X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Fcone%2Fcone.py;h=f2e8b2e9ee104e6cbd216e7473fd65dd76947d98;hb=44802773ad9e5151890ed37e7bb2463ff9fc4135;hp=23043381bca83acd6d7f17479ef4769980b9960f;hpb=012175f3a2591586099b4e955bb440958f2d63d7;p=sage.d.git diff --git a/mjo/cone/cone.py b/mjo/cone/cone.py index 2304338..f2e8b2e 100644 --- a/mjo/cone/cone.py +++ b/mjo/cone/cone.py @@ -8,6 +8,602 @@ addsitedir(abspath('../../')) from sage.all import * +def iso_space(K): + r""" + Construct the space `W \times W^{\perp}` isomorphic to the ambient space + of ``K`` where `W` is equal to the span of ``K``. + """ + V = K.lattice().vector_space() + + # Create the space W \times W^{\perp} isomorphic to V. + # First we get an orthogonal (but not normal) basis... + M = matrix(V.base_field(), K.rays()) + W_basis,_ = M.gram_schmidt() + + W = V.subspace_with_basis(W_basis) + W_perp = W.complement() + + return W.cartesian_product(W_perp) + + +def ips_iso(K): + r""" + Construct the IPS isomorphism and its inverse from our paper. + + Given a cone ``K``, the returned isomorphism will split its ambient + vector space `V` into a cartesian product `W \times W^{\perp}` where + `W` equals the span of ``K``. + """ + V = K.lattice().vector_space() + V_iso = iso_space(K) + (W, W_perp) = V_iso.cartesian_factors() + + # A space equivalent to V, but using our basis. + V_user = V.subspace_with_basis( W.basis() + W_perp.basis() ) + + def phi(v): + # Write v in terms of our custom basis, where the first dim(W) + # coordinates are for the W-part of the basis. + cs = V_user.coordinates(v) + + w1 = sum([ V_user.basis()[idx]*cs[idx] + for idx in range(0, W.dimension()) ]) + w2 = sum([ V_user.basis()[idx]*cs[idx] + for idx in range(W.dimension(), V.dimension()) ]) + + return V_iso( (w1, w2) ) + + + def phi_inv( pair ): + # Crash if the arguments are in the wrong spaces. + V_iso(pair) + + #w = sum([ sub_w[idx]*W.basis()[idx] for idx in range(0,m) ]) + #w_prime = sum([ sub_w_prime[idx]*W_perp.basis()[idx] + # for idx in range(0,n-m) ]) + + return sum( pair.cartesian_factors() ) + + + return (phi,phi_inv) + + + +def unrestrict_span(K, K2=None): + if K2 is None: + K2 = K + + _,phi_inv = ips_iso(K2) + V_iso = iso_space(K2) + (W, W_perp) = V_iso.cartesian_factors() + + rays = [] + for r in K.rays(): + w = sum([ r[idx]*W.basis()[idx] for idx in range(0,len(r)) ]) + pair = V_iso( (w, W_perp.zero()) ) + rays.append( phi_inv(pair) ) + + L = ToricLattice(W.dimension() + W_perp.dimension()) + + return Cone(rays, lattice=L) + + + +def intersect_span(K1, K2): + r""" + Return a new cone obtained by intersecting ``K1`` with the span of ``K2``. + """ + L = K1.lattice() + + if L.rank() != K2.lattice().rank(): + raise ValueError('K1 and K2 must belong to lattices of the same rank.') + + SL_gens = list(K2.rays()) + span_K2_gens = SL_gens + [ -g for g in SL_gens ] + + # The lattices have the same rank (see above) so this should work. + span_K2 = Cone(span_K2_gens, L) + return K1.intersection(span_K2) + + + +def restrict_span(K, K2=None): + r""" + Restrict ``K`` into its own span, or the span of another cone. + + INPUT: + + - ``K2`` -- another cone whose lattice has the same rank as this cone. + + OUTPUT: + + A new cone in a sublattice. + + EXAMPLES:: + + sage: K = Cone([(1,)]) + sage: restrict_span(K) == K + True + + sage: K2 = Cone([(1,0)]) + sage: restrict_span(K2).rays() + N(1) + in 1-d lattice N + sage: K3 = Cone([(1,0,0)]) + sage: restrict_span(K3).rays() + N(1) + in 1-d lattice N + sage: restrict_span(K2) == restrict_span(K3) + True + + TESTS: + + The projected cone should always be solid:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 10) + sage: K_S = restrict_span(K) + sage: K_S.is_solid() + True + + And the resulting cone should live in a space having the same + dimension as the space we restricted it to:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 10) + sage: K_S = restrict_span( intersect_span(K, K.dual()), K.dual() ) + sage: K_S.lattice_dim() == K.dual().dim() + True + + This function has ``unrestrict_span()`` as its inverse:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 10, solid=True) + sage: J = restrict_span(K) + sage: K == unrestrict_span(J,K) + True + + This function should not affect the dimension of a cone:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 10) + sage: K.dim() == restrict_span(K).dim() + True + + Nor should it affect the lineality of a cone:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 10) + sage: lineality(K) == lineality(restrict_span(K)) + True + + No matter which space we restrict to, the lineality should not + increase:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 10) + sage: J = intersect_span(K, K.dual()) + sage: lineality(K) >= lineality(restrict_span(J, K.dual())) + True + + If we do this according to our paper, then the result is proper:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 10) + sage: K_S = restrict_span(K) + sage: P = restrict_span(K_S.dual()).dual() + sage: P.is_proper() + True + + If ``K`` is strictly convex, then both ``K_W`` and + ``K_star_W.dual()`` should equal ``K`` (after we unrestrict):: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 10, strictly_convex=True) + sage: K_W = restrict_span(intersect_span(K,K.dual()), K.dual()) + sage: K_star_W_star = restrict_span(K.dual()).dual() + sage: j1 = unrestrict_span(K_W, K.dual()) + sage: j2 = unrestrict_span(K_star_W_star, K.dual()) + sage: j1 == j2 + True + sage: j1 == K + True + sage: K; [ list(r) for r in K.rays() ] + + Test the proposition in our paper concerning the duals, where the + subspace `W` is the span of `K^{*}`:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 10, solid=False, strictly_convex=False) + sage: K_W = restrict_span(intersect_span(K,K.dual()), K.dual()) + sage: K_star_W_star = restrict_span(K.dual(), K.dual()).dual() + sage: K_W.nrays() == K_star_W_star.nrays() + True + sage: K_W.dim() == K_star_W_star.dim() + True + sage: lineality(K_W) == lineality(K_star_W_star) + True + sage: K_W.is_solid() == K_star_W_star.is_solid() + True + sage: K_W.is_strictly_convex() == K_star_W_star.is_strictly_convex() + True + + """ + if K2 is None: + K2 = K + + phi,_ = ips_iso(K2) + (W, W_perp) = iso_space(K2).cartesian_factors() + + ray_pairs = [ phi(r) for r in K.rays() ] + + if any([ w2 != W_perp.zero() for (_, w2) in ray_pairs ]): + msg = 'Cone has nonzero components in W-perp!' + raise ValueError(msg) + + # Represent the cone in terms of a basis for W, i.e. with smaller + # vectors. + ws = [ W.coordinate_vector(w1) for (w1, _) in ray_pairs ] + + L = ToricLattice(W.dimension()) + + return Cone(ws, lattice=L) + + + +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 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: + + The lineality of a cone should be an integer between zero and the + dimension of the ambient space, inclusive:: + + sage: set_random_seed() + 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: set_random_seed() + sage: K = random_cone(max_dim = 10, strictly_convex = True) + sage: lineality(K) + 0 + + """ + return K.linear_subspace().dimension() + + +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: set_random_seed() + 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: set_random_seed() + 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: set_random_seed() + 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): + 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: set_random_seed() + sage: K1 = random_cone(max_dim=6) + sage: K2 = K1.dual() + sage: expected = [(x,s) for (s,x) in discrete_complementarity_set(K2)] + sage: actual = discrete_complementarity_set(K1) + sage: sorted(actual) == sorted(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: set_random_seed() + 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 + + The Lyapunov-like transformations on a cone and its dual are related + by transposition, but we're not guaranteed to compute transposed + elements of `LL\left( K \right)` as our basis for `LL\left( K^{*} + \right)` + + sage: set_random_seed() + sage: K = random_cone(max_dim=8, max_rays=10) + sage: LL2 = [ L.transpose() for L in LL(K.dual()) ] + sage: V = VectorSpace( K.lattice().base_field(), K.lattice_dim()^2) + sage: LL1_vecs = [ V(m.list()) for m in LL(K) ] + sage: LL2_vecs = [ V(m.list()) for m in LL2 ] + sage: V.span(LL1_vecs) == V.span(LL2_vecs) + True + + """ + 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. @@ -51,17 +647,21 @@ 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 +669,57 @@ 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: gs = R5.basis() + [ -r for r in R5.basis() ] + sage: K = Cone(gs) + 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: z = (0,0,0,0,0) + sage: K = Cone([e1, neg_e1, e2, neg_e2, z, z, z]) + 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)]) @@ -93,8 +727,8 @@ 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 +736,115 @@ 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 + TESTS: + + The Lyapunov rank should be additive on a product of proper cones + [Rudolf et al.]_:: + + sage: set_random_seed() + 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 [Rudolf et al.]_:: + + sage: set_random_seed() + sage: K = random_cone(max_dim=10, max_rays=10) + sage: lyapunov_rank(K) == lyapunov_rank(K.dual()) + True + + Make sure we exercise the non-strictly-convex/non-solid case:: + + sage: set_random_seed() + sage: K = random_cone(max_dim=10, strictly_convex=False, solid=False) + 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: set_random_seed() + 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: set_random_seed() + 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: set_random_seed() + sage: K = random_cone(max_dim=10) + sage: actual = lyapunov_rank(K) + sage: K_S = restrict_span(K) + sage: P = restrict_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: set_random_seed() + 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() + K_orig = K + beta = 0 - xs = [V(x) for x in K.rays()] - ss = [V(s) for s in K.dual().rays()] + m = K.dim() + n = K.lattice_dim() + l = lineality(K) - # 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] + if m < n: + # K is not solid, project onto its span. + K = restrict_span(K) - matrices = [x.column() * s.row() for (x,s) in C_of_K] + # Lemma 2 + beta += m*(n - m) + (n - m)**2 - # 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) + if l > 0: + # K is not pointed, project its dual onto its span. + # Uses a proposition from our paper, i.e. this is + # equivalent to K = restrict_span(K.dual()).dual() + K = restrict_span(intersect_span(K,K.dual()), K.dual()) + #K = restrict_span(K.dual()).dual() + + #Ks = [ list(r) for r in sorted(K.rays()) ] + #Js = [ list(r) for r in sorted(J.rays()) ] - def phi(m): - r""" - Convert a matrix to a vector isomorphically. - """ - return W(m.list()) + #if Ks != Js: + # print [ list(r) for r in K_orig.rays() ] - 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