Deriving numbers from sums Let $x_1,\dots,x_n$ be not necessarily distinct real numbers. Form a multiset $A$ by taking the $2^n-1$ sums of nonempty subsets of the numbers. For example, if we have $(x_1,x_2,x_3)=(1,2,-3)$, then by taking the seven sums, $A=\{-3,-2,-1,0,1,2,3\}$. The set $(x_1,x_2,x_3)=(-1,-2,3)$ generates the same $A$.
If we are given a multiset $A$ of size $2^n-1$, can we tell whether it is generated in this way by some $n$ numbers? Obviously the $n$ numbers are present in $A$, so one way is to try all $\binom{2^n-1}{n}$ combinations of $n$ numbers, but this is large even for modest values of $n$. Is there a more efficient way?
 A: Let $A=\{a_1,a_2\dotsb a_{2^n-1}\}$. WLOG $a_1 = x_1$, $a_2 = x_2,\dotsb a_n = x_n$, $a_{n+1}=x_1+x_2$, ...
Define the elementary symmetric polynomials 
$$\sigma_1= a_1 + a_2 \dotsb a_{2^n-1}$$
$$\sigma_2= \sum_i\sum_{j>i} a_ia_j = a_1 a_2 + a_1 a_3 \dotsb$$
etc
These polynomials can be expressed as polynomials in $x_1,x_2\dotsb x_n$ instead, and laboriously obtained by substituting. For n = 2:
$$\sigma_1 = 2(x_1+x_2)$$
$$\sigma_2 = x_1 x_2(x_1+x_2)$$
For n=3:
$$\sigma_1 = 4(x_1 + x_2 + x_3)$$
$$\sigma_2 = 6(x_1 + x_2 + x_3)^2+2(x_1x_2+x_1x_3+x_2x_3)$$
$$\sigma_3 = 4(x_1+x_2+x_3)^3+6(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)$$
$$\sigma_7 = x_1x_2x_3 (x_1+x_2+x_3)((x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)-x_1x_2x_3)$$
I'm not going to actually use $\sigma_3$ but it might come in handy for some other calculations. These are tiresome to find, so I didn't compute the other ones.
Since $\sigma_i$ is expressed as a function of the numbers we have, we can compute those and use them to find the elementary symmetric polynomials of $(x_1,x_2\dotsb x_n)$, so let $x_1 + x_2 + x_3 = a$, $x_1x_2+x_1x_3+x_2x_3 = b$ and $x_1x_2x_3 = c$, then solving this system is equivalent to finding the roots of $t^3-at^2+bt-c=0$, which can hopefully be found with the rational root test.
In your example, $\sigma_1=0$, $\sigma_2=-14$, $\sigma_7=0$, then $$x_1+x_2+x_3=0$$ $$x_1x_2+x_1x_3+x_2x_3=-7$$ $$x_1x_2x_3= ?$$ This is a degenerate case because $\sigma_1 = 0$, but the two valid solutions can be found by examining the subsets with sum 0. Also this leads to insight on the uniqueness of the solutions, if $\sigma_1\not=0$ then all of $x_1 + x_2 + x_3$, $x_1x_2+x_1x_3+x_2x_3$, $x_1x_2x_3$ are uniquely determined which wields a single solution up to permutation.
For larger n expressing $\sigma_1,\sigma_2\dotsb\sigma_n$ as a function of $x_1,x_2\dotsb x_n$ gets much more complex, but disregarding that the problem can be probably solved the same way.
