from sage.all import *
+def project_span(K):
+ r"""
+ Project ``K`` into its own span.
+
+ EXAMPLES::
+
+ 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:
+
+ The projected cone should always be solid::
+
+ sage: K = random_cone(max_dim = 10)
+ sage: K_S = project_span(K)
+ sage: K_S.is_solid()
+ True
+
+ If we do this according to our paper, then the result is proper::
+
+ 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 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: 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 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):
r"""
Compute the discrete complementarity set of this cone.
OUTPUT:
- A ``MatrixSpace`` object `M` such that every matrix `L \in M` is
- Lyapunov-like on this cone.
+ 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()
cone and Lyapunov-like transformations, Mathematical Programming, 147
(2014) 155-170.
+ .. [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.
sage: lyapunov_rank(octant)
3
+ 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: 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)])
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(max_dim=10, max_rays=10)
- sage: K2 = random_cone(max_dim=10, max_rays=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
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
+
"""
- return len(LL(K))
+ beta = 0
+
+ m = K.dim()
+ n = K.lattice_dim()
+ l = lineality(K)
+
+ if m < n:
+ # K is not solid, project onto its span.
+ K = project_span(K)
+
+ # Lemma 2
+ beta += m*(n - m) + (n - m)**2
+
+ if l > 0:
+ # K is not pointed, project its dual onto its span.
+ K = project_span(K.dual()).dual()
+
+ # Lemma 3
+ beta += m * l
+
+ beta += len(LL(K))
+ return beta
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