3 ms·
For me, it seems easy if we don't have to care about the imposed restrictions: no crossing and no straight angles. Hmmmm.... New ideas coming... Maybe we don't
by marcodiego 5y ago
For me, it seems easy if we don't have to care about the imposed restrictions: no crossing and no straight angles.
Hmmmm.... New ideas coming... Maybe we don't have to care about crossings: any crossing can be solved by reordering the vertices, so, all we have to care about is the number unique vertices positions!
I still don't know how to avoid straight angles though.
- tylerhou 5y agoIf you didn’t care about the restrictions, then the answer is N*M choose 4.