Let $\gamma(G)$ denote the domination number of a graph, and $G\,\square\,H$ denote the cartesian product of two graphs. Then $K\_8\,\square\, K\_8$ is the rook graph, whose vertices are the squares of a chessboard, with edges between squares a rook can move between. We can similarly define the rook graph of any square chessboard as $K\_n \,\square \, K\_n$, and define $d$-dimensional chessboards as $(K\_n)^{\square \,d}$, where generalized rooks move by sliding parallel to one of the coordinate axes.


Clearly, $\gamma(K\_n\,\square\, K\_n)=n$. Put another way, it is possible to place $n$ rooks on an $n\times n$ chessboard so every square contains a rook or is attacked by one, while $n-1$ rooks are insufficient. Much less easily, you can prove $\gamma(K\_n\,\square\,K\_n\,\square\,K\_n)=\lceil n^2/2\rceil$, which also has an interpretation with rooks on a 3D board. 


These nice answers for two and three dimensions gave me hope there was a nice one for 4, so my question is this:



> 
> Is anything known about $\gamma ((K\_n)^{\square \,4})$? Or, how many rooks does it takes to dominate an $n\times n\times n \times n$ chessboard?
> 
> 
> 


It's easy to show that $\gamma ((K\_2)^{\square \,4})= 4$, and not hard to show $\gamma ((K\_3)^{\square \,4})= 9$. But for a $4\times 4\times 4\times 4$ board, all I know is that the domination number is at least 23 and at most 32. Unfortunately, it is infeasible to brute force $\gamma((K\_4)^{\square \,4})$.



Four Dimensional Rook Domination
This might be an answer for n=4. There are 24 points. I think 23 is impossible.
Just a guess, however.



```
   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

```

