Timeline for An easy-to-state elusive combinatorial problem
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 30, 2013 at 17:43 | comment | added | Maaz-ul-Haq | But as you go up in dimensions the gap between the cubes gets smaller and smaller. So while the 2D case was sort of a checkerboard square lattice (with gaps of the same width as the square), the 3D case has cubes with gaps two-thirds of the cube-edge length. | |
| Jun 30, 2013 at 15:33 | comment | added | David E Speyer | I thought I could update case 1 to work in any dimension, but then I realized that I was using the fact that the gaps between the cubes are the same width as the cubes themselves in a crucial way. So I currently don't know how to do higher dimensions, although my intuition is that any fixed dimension should be doable by a case by case check like this one. | |
| Jun 30, 2013 at 1:36 | comment | added | Maaz-ul-Haq | Can Case $1$ and $2$ with $m\ne 1$ be given an inductive flavor to cover for all dimensions? Or do more cases arise when you have an increment in dimensions? | |
| Jun 28, 2013 at 1:32 | comment | added | Maaz-ul-Haq | It was inspired by the View Obstruction paper by Cusick, though this one is on a completely different route, more like scaling n-cubes to cover the entire totally positive orthant of $\mathbb{R}^n$. | |
| Jun 28, 2013 at 1:28 | comment | added | David E Speyer | Out of curiosity, where does this problem come from? | |
| Jun 28, 2013 at 1:20 | comment | added | Maaz-ul-Haq | I am actually interested in a general version of this problem where you have $a_1, a_2, a_3,... a_{\lambda-2} \in \mathbb{N}$ such that $a_i\geq 1$ for all $1\leq i\leq \lambda-2$ and it is conjectured that the bound is $x=\lambda-1$ such that $n\in[1,x]$ to ensure that all $a_i$ are contained within $\lambda-2$-dimensional hypercubes generated by $[\lambda (m-1)+1,\lambda m -1]$ in all dimensions. It would be quite a task to solve this in general as apparently the problem increases in difficulty as you increase the dimensions. | |
| Jun 28, 2013 at 1:09 | vote | accept | Maaz-ul-Haq | ||
| Jun 28, 2013 at 0:57 | comment | added | Maaz-ul-Haq | "It is easy to check that, any point in this triangle, we can rescale it by a factor of $\leq 3$ to land in the square $[9,11]\times[5,7]$." Or precisely this? | |
| Jun 27, 2013 at 23:41 | history | answered | David E Speyer | CC BY-SA 3.0 |
