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I'm looking to prove that any $k$-regular graph $G$ (i.e. a graph with degree $k$ for all vertices) with an odd number of points has edge-colouring number $>k$ ($\chi'(G) > k$).

With Vizing, I see that $\chi'(G) \leq k + 1$, so apparently $\chi'(G)$ will end up equaling $k+1$.

Furthermore, as $\#V$ is odd, $k$ must be even for $\#V\cdot k$ to be an even number (required to be even, since $\frac{1}{2}\cdot\#V\cdot k = \#E$.

Does anyone have any suggestions on what to try?

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  • $\begingroup$ For future readers, be sure to note that for multigraphs, Vizing's theorem does not hold, so the only result you can guarantee is that $\chi'(G) > k$. For example, a triangle with each edge replaced with $k$ parallel edges has $\chi'(G) = 3k > k+1$. $\endgroup$ – Quotable Jun 7 at 5:40
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Let $G=\langle V,E\rangle$ be a $k$-regular graph with $n=2m+1$ vertices; as you say, clearly $k=2\ell$ for some $\ell$, so $G$ has $$\frac{kn}2=\frac{2\ell(2m+1)}2=\ell(2m+1)$$ edges. Suppose that $c:E\to\{1,\dots,k\}$ is a coloring of the edges of $G$. $$\frac{\ell(2m+1)}k=m+\frac12\;,$$ so there is some color that is used on at least $m+1$ edges. $G$ has only $2m+1$ vertices, so two of these edges must share a vertex, and $c$ therefore cannot be a proper coloring.

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  • $\begingroup$ Excellent! Thanks! Took some thought to grasp the last bit, but now I get it I appreciate the simplicity. $\endgroup$ – FreshmanMath Oct 14 '12 at 9:44
  • $\begingroup$ @FreshmanMath: You’re welcome. That one was fun. $\endgroup$ – Brian M. Scott Oct 14 '12 at 9:45
  • $\begingroup$ I didn't quite understand the last part, "$c$ is a coloring of edges of $G$". Why is the colour used on at least $m+1$ edges? $\endgroup$ – user59036 Mar 7 '17 at 5:33
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Another way to prove this fact is to notice that in any proper edge coloring, every set of edges that share a color must form a matching. But for any given color, the matching touches an even number of vertices, so there must be one vertex missing that color. Since that vertex has $k$ edges, all of a different color, together there must be at least $k + 1$ colors.

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