Let $\alpha$ be an irrational number, $n\geq 1$ and

$ X_n=\lbrace (x,y) \in {\mathbb Z}^2 | |y| \leq n, \ x+y\alpha >0 \rbrace$

Now let $(x_n,y_n)$ minimize the quantity $x+y\alpha$ on $X_n$. This pair is unique because $\alpha$ is irrational. Now the problem is to deduce all the "location" of $\alpha$ from the three integers $n,x_n$ and $y_n$ only.

The two simplest cases are :

$x_n=0,y_n=1$. This is equivalent to $0 < \alpha < \frac{1}{n+1}$.

$x_n=1,y_n=-2$. This is equivalent to $\frac{1}{2}-\frac{1}{2m} < \alpha < \frac{1}{2}$ (where $m$ is the smallest odd integer $>n$).

More generally, it seems that for any triple $(n,x,y)$, the set $Y(n,x,y)$ of all irrationals $\alpha$ yielding $x_n=x,y_n=y$, is either empty or an interval $[A(n,x,y),B(n,x,y)] \setminus {\mathbb Q}$. Is that true, and are there recursive formulas to compute $A(n,x,y)$ and $B(n,x,y)$ ?

This problem is certainly related to continued fractions and best approximations in the usual sense, but I don't see how to make the connection effective.