TITLE: Checkmate in $\omega$ moves?
QUESTION [103 upvotes]: Is there a chess position with a finite number of pieces on the infinite chess board $\mathbb{Z}^2$ such that White to move has a forced win, but Black can stave off mate for at least $n$ moves for every $n$?

This question is motivated by a question posed here a few months ago by Richard Stanley. He asked whether chess with finitely many pieces on $\mathbb{Z}^2$ is decidable.
A compactness observation is that if Black has only short-range pieces (no bishops, rooks or queens), then the statement "White can force mate" is equivalent to "There is some $n$ such that White can force mate in at most $n$ moves". 
This probably won't lead to an answer to Stanley's question, because even if there are only short-range pieces, there is no general reason the game should be decidable. It is well-known that a finite automaton with a finite number of "counters" can emulate a Turing machine, and there seems to be no obvious reason why such an automaton could not be emulated by a chess problem, even if we allow only knights and the two kings. 
But it might still be of interest to have an explicit counterexample to the idea that being able to force a win means being able to do so in some specified number of moves. Such an example must involve a long-range piece for the losing side, and one idea is that Black has to move a rook (or bishop) out of the way to make room for his king, after which White forces Black's king towards the rook with a series of checks, finally mating thanks to the rook blocking a square for the king.
If there are such examples, we can go on and define "mate in $\alpha$" for an arbitrary ordinal $\alpha$. To say that White has a forced mate in $\alpha$ means that White has a move such that after any response by Black, White has a forced mate in $\beta$ for some $\beta<\alpha$.  
For instance, mate in $\omega$ means that after Black's first move, White is able to force mate in $n$ for some finite $n$, while mate in $2\omega + 3$ means that after Black's fourth move, White will be able to specify how many more moves it will take until he can specify how long it will take to mate.
With this definition, we can ask exactly how long-winded the solution to a chess problem can be:

What is the smallest ordinal $\gamma$ such that having a forced mate implies having a forced mate in $\alpha$ for some $\alpha<\gamma$? 

Obviously $\gamma$ is infinite, and since there are only countably many positions, $\gamma$ must be countable. Can anyone give better bounds?

REPLY [5 votes]: I have an idea for how to get up to $\omega_1^{CK}$. Consider this position: black's king is trapped and white has a mate in one. However, white's king is trapped in a perpetual check. The only way out for the white king is to go along a specific trail, emulating a finite-state machine. Along certain points on the trail, black will need to check white by moving a queen somewhere along an infinitely long line or diagonal. There will be two black queens which are far enough from the rest of the pieces that they will be able to move freely along these lines. As the white king goes along various lines the queens will be forced to emulate a two-register Minsky Machine. Note that with two queens and an infinite chess board there will be enough space to ensure that only one queen will be able to attack the king in at any one time. For the decrement operation black must be able to force white along of two paths, one of which will be obstructed for the queen if she is beyond the '0' square, the other of which will be obstructed only for a queen at the '0' square.
So now we have a Minsky machine. Next we need a source of $\omega$-power. For this, the white king will occasionally be forced to block the line white needs to checkmate black. When white is there, black will not have any checks in one move. However, one of the queens will be in square '0', and it would have been able to force a checkmate were it in any farther square. Now black moves their queen as far back as they please and threaten to checkmate white the next move. White cannot checkmate black because the line is blocked by the white king. The only way for white to avoid checkmate is to move the white king one step further and allow black to continue their perpetual checks, with an arbitrary value on one of the registers.
The way this is currently designed has a problem: If one of the registers is 0, the other must contain the state information. Then, when the first register moves to an arbitrary value, the Minsky machine must fully make use of the values in both registers. However, two-register Minsky machines are too inflexible to do this. Luckily, by moving from lines to diagonals or vice versa, a chess queen can also divide and multiply by 2. That should be enough to adequately manipulate both registers. Another possibility is to add more queens and when multiple queens can check the king along a line, make it unstrategic for all but one of the queens to do so. This approach will also be needed to modify this strategy to work in quarter-infinite and half-infinite boards.
To completely work out this strategy it is necessary to describe in full the chess positions that will make up the trail for the king and to also describe in detail how to maneuver the queens to force them to emulate a Minsky machine. I have some idea for how this will go, but not entirely. Still, I believe this idea can work.