Department of Industrial and Systems Engineering Lehigh University COR@L Seminar Series, 2005 10/27/2005
Outline 1 Duality Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation 2 Continuous Case Discrete Case LP formulation Superadditive Dual
References: Jean B. Lasserre, Duality and a Farkas lemma for integer programs. in: Optimization: Structure and Applications (E. Hunt and C.E.M. Pearce, Editors), Kluwer Academic Publishers (2004) Jean B. Lasserre, Integer programming duality and superadditive functions. Contemporary Mathematics 374, pp. 139 150. (2005)
Formulation Duality Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation I : P : LP f (b, c) := max x Ω(b) IP c x P d : f d (b, c) := max x Θ(b) Integration Summation Z ˆf (b, c) := e c x ds I d : ˆfd (b, c) := X Ω(b) x Θ(b) c x e c x where Ω(b) := {x R n : Ax = b, x 0} and Θ(b) := Ω(b) Z n.
Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation The simple relation is that: or equivalently e f (b,c) = lim r ˆf (b, rc) 1/r, 1 f (b, c) = lim r r lnˆf (b, rc), e f d (b,c) = lim r ˆfd (b, rc) 1/r 1 f d (b, c) = lim r r ln ˆf d (b, rc)
Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Let P, P d be some dual problems of P, P d. The well-known strong dual P can be obtained by Legendre-Fenchel Duality formulation. Let f : R n R, then the Fenchel conjugate: f (y) := sup x R n {x y f (x)} y R n If f is convex and lower semi-continuous f := f. Then, noting that f (y, c) is concave, P : f (b, c) := inf b f (λ, c)} = min λ A λ c} λ R m{λ λ R m{b No available transformation to get a strong P d (except the subadditive formulation, see my previous talk...) Let I, I d be the similar transforms of I, I d using Laplace-transform and Z-transform, respectively. Lasserre calls I, I d the natural duals: we can get closed form equations for ˆf (b, c) and ˆfd (b, c).
Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Laplace transform f : C m + C of f : R m + R is Z f (λ) := e λ y f (y)dy R n + Applying to ˆf (b, c): ˆF(λ, c) := := Z e λ yˆf (y, c)dy R m + ny 1 (A λ c) k k=1
Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Note that, ˆF(λ, c) is well defined when R(A λ c) > 0. To get I and hence a closed form ˆf (b, c), we solve the inverse Laplace transform problem: I : ˆf (b, c) := = 1 (2iπ) m 1 (2iπ) m Z γ+i γ i Z γ+i γ i e λ b ˆF(λ, c)dλ e λ b dλ ny (A λ c) k k=1 where γ is fixed and satisfies (A γ c) > 0. This complex integral can be solved directly by Cauchy residual techniques.
Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation In fact, the multidimensional poles of ˆF are real and solve p B A B = c B for all bases of Ω(b). Then: ˆf (b, c) = X x b.f.s of Ω(b) e c x det(b) Y ( c k + p B A k ) k NB where B is the basis of the corresponding b.f.s. In other words, ˆf (b, c) is a weighted summation over the vertices of Ω(b).
Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Z-transform f : C m + C of f : Z m + R is f (z) := X z y f (y) y Z m + Applying to ˆf d (b, c): ˆF d (z, c) := ny 1 1 e c k z A k k=1 which is well-defined if z A k > e c k k = 1,..., n
Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation To get I d and hence a closed form ˆf d (b, c), we solve the inverse Z-transform problem: Z I d : ˆf 1 d (b, c) := ˆF (2iπ) m d (z, c)z b 1 dz z =γ where γ satisfies γ A k > e c k k = 1,..., n. Cauchy residue technique can be used, however, we have complex poles! Bases of ω(b) provide these poles. Each basis B provides det(b) complex poles in the form of: v z(k) = e p B+2iπ det(b) for k = 1,..., det(b) where v {v Z m v B = 0 mod det(b)}
Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Combining these poles with discrete Brian and Vergne s formula: X ˆfd (b, c) = e c x U B (b, c) x b.f.s of Ω(b)
Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Then, 1 f d (b, c) = lim r r ln ˆf d (b, rc) = max {c 1 x + lim x b.f.s of Ω(b) r r ln U B(b, rc)} = max {c x + 1 x b.f.s of Ω(b) q (deg(p b) deg(q b ))} = c x + ρ where q is the l.c.m of det(b): B is a feasible basis, P b and Q b are some real valued polynomials. Note that, ρ is the value of the Gomory group/asymptotic problem!
Continuous Case Duality Continuous Case Discrete Case LP formulation Superadditive Dual Note that a polynomial Q R[λ 1,..., λ m] can be written Q(λ) = X α S Q α λ α = X Q α λ α 1 1...λαm m α S where S N m and Q α are real coefficients α S. Let A R m n, b R m. Then the folllowing two statements are equivalent: (a) The linear system Ax = b has a nonnegative solution x R n. (b) The polynomial b λ can be written b λ = nx Q j (λ)(a jλ), λ R m j=1 for some polynomials Q j R[λ 1,..., λ m], j = 1,..., n, all with nonnegative coefficients. Proof:
Discrete Case Duality Continuous Case Discrete Case LP formulation Superadditive Dual Let, wlog (as long as the feasible region is compact), A N m n, b N m. Then the folllowing two statements are equivalent: (a) The linear system Ax = b has a solution x N n. (b) The polynomial z b 1 can be written z b 1 = nx Q j (z)(z A j 1), z R m j=1 for some polynomials Q j R[z 1,..., z m], j = 1,..., n, all with nonnegative coefficients. Moreover, the total degree of each polynomial Q j can be bounded by b = P mx m j=1 b j min A ik. k=1,...,n Proof: i=1
Discrete Case Duality Continuous Case Discrete Case LP formulation Superadditive Dual Let q 0 be the vector of nonnegative coefficients of all polynomials Q j s. If D := {Mq = r, q 0} then, there exist such polynomials. Note that Mq = r state that the nx polynomials z b 1 and Q j (z)(z A j 1) are identical by equating the respective coefficients. j=1
Discrete Case Duality Continuous Case Discrete Case LP formulation Superadditive Dual Each Q j may be restricted to contain only monomials {z α : α b A j, α N m } Hence, in D, q R s and M R p s where p = s = my (b i + 1) i=1 nx my s j with s j = (b i A ij + 1), j = 1,..., n j=1 i=1 Note that, p is the number of monomials z α with α b and s j is the number of monomials z α with α A j b. Example: Note that M is totally unimodular.
Continuous Case Discrete Case LP formulation Superadditive Dual An LP equivalent to P d Let e sj = (1,..., 1) R s j, j = 1,..., n and let E N n s be the n-block diagonal matrix, whose each diagonal block is a row vector e sj. Then, 1 f d (b, c) = max{c Eq Mq = r, q 0} 2 If q is an optimal solution, then x := Eq is the associated optimal solution of P d. Proof:
A class of superadditive functions Continuous Case Discrete Case LP formulation Superadditive Dual Let D N m be a finite set s.t 0 D and if α D β D, β α. D be the set of functions π : N m R {+ } s.t. π(0) = 0 and π(α) < + if α D, or π(α) = +, otherwise. Given π D, f π : N m R {+ } is defined as f π(x) := inf {π(α + x) π(α)}, x α D Nm Lemma: For every π D : 1 f π D 2 f π π, and f π is superadditive 3 if π is superadditive, then π = f π. Proof:
Superadditive Dual Duality Continuous Case Discrete Case LP formulation Superadditive Dual Let D := {α N m α b}. Then the dual of max{c Eq Mq = r, q 0} is min γ s.t γ(b) γ(0) γ(α + A j ) γ(α) c j, α + A j D, j = 1,..., n Letting π(α) = γ(α) γ(0), α D and extending π to N m, the dual becomes ρ 1 = min π D s.t π(b) π(α + A j ) π(α) c j, α D, j = 1,..., n
Superadditive Dual Duality Continuous Case Discrete Case LP formulation Superadditive Dual Now, assume that f d (b, c) > and consider the following problem ρ 2 = inf π D s.t f π(b) f π(a j ) c j, j = 1,..., n where f π : N m R defined as before for every π D. Then, f d (b, c) = ρ 1 = ρ 2 = f π (b) for some π D Proof: Example:
Superadditive Dual Duality Continuous Case Discrete Case LP formulation Superadditive Dual Note the similar dual formulation of Wolsey min π s.t π(b) π(λ) + π(µ) π(λ + µ), 0 λ + µ b π(a j ) c j, j = 1,..., n π(0) = 0 where the first constraint set state that π : D R is superadditive. The number of variables are the same, however, this one has O(p 2 ) constraints whereas the introduced one has O(np).