Breaking chocolate problem and Euler characteristic. There is the following Problem:

Given an $m \times n$ chocolate bar, where you can only break it along the gridlines and only break one piece at the time. What is the minimum amout of steps needed to break it into $1\times1$ pieces?


Answer: You always need $m \times n -1$ Steps.

Now, my gut is telling me that this is related to the Euler characteristic. If that is the case, where can i find more information as to why/how that is the case?
P.S. This is not a homework problem, just a question about the more general principles and ways of seeing it.
 A: I like your thought, there is indeed a (very simple) Euler characteristic description, although I think it turns out to be a little disappointing because it's really not much different than the "square counting" solution that is linked in the comments.
But here goes anyway.
From the assumption given, each "break" is carried out on a single piece, breaking that piece along a "fault line" which follows the grid. Let's examine how the numbers of faces, edges, and vertices change from "before" the break to "after" the break.
The number of faces is the the same before and after the break (namely $m \times n$).
Before the break, assume that the fault line has $k$ edges, it therefore also has $k+1$ vertices. After the break, each of those $k$ edges and $k+1$ vertices along the fault line has doubled, producing $2k$ edges and $2k+2$ vertices.
The numbers of vertices and edges not along the fault line is the same before and after the break.
The net change in the Euler characteristic is therefore
\begin{align*}
\chi_{\text{after}} - \chi_{\text{before}} &= (\text{vertices after} - \text{vertices before}) - (\text{edges after} - \text{edges before}) \\
  &= ((2k+2)-(k+1)) - (2k - k) \\
  &= 1
\end{align*}
That is to say: after each break, the Euler characteristic has increased by $1$. Since the Euler characteristic started at $1$ and ended at $m \times n$, there are $m \times n - 1$ breaks.
