X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Fcone%2Fcone.py;h=ff7d195d134c15943dbb75c1f26b741bb4a0afba;hb=10142e85f34c47fa35df002f519d1d58a79a74f4;hp=23043381bca83acd6d7f17479ef4769980b9960f;hpb=012175f3a2591586099b4e955bb440958f2d63d7;p=sage.d.git

diff --git a/mjo/cone/cone.py b/mjo/cone/cone.py
index 2304338..ff7d195 100644
--- a/mjo/cone/cone.py
+++ b/mjo/cone/cone.py
@@ -8,6 +8,390 @@ addsitedir(abspath('../../'))
 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.
+
+    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 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.
@@ -51,17 +435,21 @@ 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 +457,57 @@ 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: 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)])
@@ -93,8 +515,8 @@ 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 +524,92 @@ 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
 
-    """
-    V = K.lattice().vector_space()
+    TESTS:
 
-    xs = [V(x) for x in K.rays()]
-    ss = [V(s) for s in K.dual().rays()]
+    The Lyapunov rank should be additive on a product of proper cones
+    [Rudolf et al.]_::
 
-    # 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]
+        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
 
-    matrices = [x.column() * s.row() for (x,s) in C_of_K]
+    The dual cone `K^{*}` of ``K`` should have the same Lyapunov rank as ``K``
+    itself [Rudolf et al.]_::
 
-    # 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)
+        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
+
+    """
+    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
 
-    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