2
$\begingroup$

Let $\mathcal{L}=\{\Bbb x\in\Bbb R^n:x_1+x_2+\dots+x_n=0\}$ be a specific hyperplane.

Let a projection of $c\in\Bbb R^n$ be $p(c)=[p_1,p_2,\dots,p_n]$ where $p_i\neq c_i\implies p_i=0$.

Let $\mathcal{P}(c)$ be set of all projections of $c=[c_1,c_2,\dots,c_n]\in {\Bbb Z^n}$.

What is the probability that $\mathcal{L}\cap\mathcal{P}(c)=\emptyset$ where $c$ is picked uniformly from ${\Bbb Z^n}$ with each $|c_i|\leq T$?

Is there an expression in terms of $T$ and $n$?

This will give probability that a given sequence of integers does not have a subset that sums to $0$.

Improve this question
$\endgroup$
6
  • $\begingroup$ Your notations are confusing. $\mathcal{L}$ is a very specific hyperplane, right? Is $p$ a orthogonal projection to a coordinate subspace? What is your random variable ($c$ ?), and what is its law? $\endgroup$
    – Benoît Kloeckner
    Dec 3, 2013 at 10:48
  • $\begingroup$ Is it clear now? $\endgroup$
    – Turbo
    Dec 3, 2013 at 11:33
  • $\begingroup$ It is a bit clearer (but a uniform random point in $B_{\mathbb{Z}^n}$ will not satisfy $c_i\neq 0$ and $c_i\neq c_j$ for all $i\neq j$, I guess you think conditionally on these). A closed form expression is certainly out of question, as even a precise asymptotic expression for the cardinal of $B_{\mathbb{Z}^n}$ is an open question. I guess that you can hope for some asymptotic information. Do you really care about taking the euclidean ball, and do you realy want the restrictions on $c$? Maybe taking all coordinates of $c$ independently uniformly between $-N$ and $N$ would be more tractable. $\endgroup$
    – Benoît Kloeckner
    Dec 3, 2013 at 16:29
  • $\begingroup$ Also, a bit of motivation would be nice, it could make people more eager to help. $\endgroup$
    – Benoît Kloeckner
    Dec 3, 2013 at 16:29
  • $\begingroup$ @BenoîtKloeckner I think a cube is also fine as long as the coordinates are different for $c$. $\endgroup$
    – Turbo
    Dec 3, 2013 at 17:59

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.