3 ms·
Although rather unfortunately a sphere can't be tiled by hexagons.
by contravariant 8y ago
Although rather unfortunately a sphere can't be tiled by hexagons.
- Waterluvian 8y agoThis reminds me of RimWorld hiding pentagons throughout its world in order to satisfy this issue.
- contravariant 8y agoKind of makes you wonder how many you'd need. Now ignoring some cheats, like just putting 6 vertices on the equator and calling both hemisphere a hexagon, a shape consisting solely of 'h' hexagons and 'p' pentagons will have (h+p) faces, (6h+5p)/2 edges and at most (6h+5p)/3 vertices (we're forbidding the construction where you just cut an edge into two and claim you've created a vertex, so each vertex is part of at least 3 faces). This gives an Euler characeristic of at most: (6h+5p)/3 - (6h+5p)/2 + (h+p) = (6h+5p)/6 - (h+p) = (1/6) p. Since a sphere has a characteristic of 2 you need at least twelve pentagons. And apparently you can achieve this lower bound by subdividing the triangular faces of an icosahedron, leaving you with 12 pentagons at the corners of the icosahedron.
- Waterluvian 8y agoApparently RimWorld uses 12 pentagons: https://rimworldwiki.com/wiki/World_generation https://rimworldwiki.com/wiki/World_generation Well done.