X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Fcone%2Fcone.py;h=ff7d195d134c15943dbb75c1f26b741bb4a0afba;hb=10142e85f34c47fa35df002f519d1d58a79a74f4;hp=3a1e190cb2ebe41f57810f04726f8123294c55cd;hpb=db71f895bcdf4550bc4212d0e2fae41c8de22d6a;p=sage.d.git diff --git a/mjo/cone/cone.py b/mjo/cone/cone.py index 3a1e190..ff7d195 100644 --- a/mjo/cone/cone.py +++ b/mjo/cone/cone.py @@ -8,56 +8,210 @@ addsitedir(abspath('../../')) from sage.all import * -def project_span(K, K2 = None): +def project_span(K): r""" - Return a "copy" of ``K`` embeded in a lower-dimensional space. - - By default, we will project ``K`` into the subspace spanned by its - rays. However, if ``K2`` is not ``None``, we will project into the - space spanned by the rays of ``K2`` instead. + Project ``K`` into its own span. EXAMPLES:: - sage: K = Cone([(1,0,0), (0,1,0)]) - sage: project_span(K) - 2-d cone in 2-d lattice N - sage: project_span(K).rays() - N(1, 0), - N(0, 1) - in 2-d lattice N + sage: K = Cone([(1,)]) + sage: project_span(K) == K + True - sage: K = Cone([(1,0,0), (0,1,0)]) - sage: K2 = Cone([(0,1)]) - sage: project_span(K, K2).rays() + 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 """ - # Allow us to use a second cone to generate the subspace into - # which we're "projecting." - if K2 is None: - K2 = K + 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: - # Use these to generate the new cone. - cs1 = K.rays().matrix().columns() + A nonnegative integer representing the dimension of the space of all + elements perpendicular to this cone. - # And use these to figure out which indices to drop. - cs2 = K2.rays().matrix().columns() + .. seealso:: - perp_idxs = [] + :meth:`dim`, :meth:`lattice_dim` - for idx in range(0, len(cs2)): - if cs2[idx].is_zero(): - perp_idxs.append(idx) + EXAMPLES: - solid_cols = [ cs1[idx] for idx in range(0,len(cs1)) - if not idx in perp_idxs - and not idx >= len(cs2) ] + The codimension of the nonnegative orthant is zero, since the span of + its generators equals the entire ambient space:: - m = matrix(solid_cols) - L = ToricLattice(len(m.rows())) - J = Cone(m.transpose(), lattice=L) - return J + 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): @@ -204,23 +358,6 @@ def LL(K): sage: sum(map(abs, l)) 0 - Try the formula in my paper:: - - sage: K = random_cone(max_dim=15, max_rays=25) - sage: actual = lyapunov_rank(K) - sage: K_S = project_span(K) - sage: J_T1 = project_span(K, K_S.dual()) - sage: J_T2 = project_span(K_S.dual()).dual() - sage: J_T2 = Cone(J_T2.rays(), lattice=J_T1.lattice()) - sage: J_T1 == J_T2 - True - sage: J_T = J_T1 - sage: l = K.linear_subspace().dimension() - sage: codim = K.lattice_dim() - K.dim() - sage: expected = lyapunov_rank(J_T) + K.dim()*(l + codim) + codim**2 - sage: actual == expected - True - """ V = K.lattice().vector_space() @@ -302,6 +439,9 @@ 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. @@ -321,6 +461,15 @@ 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: 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.]_:: @@ -334,7 +483,30 @@ 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: 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)]) @@ -360,11 +532,11 @@ 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: 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 @@ -390,5 +562,54 @@ 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: 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