X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Fcone%2Fcone.py;h=3f5a4fed4e1c49853f00eafcf6084744223ca296;hb=3e6f51aa1f2d6f300cb22281701901add3631904;hp=23043381bca83acd6d7f17479ef4769980b9960f;hpb=012175f3a2591586099b4e955bb440958f2d63d7;p=sage.d.git diff --git a/mjo/cone/cone.py b/mjo/cone/cone.py index 2304338..3f5a4fe 100644 --- a/mjo/cone/cone.py +++ b/mjo/cone/cone.py @@ -7,6 +7,270 @@ addsitedir(abspath('../../')) from sage.all import * +def rename_lattice(L,s): + r""" + Change all names of the given lattice to ``s``. + """ + L._name = s + L._dual_name = s + L._latex_name = s + L._latex_dual_name = s + +def span_iso(K): + r""" + Return an isomorphism (and its inverse) that will send ``K`` into a + lower-dimensional space isomorphic to its span (and back). + + EXAMPLES: + + The inverse composed with the isomorphism should be the identity:: + + sage: K = random_cone(max_dim=10) + sage: (phi, phi_inv) = span_iso(K) + sage: phi_inv(phi(K)) == K + True + + The image of ``K`` under the isomorphism should have full dimension:: + + sage: K = random_cone(max_dim=10) + sage: (phi, phi_inv) = span_iso(K) + sage: phi(K).dim() == phi(K).lattice_dim() + True + + The isomorphism should be an inner product space isomorphism, and + thus it should preserve dual cones (and commute with the "dual" + operation). But beware the automatic renaming of the dual lattice. + OH AND YOU HAVE TO SORT THE CONES:: + + sage: K = random_cone(max_dim=10, strictly_convex=False, solid=True) + sage: L = K.lattice() + sage: rename_lattice(L, 'L') + sage: (phi, phi_inv) = span_iso(K) + sage: sorted(phi_inv( phi(K).dual() )) == sorted(K.dual()) + True + + We may need to isomorph twice to make sure we stop moving down to + smaller spaces. (Once you've done this on a cone and its dual, the + result should be proper.) OH AND YOU HAVE TO SORT THE CONES:: + + sage: K = random_cone(max_dim=10, strictly_convex=False, solid=False) + sage: L = K.lattice() + sage: rename_lattice(L, 'L') + sage: (phi, phi_inv) = span_iso(K) + sage: K_S = phi(K) + sage: (phi_dual, phi_dual_inv) = span_iso(K_S.dual()) + sage: J_T = phi_dual(K_S.dual()).dual() + sage: phi_inv(phi_dual_inv(J_T)) == K + True + + """ + phi_domain = K.sublattice().vector_space() + phi_codo = VectorSpace(phi_domain.base_field(), phi_domain.dimension()) + + # S goes from the new space to the cone space. + S = linear_transformation(phi_codo, phi_domain, phi_domain.basis()) + + # phi goes from the cone space to the new space. + def phi(J_orig): + r""" + Takes a cone ``J`` and sends it into the new space. + """ + newrays = map(S.inverse(), J_orig.rays()) + L = None + if len(newrays) == 0: + L = ToricLattice(0) + + return Cone(newrays, lattice=L) + + def phi_inverse(J_sub): + r""" + The inverse to phi which goes from the new space to the cone space. + """ + newrays = map(S, J_sub.rays()) + return Cone(newrays, lattice=K.lattice()) + + + return (phi, phi_inverse) + + + +def discrete_complementarity_set(K): + r""" + Compute the discrete complementarity set of this cone. + + The complementarity set of this cone is the set of all orthogonal + pairs `(x,s)` such that `x` is in this cone, and `s` is in its + dual. The discrete complementarity set restricts `x` and `s` to be + generators of their respective cones. + + OUTPUT: + + A list of pairs `(x,s)` such that, + + * `x` is in this cone. + * `x` is a generator of this cone. + * `s` is in this cone's dual. + * `s` is a generator of this cone's dual. + * `x` and `s` are orthogonal. + + EXAMPLES: + + The discrete complementarity set of the nonnegative orthant consists + of pairs of standard basis vectors:: + + sage: K = Cone([(1,0),(0,1)]) + sage: discrete_complementarity_set(K) + [((1, 0), (0, 1)), ((0, 1), (1, 0))] + + If the cone consists of a single ray, the second components of the + discrete complementarity set should generate the orthogonal + complement of that ray:: + + sage: K = Cone([(1,0)]) + sage: discrete_complementarity_set(K) + [((1, 0), (0, 1)), ((1, 0), (0, -1))] + sage: K = Cone([(1,0,0)]) + sage: discrete_complementarity_set(K) + [((1, 0, 0), (0, 1, 0)), + ((1, 0, 0), (0, -1, 0)), + ((1, 0, 0), (0, 0, 1)), + ((1, 0, 0), (0, 0, -1))] + + When the cone is the entire space, its dual is the trivial cone, so + the discrete complementarity set is empty:: + + sage: K = Cone([(1,0),(-1,0),(0,1),(0,-1)]) + sage: discrete_complementarity_set(K) + [] + + TESTS: + + The complementarity set of the dual can be obtained by switching the + components of the complementarity set of the original cone:: + + sage: K1 = random_cone(max_dim=10, max_rays=10) + sage: K2 = K1.dual() + sage: expected = [(x,s) for (s,x) in discrete_complementarity_set(K2)] + sage: actual = discrete_complementarity_set(K1) + sage: actual == expected + True + + """ + V = K.lattice().vector_space() + + # Convert the rays to vectors so that we can compute inner + # products. + xs = [V(x) for x in K.rays()] + ss = [V(s) for s in K.dual().rays()] + + return [(x,s) for x in xs for s in ss if x.inner_product(s) == 0] + + +def LL(K): + r""" + Compute the space `\mathbf{LL}` of all Lyapunov-like transformations + on this cone. + + OUTPUT: + + A list of matrices forming a basis for the space of all + Lyapunov-like transformations on the given cone. + + EXAMPLES: + + The trivial cone has no Lyapunov-like transformations:: + + sage: L = ToricLattice(0) + sage: K = Cone([], lattice=L) + sage: LL(K) + [] + + The Lyapunov-like transformations on the nonnegative orthant are + simply diagonal matrices:: + + sage: K = Cone([(1,)]) + sage: LL(K) + [[1]] + + sage: K = Cone([(1,0),(0,1)]) + sage: LL(K) + [ + [1 0] [0 0] + [0 0], [0 1] + ] + + sage: K = Cone([(1,0,0),(0,1,0),(0,0,1)]) + sage: LL(K) + [ + [1 0 0] [0 0 0] [0 0 0] + [0 0 0] [0 1 0] [0 0 0] + [0 0 0], [0 0 0], [0 0 1] + ] + + Only the identity matrix is Lyapunov-like on the `L^{3}_{1}` and + `L^{3}_{\infty}` cones [Rudolf et al.]_:: + + sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)]) + sage: LL(L31) + [ + [1 0 0] + [0 1 0] + [0 0 1] + ] + + sage: L3infty = Cone([(0,1,1), (1,0,1), (0,-1,1), (-1,0,1)]) + sage: LL(L3infty) + [ + [1 0 0] + [0 1 0] + [0 0 1] + ] + + TESTS: + + The inner product `\left< L\left(x\right), s \right>` is zero for + every pair `\left( x,s \right)` in the discrete complementarity set + of the cone:: + + sage: K = random_cone(max_dim=8, max_rays=10) + sage: C_of_K = discrete_complementarity_set(K) + sage: l = [ (L*x).inner_product(s) for (x,s) in C_of_K for L in LL(K) ] + sage: sum(map(abs, l)) + 0 + + """ + V = K.lattice().vector_space() + + C_of_K = discrete_complementarity_set(K) + + tensor_products = [s.tensor_product(x) for (x,s) in C_of_K] + + # Sage doesn't think matrices are vectors, so we have to convert + # our matrices to vectors explicitly before we can figure out how + # many are linearly-indepenedent. + # + # The space W has the same base ring as V, but dimension + # dim(V)^2. So it has the same dimension as the space of linear + # transformations on V. In other words, it's just the right size + # to create an isomorphism between it and our matrices. + W = VectorSpace(V.base_ring(), V.dimension()**2) + + # Turn our matrices into long vectors... + vectors = [ W(m.list()) for m in tensor_products ] + + # Vector space representation of Lyapunov-like matrices + # (i.e. vec(L) where L is Luapunov-like). + LL_vector = W.span(vectors).complement() + + # Now construct an ambient MatrixSpace in which to stick our + # transformations. + M = MatrixSpace(V.base_ring(), V.dimension()) + + matrix_basis = [ M(v.list()) for v in LL_vector.basis() ] + + return matrix_basis + + def lyapunov_rank(K): r""" @@ -51,17 +315,21 @@ def lyapunov_rank(K): REFERENCES: - 1. M.S. Gowda and J. Tao. On the bilinearity rank of a proper cone - and Lyapunov-like transformations, Mathematical Programming, 147 + .. [Gowda/Tao] M.S. Gowda and J. Tao. On the bilinearity rank of a proper + cone and Lyapunov-like transformations, Mathematical Programming, 147 (2014) 155-170. - 2. G. Rudolf, N. Noyan, D. Papp, and F. Alizadeh, Bilinear + .. [Orlitzky/Gowda] M. Orlitzky and M. S. Gowda. The Lyapunov Rank of an + Improper Cone. Work in-progress. + + .. [Rudolf et al.] G. Rudolf, N. Noyan, D. Papp, and F. Alizadeh, Bilinear optimality constraints for the cone of positive polynomials, Mathematical Programming, Series B, 129 (2011) 5-31. EXAMPLES: - The nonnegative orthant in `\mathbb{R}^{n}` always has rank `n`:: + The nonnegative orthant in `\mathbb{R}^{n}` always has rank `n` + [Rudolf et al.]_:: sage: positives = Cone([(1,)]) sage: lyapunov_rank(positives) @@ -69,23 +337,25 @@ def lyapunov_rank(K): sage: quadrant = Cone([(1,0), (0,1)]) sage: lyapunov_rank(quadrant) 2 - sage: octant = Cone([(1,0,0), (0,1,0), (0,0,1)]) + sage: octant = Cone([(1,0,0), (0,1,0), (0,0,1)]) sage: lyapunov_rank(octant) 3 - The `L^{3}_{1}` cone is known to have a Lyapunov rank of one:: + The `L^{3}_{1}` cone is known to have a Lyapunov rank of one + [Rudolf et al.]_:: sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)]) sage: lyapunov_rank(L31) 1 - Likewise for the `L^{3}_{\infty}` cone:: + Likewise for the `L^{3}_{\infty}` cone [Rudolf et al.]_:: sage: L3infty = Cone([(0,1,1), (1,0,1), (0,-1,1), (-1,0,1)]) sage: lyapunov_rank(L3infty) 1 - The Lyapunov rank should be additive on a product of cones:: + The Lyapunov rank should be additive on a product of cones + [Rudolf et al.]_:: sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)]) sage: octant = Cone([(1,0,0), (0,1,0), (0,0,1)]) @@ -93,8 +363,8 @@ def lyapunov_rank(K): sage: lyapunov_rank(K) == lyapunov_rank(L31) + lyapunov_rank(octant) True - Two isomorphic cones should have the same Lyapunov rank. The cone - ``K`` in the following example is isomorphic to the nonnegative + Two isomorphic cones should have the same Lyapunov rank [Rudolf et al.]_. + The cone ``K`` in the following example is isomorphic to the nonnegative octant in `\mathbb{R}^{3}`:: sage: K = Cone([(1,2,3), (-1,1,0), (1,0,6)]) @@ -102,40 +372,69 @@ def lyapunov_rank(K): 3 The dual cone `K^{*}` of ``K`` should have the same Lyapunov rank as ``K`` - itself:: + itself [Rudolf et al.]_:: sage: K = Cone([(2,2,4), (-1,9,0), (2,0,6)]) sage: lyapunov_rank(K) == lyapunov_rank(K.dual()) True - """ - V = K.lattice().vector_space() + TESTS: - xs = [V(x) for x in K.rays()] - ss = [V(s) for s in K.dual().rays()] + The Lyapunov rank should be additive on a product of cones + [Rudolf et al.]_:: - # WARNING: This isn't really C(K), it only contains the pairs - # (x,s) in C(K) where x,s are extreme in their respective cones. - C_of_K = [(x,s) for x in xs for s in ss if x.inner_product(s) == 0] + sage: K1 = random_cone(max_dim=10, max_rays=10) + sage: K2 = random_cone(max_dim=10, max_rays=10) + sage: K = K1.cartesian_product(K2) + sage: lyapunov_rank(K) == lyapunov_rank(K1) + lyapunov_rank(K2) + True - matrices = [x.column() * s.row() for (x,s) in C_of_K] + The dual cone `K^{*}` of ``K`` should have the same Lyapunov rank as ``K`` + itself [Rudolf et al.]_:: - # Sage doesn't think matrices are vectors, so we have to convert - # our matrices to vectors explicitly before we can figure out how - # many are linearly-indepenedent. - # - # The space W has the same base ring as V, but dimension - # dim(V)^2. So it has the same dimension as the space of linear - # transformations on V. In other words, it's just the right size - # to create an isomorphism between it and our matrices. - W = VectorSpace(V.base_ring(), V.dimension()**2) + sage: K = random_cone(max_dim=10, max_rays=10) + sage: lyapunov_rank(K) == lyapunov_rank(K.dual()) + True - def phi(m): - r""" - Convert a matrix to a vector isomorphically. - """ - return W(m.list()) + The Lyapunov rank of a proper polyhedral cone in `n` dimensions can + be any number between `1` and `n` inclusive, excluding `n-1` + [Gowda/Tao]_. By accident, the `n-1` restriction will hold for the + trivial cone in a trivial space as well. However, in zero dimensions, + the Lyapunov rank of the trivial cone will be zero:: - vectors = [phi(m) for m in matrices] + sage: K = random_cone(max_dim=10, strictly_convex=True, solid=True) + sage: b = lyapunov_rank(K) + sage: n = K.lattice_dim() + sage: (n == 0 or 1 <= b) and b <= n + True + sage: b == n-1 + False + + In fact [Orlitzky/Gowda]_, no closed convex polyhedral cone can have + Lyapunov rank `n-1` in `n` dimensions:: + + sage: K = random_cone(max_dim=10) + sage: b = lyapunov_rank(K) + sage: n = K.lattice_dim() + sage: b == n-1 + False + + The calculation of the Lyapunov rank of an improper cone can be + reduced to that of a proper cone [Orlitzky/Gowda]_:: + + sage: K = random_cone(max_dim=15, solid=False, strictly_convex=False) + sage: actual = lyapunov_rank(K) + sage: (phi1, phi1_inv) = span_iso(K) + sage: K_S = phi1(K) + sage: (phi2, phi2_inv) = span_iso(K_S.dual()) + sage: J_T = phi2(K_S.dual()).dual() + sage: phi1_inv(phi2_inv(J_T)) == K + True + sage: l = K.linear_subspace().dimension() + sage: codim = K.lattice_dim() - K.dim() + sage: expected = lyapunov_rank(J_T) + K.dim()*(l + codim) + codim**2 + sage: actual == expected + True - return (W.dimension() - W.span(vectors).rank()) + """ + return len(LL(K))