Locality: which owners can meet a boundary segment #
Every anchor other than the listed ones is at squared distance at least 1 from every point of
the segment, so its owner (a closed triangle of side < 1) cannot contain any point of the
segment. Hexagon side i runs from vertex i to vertex (i+1); its admissible owners are the
centre, S_i, S_{i+1} and, for even i, the corner owner A_{i/2}. The corner segment from
corner j to vertex (2j+m) admits A_j, S_{2j}, S_{2j+1}. This is the content of
proof/verify_locality.py.
Admissible owners of hexagon side i.
Equations
Instances For
theorem
ConwaySoifer.side_owner_allowed
{T : Configuration}
{r : ℝ}
(hcommon : CommonSide T r)
(hr : r < 1)
(hanch : ∀ (i : Fin 10), anchor i ∈ (T i).carrier)
(i : Fin 6)
{t : ℝ}
(ht0 : 0 ≤ t)
(ht1 : t ≤ 1)
{k : Fin 10}
(hk : linePoint (vertex i) (sideDirection i true) t ∈ (T k).carrier)
:
A point of hexagon side i can only belong to an admissible owner.
theorem
ConwaySoifer.cornerSeg_owner_allowed
{T : Configuration}
{r : ℝ}
(hcommon : CommonSide T r)
(hr : r < 1)
(hanch : ∀ (i : Fin 10), anchor i ∈ (T i).carrier)
(j : Fin 3)
(m : Fin 2)
{t : ℝ}
(ht0 : 0 ≤ t)
(ht1 : t ≤ 1)
{k : Fin 10}
(hk : linePoint (corner j) (cornerDirection j m) t ∈ (T k).carrier)
:
A point of a corner segment of cell j can only belong to an admissible owner.