X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Fcone%2Fcone.py;h=ff7d195d134c15943dbb75c1f26b741bb4a0afba;hb=10142e85f34c47fa35df002f519d1d58a79a74f4;hp=507b6cea5c24d0a4eb745d9245c85a1fbf4ddb90;hpb=494fe2e8517d40e3b71a54657e4a69a83cadf423;p=sage.d.git diff --git a/mjo/cone/cone.py b/mjo/cone/cone.py index 507b6ce..ff7d195 100644 --- a/mjo/cone/cone.py +++ b/mjo/cone/cone.py @@ -8,82 +8,210 @@ addsitedir(abspath('../../')) from sage.all import * -def random_cone(min_dim=None, max_dim=None, min_rays=None, max_rays=None): +def project_span(K): r""" - Generate a random rational convex polyhedral cone. + Project ``K`` into its own span. - Lower and upper bounds may be provided for both the dimension of the - ambient space and the number of generating rays of the cone. Any - parameters left unspecified will be chosen randomly. + EXAMPLES:: - INPUT: + sage: K = Cone([(1,)]) + sage: project_span(K) == K + True + + sage: K2 = Cone([(1,0)]) + sage: project_span(K2).rays() + N(1) + in 1-d lattice N + sage: K3 = Cone([(1,0,0)]) + sage: project_span(K3).rays() + N(1) + in 1-d lattice N + sage: project_span(K2) == project_span(K3) + True + + TESTS: - - ``min_dim`` (default: random) -- The minimum dimension of the ambient - lattice. + The projected cone should always be solid:: - - ``max_dim`` (default: random) -- The maximum dimension of the ambient - lattice. + sage: K = random_cone(max_dim = 10) + sage: K_S = project_span(K) + sage: K_S.is_solid() + True - - ``min_rays`` (default: random) -- The minimum number of generating rays - of the cone. + If we do this according to our paper, then the result is proper:: - - ``max_rays`` (default: random) -- The maximum number of generating rays - of the cone. + sage: K = random_cone(max_dim = 10) + sage: K_S = project_span(K) + sage: P = project_span(K_S.dual()).dual() + sage: P.is_proper() + True + + """ + L = K.lattice() + F = L.base_field() + Q = L.quotient(K.sublattice_complement()) + vecs = [ vector(F, reversed(list(Q(r)))) for r in K.rays() ] + + newL = None + if len(vecs) == 0: + newL = ToricLattice(0) + + return Cone(vecs, lattice=newL) + + + +def lineality(K): + r""" + Compute the lineality of this cone. + + The lineality of a cone is the dimension of the largest linear + subspace contained in that cone. OUTPUT: - A new, randomly generated cone. + A nonnegative integer; the dimension of the largest subspace + contained within this cone. + + REFERENCES: + + .. [Rockafellar] R.T. Rockafellar. Convex Analysis. Princeton + University Press, Princeton, 1970. + + EXAMPLES: + + The lineality of the nonnegative orthant is zero, since it clearly + contains no lines:: + + sage: K = Cone([(1,0,0), (0,1,0), (0,0,1)]) + sage: lineality(K) + 0 + + However, if we add another ray so that the entire `x`-axis belongs + to the cone, then the resulting cone will have lineality one:: + + sage: K = Cone([(1,0,0), (-1,0,0), (0,1,0), (0,0,1)]) + sage: lineality(K) + 1 + + If our cone is all of `\mathbb{R}^{2}`, then its lineality is equal + to the dimension of the ambient space (i.e. two):: + + sage: K = Cone([(1,0), (-1,0), (0,1), (0,-1)]) + sage: lineality(K) + 2 + + Per the definition, the lineality of the trivial cone in a trivial + space is zero:: + + sage: K = Cone([], lattice=ToricLattice(0)) + sage: lineality(K) + 0 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 lineality of a cone should be an integer between zero and the + dimension of the ambient space, inclusive:: - sage: from sage.geometry.cone import is_Cone - sage: K = random_cone() - sage: is_Cone(K) # long time + sage: K = random_cone(max_dim = 10) + sage: l = lineality(K) + sage: l in ZZ + True + sage: (0 <= l) and (l <= K.lattice_dim()) True + A strictly convex cone should have lineality zero:: + + sage: K = random_cone(max_dim = 10, strictly_convex = True) + sage: lineality(K) + 0 + """ + return K.linear_subspace().dimension() - def random_min_max(l,u): - r""" - We need to handle four cases to prevent us from doing - something stupid like having an upper bound that's lower than - our lower bound. And we would need to repeat all of that logic - for the dimension/rays, so we consolidate it here. - """ - if l is None and u is None: - # They're both random, just return a random nonnegative - # integer. - return ZZ.random_element().abs() - - if l is not None and u is not None: - # Both were specified. Again, just make up a number and - # return it. If the user wants to give us u < l then he - # can have an exception. - return ZZ.random_element(l,u) - - if l is not None and u is None: - # In this case, we're generating the upper bound randomly - # GIVEN A LOWER BOUND. So we add a random nonnegative - # integer to the given lower bound. - u = l + ZZ.random_element().abs() - return ZZ.random_element(l,u) - - # Here we must be in the only remaining case, where we are - # given an upper bound but no lower bound. We might as well - # use zero. - return ZZ.random_element(0,u) - - d = random_min_max(min_dim, max_dim) - r = random_min_max(min_rays, max_rays) - - L = ToricLattice(d) - rays = [L.random_element() for i in range(0,r)] - - # We pass the lattice in case there are no rays. - return Cone(rays, lattice=L) + +def codim(K): + r""" + Compute the codimension of this cone. + + The codimension of a cone is the dimension of the space of all + elements perpendicular to every element of the cone. In other words, + the codimension is the difference between the dimension of the + ambient space and the dimension of the cone itself. + + OUTPUT: + + A nonnegative integer representing the dimension of the space of all + elements perpendicular to this cone. + + .. seealso:: + + :meth:`dim`, :meth:`lattice_dim` + + EXAMPLES: + + The codimension of the nonnegative orthant is zero, since the span of + its generators equals the entire ambient space:: + + sage: K = Cone([(1,0,0), (0,1,0), (0,0,1)]) + sage: codim(K) + 0 + + However, if we remove a ray so that the entire cone is contained + within the `x-y`-plane, then the resulting cone will have + codimension one, because the `z`-axis is perpendicular to every + element of the cone:: + + sage: K = Cone([(1,0,0), (0,1,0)]) + sage: codim(K) + 1 + + If our cone is all of `\mathbb{R}^{2}`, then its codimension is zero:: + + sage: K = Cone([(1,0), (-1,0), (0,1), (0,-1)]) + sage: codim(K) + 0 + + And if the cone is trivial in any space, then its codimension is + equal to the dimension of the ambient space:: + + sage: K = Cone([], lattice=ToricLattice(0)) + sage: codim(K) + 0 + + sage: K = Cone([(0,)]) + sage: codim(K) + 1 + + sage: K = Cone([(0,0)]) + sage: codim(K) + 2 + + TESTS: + + The codimension of a cone should be an integer between zero and + the dimension of the ambient space, inclusive:: + + sage: K = random_cone(max_dim = 10) + sage: c = codim(K) + sage: c in ZZ + True + sage: (0 <= c) and (c <= K.lattice_dim()) + True + + A solid cone should have codimension zero:: + + sage: K = random_cone(max_dim = 10, solid = True) + sage: codim(K) + 0 + + The codimension of a cone is equal to the lineality of its dual:: + + sage: K = random_cone(max_dim = 10, solid = True) + sage: codim(K) == lineality(K.dual()) + True + + """ + return (K.lattice_dim() - K.dim()) def discrete_complementarity_set(K): @@ -140,7 +268,7 @@ def discrete_complementarity_set(K): 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(0,10,0,10) + 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) @@ -158,6 +286,112 @@ 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. @@ -201,17 +435,21 @@ 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) @@ -219,23 +457,57 @@ 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 full space `\mathbb{R}^{n}` has Lyapunov rank `n^{2}` + [Orlitzky/Gowda]_:: + + sage: R5 = VectorSpace(QQ, 5) + sage: gens = R5.basis() + [ -r for r in R5.basis() ] + sage: K = Cone(gens) + sage: lyapunov_rank(K) + 25 + + 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:: + A single ray in `n` dimensions should have Lyapunov rank `n^{2} - n + + 1` [Orlitzky/Gowda]_:: + + sage: K = Cone([(1,0,0,0,0)]) + sage: lyapunov_rank(K) + 21 + sage: K.lattice_dim()**2 - K.lattice_dim() + 1 + 21 + + A subspace (of dimension `m`) in `n` dimensions should have a + Lyapunov rank of `n^{2} - m\left(n - m)` [Orlitzky/Gowda]_:: + + sage: e1 = (1,0,0,0,0) + sage: neg_e1 = (-1,0,0,0,0) + sage: e2 = (0,1,0,0,0) + sage: neg_e2 = (0,-1,0,0,0) + sage: zero = (0,0,0,0,0) + sage: K = Cone([e1, neg_e1, e2, neg_e2, zero, zero, zero]) + sage: lyapunov_rank(K) + 19 + sage: K.lattice_dim()**2 - K.dim()*codim(K) + 19 + + The Lyapunov rank should be additive on a product of proper 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)]) @@ -243,8 +515,8 @@ 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)]) @@ -252,7 +524,7 @@ 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()) @@ -260,44 +532,84 @@ 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 proper cones + [Rudolf et al.]_:: - sage: K1 = random_cone(0,10,0,10) - sage: K2 = random_cone(0,10,0,10) + sage: K1 = random_cone(max_dim=10, strictly_convex=True, solid=True) + sage: K2 = random_cone(max_dim=10, strictly_convex=True, solid=True) sage: K = K1.cartesian_product(K2) sage: lyapunov_rank(K) == lyapunov_rank(K1) + lyapunov_rank(K2) True The dual cone `K^{*}` of ``K`` should have the same Lyapunov rank as ``K`` - itself:: + itself [Rudolf et al.]_:: - sage: K = random_cone(0,10,0,10) + sage: K = random_cone(max_dim=10, max_rays=10) sage: lyapunov_rank(K) == lyapunov_rank(K.dual()) True + 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:: + + 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=10) + sage: actual = lyapunov_rank(K) + sage: K_S = project_span(K) + sage: P = project_span(K_S.dual()).dual() + sage: l = lineality(K) + sage: c = codim(K) + sage: expected = lyapunov_rank(P) + K.dim()*(l + c) + c**2 + sage: actual == expected + True + + The Lyapunov rank of a proper cone is just the dimension of ``LL(K)``:: + + sage: K = random_cone(max_dim=10, strictly_convex=True, solid=True) + sage: lyapunov_rank(K) == len(LL(K)) + True + """ - V = K.lattice().vector_space() + beta = 0 - C_of_K = discrete_complementarity_set(K) + m = K.dim() + n = K.lattice_dim() + l = lineality(K) - matrices = [x.tensor_product(s) for (x,s) in C_of_K] + if m < n: + # K is not solid, project onto its span. + K = project_span(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) + # Lemma 2 + beta += m*(n - m) + (n - m)**2 - def phi(m): - r""" - Convert a matrix to a vector isomorphically. - """ - return W(m.list()) + if l > 0: + # K is not pointed, project its dual onto its span. + K = project_span(K.dual()).dual() - vectors = [phi(m) for m in matrices] + # Lemma 3 + beta += m * l - return (W.dimension() - W.span(vectors).rank()) + beta += len(LL(K)) + return beta