from sage.all import *
-def random_cone(min_dim=None, max_dim=None, min_rays=None, max_rays=None):
+def project_span(K, K2 = None):
r"""
- Generate a random rational convex polyhedral cone.
+ Return a "copy" of ``K`` embeded in a lower-dimensional space.
- 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.
+ By default, we will project ``K`` into the subspace spanned by its
+ rays. However, if ``K2`` is not ``None``, we will project into the
+ space spanned by the rays of ``K2`` instead.
- INPUT:
+ EXAMPLES::
+
+ sage: K = Cone([(1,0,0), (0,1,0)])
+ sage: project_span(K)
+ 2-d cone in 2-d lattice N
+ sage: project_span(K).rays()
+ N(1, 0),
+ N(0, 1)
+ in 2-d lattice N
+
+ sage: K = Cone([(1,0,0), (0,1,0)])
+ sage: K2 = Cone([(0,1)])
+ sage: project_span(K, K2).rays()
+ N(1)
+ in 1-d lattice N
+
+ """
+ # Allow us to use a second cone to generate the subspace into
+ # which we're "projecting."
+ if K2 is None:
+ K2 = K
+
+ # Use these to generate the new cone.
+ cs1 = K.rays().matrix().columns()
- - ``min_dim`` (default: random) -- The minimum dimension of the ambient
- lattice.
+ # And use these to figure out which indices to drop.
+ cs2 = K2.rays().matrix().columns()
- - ``max_dim`` (default: random) -- The maximum dimension of the ambient
- lattice.
+ perp_idxs = []
- - ``min_rays`` (default: random) -- The minimum number of generating rays
- of the cone.
+ for idx in range(0, len(cs2)):
+ if cs2[idx].is_zero():
+ perp_idxs.append(idx)
- - ``max_rays`` (default: random) -- The maximum number of generating rays
- of the cone.
+ solid_cols = [ cs1[idx] for idx in range(0,len(cs1))
+ if not idx in perp_idxs
+ and not idx >= len(cs2) ]
+
+ m = matrix(solid_cols)
+ L = ToricLattice(len(m.rows()))
+ J = Cone(m.transpose(), lattice=L)
+ return J
+
+
+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 new, randomly generated cone.
+ 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:
- 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 complementarity set of the dual can be obtained by switching the
+ components of the complementarity set of the original cone::
- sage: from sage.geometry.cone import is_Cone
- sage: K = random_cone()
- sage: is_Cone(K) # long time
+ 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 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 lyapunov_rank(K):
An integer representing the Lyapunov rank of the cone. If the
dimension of the ambient vector space is `n`, then the Lyapunov rank
will be between `1` and `n` inclusive; however a rank of `n-1` is
- not possible (see the first reference).
+ not possible for any cone.
.. note::
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)
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)])
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)])
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())
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(0,10,0,10)
- sage: K2 = random_cone(0,10,0,10)
+ 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
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
- """
- V = K.lattice().vector_space()
-
- xs = [V(x) for x in K.rays()]
- ss = [V(s) for s in K.dual().rays()]
-
- # 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]
-
- matrices = [x.column() * s.row() for (x,s) in C_of_K]
+ 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 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)
-
- def phi(m):
- r"""
- Convert a matrix to a vector isomorphically.
- """
- return W(m.list())
-
- 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, max_rays=16)
+ 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, max_rays=25)
+ sage: actual = lyapunov_rank(K)
+ sage: K_S = project_span(K)
+ sage: J_T1 = project_span(K_S.dual()).dual()
+ sage: J_T2 = project_span(K, K_S.dual())
+ sage: J_T2 = Cone(J_T2.rays(), lattice=J_T1.lattice())
+ sage: J_T1 == J_T2
+ True
+ sage: J_T = J_T1
+ 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))