Dividing bars of chocolate. Say you have $n$ people and $m$ bars of chocolate (of the same size) and you want to distribute the bars evenly among these $n$ people. But you're only allowed to cut a bar one time (in any way you want). Find all $m,n\ge 1$ such that you can do this.
When $m\ge n$ we can group all the bars of chocolate in in a straight line and then cut the line by a ratio of $m/n$. This way we're guaranteed to cut every bar at most once because if a bar has been cut more than one time then the distance between these two cuts is less that $1$ which impossible since $m\ge n$.
Now the problem is when $n>m$ . I saw an approach using graph theory but I don't know the theory very well. Is there any other way?
I can solve $m=n-1$ we cut the bars in ratio of $1/n$ and $(n-1)/n$. so that every person gets $(n-1)/n$
 A: I can't find a different approach, but I can explain the graph theory approach without directly referencing graph theory. See if you can follow this explanation using minimal subsets.
Given a way to distribute the $2m$ pieces of chocolate bars $1,\cdots m$ into $n$ equal portions, call $S_i$ the set of chocolate bars from which we create portion $i$.

For your $m=n-1$ example, this could correspond to sets
$$S_1=\{1\},S_2=\{2\},\cdots,S_{n-1}=\{n-1\},S_n=\{1,2,\cdots,n-1 \}$$

Now take minimal subsets of $\{S_1,\cdots,S_n \}$ that use a nonempty subset of whole chocolate bars (these minimal subsets correspond to the connected components). Since sharing a trail of chocolate bars is an equivalence relation, if two minimal subsets intersect, they are the same subset. Thus, these minimal subsets will create a partition of the portions $\{S_1,\cdots,S_n\}$.

For your $m=n-1$ example, there would just be 1 set in the partition corresponding to all $n$ portions. This is because in order to include any whole chocolate bar, we must include $S_n$, and then we have a part of every chocolate bar, so we must include the whole of every chocolate bar.

Zooming in on a particular minimal subset, call the number of portions $n'$ and the number of whole chocolate bars $m'$. Then, we know that each one of these forms a fraction $n'/n=m'/m = a/\gcd(n,m)$ of all the chocolate, where $a\in\mathbb{N}$. Thus there are at most $\gcd(n,m)$ of these minimal subsets in the partition.
Next, we show that by minimality of these subsets, $m' \geq n'-1$ (this corresponds to the fact that a connected graph has $|E| \geq |V|-1$). Now, pick a portion $S_{i_1}$. The rest of the $S_i$ in this minimal subset must share at least one chocolate bar with $S_{i_1}$, as otherwise, we could create a smaller subset by excluding $S_{i_1}$. Call one of these portions that share a chocolate bar $S_{i_2}$.
Now again by minimality, $S_{i_1}$ and $S_{i_2}$ must share another chocolate bar with the rest, since otherwise, we could create a smaller subset by excluding $S_{i_1}$ and $S_{i_2}$. Call this new portion $S_{i_3}$. We can continue this process $n'-1$ times until we end up with all $n'$ portions, so there must be at least $m'\geq n'-1$ different chocolate bars.
Adding this up over all the ($\leq \gcd(n,m)$-many) parts, we get $m\geq n - \gcd(m,n)$. Thus, for the nontrivial case of $n>m$, we have
$$m < n \leq m + \gcd(n,m)$$
and since $m+\gcd(n,m)$ is the only multiple of $\gcd(n,m)$ on this interval, we must actually have
$$n = m + \gcd(n,m)$$
These can be constructed by repeating your $n=m+1$ construction, so we can characterize all solutions as
$$\{n,m\in \mathbb{N} \text{ s.t. } m\geq n \text{ or } n = m+\gcd(n,m)  \}$$
