0
$\begingroup$

Given $a,b\in\mathbb N$ define the set $$\chi(a,b)=\{M\in\{0,1\}^{n^a\times n^b}:\mbox{ every row of }M\mbox{ is distinct}\}.$$

Also given ${\bf{x}}=(x_1,\dots,x_{n^b})\in\mathbb Z^{n^b}$ define the quantity $$Q_{a,b}(x_1,\dots,x_{n^b})=\min\Bigg\{\|{\bf{y}}\|_\infty:\substack{{\bf{y}}^T=M{\bf{x}}^T\\ M\in\chi(a,b)\\ i\neq j\implies\mbox{ }y_i\neq y_j\\ \mbox{ }y_i-y_j=y_{i'}-y_{j'}\implies i=i'\mbox{ and }j=j'}\Bigg\}$$

If $a\leq b$ and $1\leq i\neq j\leq n\implies\mbox{ }x_i\neq x_j$ holds then is it true there is a $z\in\mathbb Z$ and a $k\in\mathbb R$ such that if $z\leq x_{min}=\min_{i\in\{1,\dots,n^b\}}x_i$ holds then we have $$Q_{a,b}(x_1,\dots,x_{n^b})\leq x_{min}^k?$$

Notes: We have $n^b$ variables $x_1,\dots,x_{n^b}$ at some $b\geq1$ such that each of $x_1,\dots,x_{n^b}$ is distinct and $z\leq\min_{i\in\{1,\dots,n^b\}}x_i$. Take the vector $$y=Mx$$ where $M$ is a matrix such that every row is distinct. If every row and pairwise difference of every row of $y=Mx$ is also distinct then is the smallest $\|{\bf{y}}\|_\infty$ that is possible always bound by a polynomial in minimum value of $(x_1,\dots,x_{n^a})$ that is valid over every $A$?

$\endgroup$
6
  • $\begingroup$ What does "minimal" in "is to be assigned a distinct minimal non-negative integer" mean? $\endgroup$ Commented Dec 2, 2019 at 13:08
  • $\begingroup$ Each $x_i$ is distinct and as low as possible. $\endgroup$
    – VS.
    Commented Dec 3, 2019 at 6:49
  • $\begingroup$ Minimal when others are arbitrary? $\endgroup$ Commented Dec 3, 2019 at 7:02
  • $\begingroup$ Well $A$ is not fixed and so minimum over all possible $A$ at every $n$ and $b\geq1$. $\endgroup$
    – VS.
    Commented Dec 3, 2019 at 7:04
  • $\begingroup$ It is all very strange. If you take minimum also by $A$, then any $x_i$ is zero for sure. $\endgroup$ Commented Dec 3, 2019 at 7:35

0

You must log in to answer this question.

Browse other questions tagged .