from sage.all import *
-def random_cone(min_dim=0, max_dim=None, min_rays=0, 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. If a
- lower bound is left unspecified, it defaults to zero. Unspecified
- upper bounds will be chosen randomly.
+ EXAMPLES::
- INPUT:
+ sage: K = Cone([(1,)])
+ sage: project_span(K) == K
+ True
- - ``min_dim`` (default: zero) -- A nonnegative integer representing the
- minimum dimension of the ambient lattice.
+ 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:
- - ``max_dim`` (default: random) -- A nonnegative integer representing
- the maximum dimension of the ambient
- lattice.
+ The projected cone should always be solid::
- - ``min_rays`` (default: zero) -- A nonnegative integer representing the
- minimum number of generating rays of the
- cone.
+ 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.
- - ``max_rays`` (default: random) -- A nonnegative integer representing the
- maximum number of generating rays of the
- 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:
- If we set the lower/upper bounds to zero, then our result is
- predictable::
+ 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: random_cone(0,0,0,0)
- 0-d cone in 0-d lattice N
+ sage: K = Cone([(1,0), (-1,0), (0,1), (0,-1)])
+ sage: lineality(K)
+ 2
- In fact, as long as we ask for zero rays, we should be able to predict
- the output when ``min_dim == max_dim``::
+ Per the definition, the lineality of the trivial cone in a trivial
+ space is zero::
- sage: random_cone(min_dim=4, max_dim=4, min_rays=0, max_rays=0)
- 0-d cone in 4-d lattice N
+ 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 # long time
- sage: K = random_cone() # long time
- 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
- Ensure that an exception is raised when either lower bound is greater
- than its respective upper bound::
-
- sage: random_cone(min_dim=5, max_dim=2)
- Traceback (most recent call last):
- ...
- ValueError: max_dim must be greater than or equal to min_dim.
+ A strictly convex cone should have lineality zero::
- sage: random_cone(min_rays=5, max_rays=2)
- Traceback (most recent call last):
- ...
- ValueError: max_rays must be greater than or equal to min_rays.
+ sage: K = random_cone(max_dim = 10, strictly_convex = True)
+ sage: lineality(K)
+ 0
"""
+ return K.linear_subspace().dimension()
- # Catch obvious mistakes so that we can generate clear error
- # messages.
- if min_dim < 0:
- raise ValueError('min_dim must be nonnegative.')
+def codim(K):
+ r"""
+ Compute the codimension of this cone.
- if min_rays < 0:
- raise ValueError('min_rays must be nonnegative.')
+ 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.
- if max_dim is not None:
- if max_dim < 0:
- raise ValueError('max_dim must be nonnegative.')
- if (min_dim > max_dim):
- raise ValueError('max_dim must be greater than or equal to min_dim.')
+ OUTPUT:
- if max_rays is not None:
- if max_rays < 0:
- raise ValueError('max_rays must be nonnegative.')
- if (min_rays > max_rays):
- raise ValueError('max_rays must be greater than or equal to min_rays.')
+ A nonnegative integer representing the dimension of the space of all
+ elements perpendicular to this cone.
+ .. seealso::
- def random_min_max(l,u):
- r"""
- We need to handle two cases for the upper bounds, and we need to do
- the same thing for max_dim/max_rays. So we consolidate the logic here.
- """
- if u is None:
- # The upper bound is unspecified; return a random integer
- # in [l,infinity).
- return l + ZZ.random_element().abs()
- else:
- # We have an upper bound, and it's greater than or equal
- # to our lower bound. So we generate a random integer in
- # [0,u-l], and then add it to l to get something in
- # [l,u]. To understand the "+1", check the
- # ZZ.random_element() docs.
- return l + ZZ.random_element(u - l + 1)
+ :meth:`dim`, :meth:`lattice_dim`
+ EXAMPLES:
- d = random_min_max(min_dim, max_dim)
- r = random_min_max(min_rays, max_rays)
+ The codimension of the nonnegative orthant is zero, since the span of
+ its generators equals the entire ambient space::
- L = ToricLattice(d)
- rays = [L.random_element() for i in range(0,r)]
+ sage: K = Cone([(1,0,0), (0,1,0), (0,0,1)])
+ sage: codim(K)
+ 0
- # The lattice parameter is required when no rays are given, so we
- # pass it just in case.
- return Cone(rays, lattice=L)
+ 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):
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.
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 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)])
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 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
The dual cone `K^{*}` of ``K`` should have the same Lyapunov rank as ``K``
- itself::
+ 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, 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
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