Which one of the following options is then true?

The problem asks whether a Hamiltonian cycle (a closed loop visiting every vertex exactly once) exists in graphs (i) and (ii).

1. Graph (i) is a $4 \times 4$ grid graph with $16$ vertices.

   • A grid graph $P_m \times P_n$ has a Hamiltonian cycle if and only if $m, n \ge 2$ and the total number of vertices $m \times n$ is even.

   • Since $4 \times 4 = 16$ is even, a Hamiltonian cycle exists (for example, traversing the perimeter and snaking through the inner rows). Thus, such a trip is possible for (i).

2. Graph (ii) has $5$ vertices:

   • Let the outer four vertices be Top-Left ($TL$), Top-Right ($TR$), Bottom ($B$), and Bottom-Left ($BL$), and the inner vertex be $I$.

   • Notice that vertex $BL$ has degree $2$, connected only to $TL$ and $B$.

   • Similarly, vertex $I$ has degree $2$, connected only to $TL$ and $B$.

   • In any Hamiltonian cycle, all vertices must have exactly degree $2$ within the cycle. Therefore, the cycle must include both incident edges of $BL$ (edges $(TL, BL)$ and $(BL, B)$) and both incident edges of $I$ (edges $(TL, I)$ and $(I, B)$).

   • This forces a smaller cycle $TL - BL - B - I - TL$ of length $4$, which visits only $4$ vertices and prematurely closes without visiting $TR$.

   • Alternatively, vertex $TL$ would require degree $4$ in the cycle to connect to $TR$ as well, which is impossible. Hence, no Hamiltonian cycle exists, and such a trip is not possible for (ii).

Therefore, the correct statement is that such a trip is possible for (i), but not for (ii) (Option A).