Consider $m$ random 0-1 vectors of length $n$. Let $L$ be the lattice spanned by them. What is the value of $m$ (as a function of $n$) for which it is true with positive probability that $L=Z^n$? More generally, let $V(L)$ be the rank of $Z^n/L$ (The volume of $L$). What is the behavior of $V(L)$ as a function of $n$ and $m$? 



The latice spanned by $m$ random 0-1 vectors of length $n$
I believe (but haven't fully checked) that you can get an upper bound of $m=cn \log^2 n$ using the second moment method. I'm including a sketched argument below. 


I will assume WLOG that $m$ is even. I will also (for now) make a parity assumption: I will assume that, modulo $2$, the sum of all $m$ vectors is equal to $(1,0,\dots,0)$. 


Consider the $\binom{m}{m/2}$ vectors 
$$v\_A := \sum\_{j \in A} v\_j - \sum\_{j \notin A} v\_j,$$
where $A$ is any subset of size $m/2$. For any given $A$, the probability $v\_A$ equals $(1,0,\dots 0)$ is 
$$\frac{\binom{m}{m/2-1}}{2^{m-1}} \left(\frac{\binom{m}{m/2}}{2^{m-1}}\right)^{n-1} = \left(\frac{4}{\pi n}+o(1)\right)^{n/2}$$
by Stirling's approximation (note that I'm dividing by $2^{m-1}$ here due to the parity assumption).


So the expected number of $v\_A$ equal to $(1,0,\dots,0)$ is (again using Stirling's approximation)
$$\frac{2^m}{\sqrt{\pi m/2}} \left(\frac{2}{\pi n}+o(1)\right)^{n/2}, $$
which tends to infinity for $m=c n \log n$ and sufficiently large $c$. We now look at the second moment. 


If $|A \cap B|=k$, then for each coordinate (except the first, which is pretty much the same), the event $v\_A(t)=v\_B(t)=0$ corresponds (after a bit of rearrangement) to the pair of events 
$$\sum\_{j \in A \cap B} v\_j (t) = \sum\_{j \in A^C \cap B^C} v\_j(t)$$
$$\sum\_{j \in A \cap B^C} v\_j (t) = \sum\_{j \in A^C \cap B} v\_j(t).$$
So the probability that both occur equals 
$$\frac{\binom{2k}{k} \binom{m-2k}{m/2-k}}{2^{m-1}}.$$
We therefore have 
$$\frac{P(v\_A=v\_B=0)}{P(v\_A=0)^2}=\left( 2^{m-1} \frac{\binom{2k}{k} \binom{m-2k}{m/2-k}}{\left(\binom{m}{m/2}\right)^2} \right)^n$$
Applying Stirling/central binomial asmyptotics again, I get that after some more algebra this becomes 
$$\left(\frac{m/4}{\sqrt{k(m/2-k)}} \left(1+O(\frac{1}{\min(k,m/2-k)})\right)\right)^n.$$


For $|k-m/4|=t\sqrt{m}$, the first fraction is $1+O\left(\frac{t^2}{m}\right)$ so for $t=o(\sqrt{\log n})$ we have 
$$\frac{P(v\_A=v\_B=0)}{P(v\_A=0)^2}= \left(1+O(\frac{t^2}{m})\right)^n = 1+o(1).$$
[The parity assumption is necessary to make this work -- otherwise the fact that $v\_A=v\_B$ modulo $2$ increases the probability by a factor of $2$ for each coordinate]. I believe (but haven't gone through the full details) that it's similarly possible to bound the tails, so by Chebyshev we will almost surely have $(1,0,0,\dots,0)$ by the time we get to $m=c n \log n$, under our parity conditioning.


By another second moment calculation, we know that any subset of size $2m$ vectors almost surely has a subset of size $m$ having the desired sum modulo $2$ (the second moment calculation's actually a lot simpler here -- for any $A \neq B$ the sums of $A$ and $B$ are independent!). So by increasing $m$ to $2cn \log n$, we can remove the parity conditioning and almost surely have a sum equal to $(1,0,\dots,0)$. Taking $\log n$ collections of this size $m$, we can almost surely hit every coordinate vector.


Effectively I lost a $\log$ in this argument when I only considered the $v\_A$ instead of more general sums, and another $\log$ in the end when I considered $\log n$ disjoint collections of vectors instead of allowing the collections to interact with each other. Both may be unncessary. 


