I have a cyclic set of $n$ unknown values $x_1 ... x_n$ that I am trying to obtain and a cyclic set of $n$ known values, each of which corresponds to $x_a + x_{a+1}$ with the n$^{th}$ value $y_{n,1}$ = $x_n + x_{1}$ wrapping around back to the start of the set.
This can be re-written as $n$ equations with $n$ unknowns:
$x_1 + x_2 = y_{1,2}\\x_2 + x_3 = y_{2,3}\\...\\x_{n-1} + x_n = y_{n-1,n}\\x_n + x_1 = y_{n,1}$
Given that the values of $y$ are known decimals, can the x values be obtained in any case other than $n=3$?
Apologies if my terminology is not perfect, I'm coming at this from a programming background.
Let $y_i = x_i + x_{i+1}$ where $x_{n+1} = x_1$.
If $n$ is even then you require that the sum of the odd $y_i$ is equal to the sum of the even $y_i$ and then every option for $x_1$ induces a solution that can be obtained in linear time easily by using $x_i = y_{i-1}-x_{i-1}$.
If $n$ is odd then note $x_i = (\sum\limits_{i=1}^n y_i)/2-y_2-y_4+\dots -y_{2\lfloor(n/2-1\rfloor}$ and every other value can be determined by using $x_i = y_{i-1}-x_{i-1}$.