+ 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
+