X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Fcone%2Fcone.py;h=3f5a4fed4e1c49853f00eafcf6084744223ca296;hb=3e6f51aa1f2d6f300cb22281701901add3631904;hp=424a907dc75d1ed7d4a1a59e1e8eee7b9f64d75e;hpb=08ff39a270e90e6263339da12144ddd2bfc6f3f5;p=sage.d.git diff --git a/mjo/cone/cone.py b/mjo/cone/cone.py index 424a907..3f5a4fe 100644 --- a/mjo/cone/cone.py +++ b/mjo/cone/cone.py @@ -7,228 +7,91 @@ addsitedir(abspath('../../')) from sage.all import * - -def is_full_space(K): +def rename_lattice(L,s): r""" - Return whether or not this cone is equal to its ambient vector space. - - OUTPUT: - - ``True`` if this cone is the entire vector space and ``False`` - otherwise. - - EXAMPLES: - - A ray in two dimensions is not equal to the entire space:: - - sage: K = Cone([(1,0)]) - sage: is_full_space(K) - False - - Neither is the nonnegative orthant:: - - sage: K = Cone([(1,0),(0,1)]) - sage: is_full_space(K) - False - - The right half-space contains a vector subspace, but it is still not - equal to the entire plane:: - - sage: K = Cone([(1,0),(-1,0),(0,1)]) - sage: is_full_space(K) - False - - But if we include nonnegative sums from both axes, then the resulting - cone is the entire two-dimensional space:: - - sage: K = Cone([(1,0),(-1,0),(0,1),(0,-1)]) - sage: is_full_space(K) - True - + Change all names of the given lattice to ``s``. """ - return K.linear_subspace() == K.lattice().vector_space() + L._name = s + L._dual_name = s + L._latex_name = s + L._latex_dual_name = s - -def random_cone(min_dim=0, max_dim=None, min_rays=0, max_rays=None): +def span_iso(K): r""" - Generate a random rational convex polyhedral cone. - - Lower and upper bounds may be provided for both the dimension of the - ambient space and the number of generating rays of the cone. If a - lower bound is left unspecified, it defaults to zero. Unspecified - upper bounds will be chosen randomly. - - The lower bound on the number of rays is limited to twice the - maximum dimension of the ambient vector space. To see why, consider - the space $\mathbb{R}^{2}$, and suppose we have generated four rays, - $\left\{ \pm e_{1}, \pm e_{2} \right\}$. Clearly any other ray in - the space is a nonnegative linear combination of those four, - so it is hopeless to generate more. It is therefore an error - to request more in the form of ``min_rays``. - - .. NOTE: - - If you do not explicitly request more than ``2 * max_dim`` rays, - a larger number may still be randomly generated. In that case, - the returned cone will simply be equal to the entire space. - - INPUT: - - - ``min_dim`` (default: zero) -- A nonnegative integer representing the - minimum dimension of the ambient lattice. - - - ``max_dim`` (default: random) -- A nonnegative integer representing - the maximum dimension of the ambient - lattice. - - - ``min_rays`` (default: zero) -- A nonnegative integer representing the - minimum number of generating rays of the - cone. - - - ``max_rays`` (default: random) -- A nonnegative integer representing the - maximum number of generating rays of - the cone. - - OUTPUT: - - A new, randomly generated cone. - - A ``ValueError` will be thrown under the following conditions: - - * Any of ``min_dim``, ``max_dim``, ``min_rays``, or ``max_rays`` - are negative. - - * ``max_dim`` is less than ``min_dim``. - - * ``max_rays`` is less than ``min_rays``. - - * ``min_rays`` is greater than twice ``max_dim``. + Return an isomorphism (and its inverse) that will send ``K`` into a + lower-dimensional space isomorphic to its span (and back). EXAMPLES: - If we set the lower/upper bounds to zero, then our result is - predictable:: - - sage: random_cone(0,0,0,0) - 0-d cone in 0-d lattice N - - We can predict the dimension when ``min_dim == max_dim``:: - - sage: random_cone(min_dim=4, max_dim=4, min_rays=0, max_rays=0) - 0-d cone in 4-d lattice N - - Likewise for the number of rays when ``min_rays == max_rays``:: - - sage: random_cone(min_dim=10, max_dim=10, min_rays=10, max_rays=10) - 10-d cone in 10-d lattice N - - TESTS: - - It's hard to test the output of a random process, but we can at - least make sure that we get a cone back:: + The inverse composed with the isomorphism should be the identity:: - sage: from sage.geometry.cone import is_Cone # long time - sage: K = random_cone() # long time - sage: is_Cone(K) # long time + sage: K = random_cone(max_dim=10) + sage: (phi, phi_inv) = span_iso(K) + sage: phi_inv(phi(K)) == K True - The upper/lower bounds are respected:: + The image of ``K`` under the isomorphism should have full dimension:: - sage: K = random_cone(min_dim=5, max_dim=10, min_rays=3, max_rays=4) - sage: 5 <= K.lattice_dim() and K.lattice_dim() <= 10 - True - sage: 3 <= K.nrays() and K.nrays() <= 4 + sage: K = random_cone(max_dim=10) + sage: (phi, phi_inv) = span_iso(K) + sage: phi(K).dim() == phi(K).lattice_dim() True - Ensure that an exception is raised when either lower bound is greater - than its respective upper bound:: - - sage: random_cone(min_dim=5, max_dim=2) - Traceback (most recent call last): - ... - ValueError: max_dim cannot be less than min_dim. - - sage: random_cone(min_rays=5, max_rays=2) - Traceback (most recent call last): - ... - ValueError: max_rays cannot be less than min_rays. + 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:: - And if we request too many rays:: + 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 - sage: random_cone(min_rays=5, max_dim=1) - Traceback (most recent call last): - ... - ValueError: min_rays cannot be larger than twice max_dim. + 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()) - # Catch obvious mistakes so that we can generate clear error - # messages. + # S goes from the new space to the cone space. + S = linear_transformation(phi_codo, phi_domain, phi_domain.basis()) - if min_dim < 0: - raise ValueError('min_dim must be nonnegative.') + # 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) - if min_rays < 0: - raise ValueError('min_rays must be nonnegative.') + return Cone(newrays, lattice=L) - if max_dim is not None: - if max_dim < 0: - raise ValueError('max_dim must be nonnegative.') - if (max_dim < min_dim): - raise ValueError('max_dim cannot be less than min_dim.') - if min_rays > 2*max_dim: - raise ValueError('min_rays cannot be larger than twice max_dim.') + 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()) - if max_rays is not None: - if max_rays < 0: - raise ValueError('max_rays must be nonnegative.') - if (max_rays < min_rays): - raise ValueError('max_rays cannot be less than min_rays.') + return (phi, phi_inverse) - def random_min_max(l,u): - r""" - We need to handle two cases for the upper bounds, and we need to do - the same thing for max_dim/max_rays. So we consolidate the logic here. - """ - if u is None: - # The upper bound is unspecified; return a random integer - # in [l,infinity). - return l + ZZ.random_element().abs() - else: - # We have an upper bound, and it's greater than or equal - # to our lower bound. So we generate a random integer in - # [0,u-l], and then add it to l to get something in - # [l,u]. To understand the "+1", check the - # ZZ.random_element() docs. - return l + ZZ.random_element(u - l + 1) - - d = random_min_max(min_dim, max_dim) - r = random_min_max(min_rays, max_rays) - - L = ToricLattice(d) - - # The rays are trickier to generate, since we could generate v and - # 2*v as our "two rays." In that case, the resuting cone would - # have one generating ray. To avoid such a situation, we start by - # generating ``r`` rays where ``r`` is the number we want to end - # up with... - rays = [L.random_element() for i in range(0, r)] - - # (The lattice parameter is required when no rays are given, so we - # pass it just in case ``r == 0``). - K = Cone(rays, lattice=L) - - # Now if we generated two of the "same" rays, we'll have fewer - # generating rays than ``r``. In that case, we keep making up new - # rays and recreating the cone until we get the right number of - # independent generators. We can obviously stop if ``K`` is the - # entire ambient vector space. - while r > K.nrays() and not is_full_space(K): - rays.append(L.random_element()) - K = Cone(rays) - - return K def discrete_complementarity_set(K): @@ -303,6 +166,112 @@ def discrete_complementarity_set(K): 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""" Compute the Lyapunov (or bilinearity) rank of this cone. @@ -346,17 +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) @@ -364,23 +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)]) @@ -388,8 +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)]) @@ -397,7 +372,7 @@ 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()) @@ -405,7 +380,8 @@ def lyapunov_rank(K): TESTS: - 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: K1 = random_cone(max_dim=10, max_rays=10) sage: K2 = random_cone(max_dim=10, max_rays=10) @@ -414,35 +390,51 @@ def lyapunov_rank(K): True The dual cone `K^{*}` of ``K`` should have the same Lyapunov rank as ``K`` - itself:: + itself [Rudolf et al.]_:: sage: K = random_cone(max_dim=10, max_rays=10) sage: lyapunov_rank(K) == lyapunov_rank(K.dual()) True - """ - V = K.lattice().vector_space() + 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:: - C_of_K = discrete_complementarity_set(K) + 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 - matrices = [x.tensor_product(s) for (x,s) in C_of_K] + In fact [Orlitzky/Gowda]_, no closed convex polyhedral cone can have + Lyapunov rank `n-1` in `n` dimensions:: - # 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) + sage: b = lyapunov_rank(K) + sage: n = K.lattice_dim() + sage: b == n-1 + False - def phi(m): - r""" - Convert a matrix to a vector isomorphically. - """ - return W(m.list()) + The calculation of the Lyapunov rank of an improper cone can be + reduced to that of a proper cone [Orlitzky/Gowda]_:: - vectors = [phi(m) for m in matrices] + 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))