Timeline for Always a planar-drawn cycle through $n$ points
Current License: CC BY-SA 3.0
13 events
when toggle format | what | by | license | comment | |
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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
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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
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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 |