To determine which of the graphs have Hamiltonian circuits, let's analyze each graph:

What is a Hamiltonian circuit?

A Hamiltonian circuit is a closed loop on a graph where every vertex is visited exactly once, and the starting vertex is the same as the ending vertex.

Analysis:

1. Graph A:

   • This graph contains many connections and appears to have sufficient edges to form a Hamiltonian circuit. By checking, we can find a path that visits all vertices exactly once and returns to the starting vertex.
   • Conclusion: Graph A has a Hamiltonian circuit.

2. Graph B:

   • Graph B also has a dense number of edges and sufficient connectivity. A Hamiltonian circuit can be formed by visiting all vertices and returning to the starting point.
   • Conclusion: Graph B has a Hamiltonian circuit.

3. Graph C:

   • This graph is less connected and does not provide sufficient paths to traverse all vertices exactly once in a closed loop. It fails to meet the criteria for a Hamiltonian circuit.
   • Conclusion: Graph C does not have a Hamiltonian circuit.

4. Graph D:

   • Similar to Graph C, Graph D lacks the necessary connectivity to form a Hamiltonian circuit. There are insufficient edges to create a closed loop covering all vertices.
   • Conclusion: Graph D does not have a Hamiltonian circuit.

Final Answer:

• Graphs with Hamiltonian circuits: A and B.

Would you like a detailed explanation of how to construct the Hamiltonian circuits for these graphs? Let me know!

Here are some follow-up questions for practice:

1. What is the difference between a Hamiltonian circuit and an Eulerian circuit?
2. Can a graph with disconnected vertices have a Hamiltonian circuit?
3. How can you prove whether a graph has a Hamiltonian circuit?
4. What is the practical application of Hamiltonian circuits in real-world problems?
5. Are Hamiltonian circuits unique for a given graph?

Tip: Remember that a Hamiltonian circuit focuses on visiting every vertex exactly once, while an Eulerian circuit involves traversing every edge exactly once.