Timeline for Inscribed parallelotope in a $d$-simplex
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 5, 2015 at 21:31 | comment | added | Martin Tancer | No. Imagine a triangle, a vertex $x$ of this triangle and $P_x$ to be a parallelogram with the faces (segments) parallel with the two edges of the triangle incident to $x$. Then $P'_x$ contains $x$ as well as almost completely the two segments of $P_x$ incident of $x$. (On the other hand, it does not contain the remaining two segments of $P_x$.) | |
| Aug 5, 2015 at 21:20 | comment | added | Roy Han | How do you do these for boundary points? Will $P'_{x_i}$ exclude $x_i$? | |
| Aug 5, 2015 at 21:09 | comment | added | Martin Tancer | @RoyHan That is why I use the sets $P'_x$ open in the simplex (and not $\mathbb{R}^d$). It is written there. | |
| Aug 5, 2015 at 20:50 | comment | added | Roy Han | I realized this is a little problematic in the sense that you'll need open covers, which simple is not true when $x_i$ is at the boundary of $\Delta^d$. | |
| Aug 5, 2015 at 6:25 | history | edited | Martin Tancer | CC BY-SA 3.0 |
added 138 characters in body
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| Aug 4, 2015 at 21:18 | comment | added | Roy Han | Yes, compactness argument will give a very easy proof. I think I may have not expressed in a clearer way of my concern: I'm concerned with the question that, how many parallelotopes do we need at least, to cover the simplex, i.e. asking for the minimum covering number, or getting an estimate of such number that scales with dimension $d$ with the correct order. | |
| Aug 3, 2015 at 21:23 | history | answered | Martin Tancer | CC BY-SA 3.0 |