Need some example graphs which are not Hamiltonian, i.e, does not admit any Hamiltonian cycle,  but which have Hamiltonian path. It has Hamiltonian paths between exactly 4 pair of vertices. I have identified one such group of graphs. Would like to see more such examples.[Redundant information deleted; thanks to Sadeq Dousti for pointing it out]
Non-hamiltonian Graphs with unique hamiltonian path between exactly 4 pair of vertices
graph theory;co.combinatorics
The idea of the constructions in this answer are from Niel de Beaudrap's answer.Assume that $H$ is an arbitrary graph with a unique Hamiltonian path between $v_l$ and $v_r$.Construction I.Assume that $G_l$ is a graph, and $U_l \subseteq V(G_l)$ s.t. there are exactly 2 Hamiltonian paths starting from vertices in $U_l$. Similarly for $G_r$. Attach $G_l$ to $H$ by adding edges between $v_l$ and vertices in $U_l$. Similarly attach $G_r$ to $H$ using $v_r$. The resulting graph has the required property.Construction II.Assume that there is no other Hamiltonian path in $H$ starting from $v_l$, and that $G$ is a graph, and there are exactly 4 Hamiltonian paths starting from vertices in $U \subseteq V(G)$. Attach $G$ to $H$ by adding edges between the vertices in $U$ and $v_l$. The resulting graph has the required property.An example for $G$ in the second construction is a $C_{4}$, taking two non-adjacent vertices as $U$.