the face cycles of the record: 20 vertices, 30 edges, 12 faces, every directed edge in exactly one face, distance profile 1,3,6,6,3,1 from every vertex: ok
graph distance of the vertices 0 and 2: 2
reachable points of length <= 18 in the full sector: 167 ; with the face word of the record: 1
crossed edges: identical with the list of the record
end vertex: 2 ; N = 16 segments; length 17.512014631639 ; slope angle alpha = 0.028007 pi
2|B|^2 = 307 + 137 sqrt 5 exactly; |B|^2 = 306.670656458736, |B| = 17.512014631639
angles of Fig. 8 for the billiard trajectory: alpha = 0.028007 pi, beta = 0.828007 pi, beta - alpha = 0.800000 pi = 2 * (2pi/5)
type by Definition 2.1 as printed (all triples): [1] ; with the rule on the parity of N: 1
last polygon = c - P_0, end point = c - X_s with s = 1 ; c == B: False (central symmetry would require s = 0)
fractions of the length at which the 15 edges are crossed: 0.059017 0.145898 0.213525 0.291796 0.368034 0.437694 0.522542 0.583592 0.669153 0.683282 0.751865 0.788854 0.834576 0.894427 0.917288
they agree with the 15 values of t printed in the record (to 1e-9)
t_8 = 0.583592 (in place of 1/2); t_i + t_(16-i) ranges from 0.9763 to 1.1917 (in place of 1)
in the unfolding: -S0 = eta^2 * S1 for S0 = 0->16, S1 = 2->10 (the relation of the record): confirmed exactly
the image in the last face of the boundary edge which leaves the last vertex of the billiard trajectory counterclockwise: 2 -> 16
the edge which leaves the vertex 2 counterclockwise in the face F1: 2 -> 10
angle between the reversed last segment and 2->10: 0.428007 pi; and 2->16: 0.171993 pi (= pi - beta)
with the angle at 2->10 taken as the terminal angle: (angle) - alpha = 0.400000 pi = 2pi/5;  pi - (angle) - alpha = 0.543986 pi, pi - (angle) + alpha = 0.600000 pi
all short geodesics from the vertex 0 in F1 with this length: (N, type, end vertex): [(16, 1, 2), (16, 1, 2), (16, 2, 10), (16, 2, 10)]
ALL CHECKS PASSED
