notes-100415 - MATH 682 Notes Combinatorics and Graph...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 682 Notes Combinatorics and Graph Theory II 1 Advanced Chromatic Properties 1.1 Color-Criticality When we were proving particular results in chromatic number, such as the five-color theorem, we frequently assumed we were looking at a minimal example of some species of graph. Very often, the minimal examples of a problem have exploitable structures we can make better use of. Such graphs are called critical : Definition 1. A graph G without isolated vertices is color-critical if ( G- e ) < ( G ) for any e E ( G ). If ( G ) = k , this property may be further described as k-criticality . We know a few color-critical graphs off the top of our heads: K n will be n-critical, since removal of any edge allows coloring with n- 1 colors, and C 2 n +1 will be 3-critical, since removing any edge admits a 2-coloring. In fact, every graph has a k-critical core, which in many cases is not a cycle or a clique. Proposition 1. If ( G ) = k , then G has a k-critical subgraph. Proof. Let us prove this by induction on k G k ; for the base case, note that k G k = 1 corresponds uniquely to a 2-critical graph. For larger G , one of two things is true: either G is k-critical, in which case it is its own k- critical subgraph, or there is an edge e such that ( G- e ) ( G ). Since any coloring of G- e is a proper coloring of G , we know that this nonstrict inequality is in fact an equality; that is, ( G- e ) = ( G ) = k . Then, by the inductive hypothesis, since k G- e k < k G k and ( G- e ) = k , G- e has a k-critical subgraph, which is in turn a k-critical subgraph of G . Since we now know k-critical graphs are all over the place, we might begin to wonder about their structure, since they are exemplars of the necessary substructures for a graph to require k colors. Proposition 2. If G is k-critical, then ( G ) k- 1 . Proof. Suppose ( G ) = k and G has a vertex v with degree less than k- 1. Let e = { u,v } for some neighbor u of v . If ( G- e ) = k- 1, then it must follow that in this coloring u and v are the same color (or this would be a ( k- 1)-coloring of G as well; however, since v has no more than k- 3 neighbors in G- e , there must be two colors k- 1 not represented in its neighborhood, so any proper coloring of G- e except for v allows at least two choices of color for v , so it can always be selected to be a different color than u , inducing a ( k- 1)-coloring of G , which contradicts the given chromatic number of G . Thus, ( G- e ) must be k , so G is not k-critical. Not only is the minimum degree at each vertex necessarily close to the chromatic number of a critical graph, but in fact, the more global concept of edge-connectivity must also be dictated by the chromatic number. We might start with a quite simple observation....
View Full Document

Page1 / 8

notes-100415 - MATH 682 Notes Combinatorics and Graph...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online