Volumes and Integrals over Polytopes

Transcription

1 Volumes and Integrals over Polytopes Jesús A. De Loera, UC Davis July 16, 2009

2 Meet Volume Why compute the volume and its cousins? The (Euclidean) volume V (R) of a region of space R is real non-negative number defined via the Riemann integral over the regions.

3 Meet Volume s Cousins Why compute the volume and its cousins? In the case when P is an n-dimensional lattice polytope (i.e., all vertices have integer coordinates) we can naturally define a normalized volume of P, NV (P) to be n!v (P). EXAMPLE: P = {(x, y) : 0 x 1, 0 y 1} NV (P) = 2! 1 = 2. Given polytopes P 1,..., P k R n and real numbers t 1,..., t k 0 the Minkowski sum is the polytope. t 1 P t k P k := {t 1 v t k v k : v i P i }

4 Why compute the volume and its cousins? EXAMPLE Theorem(H. Minkowski) There exist MV (P a 1 1,..., Pa k k ) > 0 (the mixed volumes) such that V (t 1 P 1 + ( + t k P k ) = n a 1 + +a k =n a 1,...,a k MV (P a 1 1,..., Pa k k )ta 1 1 ta 2 2 ta k k.

5 A few reasons to compute them Why compute the volume and its cousins? (for algebraic geometers) If P is an integral polytope, then the normalized volume of P is the degree of the toric variety associated to P. (for computational algebraic geometers) Let f 1,..., f n be polynomials in C[x 1,..., x n ]. Let New(f j ) denote the Newton polytope of f j, If f 1,..., f n are generic, then the number of solutions of the polynomial system of equations f 1 = 0,..., f n = 0 with no x i = 0 is equal to the normalized mixed volume n!mv (New(f 1 ),..., New(f n )). (for Combinatorialists ) Volumes count things! CR m = {(a ij ) : i a ij = 1, j a ij = 1, with a ij 0 but a ij = 0 when j > i + 1 }, then NV (CR m ) = product of first (m 2) Catalan numbers. (D. Zeilberger).

6 Computational Complexity of Volume Do we need limits to define volumes of polytopes? volume of egyptian pyramid = 1 (area of base) height 3

7 Easy and pretty in some cases... Computational Complexity of Volume

8 Computational Complexity of Volume In general, proofs seem to rely on an infinite process!

9 But not necessary in dimension two! Computational Complexity of Volume Polygons of the same area are equidecomposable, i.e., one can be partitioned into pieces that can be reassembled into the other.

10 Hilbert s Third Problem Computational Complexity of Volume Are any two convex 3-dimensional polytopes of the same volume equidecomposable?

11 Computational Complexity of Volume NOT always!!! We need calculus to define the volume of

12 high-dimensional polytopes. Computational Complexity of Volume It is hard to compute the volume of a vertex presented polytopes (Dyer and Frieze 1988, Khachiyan 1989). Number of digits necessary to write the volume of a rational polytope P cannot always be bounded by a polynomial on the input size. (J. Lawrence 1991). Theorem (Brightwell and Winkler 1992) It is #P-hard to compute the volume of a d-dimensional polytope P represented by its facets. We even know that it is hard to compute the volume of zonotopes (Dyer, Gritzmann 1998). Thus computing mixed volumes, even for Minkowski sums of line segments, is already hard! For convex bodies, deterministic approximation is already hard, but randomized approximation can be done efficiently (work by Barany, Dyer, Elekes, Furedi, Frieze, Kannan,

13 simplices Via Triangulations Via Rational Functions for Lattice Points SIMPLICES are d-dimensional polytopes with d + 1 vertices. E.g., triangles, tetrahedra, etc. The volume of a (Euclidean) simplex is given by a fast determinant calculation. To compute the volume of a polytope: divide it as a disjoint union of simplices, calculate volume for each simplex and then add them up!

14 Via Triangulations Via Rational Functions for Lattice Points Triangulations: Enough to know how to do it for simplices! Theorem: For all polytopes in fixed dimension d their whole volume can be computed in polynomial time.

15 The size of a triangulation changes! Via Triangulations Via Rational Functions for Lattice Points Triangulations of a convex polyhedron come in different sizes! i.e. the number of simplices changes. 6 8

16 Via Triangulations Via Rational Functions for Lattice Points Counting lattice points to approximate volume Lattice points are those points with integer coordinates: Z n = {(x 1, x 2,..., x n ) x i integer} We wish to count how many lie inside a given polytope! Let P be a convex polytope in R d. For each integer n 1, let np = {nq q P} P 3P

17 Via Triangulations Via Rational Functions for Lattice Points For P a d-polytope, let i(p, n) = #(np Z d ) = #{q P nq Z d } This is the number of lattice points in the dilation np. Volume of P = limit n i(p, n) n d At each dilation we can approximate the volume by placing a small unit cube centered at each lattice point:

18 Lawrence s Style Volume Formulas Via Triangulations Via Rational Functions for Lattice Points Theorem (J. Lawrence 1991) Let P be a simple d-polytope given by {x R d : b i ai t x 0, i = 1... m}. Suppose that c is a vector such that the dot produt of c with any edge of P is non-zero. Then the volume of P equals vol(p) = 1 d! v V (P) ( c, v ) d δ v γ 1 γ 2 γ d where if indices of the constraints that are binding at v are i 1,..., i d then γ i s are such c = γ 1 a i1 + γ 2 a i2 + + γ n a id and δ v = det([a i1, a i2,..., a id ]).

19 Integration of polynomials: From volume to Integration Still people need to compute integrals exactly!!! Given P be a d-dimensional rational polytope inside R n and let f Q[x 1,..., x n ] be a polynomial with rational coefficients. Compute the EXACT value of the integral P f dm?

20 Example From volume to Integration Still people need to compute integrals exactly!!! If we integrate the monomial x 17 y 111 z 13 over the three-dimensional standard simplex. Then x 17 y 111 z 13 dxdydz equals exactly

21 Why exact integration? From volume to Integration Still people need to compute integrals exactly!!! Integrals over polytopes arise in probability, statistics, algebraic geometry, combinatorics, symplectic geometry. Already computing volumes is a very important subroutine. Despite the success of APPROXIMATE integration, still EXACT integration is necessary. Example: Computation of marginal likelihood integrals in model selection. Example: Statisticians used BIC, Laplace, Montecarlo approximations in concrete 6 variable problems. They say Problems are too hard for exact methods. approximation leads to model answer. My point: Exact integration useful for calibration!!!

22 TECHNICAL DETAILS... From volume to Integration Still people need to compute integrals exactly!!! The input simplex : encoding of is given by the number of the dimension d, and the largest binary encoding size of the coordinates among vertices. For simplicity assume the polytope P is full dimension n, in R n dm is the standard Lebesgue measure, which gives volume 1 to the fundamental domain of the lattice Z n. For this dm, every integral of a polynomial function with rational coefficients will be a rational number.

23 From volume to Integration Still people need to compute integrals exactly!!! How to represent a polynomial in a computer? The input polynomial: requires that one specifies concrete data structures for reading the input polynomial and to carry on the calculations. Three main possibilities: dense representation: polynomials are given by a list of the coefficients of all monomials up to a given total degree M. sparse representation: Polynomials are specified by a list of exponent vectors of monomials with non-zero coefficients, together with their coefficients. straight-line program Φ if polynomial is a finite sequence of polynomial functions of Q[x 1,..., x n ], namely q 1,..., q k, such that each q i is either a variable x 1,..., x n, an element of Q, or either the sum or the product of two preceding polynomials in the sequence and such that q k = f.

24 Good News Bad News Best News: Fast Integration for powers of linear forms Theorem: There exists a polynomial-time algorithm for the following problem. Input: numbers d, M N. affinely independent rational vectors s 1,..., s d+1 Q d in binary encoding, a power of a linear form l, x M Output:, in binary l, x M dm.

25 Good News Bad News From fixed number of linear forms to fixed degree. We can also deal with arbitrary polynomials of fixed degree. Write a polynomial as a sum of powers of linear forms. Explicit formula with at most 2 M terms. x m 1 1 x m 2 1 m! 2 x m d 0 p i m i ( 1) m p ( m 1 ) ( p 1 md d = p d ) (p1 x p d x d ) m.

26 Good News Bad News Integration of arbitrary powers of quadratic forms is NP-hard The clique problem (does G contain a clique of size n) is NP-complete. (Karp 1972). Theorem [Motzkin-Straus 1965] G a graph with clique number ω(g). Q G (x) := 1 2 (i,j) E(G) x ix j. Function on standard simplex in R V (G). Then Q G = 1 2 (1 1 ω(g) ). Lemma Let G a graph with d vertices. For p 4(e 1)d 3 ln(32d 2 ), the clique number ω(g) is equal to Q G p. (L p -norm, Holder inequality).

27 Valuations A function S on polyhedra is a valuation. If it is a linear map from the vector space of characteristic functions χ(p i ) of any polyhedra into a field. Thus if polyhedra p i satisfy a linear relation i r iχ(p i ) = 0, then Example: r i S(p i ) = 0, i χ(p 1 p 2 ) + χ(p 1 p 2 ) χ(p 1 ) χ(p 2 ) = 0,

28 Two important valuations for polyhedra p (convex) polyhedron, rational (lattice Λ). S(p)(ξ) := x p Λ e ξ,x generating function for lattice points of p. I (p)(ξ) := e ξ,x dm. when integral and series converge. If p contains a line, then S(p) := 0 and I (p) := 0. IMPORTANT FACT: When p is a simplicial cone easy to write. p

29 Sums S(p) in dim 1 For the real line we have e nξ + e nξ = eξ 1 e ξ e ξ = 0 n>s n<s e nξ = 0 n= For the line segment [a, b] we have. e nξ + n=a χ([a, b]) = χ([, b]) + χ([a, + ]) χ(r) b n= e nξ = eaξ 1 e ξ + ebξ 1 e ξ = eaξ e (b+1)ξ 1 e ξ

30 Case of a simplicial affine cone s + c affine cone with vertex s and integral generators v 1,..., v d lattice Λ. c = R + v R + v d. I (s + c)(ξ) = det (v j) Λ j S(s + c)(ξ) = e ξ,x x (s+b) Λ where b = j [0, 1[v j, semi-closed cell. j e ξ,s ξ, v j 1 1 e ξ,v j

31 Polyhedron sum of its supporting cones at vertices Theorem(Brion-Lawrence-Varchenko) p convex polyhedron, s + c s supporting cone at vertex s. S(p) = S(s + c s ), I (p) = I (s + c s ) s vertices s Example: Let be a simplex. Let l be a linear form which is regular w.r.t., i.e., l, s i l, s j for any pair i j. Then d+1 e l e l,s i dm = d! vol(, dm) j i l, s i s j. (1) i=1 l M M! dm = d! vol(, dm) (M + d)! ( d+1 i=1 l, s i M+d ) j i l, s. (2) i s j

32 Summary Integration of arbitrary powers of linear forms can be done efficiently over simplices Not only over simplices! Also good over simple polytopes with polynomially many vertices, simplicial polytopes with polynomially many facets. Integration of polynomials of fixed degree is OK too. Integration of arbitrary powers of quadratic forms is already hard. Algorithms run nicely in practice. Have been useful to check results.

33 Thank you