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Imagine there is some set $X$ of $n$ points (different) in $\mathbb{R}^d$. The number of (linear) orders we could define for them is just $n!\approx \sqrt{2\pi n}(n/e)^n$.

Some of these orders can be defined alongside some direction: by projection on some unitary vector $v$: we take scalar products $(v\cdot x^i)_{i=1..n}$ for all these points and use the natural order for these real numbers.

It defines an interesting equivalence relation on the sphere of directions: $v\sim_X u$ iff they define the same order in $X$. Such regions should be connected (and "convex"?). Let's discard their boundaries, which are generic situations: take only directions for which projections of all points are pairwise different.

The question is what is the number of such directional orders e.g. for a $d$-dimensional cube? Upper bound? Probability distribution for points of random positions?

If all points are in one line, there are just 2 different directional orders (the regions are two half-spheres). So the main question is upper bound - denote the maximal number of directional orders as $M_{n,d}$.

Some obvious values are $M_{n,1}=2$, or for $d\geq n-1$ we have simplex: $M_{n,d}=n!$.

Is there a general formula for $M_{n,d}$? Recurrence?

Update: just made some numerics in Mathematica for $M_{n,d}$, (e.g.: "MatrixForm[Table[x = RandomReal[1, {n, d}]; CountDistinct[Table[v = RandomReal[{-1, 1}, d]; Ordering[x.v], {i, 100000}]], {n,10}, {d, 10}]]").

For $d=2$ and $n=1,\ldots,10$, we get looking certain: 1,2,6, 12, 20, 30, 42, 56, 72, 90 ... which can be guessed: $M_{n,2}=n(n-1)$ with exception of $M_{1,2}=1$. This formula suggests that we can choose e.g. first two vertices and the rest follow, but it's more complicated.

For $d=3$ we get: 1,2,6,24, 72, 172?, 352?, 642?, 1072?, 1690? (the last ones may be still increased - tested 10 sets for million directions) - the first four are factorials, but I don't see a formula for the following?

For $d=4$ we get: 1,2,6,24,120, 480, ~1404, ~3488, ~7516, ~14374.

For $d=5$ we get: 1,2,6,24,120,720,3600, ~10500, ~29000, ~62000.

For $d=6$ we get: 1,2,6,24,120,720,5040, 30240?, ~70000, ~260000.

Numerics suggest that the number varies among different sets of points with random positions, leading to further questions e.g. of characterization of sets of points maximizing the number of directional orders?

Update: Searching just below diagonal might be helpful, e.g. "adding one point to simplex": $M_{n,n-2}=(n-1)!\cdot(n-2) $ checked for $n=3,4,5,6,7$ ... (?)

Update: For cubes ($n=2^d$) and dimensions 1,2,3,4, the number of directional orders is correspondingly: 1, 8, 96, 5376.

Update: We can tell a lot about boundaries of the regions. E.g. generic $d$ points in $\mathbb{R}^d$ form unique hyperplane, and its $\pm$normal directions are two vertices of the boundary between $d!$ regions of equivalence of $\sim_X$ relation: with infinitesimal perturbation we can choose any order among these $d$ points. Analogously $d-1$ points define lines at the boundary - between $(d-1)!$ regions.

For such boundary as CW-complex, we would like to apply general Euler formula:

$$1-(-1)^d=\chi=k_0-k_1+k_2-\ldots-(-1)^d k_{d-1}$$

where $k_0$ is the number of vertices, $k_1$ edges and so on, finally allowing to calculate the number of our regions/orders as $k_{d-1}$.

In generic case, every $d$ vertices define hyperplane and so two vertices of the complex: $k_0=2{n \choose d}$.

For $k_1$ we choose $d-1$ vertices in ${n\choose d-1}$ ways - plane going through them defines 1D circle on the sphere. Projecting on two dimensional space orthogonal to such plane, these $d-1$ points become one point there, getting situation with $n-d+2$ points in 2D. Hence, the total number of arcs seems $k_1={n\choose d-1}M_{n-d+2,2}$ (?).

Analogously $k_i={n\choose d-i} M_{n-d+1+i,1+i}$ (?), what also agrees for $k_0$, and allows to get recurrence thanks to Euler formula:

$$M_{n,d}=^?(-1)^d \left((-1)^d-1 +\sum_{i=0}^{d-2} (-1)^i {n\choose d-i} M_{n-d+1+i,1+i} \right) $$

Unverified - I will probably go back to it in a few days.