Math Problem Statement

When G = (V, E) is a graph, let Δ(G) denote the maximum degree in G. Prove Brooks' Theorem: If G is connected and Δ(G) = k, then χ(G) ≤ k + 1. Furthermore, equality holds if and only if (a) k = 2 and G is an odd cycle, or (b) k ≠ 2 and G = K_{k+1}.

Solution

This problem is about proving Brooks' Theorem. Let’s break it down and go through the proof.

Problem Statement:

Given a connected graph G=(V,E)G = (V, E), where:

  • Δ(G)=k\Delta(G) = k, which means the maximum degree of any vertex in the graph is kk,
  • χ(G)\chi(G) denotes the chromatic number, the minimum number of colors needed to properly color the graph.

We need to prove Brooks' Theorem:

  1. Brooks' Theorem: If GG is connected and Δ(G)=k\Delta(G) = k, then: χ(G)k+1\chi(G) \leq k + 1 However, equality holds only if:
    • k=2k = 2 and GG is an odd cycle, or
    • G=Kk+1G = K_{k+1}, the complete graph on k+1k + 1 vertices.

Strategy for Proof:

  1. Case 1: G is a Complete Graph

    • If G=Kk+1G = K_{k+1}, all vertices are adjacent to each other, so every vertex has degree kk. In this case, we require k+1k + 1 colors because every vertex is connected to every other vertex, and thus no two vertices can share the same color. This is one of the cases where equality holds: χ(G)=k+1\chi(G) = k + 1.
  2. Case 2: G is an Odd Cycle

    • If GG is an odd cycle, the chromatic number is 33, and the maximum degree is 22. Therefore, χ(G)=3=k+1\chi(G) = 3 = k + 1, which is another case where equality holds.
  3. Case 3: G is not a Complete Graph or an Odd Cycle

    • For all other connected graphs, we will show that it’s possible to color the graph using at most kk colors, i.e., χ(G)k\chi(G) \leq k.

    To prove this:

    • Consider any vertex vv of degree kk.
    • Remove the vertex vv, and since the remaining graph is connected, by induction (assuming the graph has fewer vertices), it can be colored with kk colors.
    • When adding back vv, it has at most kk neighbors, so there will be at least one available color among the kk colors.
    • Thus, the graph can be colored with kk colors, meaning χ(G)k\chi(G) \leq k, unless GG is an odd cycle or a complete graph.

Conclusion:

Thus, the chromatic number χ(G)k\chi(G) \leq k, except when GG is an odd cycle or a complete graph, where equality χ(G)=k+1\chi(G) = k + 1 holds.

Do you want further details or more elaboration on any specific part of the proof?


Here are 5 questions to deepen your understanding:

  1. What is the chromatic number of a bipartite graph?
  2. How does the chromatic number change when adding a vertex to a cycle?
  3. Why does equality hold in Brooks' Theorem only for complete graphs and odd cycles?
  4. How would you approach proving the inductive step in Brooks' Theorem?
  5. What are some other famous theorems related to graph coloring?

Tip: Always check if your graph is complete or an odd cycle when applying Brooks' Theorem!

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Math Problem Analysis

Mathematical Concepts

Graph Theory
Graph Coloring
Chromatic Number
Maximum Degree of a Graph

Formulas

Δ(G): Maximum degree of a vertex in the graph G
χ(G): Chromatic number, the minimum number of colors needed to color G

Theorems

Brooks' Theorem

Suitable Grade Level

University Level (Discrete Mathematics, Graph Theory)