If G is a connected undirected graph such that every vertex has degree at most 8 , and the shortest path between any two vertices has length at most 2, then what is the maximum number of vertices in G? (Hint: Try to draw the BFS tree starting with any vertex)

Let's solve this problem step by step.

### Problem Restatement
We have a connected undirected graph \( G \) where:
- Every vertex has a degree of at most 8.
- The shortest path between any two vertices is at most 2.

We need to find the maximum number of vertices in \( G \).

### Step 1: Understand the Graph Properties
Given that the shortest path between any two vertices is at most 2, it means that any two vertices in \( G \) are either directly connected or have a common neighbor.

### Step 2: Consider the BFS Tree Structure
Let's consider the BFS (Breadth-First Search) tree starting from any arbitrary vertex \( v \).

1. **Vertex \( v \) (Level 0):**
   - The vertex \( v \) has at most 8 neighbors (since the degree of any vertex is at most 8).
   
2. **Neighbors of \( v \) (Level 1):**
   - There are at most 8 vertices at this level.
   - Each of these vertices can have at most 7 additional neighbors that are not \( v \) itself (since the graph is simple, and there are no self-loops).

3. **Neighbors of Neighbors (Level 2):**
   - Each vertex at Level 1 can introduce new neighbors at Level 2.
   - However, any new vertex added to Level 2 must not be connected to any other vertex in Level 1 (otherwise, the shortest path condition would violate the "at most 2" rule for some pair).

### Step 3: Counting Maximum Vertices
To maximize the number of vertices, observe that:
- The central vertex \( v \) has at most 8 neighbors (Level 1).
- Each of these 8 neighbors can have at most 7 additional neighbors, which must all be distinct from each other and from the vertex \( v \).

The maximum number of vertices \( n \) can be calculated as follows:
\[
n = 1 + 8 + 8 \times 7
\]
- 1 is the central vertex \( v \).
- 8 are the direct neighbors of \( v \).
- 8 times 7 gives the vertices at distance 2 from \( v \), which are the neighbors of the vertices at Level 1.

So, 
\[
n = 1 + 8 + 56 = 65
\]

### Conclusion
The maximum number of vertices in the graph \( G \) is \( \boxed{65} \).

Would you like further details or have any questions? Here are some related questions to consider:

1. How would the answer change if the degree of each vertex was at most 6?
2. What if the graph had different degrees, such as some vertices having degree 4 and others 8?
3. Can this concept be extended to higher path lengths, like 3 or 4?
4. How does this problem relate to the concept of a "clique" in graph theory?
5. What role does the BFS tree play in understanding graph connectivity and distances?

**Tip:** In graph problems, drawing the structure and analyzing small cases often helps in identifying patterns and generalizing the solution.

Explore the solution to finding the maximum number of vertices in a connected undirected graph where each vertex has a degree of at most 8 and the shortest path between any two vertices is at most 2. Learn how to apply Breadth-First Search (BFS) and analyze vertex degrees to determine the graph's maximum size.

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