Tomato Packing and Lettuce-Based Crypto

Transcription

1 Tomato Packing and Lettuce-Based Crypto A. S. Mosunov and L. A. Ruiz-Lopez University of Waterloo Joint PM & C&O Colloquium March 30th, 2017

2 Spring in Waterloo

3 Love is everywhere... Picture from 57f917fb6e738157a505ae3c.png.

4 I see love among people... Picture from

5 ...and among spheres! Figure: 1 Spheres kissing in R 3 1 Picture from

6 What is a kissing number?

7 What is a kissing number? It is not the number of studio albums by Kiss. Picture from http: //jenixe.com/wp-content/uploads/2015/11/382884_kiss_band.jpg.

8 What is a kissing number? The Kissing Number Problem. Given a positive integer n, what is the largest number of n-dimensional unit spheres that can kiss a unit sphere centred at the origin of R n? This number is called the kissing number τ n. It is not important what the radius is, as long as all the spheres have the same positive radius.

9 Examples for n = 1,2 Figure: 2 τ 1 = 2 Figure: 3 τ 2 = 6 2 Figure 1 in [1]. 3 Figure 1 in [3].

10 Example for n = 3 Figure: 4 τ 3 = 12 4 Figure 3 in [3].

11 Why τ 3 = 12? Intuition. How would you stack tomatoes or oranges so to occupy as much space as possible? Pictures from heap-tomatoes-shared-pyramid jpg and http: //blog.kleinproject.org/wp-content/uploads/2012/05/oranges2.png.

12 Why τ 3 = 12? Such a packing is called the face-centred cubic packing (FCC). Every sphere is kissed by 12 other spheres. Figure: 5 Face-centred cubic packing Atoms in a crystal can arrange in a form of FCC packing (nickel, crystallization of palladium or krypton). The density of FCC packing is equal to π The centres of spheres constitute a lattice in a three dimensional space. 5 Picture from Chapter1html/chapters/images/fccc.jpg.

13 What is a lattice? Let v 1,v 2,...,v n be linearly independent vectors in R n. The set Λ = {a 1 v 1 + a 2 v a n v n : (a 1,a 2,...,a n ) Z n } is said to be a lattice with basis v 1,v 2,...,v n. Example. The integers Z form a lattice in R 1 with a basis {(1)}. Example. The points of R 2 which have integer coordinates constitute a lattice with a basis {(1,0),(0,1)}.

14 More examples of lattices The hexagonal lattice in R 2 has a basis { (1,0), ( 1 2, 3 2 )}. By matching centres of spheres with the points of a hexagonal lattice, one achieves the optimal sphere kissing configuration. In this example, spheres have radius 1/2. 6 Figure 2 in [3]. Figure: 6 Hexagonal lattice

15 More examples of lattices The face-centred cubic lattice in R 3 has a basis {(0,1,1),(1,0,1),(1,1,0)}. By matching centres of spheres with the points of a hexagonal lattice, one achieves the optimal sphere kissing configuration. In this example, spheres have radius 1/ 2. This configuration is not unique! Figure: 7 Face-centred cubic lattice

16 Why τ 3 = 12? This construction gives us τ How would we prove that τ 3 = 12? The proof is not obvious. In fact, Isaac Newton and David Gregory argued about this in 1694: Newton believed that the correct answer is 12, and Gregory believed that it is 13. The equality τ 3 = 12 was proved by Schüte and van der Waerden in Related problem. In 1611, Kepler conjectured that no packing of balls of the same radius in three dimensions has density greater than the FCC packing. Kepler s conjecture was proved by Thomas C. Hales in 2000.

17 Other optimal configurations in R 3 There are optimal lattice packings in R 3 that are different from FCC packing. There are optimal irregular packings. These packings do not arise from lattices. E.g. the icosahedron configuration. Figure: 8 Icosahedron configuration / Icosahedron graph 8 Figures 3 and 10 in [3].

18 Other optimal configurations in R n Perhaps, the kissing number τ n can always be achieved by a lattice packing? No. For n = 9, the nonlattice packing known as P9a contains spheres that kiss 306 others. However, it is known that kissing numbers of all lattice packings in R 9 do not exceed 272. Same applies to the nonlattice packing P10c in R 10. The densities of (conjectured) optimal lattice packings in R 7, R 8 and R 9 are , and , respectively. Compare this to in R 2 and in R 3.

19 Known kissing numbers In the dimensions where kissing numbers are known, there exist highly symmetrical lattices. n τ n Lattice 1 2 Z 2 6 Hexagonal lattice 3 12 FCC lattice 4 24 D E Leech lattice

20 D 4 lattice in R 4 The D 4 lattice has a basis {(1,0,0,1),(1,0,1,0),(1,0,0, 1),(0,1,1,0)}. Our optimal sphere kissing configuration has 24 lattice points, namely (±1,±1,0,0), (±1,0,±1,0), (±1,0,0,±1), (0,±1,0,±1), (0,±1,±1,0), (0,0,±1,±1). In this example, spheres have radius 1/ 2, not 1. Notice that any two lattice points differ by a vector of length at least 2 = Thus the spheres may touch but they do not overlap.

21 E 8 lattice in R 8 The E 8 lattice is spanned by 8 basis vectors (see the Figure below). Our optimal sphere kissing configuration includes ( 8 2) 4 = 112 lattice points of type (0 6,±1 2 ) and 128 vectors of type ((±1/2) 8 ) with an even number of negative components. In this example, spheres have radius 1/ 2, not 1. Figure: 9 Dynkin diagram of E 8 lattice with basis vectors at vertices Diagram from C. Stewart, Lectures Notes on the Geometry of Numbers,

22 The Leech lattice R 24 The Leech lattice is spanned by 24 basis vectors (see the Figure below). Our optimal sphere kissing configuration includes three types of lattice points, with = points in total. Figure: 10 Matrix whose columns are basis vectors of the Leech lattice 10 Diagram from C. Stewart, Lectures Notes on the Geometry of Numbers, 2009.

23 Delsarte s method All of these constructions give a lower bound on τ n. What about an upper bound? In 1972, the French mathematician Phillipe Delsarte suggested the following approach. Let x 1,...,x m R n be vectors associated to kissing points. Then x i = 1 for all i, (x i,x j ) = x i x j cosθ ij = cosθ ij. Also, the matrix (cosθ ij ) i=1,...,m is positive semidefinite: j=1,...,m t 1 x t m x m 2 = (t 1 x t m x m,t 1 x t m x m ) = m m i=1 j=1 t i t j cosθ ij 0. Further, for any i j the angle θ ij between x i and x j satisfies θ ij π 3, so cosθ ij 1 2.

24 The Gagenbauer polynomials The Gagenbauer polynomials are defined recursively as follows: G (n) 0 (t) = 1, G (n) 1 (t) = t, G (n) (n) (2k+n 4)tG k 1 k (t) = (n) (t) (k 1)G k 2 (t) k+n 3. In 1943, Schoenberg proved that if (x ij ) is a positive semidefinite matrix, then so is (f (x ij )), where f (t) = d k=0 c k G (n) k (t) is a non-negative combination of Gagenbauer polynomials with c 0 > 0.

25 Main Theorem Theorem. (Delsarte, Goethals and Seidel, 1977) Let f (t) be a non-negative combination of Gagenbauer polynomials such that c 0 > 0 and f (t) 0 for all t [ 1, 1 2]. Then τ n f (1) c 0. Proof. Since the sum of the entries of (G (n) k (cosθ ij )) is non-negative, m i,j=1 f (cosθ ij ) = d c k m k=0 i,j=1 m c 0 i,j=1 G (n) k (cosθ ij ) G (n) 0 (cosθ ij ) = c 0 m 2.

26 Main Theorem On the other hand, m i,j=1 f (cosθ ij ) = mf (1) + f (cosθ ij ) mf (1). i j After combining the two estimates, we see that any sphere kissing configuration with m points must satisfy By setting m = τ n, we see that c 0 m 2 mf (1). τ n f (1) c 0.

27 Which polynomials give tight bounds? In 1973, Odlyzko and Sloane used the following polynomials to find a tight upper bound on τ 8 and τ 24 : ( f 8 (t) = t 1 ( )t 2 t (t + 1), 2 2) ( f 24 (t) = t 1 )( t + 1 ) 2 ( t 2 t + 1 ) 2 ( t (t + 1) ) For n = 4, one can prove that τ 4 25, but better estimate is impossible. Recall that D 4 lattice yields τ In 2003, Oleg Musin announced the proof of τ 4 = 24. The paper was published in Annals of Mathematics. He developed a clever modification of Delsarte s method.

28 References [1] N. Elkies, Lattices, linear codes, and invariants, part I, Notices of the AMS 47 (10), pp , [2] T. C. Hales, Cannonballs and honeycombs, Notices of the AMS 47 (4), pp , [3] F. Pfender, G. M. Ziegler, Kissing numbers, sphere packings, and some unexpected proofs, Notices of the AMS, pp , 2004.

29 But what about real applications? (Don t ask me, ask Luis)