3 ms·
I am wondering this as well. Disclaimer I did not read the paper. On the one hand, there are infinitely many geodesics on a dodecahedron, of unbounded length,
by galimaufry 6y ago
I am wondering this as well. Disclaimer I did not read the paper.
On the one hand, there are infinitely many geodesics on a dodecahedron, of unbounded length, so you can't really brute force all of them.
On the other hand, the actual solution only goes through each face at most once. There are infinitely many unfoldings to try, but for each unfolding there are only finitely many paths to try, and you can get the solution on one of the minimal unfoldings.
- blintz 6y agoYeah, I mean 12! is ~479 million, and there are 20 vertices, so that is very much within brute-force-on-a-laptop range. For some net, I think given a vertex + list of faces, there is a simple-to-get yes/no answer on whether a ray passing through the faces and returning to the vertex exists. I wonder if you can just write a Python script to do it... Edit: Even simpler - there are 43,380 nets of a dodecahedron, and each has 20 vertices. So if you just draw ~8 million straight lines and see if any are both on the net and do not intersect another vertex, you'll find (at least) one that works!