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visits | member for | 1 year |
seen | Aug 13 '13 at 17:32 | |
stats | profile views | 178 |
Jun 30 |
comment |
An easy-to-state elusive combinatorial problem
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 |
comment |
An easy-to-state elusive combinatorial problem (revisited)
If $ry$ increases by $3(y/z)\ge 3$ wouldn't that imply an increment in $rx$ by $3(x/y)\ge 3$ as well? |
Jun 30 |
asked | An easy-to-state elusive combinatorial problem (revisited) |
Jun 30 |
comment |
An easy-to-state elusive combinatorial problem
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 |
comment |
An easy-to-state elusive combinatorial problem
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 |
comment |
An easy-to-state elusive combinatorial problem
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 |
accepted | An easy-to-state elusive combinatorial problem |
Jun 28 |
comment |
An easy-to-state elusive combinatorial problem
"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 |
awarded | Commentator |
Jun 27 |
comment |
An easy-to-state elusive combinatorial problem
@domotorp exactly. |
Jun 27 |
revised |
An easy-to-state elusive combinatorial problem
added 19 characters in body |
Jun 27 |
revised |
An easy-to-state elusive combinatorial problem
added 9 characters in body |
Jun 27 |
awarded | Critic |
Jun 26 |
comment |
An easy-to-state elusive combinatorial problem
And you'd have to keep track of all the infinitely many squares in the checkerboard? Seems implausible. |
Jun 26 |
comment |
An easy-to-state elusive combinatorial problem
In the case $a_1=a_2=4$ we select $n=\frac{5}{4}$ and that would render both $a_1, a_2 \in [5,7]$. The question is that no matter what $a_1$ and $a_2$ you go for, I can always find an $1\le n\le 3$ such that $4k−3≤na_1≤4k−1$ and $4l−3≤na_2≤4l−1$ where $k,l \in \mathbb{N}$. |
Jun 26 |
comment |
An easy-to-state elusive combinatorial problem
How do you calculate unions continuously with $n\in\mathbb{R}$? Geometrically you reckon? |
Jun 26 |
asked | An easy-to-state elusive combinatorial problem |
Jun 3 |
revised |
Existence of a solution to a system of Diophantine Inequalities
added 232 characters in body |
Jun 1 |
asked | Generalizing a conjecture on interval intersection |
May 30 |
comment |
Existence of a solution to a system of Diophantine Inequalities
So you say a solution always exists if $\|b_ic\| > 2*o(1)$ which implies $a > 4$? Can you elaborate this transition a bit? |