X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Fcone%2Fcone.py;h=b9e930e6819643710b82c06faa0b72b934298d96;hb=6bd30534d5aa984c73f511121efa8fda4386c51a;hp=2e3dc8afb78cd1f4e1d5f368eb5f75733d127e82;hpb=433b47ffeac9305beb9101afa302c7e924b60744;p=sage.d.git diff --git a/mjo/cone/cone.py b/mjo/cone/cone.py index 2e3dc8a..b9e930e 100644 --- a/mjo/cone/cone.py +++ b/mjo/cone/cone.py @@ -8,6 +8,311 @@ addsitedir(abspath('../../')) from sage.all import * +def drop_dependent(vs): + r""" + Return the largest linearly-independent subset of ``vs``. + """ + result = [] + m = matrix(vs).echelon_form() + for idx in range(0, m.nrows()): + if not m[idx].is_zero(): + result.append(m[idx]) + + return result + + +def basically_the_same(K1,K2): + r""" + ``True`` if ``K1`` and ``K2`` are basically the same, and ``False`` + otherwise. + """ + if K1.lattice_dim() != K2.lattice_dim(): + return False + + if K1.nrays() != K2.nrays(): + return False + + if K1.dim() != K2.dim(): + return False + + if lineality(K1) != lineality(K2): + return False + + if K1.is_solid() != K2.is_solid(): + return False + + if K1.is_strictly_convex() != K2.is_strictly_convex(): + return False + + if len(LL(K1)) != len(LL(K2)): + return False + + C_of_K1 = discrete_complementarity_set(K1) + C_of_K2 = discrete_complementarity_set(K2) + if len(C_of_K1) != len(C_of_K2): + return False + + if len(K1.facets()) != len(K2.facets()): + return False + + return True + + + +def rho(K, K2=None): + r""" + Restrict ``K`` into its own span, or the span of another cone. + + INPUT: + + - ``K2`` -- another cone whose lattice has the same rank as this cone. + + OUTPUT: + + A new cone in a sublattice. + + EXAMPLES:: + + sage: K = Cone([(1,)]) + sage: rho(K) == K + True + + sage: K2 = Cone([(1,0)]) + sage: rho(K2).rays() + N(1) + in 1-d lattice N + sage: K3 = Cone([(1,0,0)]) + sage: rho(K3).rays() + N(1) + in 1-d lattice N + sage: rho(K2) == rho(K3) + True + + TESTS: + + The projected cone should always be solid:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8) + sage: K_S = rho(K) + sage: K_S.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_dim = 8) + sage: K_S = rho(K, K.dual() ) + sage: K_S.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_dim = 8) + sage: K.dim() == rho(K).dim() + True + + Nor should it affect the lineality of a cone:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8) + sage: lineality(K) == lineality(rho(K)) + True + + No matter which space we restrict to, the lineality should not + increase:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8) + sage: lineality(K) >= lineality(rho(K)) + True + sage: lineality(K) >= lineality(rho(K, K.dual())) + True + + If we do this according to our paper, then the result is proper:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8, strictly_convex=False, solid=False) + sage: K_S = rho(K) + sage: P = rho(K_S.dual()).dual() + sage: P.is_proper() + True + sage: P = rho(K_S, K_S.dual()) + sage: P.is_proper() + True + + :: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8, strictly_convex=True, solid=False) + sage: K_S = rho(K) + sage: P = rho(K_S.dual()).dual() + sage: P.is_proper() + True + sage: P = rho(K_S, K_S.dual()) + sage: P.is_proper() + True + + :: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8, strictly_convex=False, solid=True) + sage: K_S = rho(K) + sage: P = rho(K_S.dual()).dual() + sage: P.is_proper() + True + sage: P = rho(K_S, K_S.dual()) + sage: P.is_proper() + True + + :: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8, strictly_convex=True, solid=True) + sage: K_S = rho(K) + sage: P = rho(K_S.dual()).dual() + sage: P.is_proper() + True + sage: P = rho(K_S, K_S.dual()) + sage: P.is_proper() + True + + Test the proposition in our paper concerning the duals, where the + subspace `W` is the span of `K^{*}`:: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8, solid=False, strictly_convex=False) + sage: K_W = rho(K, K.dual()) + sage: K_star_W_star = rho(K.dual()).dual() + sage: basically_the_same(K_W, K_star_W_star) + True + + :: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8, solid=True, strictly_convex=False) + sage: K_W = rho(K, K.dual()) + sage: K_star_W_star = rho(K.dual()).dual() + sage: basically_the_same(K_W, K_star_W_star) + True + + :: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8, solid=False, strictly_convex=True) + sage: K_W = rho(K, K.dual()) + sage: K_star_W_star = rho(K.dual()).dual() + sage: basically_the_same(K_W, K_star_W_star) + True + + :: + + sage: set_random_seed() + sage: K = random_cone(max_dim = 8, solid=True, strictly_convex=True) + sage: K_W = rho(K, K.dual()) + sage: K_star_W_star = rho(K.dual()).dual() + sage: basically_the_same(K_W, K_star_W_star) + True + + """ + if K2 is None: + K2 = K + + # First we project K onto the span of K2. This can be done with + # cones (i.e. without converting to vector spaces), but it's + # annoying to deal with lattice mismatches. + span_K2 = Cone(K2.rays() + (-K2).rays(), lattice=K.lattice()) + K = K.intersection(span_K2) + + V = K.lattice().vector_space() + + # Create the space W \times W^{\perp} isomorphic to V. + # First we get an orthogonal (but not normal) basis... + W_basis = drop_dependent(K2.rays()) + W = V.subspace_with_basis(W_basis) + + # We've already intersected K with the span of K2, so every + # generator of K should belong to W now. + W_rays = [ W.coordinate_vector(r) for r in K.rays() ] + + L = ToricLattice(K2.dim()) + return Cone(W_rays, lattice=L) + + + +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: set_random_seed() + sage: K = random_cone(max_dim = 8) + 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: set_random_seed() + sage: K = random_cone(max_dim = 8, strictly_convex = True) + sage: lineality(K) + 0 + + """ + return K.linear_subspace().dimension() + + def discrete_complementarity_set(K): r""" Compute the discrete complementarity set of this cone. @@ -62,11 +367,12 @@ def discrete_complementarity_set(K): 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: set_random_seed() + sage: K1 = random_cone(max_dim=6) 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 + sage: sorted(actual) == sorted(expected) True """ @@ -87,15 +393,100 @@ def LL(K): 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] + ] + + If our cone is the entire space, then every transformation on it is + Lyapunov-like:: + + sage: K = Cone([(1,0), (-1,0), (0,1), (0,-1)]) + sage: M = MatrixSpace(QQ,2) + sage: M.basis() == LL(K) + True + + 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: set_random_seed() + sage: K = random_cone(max_dim=8) + 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 + + The Lyapunov-like transformations on a cone and its dual are related + by transposition, but we're not guaranteed to compute transposed + elements of `LL\left( K \right)` as our basis for `LL\left( K^{*} + \right)` + + sage: set_random_seed() + sage: K = random_cone(max_dim=8) + sage: LL2 = [ L.transpose() for L in LL(K.dual()) ] + sage: V = VectorSpace( K.lattice().base_field(), K.lattice_dim()^2) + sage: LL1_vecs = [ V(m.list()) for m in LL(K) ] + sage: LL2_vecs = [ V(m.list()) for m in LL2 ] + sage: V.span(LL1_vecs) == V.span(LL2_vecs) + True """ 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] + 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 @@ -171,6 +562,9 @@ def lyapunov_rank(K): 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. @@ -190,6 +584,15 @@ def lyapunov_rank(K): 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.]_:: @@ -203,7 +606,30 @@ def lyapunov_rank(K): 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: 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()*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)]) @@ -229,11 +655,12 @@ def lyapunov_rank(K): 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: set_random_seed() + sage: K1 = random_cone(max_dim=8, strictly_convex=True, solid=True) + sage: K2 = random_cone(max_dim=8, strictly_convex=True, solid=True) sage: K = K1.cartesian_product(K2) sage: lyapunov_rank(K) == lyapunov_rank(K1) + lyapunov_rank(K2) True @@ -241,7 +668,36 @@ def lyapunov_rank(K): 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: set_random_seed() + sage: K = random_cone(max_dim=8) + sage: lyapunov_rank(K) == lyapunov_rank(K.dual()) + True + + Make sure we exercise the non-strictly-convex/non-solid case:: + + sage: set_random_seed() + sage: K = random_cone(max_dim=8, strictly_convex=False, solid=False) + sage: lyapunov_rank(K) == lyapunov_rank(K.dual()) + True + + Let's check the other permutations as well, just to be sure:: + + sage: set_random_seed() + sage: K = random_cone(max_dim=8, strictly_convex=False, solid=True) + sage: lyapunov_rank(K) == lyapunov_rank(K.dual()) + True + + :: + + sage: set_random_seed() + sage: K = random_cone(max_dim=8, strictly_convex=True, solid=False) + sage: lyapunov_rank(K) == lyapunov_rank(K.dual()) + True + + :: + + sage: set_random_seed() + sage: K = random_cone(max_dim=8, strictly_convex=True, solid=True) sage: lyapunov_rank(K) == lyapunov_rank(K.dual()) True @@ -251,7 +707,8 @@ def lyapunov_rank(K): 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: set_random_seed() + sage: K = random_cone(max_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 @@ -259,5 +716,82 @@ def lyapunov_rank(K): 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_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_dim=8) + sage: actual = lyapunov_rank(K) + sage: K_S = rho(K) + sage: P = rho(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: set_random_seed() + sage: K = random_cone(max_dim=8, strictly_convex=True, solid=True) + sage: lyapunov_rank(K) == len(LL(K)) + True + + In fact the same can be said of any cone. These additional tests + just increase our confidence that the reduction scheme works:: + + sage: set_random_seed() + sage: K = random_cone(max_dim=8, strictly_convex=True, solid=False) + sage: lyapunov_rank(K) == len(LL(K)) + True + + :: + + sage: set_random_seed() + sage: K = random_cone(max_dim=8, strictly_convex=False, solid=True) + sage: lyapunov_rank(K) == len(LL(K)) + True + + :: + + sage: set_random_seed() + sage: K = random_cone(max_dim=8, strictly_convex=False, solid=False) + sage: lyapunov_rank(K) == len(LL(K)) + True + """ - return len(LL(K)) + K_orig = 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 = rho(K) + + # Lemma 2 + beta += m*(n - m) + (n - m)**2 + + if l > 0: + # K is not pointed, project its dual onto its span. + # Uses a proposition from our paper, i.e. this is + # equivalent to K = rho(K.dual()).dual() + K = rho(K, K.dual()) + + # Lemma 3 + beta += m * l + + beta += len(LL(K)) + return beta