This is a new question about Critical graph, because the previous question about it is became too big.
Let me remind. In this context the Critical graph is: graph $G = (V, E)$ is Critical, if $G$ is biconnected and $ \forall e \in E \Rightarrow G-e $ contains a point of articulation.
$G-e$ is equal to if we remove this edge from $G$.
How to prove the next properties:
- $\forall G$, where $G$ is Critical $\Rightarrow$ $\exists v \in V$ and $deg(v) = 2$. For this properties respected @EuYu gave the proof, but I can not fully understand all of it right there (regarding the first item of the proof). Maybe somebody or the author can explain me some moments of this or show another proof.
- Is it true or not (If it is true that prove otherwise give a counterexample). If $G$ is Critical graph $\Rightarrow \forall w \in V$ , $ deg(w) \ge 3$ , $\exists u \in V$ , $deg(u) = 2$, that edge $(u, w) \in E$.
- If $G$ is Critical, $|V| \ge 4$ $\Rightarrow$ $|E| \le 2 \times |V| - 4$.
Respected @EuYu also gave the incredible book which outlines some of the above properties (and other) for Critical graph. The book calls Critical graphs as Minimally 2-connected graphs.
In this book there is the proof about first property (more powerful proof "Theorem 3.6. Every $x - y$ path contains a vertex of degree 2"). But I didn't understand it (there are contradictions for me). If someone will understand this, please tell me!
In this book there is the theorem about third property ("Theorem 3.10. A minimally 2-connected graph of order $n$ $\ge$ $4$ has size at most $2 \times n - 4$. Given $n \ge 4$ the complete bipartite graph $K(2, n - 2)$ is the only minimally 2-connected graph of order $n$ and size $2 \times n - 4$."). But why we can't have a Critical graph with a large number of edges with $n$ vertices?
Probably in this book there is something helpfull for second property.
Please help with something. Thanks anyway.