proof of Vizing’s theorem (for graphs)

Proof of Vizing’s theorem (for graphs)

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

This form of stating the theorem allows us to do inductionMathworldPlanetmath on the number of edges. A graph with zero edges, for instance, certainly doesn’t need more than ρ+1 colors. Now assume (for any given ρ) 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 ρ) a graph G* with m* edges that could not be colored thus. There would then be a largest number m, boundedPlanetmathPlanetmathPlanetmathPlanetmath by 0m<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 coloringMathworldPlanetmath it after all would prove by contradictionMathworldPlanetmathPlanetmath 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 v and w1. Some terms:

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

  • Let a Kempe chainMathworldPlanetmath H(a,b) be any connected componentMathworldPlanetmathPlanetmathPlanetmathPlanetmath of the subgraphMathworldPlanetmath 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 v and at w1, use that color for the edge vw1 and we’re finished.

Else, let color a be missing at v (but present at w1), and color b1 be missing at w1 (but present at v). If the Kempe chain of colors a and b1 that terminates at v is disjoint from the one terminating at w1, swap the colors in the latter chain, then use a for edge vw1 and be done.

Else it’s the same H(a,b1) chain, in which case we proceed as follows.


Find the b1-colored edge from v, call it vw2. Steal color b1 from it and give it to edge vw1, now vw2 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 b2 that was missing at w2 (a remains the missing color at v). If v and w2 are awkward too, being in the same H(a,b2) chain


find the b2-colored edge vw3, give its color to vw2 making vw3 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 wj+1 is not some new color bj+1 but a color bi we’ve seen before (it can’t be a, that’s present at wj+1). And not only is i<j+1 but also i<j because bj is the color we just stole from wj+1.

A chain H(a,bi) runs from v to wi+1 and all its vertices except wi+1 have color bi. Now wj+1 isn’t wi+1 (because i<j) nor one of the other vertices of the chain (because it doesn’t have color bi). So wj+1 isn’t on this H(a,bi).


wk+1 does have color a so it’s on some other H(a,bi) disjoint from H(a,bi). This could be as short as a single a-colored edge or longer, it doesn’t matter.


Swap the colors in that H(a,bi). Now use a to finally color vwj+1 (merging H(a,bi) and H(a,bi) 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 entry.

Title proof of Vizing’s theorem (for graphs)
Canonical name ProofOfVizingsTheoremforGraphs
Date of creation 2013-03-22 15:10:42
Last modified on 2013-03-22 15:10:42
Owner marijke (8873)
Last modified by marijke (8873)
Numerical id 6
Author marijke (8873)
Entry type Proof
Classification msc 05C15
Related topic KempeChain