Top Ehrhart coefficients of integer partition problems

Transcription

1 Top Ehrhart coefficients of integer partition problems Jesús A. De Loera Department of Mathematics University of California, Davis Joint Math Meetings San Diego January 2013

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3 Goal: Count the solutions of the integer restricted partition problem: Given a = [α 1, α 2,..., α N ] positive integers and t is a non-negative integer, we consider the counting function E a (t) = #{x : α 1 x 1 +α 2 x 2 + +α N x N = t, x 0, x i integer}.

4 Goal: Count the solutions of the integer restricted partition problem: Given a = [α 1, α 2,..., α N ] positive integers and t is a non-negative integer, we consider the counting function E a (t) = #{x : α 1 x 1 +α 2 x 2 + +α N x N = t, x 0, x i integer}. Also known as the Sylvester s DENUMERANT.

5 Goal: Count the solutions of the integer restricted partition problem: Given a = [α 1, α 2,..., α N ] positive integers and t is a non-negative integer, we consider the counting function E a (t) = #{x : α 1 x 1 +α 2 x 2 + +α N x N = t, x 0, x i integer}. Also known as the Sylvester s DENUMERANT. We assume gcd(a) = gcd(α 1, α 2,..., α N ) = 1. E(a)(gcd(a)t) = E a/gcd(a) (t).

6 Goal: Count the solutions of the integer restricted partition problem: Given a = [α 1, α 2,..., α N ] positive integers and t is a non-negative integer, we consider the counting function E a (t) = #{x : α 1 x 1 +α 2 x 2 + +α N x N = t, x 0, x i integer}. Also known as the Sylvester s DENUMERANT. We assume gcd(a) = gcd(α 1, α 2,..., α N ) = 1. E(a)(gcd(a)t) = E a/gcd(a) (t). For a given t, one wishes to decide whether E a (t) 0, but this is NP-complete and the counting problem of lattice points is #P-complete.

7 Goal: Count the solutions of the integer restricted partition problem: Given a = [α 1, α 2,..., α N ] positive integers and t is a non-negative integer, we consider the counting function E a (t) = #{x : α 1 x 1 +α 2 x 2 + +α N x N = t, x 0, x i integer}. Also known as the Sylvester s DENUMERANT. We assume gcd(a) = gcd(α 1, α 2,..., α N ) = 1. E(a)(gcd(a)t) = E a/gcd(a) (t). For a given t, one wishes to decide whether E a (t) 0, but this is NP-complete and the counting problem of lattice points is #P-complete. E a (t) equals number of integral points in the (N 1)-dimensional simplex in R N a = { [x 1, x 2,..., x N ] : x i 0, N i=1 α ix i = t } with rational vertices s i = [0,..., 0, t α i, 0..., 0].

8 E a (b) = #{(x, y, z) 3x + 5y + 17z = b, x 0, y 0, z 0} As b changes we obtain different values for E a (b). E.g., we see that E a (100) = 25, E a (1110) = 2471, etc...

9 E a (b) = #{(x, y, z) 3x + 5y + 17z = b, x 0, y 0, z 0} As b changes we obtain different values for E a (b). E.g., we see that E a (100) = 25, E a (1110) = 2471, etc... BIG QUESTION: How does this function behave? Geometrically we are dilating the simplex as b grows...

10 For P a d-dimensional convex polytope, consider the Ehrhart function E P (n) = # {a (np Z d )} This is done for the lattice points in the dilation np. P 3P

11 Ehrhart quasipolynomials Theorem (E. Ehrhart 1962) For P a rational convex polytope on R d and n N dilatation factor. Then the function E P (n) = # {a (np Z d )}

12 Ehrhart quasipolynomials Theorem (E. Ehrhart 1962) For P a rational convex polytope on R d and n N dilatation factor. Then the function E P (n) = # {a (np Z d )} is a quasipolynomial of degree dim P: E P (n) = dim P k=0 E k (n)n k, Coefficients E P (n) are periodic modular functions Depend only on n mod M, for some integer M.

13 Ehrhart quasipolynomials Theorem (E. Ehrhart 1962) For P a rational convex polytope on R d and n N dilatation factor. Then the function E P (n) = # {a (np Z d )} is a quasipolynomial of degree dim P: E P (n) = dim P k=0 E k (n)n k, Coefficients E P (n) are periodic modular functions Depend only on n mod M, for some integer M. Its leading coefficient is the normalized volume of the simplex.

14 Ehrhart quasipolynomials Theorem (E. Ehrhart 1962) For P a rational convex polytope on R d and n N dilatation factor. Then the function E P (n) = # {a (np Z d )} is a quasipolynomial of degree dim P: E P (n) = dim P k=0 E k (n)n k, Coefficients E P (n) are periodic modular functions Depend only on n mod M, for some integer M. Its leading coefficient is the normalized volume of the simplex. When the coordinates of the vertices of P are integers, E P (n) is a polynomial in n. It is an Ehrhart polynomial.

15 Example i(p, n) = (n + 1) 2 In general for a d-dimensional unit cube i(p, n) = (n + 1) d.

16 Example Consider the Denumerant problem a = [6, 2, 3]. On each of the cosets q + 6Z, the function E a (t) coincides with a single polynomial E [q] (t)!! Here are the corresponding polynomials. E [0] (t) = 1 72 t t + 1, E [1] (t) = 1 72 t t 5 72, E [2] (t) = 1 72 t t + 5 9, E [3] (t) = 1 72 t t + 3 8, E [4] (t) = 1 72 t t + 2 9, E [5] (t) = 1 72 t t

17 Example Consider the Denumerant problem a = [6, 2, 3]. On each of the cosets q + 6Z, the function E a (t) coincides with a single polynomial E [q] (t)!! Here are the corresponding polynomials. E [0] (t) = 1 72 t t + 1, E [1] (t) = 1 72 t t 5 72, E [2] (t) = 1 72 t t + 5 9, E [3] (t) = 1 72 t t + 3 8, E [4] (t) = 1 72 t t + 2 9, E [5] (t) = 1 72 t t Warning: Hard to figure out the periodicity!!

18 Example Consider the Denumerant problem a = [6, 2, 3]. On each of the cosets q + 6Z, the function E a (t) coincides with a single polynomial E [q] (t)!! Here are the corresponding polynomials. E [0] (t) = 1 72 t t + 1, E [1] (t) = 1 72 t t 5 72, E [2] (t) = 1 72 t t + 5 9, E [3] (t) = 1 72 t t + 3 8, E [4] (t) = 1 72 t t + 2 9, E [5] (t) = 1 72 t t Warning: Hard to figure out the periodicity!! Warning: This is NOT an efficient way to represent the quasipolynomial!! Too many pieces!!!

19 Example Consider the Denumerant problem a = [6, 2, 3]. On each of the cosets q + 6Z, the function E a (t) coincides with a single polynomial E [q] (t)!! Here are the corresponding polynomials. E [0] (t) = 1 72 t t + 1, E [1] (t) = 1 72 t t 5 72, E [2] (t) = 1 72 t t + 5 9, E [3] (t) = 1 72 t t + 3 8, E [4] (t) = 1 72 t t + 2 9, E [5] (t) = 1 72 t t Warning: Hard to figure out the periodicity!! Warning: This is NOT an efficient way to represent the quasipolynomial!! Too many pieces!!! GOOD NEWS: There are other (better!!!) ways to represent quasi-polynomials.

20 Previous Algorithmic results Theorem When the number of variables is fixed, there is a polynomial-time algorithm to compute Ehrhart quasi-polynomials (shown as rational functions) (follows from Barvinok 1993).

21 Previous Algorithmic results Theorem When the number of variables is fixed, there is a polynomial-time algorithm to compute Ehrhart quasi-polynomials (shown as rational functions) (follows from Barvinok 1993). First implemented in LattE in 2000.

22 Previous Algorithmic results Theorem When the number of variables is fixed, there is a polynomial-time algorithm to compute Ehrhart quasi-polynomials (shown as rational functions) (follows from Barvinok 1993). First implemented in LattE in 2000.

23 Previous Algorithmic results Theorem When the number of variables is fixed, there is a polynomial-time algorithm to compute Ehrhart quasi-polynomials (shown as rational functions) (follows from Barvinok 1993). First implemented in LattE in But, WHAT CAN BE DONE IN NON-FIXED DIMENSION?

24 Previous Algorithmic results Theorem When the number of variables is fixed, there is a polynomial-time algorithm to compute Ehrhart quasi-polynomials (shown as rational functions) (follows from Barvinok 1993). First implemented in LattE in But, WHAT CAN BE DONE IN NON-FIXED DIMENSION? For fixed k 0, polynomial time algorithm for computing the top k Ehrhart coefficients, for the number of lattice points of a simplex. (Barvinok 2006).

25 Previous Algorithmic results Theorem When the number of variables is fixed, there is a polynomial-time algorithm to compute Ehrhart quasi-polynomials (shown as rational functions) (follows from Barvinok 1993). First implemented in LattE in But, WHAT CAN BE DONE IN NON-FIXED DIMENSION? For fixed k 0, polynomial time algorithm for computing the top k Ehrhart coefficients, for the number of lattice points of a simplex. (Barvinok 2006). Generalizations to weighted counting (PISA team 2011)

26 Previous Algorithmic results Theorem When the number of variables is fixed, there is a polynomial-time algorithm to compute Ehrhart quasi-polynomials (shown as rational functions) (follows from Barvinok 1993). First implemented in LattE in But, WHAT CAN BE DONE IN NON-FIXED DIMENSION? For fixed k 0, polynomial time algorithm for computing the top k Ehrhart coefficients, for the number of lattice points of a simplex. (Barvinok 2006). Generalizations to weighted counting (PISA team 2011) Deep Consequences in the Theory of Optimization

27 And now... THE NEW RESULTS

28 COMMERCIAL BREAK!!!

29 Are you thirsty to hear applications of Algebraic Combinatorics and Discrete Geometry?

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31 Main Theorem (2012) (Pisa Team) There is a polynomial time algorithm for the following problem. Fix k 0 positive integer. Input: a = [α 1, α 2,..., α N ] be a sequence of positive integers.

32 Main Theorem (2012) (Pisa Team) There is a polynomial time algorithm for the following problem. Fix k 0 positive integer. Input: a = [α 1, α 2,..., α N ] be a sequence of positive integers. Output: The k top degree Ehrhart coefficients of the quasi-polynomial function E a (t) = #{x : a T x = t, x 0, x integral}. This will be presented as a Step polynomials. The dimension not fixed!!!!!.

33 What are step polynomials? (i) Let {s} = s s [0, 1) for s R, where s denotes the smallest integer larger or equal to s. The function {s + 1} = {s} is a periodic function of s modulo 1.

34 What are step polynomials? (i) Let {s} = s s [0, 1) for s R, where s denotes the smallest integer larger or equal to s. The function {s + 1} = {s} is a periodic function of s modulo 1. (ii) If r is rational with denominator q, the function T {rt } is a function of T R periodic modulo q.

35 What are step polynomials? (i) Let {s} = s s [0, 1) for s R, where s denotes the smallest integer larger or equal to s. The function {s + 1} = {s} is a periodic function of s modulo 1. (ii) If r is rational with denominator q, the function T {rt } is a function of T R periodic modulo q.

36 1. Definition A periodic function φ(t ) of the form T L J l c l l=1 j=1 {r l,j T } n l,j. will be called a step polynomial.

37 1. Definition A periodic function φ(t ) of the form T L J l c l l=1 j=1 {r l,j T } n l,j. will be called a step polynomial. 2. A step polynomial φ has of degree u if j n j u for all set of indices I occurring in the formula for φ.

38 1. Definition A periodic function φ(t ) of the form T L J l c l l=1 j=1 {r l,j T } n l,j. will be called a step polynomial. 2. A step polynomial φ has of degree u if j n j u for all set of indices I occurring in the formula for φ. 3. φ is of period q if all the rational numbers r j have common denominator q.

39 Example Wish to compute E a (t) for a = [1, 2, 3, 4]. The coefficients are: t3

40 Example Wish to compute E a (t) for a = [1, 2, 3, 4]. The coefficients are: t t2

41 Example Wish to compute E a (t) for a = [1, 2, 3, 4]. The coefficients are: t t2 (1/2 1/4{t/2} + 1/4 ({t/2}) 2 )t

42 Example Wish to compute E a (t) for a = [1, 2, 3, 4]. The coefficients are: t t2 (1/2 1/4{t/2} + 1/4 ({t/2}) 2 )t 1 + 3/2 ({t/3}) 3 3/2 ({t/3}) 2 1/3 {t/4} ({t/4}) 2 + 4/3 ({t/4}) 3 7/6 {t/2} + {t/4}{t/2} + 1/2 ({t/2}) 2 {t/4} ({t/2}) 2 + 2/3 ({t/2}) 3.

43 KEY IDEAS + METHODS

44 Counting through generating functions Given a = [α 1, α 2,..., α N+1 ]. We can construct a generating function F a (z) := E a (n)z n = n=0 1 N i=1 (1 zα i )

45 Counting through generating functions Given a = [α 1, α 2,..., α N+1 ]. We can construct a generating function F a (z) := E a (n)z n = n=0 1 N i=1 (1 zα i ) EXAMPLE When a = [3, 5, 17], a short formula for E a (t) would be a generating function! E a (t)z t = t=0 1 (1 z 17 ) (1 z 5 ) (1 z 3 ).

46 Counting through generating functions Given a = [α 1, α 2,..., α N+1 ]. We can construct a generating function F a (z) := E a (n)z n = n=0 1 N i=1 (1 zα i ) EXAMPLE When a = [3, 5, 17], a short formula for E a (t) would be a generating function! E a (t)z t = t=0 1 (1 z 17 ) (1 z 5 ) (1 z 3 ). Basic complex analysis: Compute the values of E a (n), through the poles of the complex function F a(z).

47 NOTE: The poles of F a (z) are roots of unity P = N+1 i=1 { ζ C : ζα i = 1 } Lemma: Let a = [α 1, α 2,..., α N ] be a list of integers with greatest common divisor equal to 1, and let F(a)(z) := 1 N i=1 (1 zα i ). If t is a non-negative integer, then E(a)(t) = ζ P Res z=ζ z t 1 F a (z) dz (1) and the ζ-term of this sum is a quasi-polynomial function of t with degree less than or equal to p(ζ) 1.

48 IDEA 1: Only the higher-order poles matter!!! Because the α i s have greatest common divisor 1, we have ζ = 1 as a pole of order N Other poles have order strictly less than N.

49 IDEA 1: Only the higher-order poles matter!!! Because the α i s have greatest common divisor 1, we have ζ = 1 as a pole of order N Other poles have order strictly less than N. Denote by p(ζ) the order of the pole ζ for ζ P.

50 IDEA 1: Only the higher-order poles matter!!! Because the α i s have greatest common divisor 1, we have ζ = 1 as a pole of order N Other poles have order strictly less than N. Denote by p(ζ) the order of the pole ζ for ζ P. Given an integer 0 k N, we partition the set of poles P in two disjoint sets: P >N k = { ζ : p(ζ) > N k }, P N k = { ζ : p(ζ) N k }.

51 IDEA 1: Only the higher-order poles matter!!! Because the α i s have greatest common divisor 1, we have ζ = 1 as a pole of order N Other poles have order strictly less than N. Denote by p(ζ) the order of the pole ζ for ζ P. Given an integer 0 k N, we partition the set of poles P in two disjoint sets: P >N k = { ζ : p(ζ) > N k }, P N k = { ζ : p(ζ) N k }. We have E(a)(t) = E P>N k (t) + E P N k (t),

52 IDEA 1: Only the higher-order poles matter!!! Because the α i s have greatest common divisor 1, we have ζ = 1 as a pole of order N Other poles have order strictly less than N. Denote by p(ζ) the order of the pole ζ for ζ P. Given an integer 0 k N, we partition the set of poles P in two disjoint sets: P >N k = { ζ : p(ζ) > N k }, P N k = { ζ : p(ζ) N k }. We have E(a)(t) = E P>N k (t) + E P N k (t), For computing what we need it is sufficient to compute the function E P>N k (t).

53 IDEA 1: Only the higher-order poles matter!!! Because the α i s have greatest common divisor 1, we have ζ = 1 as a pole of order N Other poles have order strictly less than N. Denote by p(ζ) the order of the pole ζ for ζ P. Given an integer 0 k N, we partition the set of poles P in two disjoint sets: P >N k = { ζ : p(ζ) > N k }, P N k = { ζ : p(ζ) N k }. We have E(a)(t) = E P>N k (t) + E P N k (t), For computing what we need it is sufficient to compute the function E P>N k (t). The function E P N k (t) is a quasi-polynomial function of t of degree in t strictly less than N k.

54 IDEA 2: Posets+Groups on the higher-order poles If ζ is a pole of order p, this means that there exist at least p elements α i in the list a so that ζ α i = 1.

55 IDEA 2: Posets+Groups on the higher-order poles If ζ is a pole of order p, this means that there exist at least p elements α i in the list a so that ζ α i = 1. ζ α i = 1 for a list α i1,... α ir ζ f = 1, for f the greatest common divisor of the elements α i, i I.

56 IDEA 2: Posets+Groups on the higher-order poles If ζ is a pole of order p, this means that there exist at least p elements α i in the list a so that ζ α i = 1. ζ α i = 1 for a list α i1,... α ir ζ f = 1, for f the greatest common divisor of the elements α i, i I. I >N k be the set of sublists of a of length greater than N k. I >N k is stable by the operation of taking supersets,

57 IDEA 2: Posets+Groups on the higher-order poles If ζ is a pole of order p, this means that there exist at least p elements α i in the list a so that ζ α i = 1. ζ α i = 1 for a list α i1,... α ir ζ f = 1, for f the greatest common divisor of the elements α i, i I. I >N k be the set of sublists of a of length greater than N k. I >N k is stable by the operation of taking supersets, Define f I to be the greatest common divisor of the sublist I = [α i1,... α ir ]. Let G >N k (a) = { f I : I I >N k } be the set of greatest common divisors so obtained.

58 IDEA 2: Posets+Groups on the higher-order poles If ζ is a pole of order p, this means that there exist at least p elements α i in the list a so that ζ α i = 1. ζ α i = 1 for a list α i1,... α ir ζ f = 1, for f the greatest common divisor of the elements α i, i I. I >N k be the set of sublists of a of length greater than N k. I >N k is stable by the operation of taking supersets, Define f I to be the greatest common divisor of the sublist I = [α i1,... α ir ]. Let G >N k (a) = { f I : I I >N k } be the set of greatest common divisors so obtained. the set G >N k (a) is a set of integers stable by the operation of taking greatest common divisors. Thus, G >N k (a) is a partially ordered set, where f f if f divides f.

59 IDEA 2: Posets+Groups on the higher-order poles If ζ is a pole of order p, this means that there exist at least p elements α i in the list a so that ζ α i = 1. ζ α i = 1 for a list α i1,... α ir ζ f = 1, for f the greatest common divisor of the elements α i, i I. I >N k be the set of sublists of a of length greater than N k. I >N k is stable by the operation of taking supersets, Define f I to be the greatest common divisor of the sublist I = [α i1,... α ir ]. Let G >N k (a) = { f I : I I >N k } be the set of greatest common divisors so obtained. the set G >N k (a) is a set of integers stable by the operation of taking greatest common divisors. Thus, G >N k (a) is a partially ordered set, where f f if f divides f. Using the group G(f ) C of f -th roots of unity, G(f ) = { ζ C : ζ f = 1 }, we have thus P >N k = f G >N k (a) G(f ). Not a disjoint union!!!

60 Lemma Inclusion exclusion principle says: We can write the characteristic function of P >N k as a linear combination of characteristic functions of the groups G(f ): [P >N k ] = f G >N k (a) where µ(f ) is the Möbius function. µ(f )[G(f )],

61 Lemma Inclusion exclusion principle says: We can write the characteristic function of P >N k as a linear combination of characteristic functions of the groups G(f ): [P >N k ] = f G >N k (a) where µ(f ) is the Möbius function. Lemma If f is a positive integer define µ(f )[G(f )], E(a, f )(t) = ζ f =1 Res z=ζ z t 1 F(a)(z) dz. Let k be a fixed integer. Then E P>N k (t) = µ(f )E(a, f )(t). f G >N k (a)

62 IDEA 3: Use polyhedral cones and special lattices! Use convex geometry to compute E(a, f )(t) = ζ f =1 Res z=ζ z t 1 F(a)(z) dz.

63 IDEA 3: Use polyhedral cones and special lattices! Use convex geometry to compute E(a, f )(t) = ζ f =1 Res z=ζ z t 1 F(a)(z) dz. SUPER COOL Lemma For each f G >N k (a), the function E(a, f )(t) is the generating functions for lattice points of cones of fixed dimension k but on a different lattice depending on f.

64 IDEA 3: Use polyhedral cones and special lattices! Use convex geometry to compute E(a, f )(t) = ζ f =1 Res z=ζ z t 1 F(a)(z) dz. SUPER COOL Lemma For each f G >N k (a), the function E(a, f )(t) is the generating functions for lattice points of cones of fixed dimension k but on a different lattice depending on f. NEWS Very Nice paper with experiments coming soon!!!