Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show that every graph $G$, such that $|G| \ge 2$ has at least two vertices which are not its cut-vertices.

share|cite|improve this question

Let $P$ be a maximal path in $G$. I claim that the end points of $P$ are not cut vertices.

Suppose that an end point $v$ of $P$ was a cut vertex. Let $G$ be separated into $G_1,\ G_2,\ \cdots,\ G_k$. It follows that any path from one component to another must pass through $v$ and namely such a path does not end on $v$ and therefore cannot be $P$. Therefore $P$ is contained entirely within some $G_i\cup\{v\}$. But this contradicts the fact that $P$ is maximal for there exists at least one vertex in $G_j$ for $i\neq j$ which connects to $v$ and extends $P$. Therefore $v$ must not be a cut vertex.

share|cite|improve this answer

Firstly, this question only makes sense if $G$ is connected. So, assuming that $G$ is connected...

Let $H$ be a spanning tree of $G$ and let $l_1$ and $l_2$ be two leaf nodes of $H$. Then $G \setminus \{l_1,l_2\}$ is connected.

We need to check:

  • $G$ indeed has a spanning tree (since it's connected).

  • $H$ has two leaf nodes (in fact, all trees on $\geq 2$ vertices have $\geq 2$ leaf nodes; this can be shown by induction).

  • $G \setminus \{l_1,l_2\}$ is indeed connected. This follows since $H \setminus \{l_1,l_2\}$ is a connected spanning subgraph of $G$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.