Chromatic index of d-regular 1-connected graph Let $G$ be a 1-vertex-connected $d$-regular graph with $d\geq2$. Find the chromatic index $\chi'(G)$.
Brooks' theorem give $\Delta(G)\leq\chi'(G)\leq\Delta(G)+1$.
I'm trying to look at $G'-v$, with $v$ being a disconnecting vertex in $V$, and defining $c_i$ to be a coloring of one of its connected components, but so far I haven't been able to rule out $\Delta(G)$ or $\Delta(G)+1$.  
 A: Assume by contradiction there is a $d$ coloring of the edges of $G$, and $v$ a disconnecting vertex. Let $A$ be a connected component of $G-v$, and $B$ be the union of all the other connected components. At least one of the edges incident to $v$ goes to $A$ (and the same for $B$), so there is an edge of color 1 not going from $v$ to $A$ and a color 2 not going to $B$ from $v$. Each vertex in $A$ is incident to exactly one edge of color 1 (because the coloring is the size of the degree), and the edges from $A$ all go into $A$ from connectivity. So if we take the graph induced by $A$, its subgraph of edges of color 1 has a sum of degrees equal to $|A|$, meaning $|A|$ needs to be even. However, in the graph induced by $A\cup {v}$, there is an edge incident to $v$ that is of color 2 (because that edge doesn't go to $B$), and of course an edge colored 2 incident to all the vertices of $A$. Therefore by the same argument $A\cup {v}$ must be even, which leads to a contradiction. Therefore by Vizing the chromatic index is $d+1$.
EDIT: By the way, the relevant theorem is Vizing's, not Brooks'.
