What is the most ``diverse'' $k$-subset of $[0, 1]^m$? Given a non-negative integer $m$, let $\Omega_m$ denote the set of vectors $\omega = (\omega_1, \dots, \omega_m) \in [0, 1]^m$ such that $\sum_i{\omega_i} = 1$.
The set $\Omega_m$ is together with a metric $d_M$, that is the Manhattan distance between vectors, i.e., $\forall \omega, \omega' \in \Omega_m$, $d_M(\omega, \omega') = \sum_i{|\omega_i - \omega'_i|}$. Given $S \subseteq \Omega_m$, we denote $\min(d_M, S) = \min_{\omega, \omega' \in S}{d_M(\omega, \omega')}$.
I am interested in the following ``most diverse $k$-subset'' problem: given two non-negative integers $m, k$, return a set $S \subset \Omega_m$ of size $k$ that maximizes $\min(d_M, S)$ (i.e., $S$ is such that $\forall S' \subset \Omega_m$ with $|S'| = k$, we have $\min(d_M, S') \leq \min(d_M, S)$).
For example, if $m = 3$ and $k = 4$, it seems that the only possible resulting set is $S = \{(0, 0, 1), (0, 1, 0), (1, 0, 0), (1/3, 1/3, 1/3)\}$, with $\min(d_M, S) = 4/3$.
What would be the general algorithm to compute $S$, for all non-negative integers $m, k$?
 A: In the case $m=3$, I've used an SMT-solver (MathSAT 5) to find the optimal solutions for $k = 4$ to $8$.  If my programming is correct, here they are:
$k = 4$: $\min(d_M,S) = 4/3$ for $[0,0,0],[0,1,0],[1,0,0],[1/3,1/3,1/3]$:

$k = 5$: $\min(d_M,S) = 1$ for $[0,1/2,1/2],[1/2,1/2,0],[1,0,0],[1/2,0,1/2],[0,0,1]$

$k = 6$: $\min(d_M,S) = 1$ for $[1/2,0,1/2],[1,0,0],[0,0,1],[1/2,1/2,0],[0,1/2,1/2],[0,1,0]$

$k = 7$: $\min(d_M,S) = 4/5$ for $[1/5,3/5,1/5],[0,1,0],[0,2/5,3/5],[1,0,0],[3/5,11/40,1/8],[9/20,1/40,21/40],[1/20,0,19/20]$

$k= 8$: $\min(d_M,S) = 3/4$ for $[5/8,3/8,0],[5/8,0,3/8],[1/4,1/8,5/8],[0,1/2,1/2],[1/4,5/8,1/8],[0,1,0],[1,0,0],[0,0,1]$

A: To expand upon the commentary, consider placing k instances of a certain kind of region in a unit square.  These are called packing problems, with two popular instances involving circles in a square and another involving (non axis-aligned) congruent small squares in a square.  Exact solutions for small k and proofs of optimlity are elusive, but k times the volume of a region cannot exceed the volume of the unit square, and there is often a scaling factor s (and accompanying proof) that says k times the volume of a scaled down version of the region is feasible, and s is often not far from 1.
In your case you have a large simplex to pack with regions I call Manhattan spheres, which I imagine for high dimensions are generalized octahedra.  They aren't rotatable as in the above packing problem with square regions, but you can perform the volume computations just the same. This will help you figure the radius of the sphere to within the scaling factor, which leaves the problem of placement.  From an initial placement with small radius of k regions, some dynamic algorithms based on charge-repulsion (think of simultaneously blowing up k balloons in a box and watch them move around) often lead to near optimal placements.
A: It seems that if $m<<k$ ( $k$ is so larger than $m$), the minimum is between $1/n$ and $1/(n+1)$, when $n$ is a number that satisfies 
$n+m-1 \choose n$$\leq k \leq$$ n+m \choose n+1$.
