R n convex, either 1. covering radius μ K, Z n 1/2, or 2. y Z n {0} such that K has width O(n Τ 4 3 ) w.r.t. y: K K Z 2

56 Khinchine s Flatness Theorem Theorem: For K R n convex, either 1. covering radius μ K, Z n 1/2, or 2. y Z n 0 such that K has width O(n Τ 4 3 ) w.r.t. y: y T x = y T x =.8 K Z 2 y

57 Khinchine s Flatness Theorem Theorem: For K R n convex, either 1. covering radius μ K, Z n 1/2, or 2. y Z n 0 such that K has width O(n Τ 4 3 ) w.r.t. y. y T x = 3 y T x = 2 y T x = 1 y T x = 0 K Z 2 y

58 Khinchine s Flatness Theorem Theorem: 1 min y Z n K y 0 μ K, Z n O(n ) To cover space, suffices to scale K to have width somewhere between 1 and O(n 4Τ3 ).

59 Khinchine s Flatness Theorem Bounds improvements for general bodies: Khinchine `48: (n + 1)! Babai `86: 2 O(n) Lenstra-Lagarias-Schnorr `87: n 5/2 Kannan-Lovasz `88: n 2 Banaszczyk et al `99: n 3/2 Rudelson `00: n 4/3 log c n [Khinchine `48, Babai `86, Hastad `86, Lenstra-Lagarias- Schnorr `87, Kannan-Lovasz `88, Banaszczyk `93-96, Banaszczyk-Litvak-Pajor-Szarek `99, Rudelson `00]

60 Best known bounds: Ellipsoids: Θ(n) [Banaszczyk `93] Centrally Symmetric: O(n log n) [Banaszczyk `96] General: Khinchine s Flatness Theorem O(n Τ 4 3 ) [Banaszczyk `96, Rudelson `00] Bound conjectured to be O(n) (best possible). K Z n K = {x: σ i x i n, x 0}

61 How can we improve decompositions? Recursion only uses integrality of one variable at a time. Why? y T x = 3 y T x = 2 y T x = 1 y T x = 0 K Z y.8 Reason: It s very efficient to enumerate integer points in an interval.

62 How can we improve decompositions? When else is it efficient to enumerate integer points? Theorem [D. Peikert Vempala `11, D. `12]: For K R d convex, can enumerate K Z d in time 2 O(d) max K + t. t Rd K

63 How can we improve decompositions? When else is it efficient to enumerate integer points? Theorem [D. Peikert Vempala `11, D. `12]: For K R d convex, can enumerate K Z d in time 2 O(d) max K + t. t Rd K+t

64 How can we improve decompositions? Generalized Decomposition Strategy for K: Choose d 1, P Z d n rank d such that max P K + t 1/d is small t R d P(K)

65 How can we improve decompositions? Generalized Decomposition Strategy for K: Choose d 1, P Z d n rank d such that max P K + t 1/d is small t R d Enumerate P K Z d and recurse on subproblems. P(K) subproblems

66 How can we improve decompositions? How should we choose d 1, P Z d n? Hardest bodies to decompose satisfy μ K, Z n = 1/2, i.e. as fat as possible without containing deep points. K

67 How can we improve decompositions? How should we choose d 1, P Z d n? If μ K, Z n = 1/2 then also μ P K, Z d 1/2, so integer projections of K are only fatter. K P(K)

68 How can we improve decompositions? How should we choose d 1, P Z d n? If μ P K, Z d 1/2, then max P K + t 1/d = O 1 vol d P K 1/d. t R d Good choice of d 1, P Z d n should approximately minimize min d 1 min P Z d n rank d vol d P K 1/d.

69 Volumetric bounds on the Covering Radius Lemma: If K + Z n = R n, then vol n K 1. K Z 2

70 Volumetric bounds on the Covering Radius Lemma: If K + Z n = R n, then vol n K 1. Pf: Let C = 0,1 n be the cube. vol n K = σ y Z n vol n K C + y = σ y Z n vol n ( K y C) vol n K + Z n C = vol C = 1. C K Z 2

71 Volumetric bounds on the Covering Radius Lemma: If K + Z n = R n, then vol n K 1. Lemma: Let P Z k n have rank k. If K + Z n = R n, then vol k PK 1. Pf: R k = P K + Z n = P K + P Z n P K + Z k.

72 Volumetric bounds on the Covering Radius Lemma: Let P Z d n have rank d. If K + Z n = R n, then vol d PK 1. Lemma: For a convex body K R n μ K, Z n max d [n] max P Z d n rank d 1 vol d PK 1/d Pf: Must scale K to have volume at least 1 in each integer projection.

73 Kannan Lovász Conjecture Theorem [Kannan-Lovász `88]: For K R n convex, either 1. Covering radius μ K, Z n 1/2, or 2. d 1, P Z d n rank d such that vol d PK 2n d. In particular, 1 μ K, Z n min d n min P Z d n rank d vol d PK 1 d 2n Proof is constructive. Enables a 2 O(n) n n time IP algorithm [D. `12].

74 Kannan Lovász Conjecture Theorem [Kannan-Lovász `88]: For K R n convex, either 1. Covering radius μ K, Z n 1/2, or 2. d 1, P Z d n rank d such that vol d PK 2n d. ex: d = 2, P = K Z 2

75 Kannan Lovász Conjecture Theorem [Kannan-Lovász `88]: For K R n convex, either 1. Covering radius μ K, Z n 1/2, or 2. d 1, P Z d n rank d such that vol d PK 2n d. ex: d = 2, P = subproblems K Z 2

76 Kannan Lovász Conjecture Theorem [Kannan-Lovász `88]: For K R n convex, either 1. Covering radius μ K, Z n 1/2, or 2. d 1, P Z d n rank d such that vol d PK 2n d. ex: d = 1, P = 0 1 subproblems Same as standard flatness. K Z 2

77 Kannan Lovász Conjecture Conjecture [Kannan-Lovász `88]: For K R n convex, either 1. Covering radius μ K, Z n 1/2, or 2. d 1, P Z d n rank d such that vol d PK O log n d. In particular, 1 μ K, Z n min d n min P Z d n rank d vol d PK 1 d O(log n) An efficient enough construction would imply a 2O n log n O(n) time IP algorithm!

78 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.

79 THANK YOU!