+ # 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 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: K.lattice_dim()
+ 0
+ sage: codim(K)
+ 0