Wednesday, January 20, 2021

[tocffidt] cube star

first, consider a 3x3 grid of points.  delete the center point.  one can draw a pretty 8-pointed star through the remaining points.  it happens to be a knight's tour.

next, we consider a 3D analogue, concentrating on making it a pretty star, not a Hamiltonian circuit.  26 points of a 3x3x3 grid, omitting the internal center point.  the remaining points fall into 3 categories: 8 corners, 12 edges, 6 centers (terminology from Rubik's cube).

for a center, consider the 9 points on the opposite face.  don't connect to the opposite center, so 8 possibilities: 4 corners and 4 edges.

for a corner, consider its opposite corner.  the opposite corner is adjacent to 3 edges.   there are also 3 centers diagonally adjacent.

for an edge, consider its opposite edge.  adjacent to it are 2 corners and 2 faces.  diagonally adjacent are 4 edges.

graph edge lengths (line segments connecting points) can be square roots of 4+1+{0, 1, 4} = sqrt {5, 6, 9}.

many possibilities to form the star.  one that seems elegant: centers connect to edges.  corners connect to edges.  (and vice versa.)

fairly straightforward to generalize to larger cubic surface grids of points: neighbors of the diametrically opposite point.  perhaps project surface points to sphere.

vertices of a cube, i.e., 2x2x2 grid of points: connect each vertex to the 3 that are neither diametrically opposite nor share an edge.  result is stella octangula.

it might be possible to argue that no pretty symmetric Hamiltonian circuit exists because the symmetry group of a cube (isomorphic to permutations of 4 items) is not a cyclic group.  however, if we really want a circuit, we can adjust the definition of pretty.

icosahedron vertices (12 points): each vertex connected to 5 adjacent to diametrically opposite: small stellated dodecahedron.

dodecahedron vertices (20 points): 3 adjacent to diametrically opposite: great stellated dodecahedron.

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