proof of Vizing’s theorem (for graphs)

Proof of Vizing’s theorem (for graphs)

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

This form of stating the theorem allows us to do induction on thenumber of edges. A graph with zero edges, for instance, certainly doesn’tneed more than $\rho +1$ colors. Now assume (for any given $\rho$) thatall 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, andtry putting the edge back.

Another way of looking at this: suppose there was (for a certain $\rho$) agraph $G^{*}$ with $m^{*}$ edges that could not be colored thus. There would thenbe a largest number $m$, bounded by , 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 therewas no such $m$ and hence no such $G^{*}$.

Either way we have our $G$ where after picking an edge we can color everythingexcept that edge, and fitting it in too proves the theorem. Let the edge tofit run between v and ${\mathrm{\text{w}}}_1$. Some terms:

• With a palette of $\rho +1$ edge colors, and no more than $\rho$ edgesat any vertex, each vertex has at least one missing color.

• Let a Kempe chain $H(a,b)$ be any connected component of thesubgraph formed by all edges colored $a$ or $b$. This is either acycle of even length, or an open path terminating at two verticeswhere one of $a$ or $b$ is missing.

Step 1. If there is a color missing both at v and at ${\mathrm{\text{w}}}_1$, usethat color for the edge ${\mathrm{\text{v}}\mathrm{\text{w}}}_1$ and we’re finished.

Else, let color $a$ be missing at v (but present at ${\mathrm{\text{w}}}_1$), and color$b_1$ be missing at ${\mathrm{\text{w}}}_1$ (but present at v). If the Kempe chain ofcolors $a$ and $b_1$ that terminates at v is disjoint from the oneterminating at ${\mathrm{\text{w}}}_1$, swap the colors in the latter chain, then use$a$ for edge ${\mathrm{\text{v}}\mathrm{\text{w}}}_1$ and be done.

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

Find the $b_1$-colored edge from v, call it${\mathrm{\text{v}}\mathrm{\text{w}}}_2$. Steal color $b_1$ from it and give it to edge ${\mathrm{\text{v}}\mathrm{\text{w}}}_1$, now${\mathrm{\text{v}}\mathrm{\text{w}}}_2$ is the problem edge that needs a color.

Go back to Step 1 calling it Step 2, replacing subscripts ${}_1$ by ${}_2$in all text. This involves a color $b_2$ that was missing at ${\mathrm{\text{w}}}_2$ ($a$remains the missing color at v). If v and ${\mathrm{\text{w}}}_2$ are awkward too,being in the same $H(a,b_2)$ chain

find the $b_2$-colored edge ${\mathrm{\text{v}}\mathrm{\text{w}}}_3$, give its color to ${\mathrm{\text{v}}\mathrm{\text{w}}}_2$making ${\mathrm{\text{v}}\mathrm{\text{w}}}_3$ 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 runout of colors, in the following sense: the color missing at some ${\mathrm{\text{w}}}_{j+1}$is not some new color $b_{j+1}$ but a color $b_i$ we’ve seen before (it can’tbe $a$, that’s present at ${\mathrm{\text{w}}}_{j+1}$). And not only is but also because $b_j$ is the color we just stole from ${\mathrm{\text{w}}}_{j+1}$.

A chain $H(a,b_i)$ runs from v to ${\mathrm{\text{w}}}_{i+1}$ and all its verticesexcept ${\mathrm{\text{w}}}_{i+1}$ have color $b_i$. Now ${\mathrm{\text{w}}}_{j+1}$ isn’t ${\mathrm{\text{w}}}_{i+1}$(because ) nor one of the other vertices of the chain (becauseit doesn’t have color $b_i$). So ${\mathrm{\text{w}}}_{j+1}$ isn’t on this $H(a,b_i)$.

${\mathrm{\text{w}}}_{k+1}$ does have color $a$ so it’s on some other $H^{\prime }(a,b_i)$ disjointfrom $H(a,b_i)$. This could be as short as a single $a$-colored edge orlonger, it doesn’t matter.

Swap the colors in that $H^{\prime }(a,b_i)$. Now use $a$ to finally color${\mathrm{\text{v}}\mathrm{\text{w}}}_{j+1}$ (merging $H(a,b_i)$ and $H^{\prime }(a,b_i)$ 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 thehttp://planetmath.org/node/6930parent entry.