1

Prove König's theorem: edge coloring of a bipartite graph with a maximum degree of $D$ requires only $D$ colors.

enter image description here

Induction after $n$ for a fixed $D$:
Let $G_{n+1}$ which have such vertex that $\deg(w)=\max \left\{ deg(v): v \in V[G]\right\}$.
We know that $\deg(w) \le D$. We delete $w$ and have $G_n$.
By induction, we can paint the edges with $D$ colors. We are restoring $ w $. Since we had in $G_{n+1}$ for every $v$: $\deg (v) \le D$ then in $G_N$ (so without $w$) we have that for every vertex $v$ which in $G_{n+1}$ is with $w$, for $G_n$ we have $deg(v)\le D-1$.
That is why we paint each of the added edges with $1$ free of $D$ colors (because at most $D-1$ is occupied).

Is it correct?

Community
  • 1
MP3129
  • 3,215

1 Answers1

4

No, it's not.

For each edge $vw$, we have up to $D-1$ colors that are already used on existing edges out of $v$ (in $G_n$). So you're right that there is some color we can give $vw$ that does not conflict with other edges out of $v$.

But this is not enough, by itself. We have many such edges $v_1w, \dots, v_Dw$ out of $w$, and even if we can give them colors that don't conflict at $v_1, \dots, v_D$, there's no reason to think that they will all be different. In all likelihood, this strategy will give several of these edges the same color, and then you have a conflict at $w$.

A "greedy" proof of König's theorem will not work. Instead, induct on $D$, proving the auxiliary theorem: in a bipartite graph with maximum degree $D$, there is a matching saturating all vertices of degree $D$. (Make this matching one of your color classes, and repeat.) This problem, in turn, can be reduced to the problem of finding a perfect matching in a $D$-regular graph.

Misha Lavrov
  • 159,700