Rooks in three dimensions Given is an infinite 3-dim chess board and a black king. What is the minimum number of white rooks necessary that can guarantee a checkmate in a finite number of moves?
(In 3-dimensional chess rooks move in straight lines, entering each cube through a face and departing through the opposite face.  Kings can move to any cube which shares a face, edge, or corner with the cube that the king starts in.
See Wikipedia).
Update: Comments (below the line) give interesting information on this problem including a connection to Conway's angel problem with 2-angel (Zare) and interesting comments towards positive answer and connection with "kinggo" (Elkies). Also a link to an identical SE question is provided MSE Question 155777 (Snyder).
 A: The following is for a finite board (the question actually assumes an infinite board). 
For an $N\times N \times N$ board, wouldn't $2N$ rooks suffice? The idea comes from adapting the checkmate with two rooks for the two dimensional case in which the two rooks alternate rows and force the king to the last rank.
For the three dimensional case, for each $y$ with $1\leq y \leq N$, place one rook at $(1,y,k)$ and another at $(2,y,k+1)$. Then, for $y$ going from $1$ to $N$, move the rook at $(1,y,k)$ to the square $(1,y,k+2)$. Again, for $y$ going from $1$ to $N$, move the rook at $(2,y,k+1)$ to $(2,y,k+3)$. Each for loop over $y$ involves moving, alternately, the rooks with $x$ coordinate $1$ by increasing their $z$ coordinate by $2$ units, or the rooks with $x$ coordinate $2$ by increasing their $z$ coordinate by $2$ units. We alternate, so that a for loop in which the rooks with $x$ coordinate $1$ are moved is followed by a for loop in which the rooks with $x$ coordinate $2$ are moved, and vice versa. Eventually, either the rooks with $x$ coordinate $1$ or the rooks with $x$ coordinate $2$ will have $z$ coordinate $N$.
The effect of this is that a subset of the squares guarded by the rooks form a "floor" of two layers that keeps moving upward. So if the black king is between this "floor" and the top of the $N\times N\times N$ cube, it gets pushed to the top face. The "floor" must be moved upward in such a way that it never becomes disconnected, so that the black king can never escape to beneath the "floor" through some gap.
For an infinite board, @Noam Elkies has already mentioned $5N$, so the following is not an improvement: the "ceiling" (and the other walls of the $N\times N \times N$ cube) can be formed by placing $N$ more rooks at $(N,y,N)$, for each $y$ with $1\leq y\leq N$, and $N-1$ more rooks at $(x,1,N)$, for each $x$ with $1 \leq x \leq N-1$, and $N-1$ more rooks at $(x,N,N)$ for each $x$ with $1 \leq x \leq N-1$.
A: As far as I know, it is still open whether or not there is any finite number of rooks which can force checkmate. However, this is possible. I answered the question over at math.stackexchange describing a strategy which forces checkmate with 96 rooks. This is certainly not optimal. The strategy I describe is very wasteful, keeping some rooks in place even when they are not actually needed, so it can be improved at the expense of becoming more difficult to describe.
