1
$\begingroup$

I need help with that question.

Suppose that G has a perfect matching M such that G\M is bipartite.

Determine χ'(G) in terms of Δ, where χ'(G) is the edge chromatic number (ie the minimum number of colours we need to colour each edge of G such that no 2 adjacent edges have the same colour) and Δ is the maximal degree of G (ie the maximum degree of its vertices). Justify your answer.

I managed to prove earlier that χ'(G) ≤ 2Δ - 1 . Maybe I can try to find a lower bound ?

Thank you.

1 Answers 1

0

Vizing's theorem states that all graphs $G$ have $\chi'(G)=\Delta$ or $\chi'(G)=\Delta+1$. The former are called class $1$ graphs and the latter are class $2$ graphs. So $\chi'(G\setminus M)=\Delta - 1$ because removing a perfect matching reduces the degree of every vertex by $1$ and all bipartite graphs are of class $1$.

When we add the matching back in, we increase the degree of every vertex by $1$ and we increase the number of required colors on the edges incident with each vertex by at most $1$. Thus $\chi'(G)\ne \Delta +1$ since the number of required colors is no more than $\Delta$, thus $\chi'(G)=\Delta$.

ALTERNATE PROOF (without Vizing's theorem): Let $\Delta(G)=\Delta$ and observe that $\Delta(G\setminus M)=\Delta -1~$ since removing a perfect matching reduces the degree of every vertex by $1$. Since $G\setminus M$ is bipartite, it is also easily seen that $\chi'(G\setminus M)=\Delta-1$. So when we add $M$ back in, there is some edge incident with a vertex of maximum degree in $G\setminus M$. This edge will force us to require an additional color, thus $\chi(G)=(\Delta-1)+1=\Delta$.

@bof: Thanks for pointing out that Vizing's theorem is unnecessary here.

  • 0
    Fine, but Vizing's theorem (which the OP evidently has not learned about yet) is overkill. All you need to answer this question is the simpler fact, that the edge chromatic number of a bipartite graph is equal to the maximum degree. – 2017-01-26
  • 0
    I agree, but without Vizing's theorem, how can we be sure that $\chi'(G)\ne\Delta-1$ since the number of colors at each vertex increases by $at~most~1$ when we add the matching back in? – 2017-01-26
  • 0
    $\chi'(G)\ge\Delta(G)$ is pretty obvious, isn't it? The edges incident with a maximum-degree vertex all have to get different colors. – 2017-01-26
  • 0
    I agree. See my edits. – 2017-01-26
  • 0
    I still find your concluding argument (starting with "So when" a little confusing. I don't see why we have to talk about the degree of some vertex increasing from $\Delta-1$ to $\Delta$ when $M$ is added back; we knew from the start that $G$ has a vertex of degree $\Delta$ and therefore $\chi'(G)\ge\Delta.$ We do have to explain why $\chi'(G)\le\chi'(G\setminus M)+1$ and that's because we can use one color for all the edges of $M.$ – 2017-01-26