Consider the following graph, $G$. I want to determine the location number of $G$.
From the source I am using to learn graph theory, an ordered set $S=\{w_1, w_2, ..., w_k\}$ of vertices in a connected graph $G$ such that for every two vertices $u$ and $v$ of $G$, there is some vertex $w_i \in S$ whose distance to $u$ is different from its distance to $v$ is a locating set. In addition, the minimum number of vertices in a locating set is the location number of $G$.
My approach has been to test various possibilities of $S$ and see whether or not $S$ coveres the whole of $G$. For example, if I select two vertices of $G$ at random, I can define a collection or ordered pairs $(x, y)$ for every other vertex in $G$ consisting of the distance from a said vertex to both of the vertices in $S$. If any number of ordered pairs are the same, then $S$ is not a locating set. Other than this, I am not sure of any good approached for a problem of this sort, as this has not been very successful. All I have proven is that the location number is greater than 1.
That being said, any redirection to a text or reading to help with my understanding would be appreciated!
You can solve this set covering problem via integer linear programming as follows. Let $N$ be the node set, let $P=\{u\in N, v\in N: u < v\}$ be the set of node pairs, and let $d_{uv}$ be the shortest path distance from $u$ to $v$. For $(u,v)\in P$, let $N_{uv}=\{i\in N: d_{iu} \not= d_{iv}\}$. For $i\in N$, let binary decision variable $x_i$ indicate whether $i\in S$. The problem is to minimize $$|S|=\sum_{i\in N} x_i$$ subject to linear constraints $$\sum_{i\in N_{uv}} x_i \ge 1 \quad \text{for all $(u,v)\in P$}.$$ These constraints force at least one $i\in S \cap N_{uv}$ for each node pair $(u,v)$.
For your sample graph, if you number the nodes from left to right and top to bottom so that the top left node is $2$ and the bottom left node is $4$, we have $N_{2,4}=\{2,4,5,7,8,10,11,13,14,16\}$, which is the set of nodes that do not appear in the middle row. The graph has $17$ nodes, so $|P|=\binom{17}{2}=136$, and there are $136$ constraints. For example, the constraint for $(u,v)=(2,4)$ is $$x_2+x_4+x_5+x_7+x_8+x_{10}+x_{11}+x_{13}+x_{14}+x_{16} \ge 1.$$ The minimum $|S|$ turns out to be $2$, and one optimal solution is $x_1=x_{10}=1$, with all other $x_i=0$. There are $20$ such optimal solutions, each consisting of either node $1$ or node $17$, together with any node from the top or bottom row.