Timeline for Peeling a polygonal vegetable
Current License: CC BY-SA 3.0
14 events
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Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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Nov 23, 2014 at 10:16 | history | edited | Erel Segal-Halevi | CC BY-SA 3.0 |
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Nov 11, 2014 at 4:50 | answer | added | S. Carnahan♦ | timeline score: 5 | |
Nov 11, 2014 at 1:14 | answer | added | Joseph O'Rourke | timeline score: 1 | |
Nov 10, 2014 at 22:07 | comment | added | André Henriques | Unrelated to the question, but related to vegetable peeling: arxiv.org/abs/1202.3033 | |
Nov 10, 2014 at 19:45 | history | edited | Erel Segal-Halevi | CC BY-SA 3.0 |
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Oct 4, 2014 at 0:36 | comment | added | Joseph O'Rourke | In any individual case the task seems trivial. What is challenging is finding a scheme that handles all cases, especially ensuring that $V \setminus P(t)$ is connected at all times. | |
Oct 3, 2014 at 7:46 | comment | added | Per Alexandersson | The floodfill algorithm comes to my mind... | |
Oct 3, 2014 at 0:11 | answer | added | Joseph O'Rourke | timeline score: 2 | |
Oct 2, 2014 at 22:56 | comment | added | Lucian | This question brings tears to my eyes. ;-) | |
Oct 2, 2014 at 19:43 | comment | added | Joonas Ilmavirta | If you construct the peeling function by expanding $P(0)$ step by step, $P(t)$ is always connected. It seems intuitively possible to first expand $P(t)$ so that $P(1)\setminus P(\frac12)$ is simply connected (cutting the loops in $P(1)\setminus P(0)$) and then shrink the simply connected set away. I'm not sure I can formalize this two-step procedure properly, though. | |
Oct 2, 2014 at 18:50 | comment | added | Erel Segal-Halevi | @JoonasIlmavirta This sounds promising, but how do you guarantee the connectivity? | |
Oct 2, 2014 at 18:26 | comment | added | Joonas Ilmavirta | A naive suggestion: Triangulate $P(1)\setminus P(0)$ (the complement of the union of the pulp and the head) and proceed triangle by triangle. Have you tried anything along these lines? | |
Oct 2, 2014 at 18:20 | history | asked | Erel Segal-Halevi | CC BY-SA 3.0 |