Computing upper bounds for densest packings with congruent copies of - PowerPoint PPT Presentation

Computing upper bounds for densest packings with congruent copies of a convex body Fernando Mario de Oliveira Filho (FU Berlin) Frank Vallentin (TU Delft, CWI Amsterdam) ERC Workshop: High complexity discrete geometry, FU Berlin October 26, 2011

The beginning of the story tetrahedra tetris image credit: Torquato, Jiao the search for the densest packing of tetrahedra

This talk: Let’s start the race!

Overview 1. Packing problems in discrete geometry 2. Spectral bounds & convex optimization 3. Computational strategies

1. Packing problems α ( G ) = 4

vertices: S n − 1 = { x ∈ R n : x · x = 1 } edges: x ∼ y if α < x · y < 1 pack spherical caps into S n − 1 to maximize fraction of covered space

vertices: R n edges: x ∼ y if 0 < � x − y � < 2 pack balls into R n to maximize fraction of covered space

vertices: R n � SO( n ) edges: ( x, A ) ∼ ( y, B ) if ( x + A K o ) ∩ ( y + B K o ) � = ∅ image credit: Torquato, Jiao pack bodies into R n to maximize fraction of covered space

2. Spectral bounds & convex optimization

G = ( V, E ) be finite graph, K : V × V → R be a symmetric matrix such that (1) K is positive semidefinite (2) K ( x, y ) ≤ 0 if { x, y } �∈ E Hoffman (1970) (3) K 1 = λ 1 Delsarte (1973) Lov´ asz (1979) Then, α ( G ) ≤ | V | λ max x ∈ V K ( x, x ) . K − λ | V | 11 T � 0 Spectral decomposition: I ⊆ V independent set λ | V || I | 2 ≤ � K ( x, y ) ≤ | I | max x ∈ V K ( x, x ) x,y ∈ I

G = ( V, E ) be finite graph, K : V × V → R be a symmetric matrix such that (1) K is positive semidefinite (2) K ( x, y ) ≤ 0 if { x, y } �∈ E Hoffman (1970) (3) K 1 = λ 1 Delsarte (1973) Lov´ asz (1979) Then, α ( G ) ≤ | V | λ max x ∈ V K ( x, x ) . � Finding optimal K can be done by SDP � Also works if V is compact measure space � If V is not compact, then (3) makes trouble. � Symmetries of G and harmonic analysis on V help to simplify the SDP

packing harmonic vertex set authors problem analysis compact, non- Delsarte, Goethals, Seidel (1977), S n − 1 spherical caps abelian Kabatiansky, Levenshtein (1978) non-compact, R n sphere packing Cohn, Elkies (2003) abelian R n � SO( n ) non-compact, body packing Oliveira, V. (2011) non-abelian

Main theorem Let P be a packing with body K , and f ∈ L 1 ( R n � SO( n )) be continuous such that (1) f is of positive type: ∀ N ∈ N ∀ ( x 1 , A 1 ) , . . . , ( x N , A N ) ∈ R n � SO( n ) : f (( x i , A i )( x j , A j ) − 1 ) � � i,j � 0 (2) f ( x, A ) ≤ 0 if K o ∩ ( x + A K o ) = ∅ � (3) λ = R n � SO( n ) f ( x, A ) d ( x, A ) > 0 . Then, δ ( P ) ≤ f (0 ,I ) · vol K λ

Approximate P by periodic packing P � N P � = � � v + x i + A i K v ∈ L i =1 for some lattice L ⊆ R n and ( x i , A i ) ∈ R n � SO( n ) compact Define K : ( R n � SO( n )) /L × ( R n � SO( n )) /L → R by � f (( y − v, B ) − 1 ( x, A )) K (( x, A ) , ( y, B )) = v ∈ L and ”apply” previous theorem.

3. Computational strategies image credit: wikipedia

Question: How to find a good function f ? Main technical step: Parametrize functions using its Fourier coefficients. Here: n = 2 , other n only on demand. . . Fourier coefficients are Hilbert Schmidt operators �� a cos φ � � cos θ �� − sin θ f , = a sin φ sin θ cos θ Fourier basis � ∞ � ∞ � f rs ( p ) i s − r e − i ( sθ +( r − s ) φ ) J s − r ( pa ) pdp 0 � � r,s = −∞ � ∀ p ≥ 0 : rs � 0 f rs ( p ) Answer: Finding an optimal f is an ∞ -dimensional SDP.

Use tools from polynomial optimization Now to approximate optimal solution of this ∞ -dimensional SDP. Main trick n � ( q i ( x )) 2 p ( x ) = p ∈ R [ x ] is sum of squares (SOS) i =1 T     1 1 x x     for Q � 0 . iff p ( x ) = Q      .   .  . .     . . x d x d

Case K = B (0 , 1)

Case K = B (0 , 1) ; reproving Cohn-Elkies � � � � rs � 0 f rs ( p ) Instead of f 00 ( p ) ≥ 0 use d � Laguerre polynomials f 2 k p 2 k e − p 2 � f 00 ( p ) = with k =0 then � ∞ d � k ! 2 L k ( a 2 / 4) e − a 2 � f ( a ) = f 00 ( p ) J 0 ( pa ) pdp = f 2 k 0 k =0 Conditions ∀ p ∈ R : � (1) : f 00 ( p ) ≥ 0 (2) : ∀ a ∈ R ≥ 2 : f ( a ) ≤ 0 are equivalent to SOS conditions. (Differs from numerical scheme of Cohn-Elkies and gives global optima)

Numerical results 0 . 28867493155 . . . 0 . 18615073311 . . . 0 . 13125261809 . . . 0 . 09973444216 . . . 0 . 08082433040 . . . 0 . 06930754913 . . . 0 . 06250430042 . . . 0 . 05898945087 . . . with d = 40 and high accuracy SDP solver and numerical help from Hans Mittelmann

Case K = C 5

Step 1: Making things of positive type Restrict to ”polynomial” Fourier coefficients d � f rs, 2 k t 2 k e − t 2 � f rs ( t ) = k =0 Then � � ∀ t ∈ R � P ( t ) = − N ≤ r,s ≤ N � 0 f rs ( t ) is a univariate PSD matrix inequality. This is equivalent to: Multivariate polynomial p ( t, y ) = y T P ( t ) y ∈ R [ t, y − N , . . . , y N ] is a sum of squares

Step 2: Evaluating the function � ∞ N d t 2 k +1 e − t 2 J s − r ( at ) dt � � f rs, 2 k i s − r e − i ( sθ +( r − s ) φ ) f ( a, φ, θ ) = 0 r,s = − N k =0 if 2 k + 2 + s − r > 0 AAR, page 222 (4.11.24) = Γ((2 k + 2 + s − r ) / 2)( a/ 2) s − r e − a 2 / 4 2Γ( s − r + 1) � ( s − r − 2 k − 2) / 2 + 1 ; a 2 � 1 F 1 If s − r even and ( s − r − 2 k − 2) / 2 + 1 ≤ 0 , then polynomial in a : s − r + 1 4 � − n � n ( a ) = ( α + 1) n 1 F 1 α ; a L α n ! Laguerre polynomial, orthogonal on [0 , ∞ ) wrt a α e − a dx So f is a polynomial in a and trigonometric in φ , θ . linear combination of ”monomials”: a k sin r φ cos t θ

Step 3: Dealing with the geometry (2) f ( x, A ) ≤ 0 if K o ∩ ( x + A K o ) = ∅ Fix A ∈ SO( n ) . x ∈ R n with K o ∩ x + AK o � = ∅ is Minkowski difference K o − AK o If K is a polytope, this is a linear condition in x . K o − A K o is an open 10 -gon

x ∈ R 2 , θ ∈ [ − 2 π/ 10 , 2 π/ 10] with x ∈ K o − A ( θ ) K o (If K is the ball, then shape is a round cylinder.) f ≤ 0 on the ten ”semialgebraic” sets { ( a, φ, θ ) : g i ( a, φ, θ ) ≥ 0 } g i ( a, φ, θ ) < 0 ”facet” defining ”polynomial” Can be relaxed as SOS condition.

Numerical results image credit: WWW

Conclusions We are developing the first step of an algorithmic solution � for a large class of packing problems Complexity of body K is reflected in the complexity � of the computation Numerical calculations are challenging � but seem to be in reach (in dimensions 2 , 3 )