Let G be a simple graph with an odd number of vertices. Prove that if G is regular of degree Δ,

Let G be a simple graph with an odd number of vertices....

Key Concepts

Regular Graph

A regular graph is one in which every vertex has the same degree, which means each vertex is connected to the same number of edges. This uniformity in structure is significant because many properties and proofs in graph theory, including edge-coloring arguments, rely on this consistent degree across all vertices.

Chromatic Index

The chromatic index of a graph is the smallest number of colors needed to color its edges such that no two adjacent edges (edges sharing a common vertex) have the same color. It is a key concept in edge-coloring problems and has a close relationship with the graph's maximum degree.

Vizing's Theorem

Vizing's Theorem is a fundamental result in graph theory that establishes that the chromatic index of any simple graph is either equal to its maximum degree (?) or to ?+1. This theorem classifies graphs into two categories (Class 1 or Class 2) and is essential for determining the required number of colors in edge-coloring problems.

Parity Argument

A parity argument involves reasoning based on the evenness or oddness of a number, such as the number of vertices or edges in a graph. In the context of a regular graph with an odd number of vertices, parity plays a crucial role in demonstrating that a perfect matching (or equal partitioning) cannot cover all vertices, which in turn implies the necessity of using more colors in an edge-coloring scheme.

Matching and Perfect Matching

A matching in a graph is a set of edges without common vertices, and a perfect matching is a matching that covers every vertex of the graph. In edge-coloring, each color class forms a matching. When a graph has an odd number of vertices, perfect matchings cannot exist, which forces at least one color class to be incomplete and necessitates the use of an additional color to achieve a proper edge-coloring.

Transcript

00:01 All right, so we want to suppose that we have a connected graph with n vertices, with n greater than one, and n minus one edges, that is one fewer edge than it has vertices.

00:10 We wish to show that g has a vertex of degree one.

00:15 Well, let's say that we have such a vertex, such a graph, so g has n vertices, n minus one edges, and we're going to assume for the sake of contradiction that every vertex is of degree greater than or equal to two.

00:42 It's a connected graph, so no vertex has degree zero, and no vertex has degree one, so they all must have degree at least two.

00:53 We're going to let x be a vertex of g, such that g without x is still connected.

01:07 A vertex is always going to exist, but it won't necessarily be every vertex.

01:16 For instance, if you imagine removing the center vertex from this graph, no longer a connected graph.

01:23 That's fine, we're going to let d, which we know is greater than or equal to two, be the degree of x.

01:32 We consider the graph now g minus x, which has n minus one nodes, vertices, and n minus one minus d edges, remember where d was the degree of x.

02:02 It also has no more than d vertices with degree.

02:16 If you imagine, for instance, you have our node here, it's connected like this, this is our node x, this connects to the rest of the graph.

02:33 By removing x and its edges, the two nodes it was connected to lost a neighbor, and if they were of degree two, may no longer be a degree two.

02:44 That'd be a degree one, but also if they had degree three, they still have to be greater than two, and that's fine.

02:54 This is cool though, because now we have our d edges...

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