Rooks on cube surface Given an $n\times n\times n$ cube, each of the six faces is divided into $n^2$ squares. So there are $6n^2$ squares in total. A rook placed on a square can attack any square, including squares on the other faces, that can be reached vertically or horizontally from its square. In particular, any rook can choose one of the two directions "vertical" or "horizontal", and walk in that direction across other faces until it comes back, spanning a total of $4n$ squares in each direction.  What is the maximum number of non-attacking rooks that can be placed?
We can place $n$ rooks on the diagonal of one face -- this already leaves no square unattacked. Unlike the $2$-dimensional case, however, it is unclear if this is the best we can do.
 A: Imagine the cube is made of square-shaped rings. The rings go $3$ ways. When a rook is placed on a square, it rules one ring one way and another ring the other way. This means every rook rules two rings. To find the amount of rooks you can place, all you have to do is find how much rings there are and divide by $2$.
There will be $n$ rings going in three directions, up, horizontal and sideways to cover the whole square with rings (This is hard to describe with words, see image). This means $\frac{3n}{2} = 1.5n$, meaning that the diagonal is not the best you can do.
Here's an image of the cube covered in rings in one direction to help you visualize it. You can tell that if you put rings the other way the cube will be covered.
Edit:
Thanks to @6005 for correcting me. I realized that the rings go in 3 ways instead of 2.
A: The maximum number of rooks that can be placed is
$\left\lfloor \frac{3n}{2} \right\rfloor$.
First, we can prove that you can't do better than $\left\lfloor \frac{3n}{2} \right\rfloor$.
Consider "rings" in three directions as described by u8y7541, (each ring contains $4n$ squares). Then there are $3n$ rings, and each rook covers two rings; no other rook may be placed on those rings. So if there are $r$ rooks, then $$2r \text{ (number of rings covered)} \le 3n\text{ (number of rings total)}$$
so $r \le \frac{3n}{2}$.
Sine $r$ is an integer, $r \le \left\lfloor \frac{3n}{2} \right\rfloor$.
Now to achieve $r = \left\lfloor \frac{3n}{2} \right\rfloor$, just pick $r$ pairs of rings, such that no pair consists of two parallel (i.e. non-intersecting) rings. Convince yourself that it's always possible to pick the maximum number of pairs of non-parallel rings. Now place a rook at one of the two intersections of each pair, and we have a valid arrangement.
