Solving for a set of points from projections Consider a set of $N$ points in $n$-dimensional space, i.e.
\begin{align*}
   S = \{x_1, \dots, x_N\} \subset \mathbb R^n.
\end{align*}
Let $v \in \mathbb R^n$ and consider the image set (not counting multiplicities (!))
  $$ P_v S := \{y \; | \;\exists x \in S \text{ such that } \langle v , x \rangle = y \} $$
which can be seen as a projection of the set $S=\{x_1, \dots, x_N\}$.
Consider the problem of reconstructing the set $S$ from projections $P_v S$ where $v \in V \subset \mathbb R^n$. 
My question is: Under which conditions on the family of vectors $V$ we can uniquely reconstruct the set $S=\{x_1, \dots, x_N\}$? Clearly it is necessary that $V$ contains a basis of $\mathbb R^n$. But this is not sufficient as one easily sees in simple examples (e.g. $n=2$, $N=3$). Loosely speaking, the set $V$ has to be large enough.
 A: $n+1$ generic vectors are enough; on the other hand, $n$ never suffice.
Take the canonical basis $\{e_1,\dots,e_n\}$ of $\mathbb{R}^n$, and let $E_i = P_{e_i}S$. The set $S$ lies in $E = E_1\times E_2\times\cdots\times E_n$, which is a finite set.
No set of $n$ vectors can get you further than this "discrete cube", so you can never tell the difference between -say- $S=E$ and $S = E\setminus \{p\}$ (so long as $E$ has at least 4 points); if the $n$ vectors are not linearly independent, you don't even get any discrete set, but you get a union of affine subspaces of positive dimension, so that's hopeless.
We are looking for a vector $v$ such that no (affine) hyperplane orthogonal to $v$ contains two points in $E$. The condition that a (linear) hyperplane doesn't contain a fixed linear subspace is dense and open, hence so is the condition that it doesn't contain a finite set of linear subspaces. Since $E$ is finite, so is the set of lines going through pairs of points of $E$. Therefore, any generic $v$ separates points in $E$, and $P_v$ is guaranteed to be injective on $S$; so we are able to decide which points of $E$ are actually in $S$.
On the "practical" level, I think that choosing coordinates of $v$ so that they give a set of trascendece $n$ over the subfield of $\mathbb{R}$ generated by $E_1\cup\dots\cup E_n$ is abundantly sufficient. (e.g. if all coordinates of points in $E$ are rational, and $n=2$, taking $\pi$ and $e^\pi$ should be sufficient).
