Question

use the graph on the right to find a hamilton circuit that begins as a, b,.... (use a comma to separate vertices as needed )

Explanation:

Step1: Recall Hamilton Circuit Definition

A Hamilton circuit is a path that visits each vertex exactly once and returns to the start. The graph has vertices \( A, B, C, D, E, F, G \)? Wait, no, looking at the graph: vertices are \( A, B, C, D, E, G, F \)? Wait, the graph shows \( A, B, C, D, E, G \) and \( F \) in the center. Wait, let's list connections: \( A \) connected to \( B, C, E, F \); \( B \) connected to \( A, D, G, F \); \( C \) connected to \( A, D, E, F \); \( D \) connected to \( B, C, G, F \); \( E \) connected to \( A, C, F \); \( G \) connected to \( B, D, F \); \( F \) connected to all.

We start with \( A, B \). From \( B \), possible next: \( D, G, F \), but we need to cover all vertices. Let's try \( A, B, D \). From \( D \), next: \( C, G, F \). Let's take \( D, C \). From \( C \), next: \( E, F \). Take \( C, E \). From \( E \), next: \( F \) (but \( F \) is center, maybe later). Wait, no, let's count vertices: \( A, B, D, C, E, F, G \)? Wait, no, maybe the vertices are \( A, B, C, D, E, G, F \)? Wait, maybe the graph has 7 vertices? Wait, no, looking at the diagram: \( A \) top-left, \( B \) bottom-left, \( C \) top-right, \( D \) bottom-right, \( E \) top-center, \( G \) bottom-center, \( F \) center. So vertices: \( A, B, C, D, E, G, F \) (7 vertices). Wait, but a Hamilton circuit must visit each once. Wait, maybe I miscounted. Wait, \( A \)-\( B \), \( B \)-\( D \), \( D \)-\( C \), \( C \)-\( E \), \( E \)-\( A \)? No, that's a square with diagonals. Wait, no, the graph is a complete-like graph with center \( F \). Wait, maybe the vertices are \( A, B, C, D, E, G \) and \( F \), but maybe \( F \) is a vertex? Wait, the problem says "a Hamilton circuit", so we need to visit each vertex once and return to \( A \).

Start with \( A, B \). From \( B \), let's go to \( G \) (bottom-center). Then \( G \) to \( D \) (bottom-right). \( D \) to \( C \) (top-right). \( C \) to \( E \) (top-center). \( E \) to \( A \)? No, that skips \( F \). Wait, \( F \) is a vertex, so we must include \( F \). So start \( A, B, F \)? No, we start with \( A, B \). So \( A, B, D \) (from \( B \) to \( D \)), \( D, C \) (from \( D \) to \( C \)), \( C, E \) (from \( C \) to \( E \)), \( E, F \) (from \( E \) to \( F \)), \( F, G \) (from \( F \) to \( G \)), \( G, A \)? Wait, no, \( G \) to \( A \)? No, \( G \) is connected to \( B, D, F \), not \( A \). Wait, \( A \) is connected to \( B, C, E, F \). So \( G \) can't go to \( A \). So backtrack.

Alternative: \( A, B, G \) (from \( B \) to \( G \)). Then \( G, D \) (from \( G \) to \( D \)). \( D, C \) (from \( D \) to \( C \)). \( C, E \) (from \( C \) to \( E \)). \( E, F \) (from \( E \) to \( F \)). \( F, A \)? No, \( F \) to \( A \) is allowed, but we missed \(... \) Wait, vertices: \( A, B, G, D, C, E, F \). Let's check: \( A \) (1), \( B \) (2), \( G \) (3), \( D \) (4), \( C \) (5), \( E \) (6), \( F \) (7). Now return to \( A \). So the circuit would be \( A, B, G, D, C, E, F, A \)? Wait, but \( F \) to \( A \) is a connection. Let's verify each step:

- \( A \) to \( B \): connected.
- \( B \) to \( G \): connected.
- \( G \) to \( D \): connected.
- \( D \) to \( C \): connected.
- \( C \) to \( E \): connected.
- \( E \) to \( F \): connected.
- \( F \) to \( A \): connected.

Wait, but did we miss any vertices? \( A, B, G, D, C, E, F \) – that's 7 vertices, each visited once, then back to \( A \). But wait, maybe the graph has 6 vertices? Wait, maybe \( F \) is not a vertex? No, the diagram shows \( F \) as a vertex. Wait, maybe the problem's graph has vertices \( A, B, C, D, E, G \) (6 vertices) and \( F \) is a center, but maybe \( F \) is not a vertex? No, the problem says "vertices", so \( F \) is a vertex.

Wait, maybe a simpler path: \( A, B, D, C, E, F, G, A \). Let's check connections:

- \( A-B \): yes.
- \( B-D \): yes.
- \( D-C \): yes.
- \( C-E \): yes.
- \( E-F \): yes.
- \( F-G \): yes.
- \( G-A \): no, \( G \) is connected to \( B, D, F \), not \( A \). Oops, mistake. So \( G \) to \( B \) is connected, but we already visited \( B \). So from \( G \), after \( F \), we need to go to \( B \) or \( D \), but they are visited. Wait, maybe my vertex list is wrong. Let's re-examine the graph:

- \( A \) connected to \( B, C, E, F \)
- \( B \) connected to \( A, D, G, F \)
- \( C \) connected to \( A, D, E, F \)
- \( D \) connected to \( B, C, G, F \)
- \( E \) connected to \( A, C, F \)
- \( G \) connected to \( B, D, F \)
- \( F \) connected to \( A, B, C, D, E, G \)

So vertices: \( A, B, C, D, E, G, F \) (7 vertices). So we need a circuit of 7 edges (visiting 7 vertices, returning to start).

Start with \( A, B \). Next: \( B \) to \( D \) (since \( B \) to \( A \) is start, \( B \) to \( G \) or \( F \) or \( D \)). Let's take \( B, D \). Then \( D \) to \( C \) ( \( D \) to \( B \) visited, \( D \) to \( G \) or \( F \) or \( C \) ). \( D, C \). Then \( C \) to \( E \) ( \( C \) to \( A \) visited, \( C \) to \( D \) visited, \( C \) to \( E \) or \( F \) ). \( C, E \). Then \( E \) to \( F \) ( \( E \) to \( A \) or \( C \) visited, \( E \) to \( F \) ). \( E, F \). Then \( F \) to \( G \) ( \( F \) to \( A, B, C, D, E \) visited, \( F \) to \( G \) ). \( F, G \). Then \( G \) to \(... \) Wait, \( G \) is connected to \( B, D, F \), all visited except none? Wait, no, we have visited \( A, B, D, C, E, F, G \) – that's 7 vertices. Now from \( G \), we need to return to \( A \), but \( G \) and \( A \) are not connected. Oh no, mistake! So my path is wrong.

Alternative: Start \( A, B, G \). Then \( G, D \). Then \( D, C \). Then \( C, E \). Then \( E, F \). Then \( F, A \)? No, \( F \) to \( A \) is connected, but we missed \( D \)? Wait, no, \( A, B, G, D, C, E, F \) – that's 7 vertices. Wait, \( G \) to \( D \) is connected, \( D \) to \( C \), \( C \) to \( E \), \( E \) to \( F \), \( F \) to \( A \). Wait, \( F \) to \( A \) is connected. So the circuit: \( A, B, G, D, C, E, F, A \). Wait, but \( G \) to \( A \) is not connected, but \( F \) to \( A \) is. Wait, no, the last step is from \( F \) to \( A \), not \( G \) to \( A \). Wait, let's list the order:

1. \( A \)
2. \( B \) (from \( A \))
3. \( G \) (from \( B \))
4. \( D \) (from \( G \))
5. \( C \) (from \( D \))
6. \( E \) (from \( C \))
7. \( F \) (from \( E \))
8. \( A \) (from \( F \))

Yes! Because \( F \) is connected to \( A \). So the path is \( A, B, G, D, C, E, F, A \). Wait, but let's check each connection:

- \( A-B \): yes.
- \( B-G \): yes.
- \( G-D \): yes.
- \( D-C \): yes.
- \( C-E \): yes.
- \( E-F \): yes.
- \( F-A \): yes.

And each vertex is visited once: \( A, B, G, D, C, E, F \) (7 vertices), then back to \( A \).

Step2: Verify All Vertices Visited

Check vertices: \( A \) (1), \( B \) (2), \( G \) (3), \( D \) (4), \( C \) (5), \( E \) (6), \( F \) (7). All 7 vertices are visited, each once, and returns to \( A \). So this is a Hamilton circuit.

Answer:

\( A, B, G, D, C, E, F, A \)