# Module Independence Graphs

Given $n, k \in \mathbb{N}$, we create a set $S \subseteq \mathbb{Z}_m^k$ by treating it as an $n\times k$ grid and choosing each $S_{ij}$ by sampling from $\mathbb{Z}_m$, such that $P(S_{ij} = z) = p_z$. We'll define the occupant count $o$ to be the total number of non-zero elements from every vector, and we'll call the quantity $o/nk = 1 - \alpha$. We can think of $\alpha$ as the probability, if we were to probe our grid at random, that we would find a zero. I want to constrain my sets to some known $\alpha$ and then have $p_{z\neq 0} = p_{0\neq x\neq z}$. Clearly if $\forall x, y \in \mathbb{Z}_m, p_z = p_x$, this condition is met, but more generally, we can reason that

$\sum_{z \neq 0}p_z + \alpha$ = 1\sum_{z \neq 0}p_z = (n - 1)p_z(n - 1)p_z + \alpha = 1p_z = \frac{1 - \alpha}{n-1}$Then I can construct a graph$G$where$V = S$and the edge relation is defined as:$E = \{ (x, y) \mid \langle x,y\rangle = 0 \}$What I want to know is if we choose uniformly$x, y \in S$, what is$P((x, y) \in E)$? A question I'd be quite willing to hear answered as well is the case when$S = \mathbb{Z}_n^k$. I suspect that this probability will rise with$n$'s compositeness where$k > n$. It'd be curious if you could compute the number. • What is m in the above question, and how does it relate to n? Mar 23, 2017 at 18:23 • That was a screwup on my part, there is only$n\$. I'll fix it when I get back to a computer or someone else could. Mar 23, 2017 at 19:57