Timeline for Reordering vertices of a polygon
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| S Jun 9, 2018 at 19:03 | history | bounty ended | CommunityBot | ||
| S Jun 9, 2018 at 19:03 | history | notice removed | CommunityBot | ||
| Jun 3, 2018 at 8:45 | comment | added | Wlod AA | Otherwise, when $Q"$ has more vertices than $Q$ then it seems clear to me that the answer in general is NO. | |
| Jun 3, 2018 at 8:22 | comment | added | user101163 | @WlodAA yes, you can suppose $Q$ and $Q'$ have the same number of vertices | |
| Jun 3, 2018 at 5:01 | comment | added | Wlod AA | When $Q'$ has more vertices than $Q$ then it can be messy even for the case of 4-vertice $Q$. Is there an assumption about $Q'$ having the same number of vericies as $Q$? | |
| Jun 3, 2018 at 4:57 | comment | added | Wlod AA | It seems that the 4-vertice case is obvious. | |
| Jun 3, 2018 at 4:56 | comment | added | Wlod AA | It seems that the 4-vertice case is obvious. | |
| Jun 3, 2018 at 4:53 | comment | added | Wlod AA | What about the simplest cases with 4 or 5 vertices? | |
| Jun 3, 2018 at 4:11 | comment | added | Wlod AA | Thank you for your kind answer. This is a nice question. | |
| Jun 2, 2018 at 8:41 | comment | added | user101163 | @j.c. Yes, $Q,Q'$ are pseudotriangles. I am sorry, the point you cited was not clear, I hope it is clear now. I also means that $v_i',i=1,2,3$ must be vertices of $Q''$ | |
| Jun 2, 2018 at 8:39 | history | edited | user101163 | CC BY-SA 4.0 |
added 188 characters in body
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| Jun 2, 2018 at 8:05 | comment | added | user101163 | @WlodAA I want to prove that for every $Q$ and $Q'$ there is a $Q''$ | |
| Jun 2, 2018 at 8:05 | history | edited | user101163 | CC BY-SA 4.0 |
added 22 characters in body
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| Jun 2, 2018 at 3:48 | comment | added | j.c. | To check my understanding, $Q,Q'$ are pseudotriangles en.wikipedia.org/wiki/Pseudotriangle . I do not understand the first bullet point in your conditions on $Q''$: "there is a correspondence between vertices of $Q$ and $Q''$ and it results $v_i''=v_i'$ "? Does the first part just amount to the condition that $Q''$ must have $n$ vertices? | |
| Jun 1, 2018 at 22:00 | comment | added | Wlod AA | Do you mean $\ \forall_{Q\ Q'}\ $ (etc.) or $\ \exists_{Q\ Q'}\ $ (etc.) ? | |
| S Jun 1, 2018 at 17:50 | history | bounty started | CommunityBot | ||
| S Jun 1, 2018 at 17:50 | history | notice added | user101163 | Draw attention | |
| May 31, 2018 at 1:02 | answer | added | Joseph O'Rourke | timeline score: 4 | |
| May 30, 2018 at 20:23 | history | edited | user101163 |
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| May 30, 2018 at 17:20 | history | asked | user101163 | CC BY-SA 4.0 |