from sage.all import *
+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 ``MatrixSpace`` object `M` such that every matrix `L \in M` is
+ Lyapunov-like on this cone.
+
+ """
+ V = K.lattice().vector_space()
+
+ C_of_K = discrete_complementarity_set(K)
+
+ matrices = [x.tensor_product(s) 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 matrices ]
+
+ # 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())
+
+ matrices = [ M(v.list()) for v in LL_vector.basis() ]
+
+ return matrices
+
+
+
def lyapunov_rank(K):
r"""
Compute the Lyapunov (or bilinearity) rank of this cone.
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
+ .. [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())
True
+ TESTS:
+
+ 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)
+ 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: 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, this holds for the trivial cone in a
+ trivial space as well)::
+
+ sage: K = random_cone(max_dim=10, strictly_convex=True, solid=True)
+ sage: b = lyapunov_rank(K)
+ sage: n = K.lattice_dim()
+ sage: 1 <= b and b <= n
+ True
+ sage: b == n-1
+ False
+
"""
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]
+ C_of_K = discrete_complementarity_set(K)
- matrices = [x.column() * s.row() for (x,s) in C_of_K]
+ matrices = [x.tensor_product(s) 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
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