TITLE: Dominating a Four Dimensional Chessboard with Rooks
QUESTION [9 upvotes]: There is a family of chess problems where you try to dominate a board with as few copies of a given piece as possible. The chessboard is dominated if every square either contains a piece, or is attacked by one.
For example, to dominate a usual chessboard with $8$ rooks, simply place a rook on every square of a diagonal. You can easily verify this is optimal, that is, no $7$ rooks can dominate a chessboard. This result easily extends to an $n\times n$ board, for any $n$.
You can also ask about rook domination in three dimensions, where rooks can slide left/right, back/front, or up/down. This problem is harder, but there is a nice solution: $\lceil n^2/2\rceil$ rooks are necessary and sufficient to dominate an $n\times n\times n$ board. See here for a proof.
My question is about the four dimensional case. Here, there are four axes, and rooks can slide along any axis. Specifically, 

How many rooks does it take to dominate a $4\times4\times4\times4$ board?

I was wondering if anyone had studied this problem before, or had any good ideas of an attack.
I know that $32$ rooks are sufficient, while $22$ rooks are insufficient. I've filled pages trying to succeed with $31$ rooks, to no avail. If your curious, the optimal number for a 2 x 2 x 2 x 2  board is 4 rooks, and for a 3 x 3 x 3 x 3 board is 9 rooks.

REPLY [2 votes]: I asked this question on Math Overflow as well, and got an answer which may be of interest. One of the answers provides a $24$ rook solution, and the other provides a source saying that $24$ is the minimum number required.
Viewing the hypercube as the set of points $(x,y,z,t)$, with $0\le x,y,z,t \le 3$ all integers, below is the locations of the rooks in a $24$ rook solution.
(0, 0, 0, 0)   (0, 0, 0, 1)   (0, 1, 1, 2)   (0, 2, 2, 3)   (0, 2, 3, 3)   (0, 3, 1, 2)
(1, 0, 1, 3)   (1, 1, 2, 0)   (1, 1, 3, 1)   (1, 2, 0, 2)   (1, 3, 2, 1)   (1, 3, 3, 0)
(2, 0, 2, 2)   (2, 0, 3, 2)   (2, 1, 0, 3)   (2, 2, 1, 0)   (2, 2, 1, 1)   (2, 3, 0, 3)
(3, 0, 1, 3)   (3, 1, 2, 1)   (3, 1, 3, 0)   (3, 2, 0, 2)   (3, 3, 2, 0)   (3, 3, 3, 1)