Covering a uniform hypergraph with complete $r$-partite hypergraphs In combinatorial terms, I was wondering how many complete $r$-partite $k$-uniform hypergraphs are needed to cover the edges of the complete $n$-vertex $k$-uniform hypergraph $\binom{[n]}{k}$.  An $r$-partite $k$-uniform hypergraph is one where the vertices are partitioned into $r$ parts, with every edge (an edge is a set of $k$ vertices) having at most $1$ vertex from each part.
In non-combinatorial terms, for $n \ge r \ge t$, what is the minimum $t$ such that we can have $n$ vectors in $[r]^t$ with the property that for any subset of $k$ of the $n$ vectors, there is some coordinate where they have pairwise-distinct entries?
We have a lower bound of $t \ge \log_r n$, since otherwise there will be a pair of vectors (or vertices) that have the same coordinates (or are always in the same part).  Random vectors give an upper bound of the form $t \le \log \binom{n}{k} / \log \left( \frac{r}{k^2} \right)$.
I am most interested in the case where $r = 2^k - 1$ and $n$ is large.  In this case the above bounds give $t = \Omega \left( \frac{\log n}{k} \right)$ and $t = O \left( \log n \right)$.  What is the correct dependence on $k$?
 A: So this problem didn't attract a whole lot of attention (yay for tumbleweeds!), but for the future reader:
I asked my PhD adviser about it, and we discovered that this problem is known in the literature as perfect hashing, and remains open.  In computer science, the preferred point of view is to think of the coordinates of the vectors as functions, so we have functions $f_1, f_2, ..., f_t : [n] \rightarrow [r]$, with the property that for every $k$-set $X \in \binom{[n]}{k}$, there is some $i$ such that $f_i$ is injective on $X$.
It would appear that the best-known bounds on the minimum possible value of $t$, as given in Jukna's "Extremal Combinatorics", are
$$\frac{\log n}{\min_{1 \le s \le k-1} g(r,s) \log \frac{r - s + 1}{k - s} } \lesssim t \lesssim \frac{(k-1) \log n}{\log \frac{1}{1 - g(r,k)}}, $$
where $g(r,k) = \frac{(r)_k}{r^k} = \prod_{j = 0}^{k-1} \left( 1 - \frac{j}{r} \right)$.
The upper bound comes from defining the functions uniformly at random.  The lower bound is more complicated than the simple pigeonhole argument I'd given above (Alon has a probabilistic argument; others use information theoretic tools).  However, when $r \approx 2^k$, this lower bound is essentially the $\log_r n$ we had earlier.
