Let $G$ be a graph. A subset $C \subseteq V(G)$ is a vertex cover of $G$ if for each $e \in E(G)$, $e\cap C \neq \phi$. If $C$ is minimal with respect to inclusion, then $C$ is called minimal vertex cover of $G$.
Let $G=C_{2n+1}$ be an odd cycle. What are the minimal vertex covers of $G$
Say a set of vertices is $V = \{1,2,3...,2n,2n+1\}$.
Let $\mathcal{M}$ be such a subset of $V$ that a following is valid:
a) for each pair $\{k,k+1\}$ (modulo $2n+1$) at least $1$ number is in $\mathcal{M}$ and
b) for each triple $\{k-1,k,k+1\}$ (modulo $2n+1$) at most $2$ numbers are in $\mathcal{M}$.
What can we say about $\mathcal{M}$?
Clearly from a) we see $\mathcal{M}$ is a vertex cover and from b) $\mathcal{M}$ is minimal. Say that second is not true. Then there exist vertex cover $\mathcal{M}'$ such that $\mathcal{M}' \subset \mathcal{M}$. So there exist $k\in \mathcal{M}\setminus \mathcal{M}'$. So $k-1$ and $k+1$ are in $\mathcal{M}'$ else pair $\{k-1,k\}$ or $\{k,k+1\}$ would not be covered with the verices in $\mathcal{M}'$. But then $k-1$ and $k+1$ are also in $\mathcal{M}$ which contradicts with b).
So we have explicit description on minimal vertex cover in $C_{2n+1}$.