Skip to main content
13 events
when toggle format what by license comment
Dec 19, 2015 at 15:29 answer added srmusawi timeline score: 2
Dec 19, 2015 at 2:24 answer added Jan Kyncl timeline score: 15
Dec 18, 2015 at 20:34 history edited JMP CC BY-SA 3.0
removed 'tree' bit
Dec 18, 2015 at 20:24 answer added Joseph O'Rourke timeline score: 14
Dec 18, 2015 at 20:18 vote accept JMP
Dec 18, 2015 at 20:16 comment added Joseph O'Rourke You have to exclude all $n$ points collinear, i.e., lying on one line.
Dec 18, 2015 at 20:03 comment added Fedor Petrov @PietroMajer even if this works, we must choose removing edges carefully.
Dec 18, 2015 at 20:02 comment added Pietro Majer Ops, I was thinking to a simple arc. It could be modified to get a cycle, but FP's construction below is much better.
Dec 18, 2015 at 20:00 history edited JMP CC BY-SA 3.0
added 30 characters in body
Dec 18, 2015 at 19:59 comment added JMP @PietroMajer; does this guarantee a cycle?
Dec 18, 2015 at 19:58 comment added Pietro Majer What about this: first draw the boundary of the convex hull of these points. Then remove the points in this boundary, and repeat. This way you get a nested family of convex polygons. Lastly, remove an edge in each polygon and connect them in a spiral.
Dec 18, 2015 at 19:56 answer added Fedor Petrov timeline score: 15
Dec 18, 2015 at 19:46 history asked JMP CC BY-SA 3.0