# Sage doesn't load ~/.sage/init.sage during testing (sage -t), so we # have to explicitly mangle our sitedir here so that "mjo.cone" # resolves. from os.path import abspath from site import addsitedir addsitedir(abspath('../../')) from sage.all import * def _basically_the_same(K1, K2): r""" Test whether or not ``K1`` and ``K2`` are "basically the same." This is a hack to get around the fact that it's difficult to tell when two cones are linearly isomorphic. We have a proposition that equates two cones, but represented over `\mathbb{Q}`, they are merely linearly isomorphic (not equal). So rather than test for equality, we test a list of properties that should be preserved under an invertible linear transformation. OUTPUT: ``True`` if ``K1`` and ``K2`` are basically the same, and ``False`` otherwise. EXAMPLES: Any proper cone with three generators in `\mathbb{R}^{3}` is basically the same as the nonnegative orthant:: sage: K1 = Cone([(1,0,0), (0,1,0), (0,0,1)]) sage: K2 = Cone([(1,2,3), (3, 18, 4), (66, 51, 0)]) sage: _basically_the_same(K1, K2) True Negating a cone gives you another cone that is basically the same:: sage: K = Cone([(0,2,-5), (-6, 2, 4), (0, 51, 0)]) sage: _basically_the_same(K, -K) True TESTS: Any cone is basically the same as itself:: sage: K = random_cone(max_ambient_dim = 8) sage: _basically_the_same(K, K) True After applying an invertible matrix to the rows of a cone, the result should be basically the same as the cone we started with:: sage: K1 = random_cone(max_ambient_dim = 8) sage: A = random_matrix(QQ, K1.lattice_dim(), algorithm='unimodular') sage: K2 = Cone( [ A*r for r in K1.rays() ], lattice=K1.lattice()) sage: _basically_the_same(K1, K2) True """ if K1.lattice_dim() != K2.lattice_dim(): return False if K1.nrays() != K2.nrays(): return False if K1.dim() != K2.dim(): return False if K1.lineality() != K2.lineality(): return False if K1.is_solid() != K2.is_solid(): return False if K1.is_strictly_convex() != K2.is_strictly_convex(): return False if len(K1.LL()) != len(K2.LL()): return False C_of_K1 = K1.discrete_complementarity_set() C_of_K2 = K2.discrete_complementarity_set() if len(C_of_K1) != len(C_of_K2): return False if len(K1.facets()) != len(K2.facets()): return False return True def _restrict_to_space(K, W): r""" Restrict this cone a subspace of its ambient space. INPUT: - ``W`` -- The subspace into which this cone will be restricted. OUTPUT: A new cone in a sublattice corresponding to ``W``. EXAMPLES: When this cone is solid, restricting it into its own span should do nothing:: sage: K = Cone([(1,)]) sage: _restrict_to_space(K, K.span()) == K True A single ray restricted into its own span gives the same output regardless of the ambient space:: sage: K2 = Cone([(1,0)]) sage: K2_S = _restrict_to_space(K2, K2.span()).rays() sage: K2_S N(1) in 1-d lattice N sage: K3 = Cone([(1,0,0)]) sage: K3_S = _restrict_to_space(K3, K3.span()).rays() sage: K3_S N(1) in 1-d lattice N sage: K2_S == K3_S True TESTS: The projected cone should always be solid:: sage: set_random_seed() sage: K = random_cone(max_ambient_dim = 8) sage: _restrict_to_space(K, K.span()).is_solid() True And the resulting cone should live in a space having the same dimension as the space we restricted it to:: sage: set_random_seed() sage: K = random_cone(max_ambient_dim = 8) sage: K_P = _restrict_to_space(K, K.dual().span()) sage: K_P.lattice_dim() == K.dual().dim() True This function should not affect the dimension of a cone:: sage: set_random_seed() sage: K = random_cone(max_ambient_dim = 8) sage: K.dim() == _restrict_to_space(K,K.span()).dim() True Nor should it affect the lineality of a cone:: sage: set_random_seed() sage: K = random_cone(max_ambient_dim = 8) sage: K.lineality() == _restrict_to_space(K, K.span()).lineality() True No matter which space we restrict to, the lineality should not increase:: sage: set_random_seed() sage: K = random_cone(max_ambient_dim = 8) sage: S = K.span(); P = K.dual().span() sage: K.lineality() >= _restrict_to_space(K,S).lineality() True sage: K.lineality() >= _restrict_to_space(K,P).lineality() True If we do this according to our paper, then the result is proper:: sage: set_random_seed() sage: K = random_cone(max_ambient_dim = 8) sage: K_S = _restrict_to_space(K, K.span()) sage: K_SP = _restrict_to_space(K_S.dual(), K_S.dual().span()).dual() sage: K_SP.is_proper() True sage: K_SP = _restrict_to_space(K_S, K_S.dual().span()) sage: K_SP.is_proper() True Test the proposition in our paper concerning the duals and restrictions. Generate a random cone, then create a subcone of it. The operation of dual-taking should then commute with _restrict_to_space:: sage: set_random_seed() sage: J = random_cone(max_ambient_dim = 8) sage: K = Cone(random_sublist(J.rays(), 0.5), lattice=J.lattice()) sage: K_W_star = _restrict_to_space(K, J.span()).dual() sage: K_star_W = _restrict_to_space(K.dual(), J.span()) sage: _basically_the_same(K_W_star, K_star_W) True """ # First we want to intersect ``K`` with ``W``. The easiest way to # do this is via cone intersection, so we turn the subspace ``W`` # into a cone. W_cone = Cone(W.basis() + [-b for b in W.basis()], lattice=K.lattice()) K = K.intersection(W_cone) # We've already intersected K with the span of K2, so every # generator of K should belong to W now. K_W_rays = [ W.coordinate_vector(r) for r in K.rays() ] L = ToricLattice(W.dimension()) return Cone(K_W_rays, lattice=L) def lyapunov_rank(K): r""" Compute the Lyapunov rank (or bilinearity rank) of this cone. The Lyapunov rank of a cone can be thought of in (mainly) two ways: 1. The dimension of the Lie algebra of the automorphism group of the cone. 2. The dimension of the linear space of all Lyapunov-like transformations on the cone. INPUT: A closed, convex polyhedral cone. OUTPUT: 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 [Orlitzky/Gowda]_). ALGORITHM: The codimension formula from the second reference is used. We find all pairs `(x,s)` in the complementarity set of `K` such that `x` and `s` are rays of our cone. It is known that these vectors are sufficient to apply the codimension formula. Once we have all such pairs, we "brute force" the codimension formula by finding all linearly-independent `xs^{T}`. REFERENCES: .. [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. .. [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` [Rudolf et al.]_:: sage: positives = Cone([(1,)]) sage: lyapunov_rank(positives) 1 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: lyapunov_rank(octant) 3 The full space `\mathbb{R}^{n}` has Lyapunov rank `n^{2}` [Orlitzky/Gowda]_:: sage: R5 = VectorSpace(QQ, 5) sage: gs = R5.basis() + [ -r for r in R5.basis() ] sage: K = Cone(gs) 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 [Rudolf et al.]_:: sage: L3infty = Cone([(0,1,1), (1,0,1), (0,-1,1), (-1,0,1)]) sage: lyapunov_rank(L3infty) 1 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: z = (0,0,0,0,0) sage: K = Cone([e1, neg_e1, e2, neg_e2, z, z, z]) sage: lyapunov_rank(K) 19 sage: K.lattice_dim()**2 - K.dim()*K.codim() 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: K = L31.cartesian_product(octant) sage: lyapunov_rank(K) == lyapunov_rank(L31) + lyapunov_rank(octant) True 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)]) sage: lyapunov_rank(K) 3 The dual cone `K^{*}` of ``K`` should have the same Lyapunov rank as ``K`` 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 proper cones [Rudolf et al.]_:: sage: set_random_seed() sage: K1 = random_cone(max_ambient_dim=8, ....: strictly_convex=True, ....: solid=True) sage: K2 = random_cone(max_ambient_dim=8, ....: strictly_convex=True, ....: solid=True) sage: K = K1.cartesian_product(K2) sage: lyapunov_rank(K) == lyapunov_rank(K1) + lyapunov_rank(K2) True The Lyapunov rank is invariant under a linear isomorphism [Orlitzky/Gowda]_:: sage: K1 = random_cone(max_ambient_dim = 8) sage: A = random_matrix(QQ, K1.lattice_dim(), algorithm='unimodular') sage: K2 = Cone( [ A*r for r in K1.rays() ], lattice=K1.lattice()) sage: 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: set_random_seed() sage: K = random_cone(max_ambient_dim=8) 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: set_random_seed() sage: K = random_cone(max_ambient_dim=8, ....: 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: set_random_seed() sage: K = random_cone(max_ambient_dim=8) 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: set_random_seed() sage: K = random_cone(max_ambient_dim=8) sage: actual = lyapunov_rank(K) sage: K_S = _restrict_to_space(K, K.span()) sage: K_SP = _restrict_to_space(K_S.dual(), K_S.dual().span()).dual() sage: l = K.lineality() sage: c = K.codim() sage: expected = lyapunov_rank(K_SP) + K.dim()*(l + c) + c**2 sage: actual == expected True The Lyapunov rank of any cone is just the dimension of ``K.LL()``:: sage: set_random_seed() sage: K = random_cone(max_ambient_dim=8) sage: lyapunov_rank(K) == len(K.LL()) True We can make an imperfect cone perfect by adding a slack variable (a Theorem in [Orlitzky/Gowda]_):: sage: set_random_seed() sage: K = random_cone(max_ambient_dim=8, ....: strictly_convex=True, ....: solid=True) sage: L = ToricLattice(K.lattice_dim() + 1) sage: K = Cone([ r.list() + [0] for r in K.rays() ], lattice=L) sage: lyapunov_rank(K) >= K.lattice_dim() True """ beta = 0 m = K.dim() n = K.lattice_dim() l = K.lineality() if m < n: # K is not solid, restrict to its span. K = _restrict_to_space(K, K.span()) # Non-solid reduction lemma. beta += (n - m)*n if l > 0: # K is not pointed, restrict to the span of its dual. Uses a # proposition from our paper, i.e. this is equivalent to K = # _rho(K.dual()).dual(). K = _restrict_to_space(K, K.dual().span()) # Non-pointed reduction lemma. beta += l * m beta += len(K.LL()) return beta def is_lyapunov_like(L,K): r""" Determine whether or not ``L`` is Lyapunov-like on ``K``. We say that ``L`` is Lyapunov-like on ``K`` if `\left\langle L\left\lparenx\right\rparen,s\right\rangle = 0` for all pairs `\left\langle x,s \right\rangle` in the complementarity set of ``K``. It is known [Orlitzky]_ that this property need only be checked for generators of ``K`` and its dual. INPUT: - ``L`` -- A linear transformation or matrix. - ``K`` -- A polyhedral closed convex cone. OUTPUT: ``True`` if it can be proven that ``L`` is Lyapunov-like on ``K``, and ``False`` otherwise. .. WARNING:: If this function returns ``True``, then ``L`` is Lyapunov-like on ``K``. However, if ``False`` is returned, that could mean one of two things. The first is that ``L`` is definitely not Lyapunov-like on ``K``. The second is more of an "I don't know" answer, returned (for example) if we cannot prove that an inner product is zero. REFERENCES: .. [Orlitzky] M. Orlitzky. The Lyapunov rank of an improper cone (preprint). EXAMPLES: The identity is always Lyapunov-like in a nontrivial space:: sage: set_random_seed() sage: K = random_cone(min_ambient_dim = 1, max_rays = 8) sage: L = identity_matrix(K.lattice_dim()) sage: is_lyapunov_like(L,K) True As is the "zero" transformation:: sage: K = random_cone(min_ambient_dim = 1, max_rays = 5) sage: R = K.lattice().vector_space().base_ring() sage: L = zero_matrix(R, K.lattice_dim()) sage: is_lyapunov_like(L,K) True Everything in ``K.LL()`` should be Lyapunov-like on ``K``:: sage: K = random_cone(min_ambient_dim = 1, max_rays = 5) sage: all([is_lyapunov_like(L,K) for L in K.LL()]) True """ return all([(L*x).inner_product(s) == 0 for (x,s) in K.discrete_complementarity_set()]) def random_element(K): r""" Return a random element of ``K`` from its ambient vector space. ALGORITHM: The cone ``K`` is specified in terms of its generators, so that ``K`` is equal to the convex conic combination of those generators. To choose a random element of ``K``, we assign random nonnegative coefficients to each generator of ``K`` and construct a new vector from the scaled rays. A vector, rather than a ray, is returned so that the element may have non-integer coordinates. Thus the element may have an arbitrarily small norm. EXAMPLES: A random element of the trivial cone is zero:: sage: set_random_seed() sage: K = Cone([], ToricLattice(0)) sage: random_element(K) () sage: K = Cone([(0,)]) sage: random_element(K) (0) sage: K = Cone([(0,0)]) sage: random_element(K) (0, 0) sage: K = Cone([(0,0,0)]) sage: random_element(K) (0, 0, 0) TESTS: Any cone should contain an element of itself:: sage: set_random_seed() sage: K = random_cone(max_rays = 8) sage: K.contains(random_element(K)) True """ V = K.lattice().vector_space() F = V.base_ring() coefficients = [ F.random_element().abs() for i in range(K.nrays()) ] vector_gens = map(V, K.rays()) scaled_gens = [ coefficients[i]*vector_gens[i] for i in range(len(vector_gens)) ] # Make sure we return a vector. Without the coercion, we might # return ``0`` when ``K`` has no rays. v = V(sum(scaled_gens)) return v def positive_operators(K): r""" Compute generators of the cone of positive operators on this cone. OUTPUT: A list of `n`-by-``n`` matrices where ``n == K.lattice_dim()``. Each matrix ``P`` in the list should have the property that ``P*x`` is an element of ``K`` whenever ``x`` is an element of ``K``. Moreover, any nonnegative linear combination of these matrices shares the same property. EXAMPLES: The trivial cone in a trivial space has no positive operators:: sage: K = Cone([], ToricLattice(0)) sage: positive_operators(K) [] Positive operators on the nonnegative orthant are nonnegative matrices:: sage: K = Cone([(1,)]) sage: positive_operators(K) [[1]] sage: K = Cone([(1,0),(0,1)]) sage: positive_operators(K) [ [1 0] [0 1] [0 0] [0 0] [0 0], [0 0], [1 0], [0 1] ] Every operator is positive on the ambient vector space:: sage: K = Cone([(1,),(-1,)]) sage: K.is_full_space() True sage: positive_operators(K) [[1], [-1]] sage: K = Cone([(1,0),(-1,0),(0,1),(0,-1)]) sage: K.is_full_space() True sage: positive_operators(K) [ [1 0] [-1 0] [0 1] [ 0 -1] [0 0] [ 0 0] [0 0] [ 0 0] [0 0], [ 0 0], [0 0], [ 0 0], [1 0], [-1 0], [0 1], [ 0 -1] ] TESTS: A positive operator on a cone should send its generators into the cone:: sage: K = random_cone(max_ambient_dim = 6) sage: pi_of_K = positive_operators(K) sage: all([K.contains(p*x) for p in pi_of_K for x in K.rays()]) True """ # 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. V = K.lattice().vector_space() W = VectorSpace(V.base_ring(), V.dimension()**2) tensor_products = [ s.tensor_product(x) for x in K for s in K.dual() ] # Turn our matrices into long vectors... vectors = [ W(m.list()) for m in tensor_products ] # Create the *dual* cone of the positive operators, expressed as # long vectors.. L = ToricLattice(W.dimension()) pi_dual = Cone(vectors, lattice=L) # Now compute the desired cone from its dual... pi_cone = pi_dual.dual() # And finally convert its rays back to matrix representations. M = MatrixSpace(V.base_ring(), V.dimension()) return [ M(v.list()) for v in pi_cone.rays() ]