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

### Problem Statement:
Given a connected graph \( G = (V, E) \), where:
- \( \Delta(G) = k \), which means the maximum degree of any vertex in the graph is \( k \),
- \( \chi(G) \) 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 \( G \) is connected and \( \Delta(G) = k \), then:
   \[
   \chi(G) \leq k + 1
   \]
   However, **equality holds** only if:
   - \( k = 2 \) and \( G \) is an odd cycle, or
   - \( G = K_{k+1} \), the complete graph on \( k + 1 \) vertices.

### Strategy for Proof:

1. **Case 1: G is a Complete Graph**
   - If \( G = K_{k+1} \), all vertices are adjacent to each other, so every vertex has degree \( k \). In this case, we require \( k + 1 \) 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: \( \chi(G) = k + 1 \).

2. **Case 2: G is an Odd Cycle**
   - If \( G \) is an odd cycle, the chromatic number is \( 3 \), and the maximum degree is \( 2 \). Therefore, \( \chi(G) = 3 = k + 1 \), 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 \( k \) colors, i.e., \( \chi(G) \leq k \).

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

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

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

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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!

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}.

Explore the proof of Brooks' Theorem, which relates the chromatic number χ(G) and the maximum degree Δ(G) of a connected graph G. Learn when equality holds for odd cycles and complete graphs, with detailed explanations.

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