Clever proof for showing that if a graph G is critically k-colorable then &#x03B4;<!-- δ --> (

Mohamed Mooney 2022-06-26 Answered
Clever proof for showing that if a graph G is critically k-colorable then δ ( G ) k 1
While reading for my graph theory class, I came across a short - yet curious - proof for the following theorem: if a graph G is critically k−colorable then δ ( G ) k 1. Here is the proof to the claim:
Suppose (for a contradiction) that G is k-critical and that v V ( G ) satisfies deg ( G ) < k 1. Then G v has a ( k 1 ) coloring, and this coloring extends to a k 1- coloring of G. This yields a contradiction.
Everything in this proof makes sense besides one particular item: in the second line, what does the author mean by extending a coloring from a subgraph of G to the whole graph? In addition, why does G v have a ( k 1 ) coloring that extends to all of G (if that makes any sense)? I understand this may seem like an easy Google search but to be honest I can't find anything helpful and figured someone could provide some insight.
Note that this is not a homework question but simply for going beyond what I am learning in class.
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Answers (1)

Cristian Hamilton
Answered 2022-06-27 Author has 23 answers
Step 1
In the second line, what does the author mean by extending a coloring from a subgraph of G to the whole graph?
Recall that a coloring formally is a function f : V ( G ) { 1 , 2 , , k } for some integer k. So the notion of extension is basically the same as it would be for functions in other cases (extending a function from some substructure to a larger one).
In addition, why does G v have a ( k 1 ) coloring that extends to all of G (if that makes any sense)?
We have a coloring f : V ( G v ) { 1 , 2 , , k 1 } from k-criticality. We then define a coloring f ~ : V ( G ) { 1 , 2 , , k 1 } where f ~ = f on V ( G v ). To complete the extension, we need to define f ~ ( v ).
Step 2
Notice that deg G ( v ) < k 1 by hypothesis (we can choose such a v where deg G ( v ) = δ ( v ) < k 1 in this approach by contradiction). Hence, there are at most k 2 nodes in V(G) adjacent to v, which does not exhaust all of the possible k 1 colors in f , f ~ . Consequently, let f ~ ( v ) be any of the colors not taken up by a neighbor of v.
This ensures that f ~ is indeed a k 1 vertex-coloring. However, this contradicts the k-criticality of G, in the sense that deleting v does not change the chromatic number of G.

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