Consider a more general version of the problem: a connected graph of \(n\) vertices/rooms. Remove edges until the graph is a tree, that is, until all cycles are removed. By König’s theorem a bipartition of the graph exists and putting custodians in the smaller of the two groups solves the problem. Therefore at most \(\lfloor\frac{n}{2}\rfloor\) custodians are necessary. Constructing an example showing this is the optimal bound is trivial - for instance, consider a straight line of rooms.

Does one move of a chess queen mean only the adjacent rooms in the line of sight of the Custodian, or all the rooms in the line of sight of the Custodian? If it's the latter, then the example of answer by @above is wrong since a straight line of rooms can be covered by a single Custodian!

I’m not entirely sure what I was thinking in my previous post, but here’s a (hopefully) fixed solution. I'll assume "adjacent room" to mean adjacent by side. Claim: The minimal number of custodians sufficient for any gallery of \(n>2\) rooms is \(\left \lfloor{ \frac{n}{3}}\right \rfloor\). Proof: Let’s start with proving \(\left \lfloor{ \frac{n}{3}}\right \rfloor\) is necessary. Indeed, consider the family of galleries shown in the attached picture - given \(n\), the gallery consists of the rooms enumerated \(1,2,\dots,n.\) Note each of the blue rooms requires a queen in its column, so our bound is achieved. Now let's prove \(\left \lfloor{ \frac{n}{3}}\right \rfloor\) is sufficient. Observe first that a chess queen can move to any square within two adjacencies of its own. Consider now the largest anticlique of rooms in the gallery - that is, the largest set \(A_1\) of pairwise non-adjacent rooms. Note any room not in \(A_1\) is adjacent to a room in \(A_1\) - otherwise \(A_1\) wouldn't be of maximal size. And of course any room in $A_1$ is adjacent to a room not in $A_1,$ as it must have some neighbor. Further consider the set of rooms not in \(A_1\). Call the largest anticlique of rooms in this set \(A_2\), and the remaining rooms \(A_3\). By the same argument as before, any room in $A_3$ is adjacent to a room in $A_2$, as otherwise $A_2$ wouldn't be the largest anticlique. So, to recap: -any room in $A_1$ is adjacent to a room in $A_2$ or $A_3$, -any room in $A_2$ or $A_3$ is adjacent to a room in $A_1$, -any room in $A_3$ is adjacent to a room in $A_2$. So any room is within two adjacencies of rooms in either of the two other sets. But this means that putting queens in all rooms of $A_i$ guarantees the entire gallery is watched by the observation above, regardless of $i$. Choose the $A_i$ with fewest rooms. $\square$

Attachments:

First post! As the graph is connected horizontally or vertically, we will consider the graph interpretation with edges connecting adjacent vertices. As the graph is connected, a natural first step would be to consider its spanning tree. Notice that a custodian can travel to all vertices with distance $\leq 2$ to it. We will prove that the number of custodians needed is $\lfloor \frac{n}{3} \rfloor$. Use the above config for necessity. The main idea is by removing a "middle" edge, it's possible to create two connected components of $\textit{not miniscule}$ size ($\geq 3$) -- so that we can induct downwards easily. We will divide into two cases. 1. If there exists a vertex $v$ and edge $e$ incident to $v$ so that two connected components $C_1$ and $C_2$ are formed so that $\min(|V(C_1)|,|V(C_2)| \geq 3$. This implies $|V(C_i)| = c_i \geq 3$ ($i \in \{1,2\}$), and $n = c_1+c_2$. We induct strongly in $n$ downwards; so that we can cover $c_i$ vertices with $\lfloor \frac{c_i}{3} \rfloor$ ($i \in \{1,2\}$) as $3 \leq |V(C_i)| < n$. By adjoining the custodians in both components, we can cover $n$ vertices with $\lfloor \frac{c_1}{3} \rfloor$ + $\lfloor \frac{c_2}{3} \rfloor$ $\leq$ $\lfloor \frac{n}{3} \rfloor$. 2. If there does not exist a vertex and edge with above conditions. $\textbf{Claim.}$ There does not exist a path of length 5 or more. $\textbf{Proof.}$ Let the path's edges consist of $e_1,e_2,\ldots,e_k$, and vertices consist of $v_1,v_2,\ldots,v_{k+1}$, $k\geq 5$. However, picking $v = v_3$ and $e = e_3$ satisfy the conditions in (1) -- as $C_1$ consists of at least the set $\{v_1,v_2,v_3\}$ and $C_2$ at least $\{v_4, \ldots, v_{k+1}\}$. $\blacksquare$ Now pick a longest path $\mathcal{P}$, let it be $e_1,e_2,\ldots,e_k$ with $k \leq 4$. Let its vertices be $v_1, v_2, \ldots, v_{k+1}$, with $e_i$ connecting $v_i$ and $v_{i+1}$. We will prove that we can place $\textbf{one}$ custodian on such graphs, which is clearly less than $\lfloor \frac{n}{3} \rfloor$. Case 1. $k = 1$. Then, $|V_G| = 2$, a contradiction. Case 2. $k = 2$. Then, $G$ is a star graph, and we can resolve the issue by putting a custodian anywhere. Case 3. $k = 3,4$. Now place the custodian on vertex $v_3$ regardless of $k$. Let there be a vertex $a$ with dist.($a,v_3$) $\geq 3$, and let its (unique) path to $a$ be $\mathcal{A}: a = a_0 - a_1 - \ldots - a_p = v_3$. We will prove there is a path of length 4 if $k = 3$, and a path of length 5 if $k = 4$. 3-1. Vertices of $\mathcal{A}$ and $\mathcal{P}$ are mutually exclusive, excluding $v_3$. There will exist a path of length $k+2$: $a_0 - a_1 - \ldots - a_p - v_2 - v_1$. 3-2. There exist a vertex excluding $v_3$ which belong to $\mathcal{A}$ and $\mathcal{P}$. Call it $X = a_i = v_j$ (if there are multiples pick one at random.) If $j > 3$: Extend $\mathcal{A}$ to $a_0 - a_1 - \ldots - v_3 - v_2 - v_1$, where a path of length $k+2$ is formed. If $j < 3$: For $k = 3$, extend $\mathcal{A}$ to $a_0 - a_1 - \ldots - v_3 - v_4$, creating a path of length 4, and for $k = 4$, push it beyond $v_4$ and add $v_5$ to it, creating a path of length 5. We are done.