is_inscribed depends on order of vertices #29125
Description
Currently the inscription test for polyhedra depends on the order of vertices:
sage: Polyhedron(vertices=[[-2,-1], [-2,1], [0,-1], [0,1]]).is_inscribed()
True
sage: P = Polyhedron(vertices=[[-2,-1], [-2,1], [0,-1], [0,1]], backend='field')
sage: P.is_inscribed()
False
sage: V = P.Vrepresentation()
sage: H = P.Hrepresentation()
sage: parent = P.parent()
sage: dic = {True: 0, False: 0}
sage: for V1 in Permutations(V):
....: P1 = parent._element_constructor_(
....: [V1, [], []], [H, []], Vrep_minimal=True, Hrep_minimal=True)
....: dic[P1.is_inscribed()] += 1
....:
sage: dic
{True: 18, False: 6}
The algorithm constructs a sphere around dim + 1 vertices in general position. The circumcenter is computed up to sign. Then, one vertex is taken to determine, which sign to choose. However, up to dim vertices might lie on the intersection of both spheres.
We fix this by checking distance from the circumcenter for all vertices of that simplex.
Component: geometry
Keywords: polytopes, is inscribed
Author: Jonathan Kliem
Branch/Commit: 3f9cc75
Reviewer: Jean-Philippe Labbé
Issue created by migration from https://trac.sagemath.org/ticket/29125