Consider the weighted undirected graph with 4 vertices, where the weight of edge $\{i, j\}$ is given by the entry $W_{ij}$ in the matrix $W$. $$W = \begin{bmatrix} 0 & 2 & 8 & 5 \\ 2 & 0 & 5 & 8 \\ 8 & 5 & 0 & x \\ 5 & 8 & x & 0 \end{bmatrix}$$ The largest possible integer value of $x$, for which at least one shortest path between some pair of vertices will contain the edge with weight $x$ is _______.

Let the 4 vertices be $1, 2, 3, 4$.

The edge with weight $x$ connects vertices $3$ and $4$, i.e., $e = \{3, 4\}$.

For the edge $\{3, 4\}$ to be part of at least one shortest path between some pair of vertices, it must be part of a shortest path between its endpoints $3$ and $4$ (since all edge weights are positive, if it is used in any shortest path, the direct path between $3$ and $4$ will use it at least when $x \le \text{dist}(3, 4)$ via alternative paths).

Let us compute all possible alternative paths between vertex $3$ and vertex $4$ not using the direct edge $\{3, 4\}$:

1. Path $3 - 1 - 4$: weight $= W_{31} + W_{14} = 8 + 5 = 13$

2. Path $3 - 2 - 4$: weight $= W_{32} + W_{24} = 5 + 8 = 13$

3. Path $3 - 2 - 1 - 4$: weight $= W_{32} + W_{21} + W_{14} = 5 + 2 + 5 = 12$

4. Path $3 - 1 - 2 - 4$: weight $= W_{31} + W_{12} + W_{24} = 8 + 2 + 8 = 18$

The shortest alternative path between vertex $3$ and vertex $4$ has weight $12$.

For edge $\{3, 4\}$ to be part of at least one shortest path between $3$ and $4$, its weight $x$ must satisfy:

When $x = 12$, the direct edge $\{3, 4\}$ provides a shortest path of length $12$ (tied with $3-2-1-4$). If $x > 12$, the path $3-2-1-4$ is strictly shorter, so the edge of weight $x$ would not belong to any shortest path.

Thus, the largest possible integer value of $x$ is 12.