co.combinatorics - efficient way to count hamiltonian paths in a grid graph for a given pair of vertices

Here is Mathematica code that finds all the Hamiltonian paths between opposite
corners of a $5 times 5$ grid graph:

<< Combinatorica`;
n = 5;
G = GridGraph[n, n];
(* Add dangling edges to corners to force start/end vertices *)
Gplus = AddVertex[G, {0, 0}];
Gplus = AddVertex[Gplus, {n + 1, n + 1}];
Gplus = AddEdge[Gplus, {1, n^2 + 1}];
Gplus = AddEdge[Gplus, {n^2, n^2 + 2}];
ShowGraph[Gplus]
H = HamiltonianPath[Gplus, All];
Print["Number of paths=", Length[H]];
Print["Paths=", H];
Number of paths=208
Paths={{26,1,2,3,4,5,10,9,8,7,6,11,12,13,14,15,20,19,18,17,16,21,22,23,24,25,27}, [etc.]}

---

Addendum. Setting $n=7$ to compute the comparable number for a $7 times 7$ grid
returns 223,424 Hamiltonian paths between opposite corners. [5 hrs computation time on a 2.5GHz laptop.]
The first one returned is:

{50, 1, 2, 3, 4, 5, 6, 7, 14, 13, 12, 11, 10, 9, 8, 15, 16, 17, 18,
19, 20, 21, 28, 27, 26, 25, 24, 23, 22, 29, 30, 31, 32, 33, 34, 35,
42, 41, 40, 39, 38, 37, 36, 43, 44, 45, 46, 47, 48, 49, 51}