Question 23

1. Graph Structure and Vertex Count:

   • Left component: $3 \times 3$ grid of $9$ vertices.
   • The bottom-right vertex of the left grid connects via an edge to a bridge path.
   • Bridge path: a bottom vertex with a degree-1 pendant vertex above it ($2$ vertices).
   • Right component: $3 \times 3$ grid of $9$ vertices.
   • Pendant vertex on the right: degree-1 vertex connected to the middle-right vertex of the right grid ($1$ vertex).
   $$\text{Total Vertices} = 9 + 2 + 9 + 1 = 21\text{ vertices}.$$

2. DFS Recursion Depth and Simple Paths:
   The maximum possible recursion depth in a DFS traversal without backtracking corresponds to the number of vertices in the longest simple path in the graph.

3. Analyzing Endpoints of a Simple Path:

   • A simple path has exactly two endpoints (start and finish).
   • The left $3 \times 3$ grid has only ONE edge connecting it to the rest of the graph (a bridge edge at its bottom-right vertex). Once a simple path enters the left grid, it cannot exit without repeating vertices. Therefore, any simple path visiting the left grid must either start or end in the left grid.
   • This leaves only ONE endpoint available for the remainder of the graph.
   • The remainder of the graph contains two distinct degree-1 pendant vertices:
     1. The middle upright pendant vertex.
     2. The rightmost pendant vertex.
   • To visit a degree-1 vertex in a simple path, that vertex must be an endpoint of the path. Since there is only one remaining endpoint, a simple path can visit at most ONE of these two degree-1 vertices.
   • Furthermore, if the path visits the middle pendant vertex as an endpoint, it cannot reach the right $3 \times 3$ grid. Thus, to maximize vertices visited, the path must skip the middle pendant vertex and continue to the right grid, ending at the rightmost pendant vertex.
   • In the right $3 \times 3$ grid, a Hamiltonian path entering at the bottom-left and exiting to the right pendant at the middle-right vertex can visit 8 of the 9 grid vertices (or all 9 if properly routed), but at least two vertices overall in the graph cannot be included in any single simple path.

4. Therefore, the longest simple path can visit at most $21 - 2 = 19$ vertices.

Hence, the maximum recursion depth is $19$.