Proof of Vizing's theorem (for graphs)

We must prove, for any integer $\rho$ , the following: if $G$ is a graph all of whose vertices have valency $\le \rho$ , then its edges can be colored (with adjacent edges receiving different colors) in no more than $\rho+1$ colors.

This form of stating the theorem allows us to do induction on the number of edges. A graph with zero edges, for instance, certainly doesn't need more than $\rho+1$ colors. Now assume (for any given $\rho$ ) that all graphs with $m$ edges can be thus colored. For any graph $G$ with $m+1$ edges we choose an edge to remove, color the remaining $m$ edges, and try putting the edge back.

Another way of looking at this: suppose there was (for a certain $\rho$ ) a graph $G^*$ with $m^*$ edges that could not be colored thus. There would then be a largest number $m$ , bounded by $0\le m\lt m^*$ , such that all graphs with $m$ edges can still be colored thus. Then there would also be a $G$ with $m+1$ edges that cannot. Choose one of its edges and proceed as above. This time, succeeding in coloring it after all would prove by contradiction that there was no such $m$ and hence no such $G^*$ .

Either way we have our $G$ where after picking an edge we can color everything except that edge, and fitting it in too proves the theorem. Let the edge to fit run between $\0v$ and $\0w_1$ . Some terms:

• With a palette of $\rho+1$ edge colors, and no more than $\rho$ edges at any vertex, each vertex has at least one missing color.
• Let a Kempe chain $H(a,b)$ be any connected component of the subgraph formed by all edges colored $a$ or $b$ . This is either a cycle of even length, or an open path terminating at two vertices where one of $a$ or $b$ is missing.

Step 1. If there is a color missing both at $\0v$ and at $\0w_1$ , use that color for the edge $\0v\0w_1$ and we're finished.

Else, let color $a$ be missing at $\0v$ (but present at $\0w_1$ ), and color $b_1$ be missing at $\0w_1$ (but present at $\0v$ ). If the Kempe chain of colors $a$ and $b_1$ that terminates at $\0v$ is disjoint from the one terminating at $\0w_1$ , swap the colors in the latter chain, then use $a$ for edge $\0v\0w_1$ and be done.

Else it's the same $H(a,b_1)$ chain, in which case we proceed as follows.

Find the -colored edge from , call it . Steal color from it and give it to edge , now is the problem edge that needs a color.

Go back to Step 1 calling it Step 2, replacing subscripts by in all text. This involves a color that was missing at ( remains the missing color at ). If and are awkward too, being in the same chain

find the -colored edge , give its color to making the problem:

Next time round (Step 3) increase the subscripts again. And if still necessary, transfer the color again.

Sooner or later (if we don't get to bail out at one of the Steps) we will run out of colors, in the following sense: the color missing at some is not some new color but a color we've seen before (it can't be , that's present at ). And not only is but also because is the color we just stole from .

A chain runs from to and all its vertices except have color . Now isn't (because ) nor one of the other vertices of the chain (because it doesn't have color ). So isn't on this .

does have color so it's on some other disjoint from . This could be as short as a single -colored edge or longer, it doesn't matter.

Swap the colors in that . Now use to finally color (merging and in the process).

This is in essence the proof usually given, as taught to students and found for instance in references [FW77] and [Wil02] of the parent entry.