Mounting A Defense Against Sheldon

Assume there are n point on each plane. Each point P can dominate 36-2=34 squares on chessboard. Two points on each plane have at least two common squares. Hence n points have at least 2 n!/[2! (n-2)!] common squares. On 12 planes there have 12n points that dominate 34 12 n-12 2 n*(n-1). Number of rooks that dominate an 12 x 12 x 12 chessboard, satisfy the equation: $34 \times 12n-12 \times 2 \times n \times (n-1)=12^3 \Rightarrow n=\boxed{72}$

There are 6 faces on a cube. Each face requires 12 rooks to dominate the face. 6 x 12 = 72!

However,you can also generalize the problem using extreme principle.I am giving such a solution.

We place $n$ layers of size $n$ x $n$ x $1$ over an $n$ x $n$ square and number them $1,2,...,n$

Suppose $R$ rooks are so placed on the $n^{3}$ cubes of the board such that they dominate all cubes.We choose a layer $L$ that contains the minimum number of rooks. We may assume it is parallel to $x_{1}x_{2}$ -plane.Suppose $L$ contains t rooks.Suppose these t rooks dominate $t_{1}$ rows in the $x_{1}$ - direction and $t_{2}$ rows in the $x_{2}$ -direction.We may further assume that $t_{1}≥t_{2}$ .Obviously $t≥t_{1}$ and $t≥t_{2}$ .In the layer $L$ , these rooks fail to dominate ( $n-t_{1})(n-t_{2})$ cubes,which must be dominated in the $x_{3}$ -direction.We consider all $n$ layers parallel to the $x_{1}x_{3}$ -plane.In $n-t_{1}$ of these not containing a rook from $L$ , there must be $(n-t_{1})(n-t_{2})$ rooks.In each of the remaining $t_{1}$ layers there are at least $t$ rooks(by the choice of $t$ ).Hence,we have

$R≥(n-t_{1})(n-t_{2})+tt_{1}≥(n-t_{1})^{2}+t_{1}^{2}=\frac{n^{2}}{2}+\frac{(2t_{1}-n)^{2}}{2}$

The right hand side assumes its minimum $\frac{n^{2}}{2}$ if n is even and $\frac{(n^{2}+1)}{2}$ if n is odd

(This is a constructive solution: the proof that you need at least 72 rooks is given in another, very neat, solution on this thread and here is one way to see it's possible fairly quickly.) Think of a 6x6x6 cube first: split it into eight 3x3x3 cubes, and consider two of these 3x3x3's that are diagonally opposite of each other. If we can fill these cubes in such a way that the rooks in each cube attack all positions in the three adjacent 3x3x3 cubes, then all eight 3x3x3 cubes will be fully attacked and we're done. Now draw a 3x3 Latin square: that is, a 3x3 grid filled in with the numbers 1, 2, 3 so that each row and each column has exactly one copy of each number (there will be three 1's in total, three 2's, three 3's). These will be the rooks of my 3x3x3 cube, where the 1 squares will be the rooks of the upper level, the 2 squares the rooks of the mid level, and the 3 squares the rooks of the bottom level. You can now imagine that when I place an empty 3x3x3 cube next to this one, that wherever I place the cube each of its cells will be attacked by a rook from my Latin square cube. Now you can do the exact same thing with the 12x12x12 cube: we only need to fill two 6x6 Latin squares and then we can use that to build our solution. :)