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An $r$-regular graph with $r\ge 2$ and a bridge in a graph separating an odd number of vertices from the rest cannot have an $r$-colour proper edge colouring. Such a colouring would partition the edges into $r$ perfect matchings. Each perfect matching must cross the odd-sided cut of a graph, so each would have to contain its unique crossing edge, the bridge in a graph. An edge cannot belong to two colour classes.

For odd $r$, every bridge in a graph in an $r$-regular graph automatically has odd-sized sides: if $S$ is either resulting connected component of a graph, double counting gives $r|S|=2|E(S)|+1$, so parity forces $|S|$ odd. Thus an odd-$r$ regular graph with a bridge in a graph is a class two regular graph.

For even $r\ge 2$, take the complete graph $K_{r+1}$. It is an $r$-regular graph on an odd number of vertices. Every matching in a graph has at most $r/2$ edges, so $r$ colours in a proper edge colouring could cover at most $r^2/2$ edges. ButTherefore ${\chi }^{\prime }(K_{r+1})>r$; Vizing's theorem gives ${\chi }^{\prime }(K_{r+1})=r+1$. This includes the graph triangle when $r=2$.

For odd $r\ge 3$, first construct a graph $H_r$ on $r+2$ vertices. Start with the complete graph onand delete the edges $va$, $vb$, and $c_1{c}_2,c_3{c}_4,\dots ,c_{r-2}{c}_{r-1}$. The pairing is possible because $r-1$ is even. In the initial complete graph, every vertex has vertex degree $r+1$. The distinguished vertex $v$ loses two edges and every other vertex loses exactly one, givingTake two disjoint copies of $H_r$, with distinguished vertices $v,v^{\prime }$, and add the edge $v{v}^{\prime }$. The resulting simple graph $G_r$ is an $r$-regular graph. The added edge is the only edge between the two copies; each copy has the odd number $r+2$ of vertices.

Suppose an $r$-colour proper edge colouring existed. At each vertex, the $r$ incident edges would use all $r$ colours. Thus each colour would form a perfect matching. Any perfect matching must use $v{v}^{\prime }$: without it, each odd-sized copy would have to be covered by internal pairs, which is impossible by parity. Therefore all $r$ colour classes would have to contain the same edge $v{v}^{\prime }$, a contradiction. Hence Vizing's theorem again givesThese constructions give class two regular graphs for every $r\ge 2$. The odd-$r$ construction exhibits the bridge obstruction to regular edge colouring; $H_r$ is a connected graph because every vertex other than $a,b,v$ is adjacent to $v$, while $a,b$ are adjacent to those vertices.