Wikipedia:Reference desk/Archives/Mathematics/2015 December 31

= December 31 =

Spherical polyhedron with biggest all-hexagon radius?
In a spherical polyhedron made of many hexagons and 12 pentagons, such as an icosahedral Goldberg polyhedron, how far away can all the pentagonal faces be from a chosen point, in order to help give the illusion of a sphere tiled with hexagons alone? Also, if a map is projected onto such a polyhedron, is it possible to rotate it such that the hexagon-for-hexagon matching only ever breaks on the back side when one is facing that point? Neon Merlin  13:25, 31 December 2015 (UTC)
 * Note that most Goldberg polyhedra do not use perfectly regular hexagons, so for a well-posed problem you need to specify bounds on acceptable shape distortion and size variation. The inverse stereographic projection of the hexagonal tiling will be hexagon-only with only one defect. It should look like . In the figure, the largest hexagon is at the bottom. Making it face the viewer would hide most of the size variation, with a ratio of about 2 between the largest and the smallest visible hexagons. Hexagons should be perfectly flat because circle are projected to circles. Egnau (talk) 17:00, 31 December 2015 (UTC)
 * Perhaps a more precise formulation of the problem is this: given an array of 1, 7, 19, 37, ... hexagons, arranged in a hexagon, can it be extended to a tiling of the sphere with 12 pentagons and fewer than 1, 7, 19, 37, ... additional hexagons. The array is to be considered as a planar graph so distortions are allowed, but edges are not allowed to cross. My pencil and paper experiments suggest that it's possible to do it with the same number of hexagons (at least up to 37), but fewer seems to be impossible. You can ask the same question but with 6 squares or 4 triangles, but in this case it does seem to be possible with fewer hexagons. --RDBury (talk) 01:35, 1 January 2016 (UTC)


 * PS. Actually I was able to do the 19 hexagon case with 14 additional hexagons (and 12 pentagons), so it seems that "lopsided" pentagon distribution is possible. Also, the name for this type of graph (planar, trivalent with only hexagon and pentagon faces) is fullerine, after the allotrope of carbon. These have been enumerated up to a large number of vertices so it may be possible to find other examples by searching through them. --RDBury (talk) 03:05, 1 January 2016 (UTC)

Inward-biconnected, outward-biconnected components
Is there a name for a decomposition of a directed graph into components such that each component has exactly one exit vertex and/or contains leaves, and has exactly one entrance vertex and/or contains roots -- a decomposition more or less into flow graphs? If so, what other restrictions on the components are common, to avoid choosing trivial ones (such as any single vertex, or any entire weakly-connected component that is directed-acyclic) unnecessarily? I figure this type of decomposition must be useful, at least for directed-acyclic graphs, because any shortest or longest path from a root to a leaf must consist of the shortest/longest paths through some subset of these components. Neon Merlin  23:26, 31 December 2015 (UTC)