1. **State the problem:** We need to show that the graphs in problems 31 and 32 do not have a Hamiltonian circuit. A Hamiltonian circuit is a cycle that visits every vertex exactly once and returns to the starting vertex. 2. **Recall the necessary condition:** A Hamiltonian circuit must include every vertex exactly once and form a closed loop. 3. **Analyze Graph 31:** - Vertices: $\{a,b,c,d,e,f,g\}$ - The graph consists of two rhombus-like shapes connected at vertex $c$. - Check the degree of vertices and connectivity: - Vertex $c$ connects the two rhombuses. - The structure forces any path visiting all vertices to pass through $c$ twice to cover both rhombuses, which is impossible in a Hamiltonian circuit. - Therefore, no Hamiltonian circuit exists in Graph 31. 4. **Analyze Graph 32:** - Vertices: $\{a,b,c,d,e,f,g,h,i,j\}$ arranged in a grid with edges connecting adjacent vertices. - The graph is bipartite with two sets of vertices. - A necessary condition for a Hamiltonian circuit in bipartite graphs is that both sets have the same number of vertices. - Here, the bipartition sets differ in size (for example, counting vertices in a checkerboard pattern). - Hence, no Hamiltonian circuit exists. 5. **Conclusion:** Both graphs 31 and 32 fail to satisfy necessary conditions for Hamiltonian circuits, so none of them has a Hamiltonian circuit.