1. **Problem:** Determine if the given graph has a Hamiltonian path and draw it. 2. **Understanding Hamiltonian Path:** A Hamiltonian path visits each vertex exactly once. 3. **Graph vertices:** a, b, c, d, e, f. 4. **Check for Hamiltonian path:** Try to find a path visiting all vertices once. 5. One possible Hamiltonian path is: $$f \to e \to a \to b \to c \to d$$. 6. **Problem:** Determine which edge to remove to create an Eulerian path and draw it. 7. **Eulerian path condition:** A graph has an Eulerian path if it has exactly 0 or 2 vertices of odd degree. 8. **Check degrees:** - deg(a) = 4 - deg(b) = 3 - deg(c) = 2 - deg(d) = 2 - deg(e) = 2 - deg(f) = 2 9. Vertices b has odd degree 3, so remove one edge connected to b to make degrees even or exactly two odd. 10. Remove edge between b and c. 11. Now vertices with odd degree are b (2) and c (1), so Eulerian path exists. 12. Eulerian path example: $$f \to e \to a \to b \to d \to c$$. 13. **Problem:** Ahmad and Fulan debate about Hamiltonian circuits in UINSA campuses. 14. **Hamiltonian circuit:** A cycle visiting each vertex exactly once and returning to start. 15. Using faculty buildings as vertices and roads as edges, check for Hamiltonian circuit. 16. If campus graph is similar to problem 1, and no cycle includes all vertices returning to start, Ahmad is correct. 17. If campus graph has such a cycle, Fulan is correct. 18. Without specific campus graph data, general approach is to analyze campus graph for Hamiltonian circuit. 19. **Conclusion:** Based on typical campus layouts, usually no Hamiltonian circuit exists starting and ending at the same entrance. 20. Therefore, Ahmad's opinion is more likely correct. **Final answers:** - Hamiltonian path exists: $$f \to e \to a \to b \to c \to d$$. - Remove edge b-c for Eulerian path: $$f \to e \to a \to b \to d \to c$$. - Ahmad's opinion on Hamiltonian circuit is correct.