Checkmate in $\omega$ moves?

Question

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?

Answer 1

Here is my first try at a solution. Your idea was a good one, but bishops are better than rooks, I surmise.

The two pictures here are placed in some distinct parts of the infinite board. The first just ensures it is White to move (in check), and that White's king will never play a role, as capturing a black unit, which are nearly stalemated as is, will release heavy pieces.

alt text http://www.freeimagehosting.net/uploads/3c8e277e7d.jpg alt text http://www.freeimagehosting.net/uploads/72ef1c9b7e.jpg

So White is left to checkmate with the four bishops and pawns. White threatens checkmate via a check from below on the northwest diagonal, and Black can only avoid this by moving the bishop northeast some amount. Upon Black moving this bishop, White then makes the bishop check anyways, the Black king moves where the Black bishop was, the pawn moves with check, the Black king again retreats northeast along the diagonal, and then White alternately moves the dark-square bishops, giving checks until the Black bishop is reached when it is mate.

The point of this second picture is that White cannot checkmate Black unless the Black bishop plays a role. Four bishops are not enough to checkmate a king on an infinite board, and hopefully I have set it up so that the White pawns play no part once Black starts the king running northeast. Pawns are not too valuable when they cannot become queens.

In extended chess notation, White plays 1. Ke5 on board A, then Black plays 1...Bz26 on board B, followed by 2. Bg3+ Kf6 3. e5+ Kg7 3. Bi5+ Kh8 4. Bf10+ Ki9 5. Bk7+ Kj10 6. Bh12+ ..., as White successively cuts off NW-SE diagonals until the Black bishop is reached. By moving the bishop X squares northeast on move 1, Black can delay the checkmate for X moves, if I set this up proper.

Other plans by White should be beatable by moving the Black king off the long diagonal or capturing the light White bishop with the pawn. Once Black's king exits the area with the pawns, the Black bishop must be a part of the mating pattern. I don't think the Black king can be forced back to that area.

Well, this is a first try.

Answer 2

Thanks to Richard Stanley and Kevin Buzzard for independently drawing my attention to this thread.

Such constructions are often easier on a half- or quarter-infinite board: the board edges are useful and also let us adapt more patterns known from the orthodox $8 \times 8$ game. I'll show that a known theoretical position with only two men on each side becomes a "checkmate in $\omega$" on a quarter-infinite board. I'll also show for each natural number $N$ two routes to "checkmate in $N\omega$" on a half-infinite board. I think one of them should adapt with some more work to chess on the edgeless square lattice.

On an infinite board even K+Q vs. K is not sufficient mating material against a lone King, while on a quarter-infinite board it is well known that K+R still suffice, with a mate of bounded length given the positions of the Kings (I think this is even in Winning Ways). Since mate in $\omega$ also requires Black to have a long-range piece, the minimum conceivable material is K+R vs. K+B. I claim that this is sufficient!

In the orthodox game K+R vs. K+B is usually an easy draw, but there are some known nontrivial wins. One standard example is Kb3,Rc2 / Kb1,Bc1. I claim that if we set this up on a quarter-infinite board with Black to move then White forces checkmate in $\omega$ moves.

White's winning plan is to play something like Rh2, Rh1, and then a waiting move like Rf1 to force Black to play Ka1 when Rxc1 is mate. (That's why this wouldn't work shifted one square left.) On the $8 \times 8$ board Black can postpone this for only a few moves. For example, if Bf4 then Rf2 and if Black saves the Bishop then Rf1 etc. (best is Kc1 but we know that after Rxf4 White wins in $O(1)$ moves). Black does better with Bg5, so after Rg2 Black can play Be3 to prevent Rg1; but White continues with Re2 and next move either takes the Bishop or initiates the mating pattern with Re1. Note that if White went to a "random" spot on the second row Black would escape with Kc1; that's why it's important to move to the file the Bishop is on.

I observed some years ago that on an $n \times n$ board the same position is checkmate in $\log_2(n) + O(1)$ moves, which seems to be the maximum for K+R against K+B. For example, with at least 11 columns and 9 rows, Black could hold on to his Bishop for an extra move by starting Bk9, so that Rk2 can be answered with Bg5 holding k1. But then Rg2 reduces to a previously solved problem. On our larger board Black can answer with either Be3 or Bi3, but Re2/Bi2 etc. wins as before. To survive one more move than that, Black would have to start by moving the Bishop 16 squares out, etc.; in general if Black moves to row $k+1$ then White checkmates in $v_2(k)+O(1)$ moves (where $v_2$ is the 2-adic valuation). So on a quarter-infinite board we get checkmate in $\omega$ as claimed. With some more effort (and a lot of added passive pieces) I think one can make this work on the edgeless board by contriving an artificial corner around a1.

EDIT See my subsequent answer for a variant of this position with K+R vs. K+B+P on a quarter-infinite board thats mate in $2\omega$, and might be extended to $3\omega$, $4\omega$, etc. with more pawns. TIDE

(I think the theoretical position Kc3,Qd1/Ka2,Rb2 is likewise a White win in $\log_2(n) + O(1)$ on an $n \times n$ board, and thus in $\omega$ on a quarter-infinite board, but the analysis is harder and it might be harder to adapt to an edgeless board.)

To get checkmate in $N\omega$ for arbitrarily large $N$ on a half-infinite board, set up something like the following, suggested by K.Buzzard's e-mail. I assume the board edge is horizontal, but much the same works with a vertical edge. Give Black Ka3 and Rb2 and White Ka1 plus a few Queens and about 3N pawns: use the pawns to fill a rectangle of 3 columns and about $N$ rows starting somewhere above the third row, and in the middle column replace each of (say) the second, third, and fourth pawns with a Queen. White will win after moving $N + O(1)$ pawns in one of the outer columns, after which the bottled-up Queens escape and finish Black off. After each pawn move, Black gets to move his Rook arbitrarily far along the second row, threatening mate; White will have to move his King one step at a time, pursued by Black's, until reaching the Rook to get a "tempo" for the next pawn move: 1...Rz2 2 Kb1 Kb3 3 Kd1 Kd3 4 Ke1 Ke3 ... Ky1 Ky3 and now another pawn move.

I don't know how to adapt this construction to an edgeless board. So here's another approach. By the vertical edge of the board, set up a position with the Black King and some White and Black pawns, none of which can move except for one White pawn that will give checkmate in $N$ moves. Surround this with a Black shell of pieces surrounded by pawns that the White King cannot penetrate and that cannot unravel within $N$ moves to either escape or stop the mate. Outside that shell put the White King and a Black Rook. $N$ times Black will choose how far out to play the Rook to harass the White King with horizontal checks.

EDIT See below for an explicit construction of mate in $N\omega$ with a fixed number of pieces on a ${\bf Z}^2$ board. TIDE

This doesn't work as it stands on an edgeless board because the White King can hide around the shell in $O(1)$ moves rather than go after the Rook. But I think something similar should succeed, using a protected but pinned Black rook to substitute for the vertical edge.

NDE

Answer 3

Update. (Oct 28, 2015) See below, for a position with game value $\omega^4$.

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This is a great question, which I have been pondering for some time.

I have just completed a joint article Transfinite game values in infinite chess with C. D. A. Evans, which describes several new positions exhibiting high transfinite game values in infinite chess. (Follow the link through to the arxiv for a pdf preprint.)

Because we found interesting positions with infinitely many pieces, we took the liberty of abandoning the finiteness requirement of the original question, considering the finite positions merely as a special case.

C. D. A. Evans and Joel David Hamkins, Transfinite game values in infinite chess, under review.

Abstract. We investigate the transfinite game values arising in infinite chess, providing both upper and lower bounds on the supremum of these values---the omega one of chess---denoted by $\omega_1^{\mathfrak{Ch}}$ in the context of finite positions and by $\omega_1^{\mathfrak{Ch}_{\hskip-1.5ex{\ \atop\sim}}}$ in the context of all positions, including those with infinitely many pieces. For lower bounds, we present specific positions with transfinite game values of $\omega$, $\omega^2$, $\omega^2\cdot k$ and $\omega^3$. By embedding trees into chess, we show that there is a computable infinite chess position that is a win for white if the players are required to play according to a deterministic computable strategy, but which is a draw without that restriction. Finally, we prove that every countable ordinal arises as the game value of a position in infinite three-dimensional chess, and consequently the omega one of infinite three-dimensional chess is as large as it can be, namely, true $\omega_1$.

The paper has 38 pages and 18 figures, detailing several positions. We also included an elementary discussion of the game-theoretic meaning of the smallish ordinal games values, such as $\omega^2$ and $\omega^3$.

Let's display here a few of the positions.

First, a simple position with value $\omega$. The main line of play here calls for black to move his center rook up to arbitrary height, and then white slowly rolls the king into the rook for checkmate. For example, 1...Re10 2.Rf5+ Ke6 3.Qd5+ Ke7 4.Rf7+ Ke8 5.Qd7+ Ke9 6.Rf9#. By playing the rook higher on the first move, black can force this main line of play have any desired finite length. We have further variations with more black rooks and a white king.

Next, consider an infinite position with value $\omega^2$. One should imagine here that the wall of pawns continues infinitely upward and downward. The central black rook, currently attacked by a pawn, may be moved up by black arbitrarily high, where it will be captured by a white pawn, which opens a hole in the pawn column. White may systematically advance pawns below this hole in order eventually to free up the pieces at the bottom that release the mating material. But with each white pawn advance, black embarks on an arbitrarily long round of harassing checks on the white king.

Here is a similar position with value $\omega^2$, which we call, "releasing the hordes", since white aims ultimately to open the portcullis and release the queens into the mating chamber at right. The black rook ascends to arbitrary height, and white aims to advance pawns, but black embarks on arbitrarily long harassing check campaigns to delay each white pawn advance.

Next, by iterating this idea, we produce a position with value $\omega^2\cdot 4$. We have in effect a series of four such rook towers, where each one must be completed before the next is activated, using the "lock and key" concept explained in the paper.

We can arrange the towers so that black may in effect choose how many rook towers come into play, and thus he can play to a position with value $\omega^2\cdot k$ for any desired $k$, making the position overall have value $\omega^3$.

Please see the article for further explanation of these positions and others.

Another interesting thing we noticed is that there is a computable position in infinite chess, such that in the category of computable play, it is a win for white---white has a computable strategy defeating any computable strategy of black---but in the category of arbitrary play, both players have a drawing strategy. Thus, our judgment of whether a position is a win or a draw depends on whether we insist that players play according to a deterministic computable procedure or not.

The basic idea for this is to have a computable tree with no computable infinite branch. When black plays computably, he will inevitably be trapped in a dead-end.

In the paper, we conjecture that the omega one of chess is as large as it can possibly be, namely, the Church-Kleene ordinal $\omega_1^{CK}$ in the context of finite positions, and true $\omega_1$ in the context of all positions.

We had an idea for proving this conjecture, but unfortunately, it does not quite fit into two-dimensional chess geometry. But we were able to make the idea work in infinite three-dimensional chess. In the last section of the article, we prove:

Theorem. Every countable ordinal arises as the game value of an infinite position of infinite three-dimensional chess. Thus, the omega one of infinite three dimensional chess is as large as it could possibly be, true $\omega_1$.

Here is one component of the three-dimension position, used to allow white to force the black king from one layer to a higher layer. Imagine the layers stacked atop each other, with $\alpha$ at the bottom and further layers below and above. The black king had entered at $\alpha$e4, was checked from below and has just moved to $\beta$e5. Pushing a pawn with check, white continues with 1.$\alpha$e4+ K$\gamma$e6 2.$\beta$e5+ K$\delta$e7 3.$\gamma$e6+ K$\epsilon$e8 4.$\delta$e7+, forcing black to climb the stairs (the pawn advance 1.$\alpha$e4+ was protected by a corresponding pawn below, since black had just been checked at $\alpha$e4).

The argument works in higher dimensional chess, as well as three-dimensional chess that has only finite extent in the third dimension $\mathbb{Z}\times\mathbb{Z}\times k$, for $k$ above 25 or so.

My co-author Cory Evans holds the chess title of U.S. National Master.

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Update. In new joint work, we've found a position with game value $\omega^4$.

• C. D. A. Evans, J. D. Hamkins, and N. L. Perlmutter, A position in infinite chess with game value $\omega^4$.

In this position, the kings sit facing each other in the throne room, an uneasy détente, while white makes steady progress in the rook towers. Meanwhile, at every step black, doomed, mounts increasingly desperate bouts of long forced play using the bishop cannon battery, with bishops flying with force out of the cannons, and then each making a long series of forced-reply moves in the terminal gateways. Ultimately, white wins with value $\omega^4$.

The position is fully explained in the article (click through to the arxiv for the pdf).

Answer 4

Here's an example on the edgeless ${\bf Z}^2$ board that shows "mate in $3\omega$ moves" and, I think, can be extended to arbitrarily large multiples $N\omega$ by moving the outlying Knight about $N/2$ squares to the left. FEN = 14/5p4/5Pp3/4p1B1p1/N3p1Prrp/2p1p1prkp/2p1p1prpp/2PpN1B1np/3P1B3p/4P4p/9p/9p/9Pr/11K/1:

_{(source: harvard.edu)}

Black to move. White will play Na11-b13-c11/d12, then capture the pawn on e10 (four squares left of the Black King), and checkmate with N(either)xg9+, Nxg9; Nxg9 exploiting the pin on the Rook at h9. Black gets three chances to move the Rook arbitrarily far to the right and harass the White King with horizontal checks until the King reaches the Rook.

Other Black defenses are no better, as long as White takes care not to move the King to one of the few squares where it could be checked with a move of the Black Knight. If that Knight moves without giving check then Nxg9 is mate immediately (which in turn means the White Knight on e7 must not move, else Black can unwind by moving the Knight, pushing the pawn to its square, and escaping with the King). None of the other Black pieces have a legal move except the Rook now on k3 (near the White King). If that Rook captures a pawn, White retakes with a pawn or Bishop and Black must soon move the Knight and get mated. Likewise after ...RxBf7; Pxf7, or ...RxBg8; Bxg8, or ...Rh12 (attacking Bg11); Pxh12. Finally if Black moves the Rook up and around to the d-file to answer Nc11 with Rd10 then White plays a random "waiting" move with his King and Black must move the Rook and allow Nxe10 and mate in two more moves.

Except for the Rook captures of the previous paragraph, White in turn must not move any piece except the King, the roving Knight, and possibly the Bf7 which must still be ready to return to f7 to defend the Bg8 and block the Rook's path to f8-f11. For example, after ...RxBf7 White must not play Bxf7? lest Black escape with ...g8; Bxg8, g9, followed by Rf10 (or g10), Kh10, etc.

To construct larger multiples of $\omega$, move the Knight from a11 far enough to the left. Black still has nothing better than to delay the King with horizontal checks. For instance, if Black could answer Nb13 with Rc12/d11 then White would play a random King move to force the Rook to relinquish its control over either c11 or d12 so the Knight could advance further. I think the Knight has enough freedom to reach its goal with the assistance of such waiting moves no matter what the Rook does. But if I'm wrong then a cavalry of $O(1)$ Knights would certainly suffice.

Answer 5

alt text http://www.freeimagehosting.net/uploads/d2ac857a24.jpg

Here is another one, hopefully it fits on one(!) board with no more modifiers. White (in check) plays Kh3xQg3, and Black threatened by Rxb7#, moves the Rf4 arbitrarily far to the east, uncovering check from the Bd6. White just takes the bishop (any way), and Black has no defense but to keep on checking White horizontally with the eastern rook, with the White king heading east until it (finally!) attacks the rook, when then White will win via Rxb7.

Notes: White has no other way to avoid the annoying rook checks, for the self-guarding Black rooks on the e-file prevent king movement to the west, and no interposes are possible by geometry. White's king can simply move east on ranks 2 and 3, but it doesn't matter too much. Black's moving the rook north on move 1 (when uncovering the bishop check) is not effective, for then White can interpose a rook on future vertical checks. The double check Rg4+ on Black's move 1 is also easily defeated by capturing that rook with the king. The only loose end is then whether by White's 1. Ki2 (not taking the queen) a faster checkmate is possible. The answer is no, for White doesn't even win, for Black can check forever with the queen on the g-file, the White king restricted to rank 3 and below.

Unless there is something missing, this seems to work also.

Answer 6

Since you were nice enough to ask for any bounds on $\gamma$ better than countable: $\gamma$ is a recursive ordinal, because the game tree, starting from any finite position, is recursive.

Correction: As pointed out by Joel David Hamkins in the comments (see also my subsequent comment), the recursiveness of the game tree implies only that, for every position $p$ from which White has a forced win, there is a recursive ordinal $\alpha$ such that White wins in $\alpha$. A uniform bound $\gamma$ that works for all such $p$ simultaneously would thus be the first non-recursive ordinal $\omega_1^{CK}$.

Answer 7

Earlier I showed how to interpret a theoretical position of K+R vs. K+B as a "mate in $\omega$" on a quarter-infinite board. If I'm doing this right, it takes only an additional Black pawn on the quarter-infinite board to get $2\omega$.

Reflect the board about the diagonal, to get White Kc2, Rb3 vs. Black Ka2, Ba3, and add a Black pawn on h4:

_{(source: harvard.edu)}

Black to move. White can't go after the pawn at once, because then the Black King escapes. White's plan is to pin the Bishop on a3 as before, forcing the pawn to advance to h3, so that when the Rook attacks and captures it the Rook will also prevent the Black King's escape: eventually ...Bc5; Rb5, Bq19; Ra5+, Ba3; Ra4, h3; Rh4, B-any; Rxh3. But then Black will have a second chance to make the game arbitrarily long.

[In case you've not seen this before: the Rook doesn't need to be on the b-file for this to work -- e.g. Rxh3, Bc5; Rh5, Bb4; and now not Rh4?, Ka3 = draw, but (say) Rh8 and if Ka3 then Ra8+ still wins because the Bishop blocks its own King's escape!]

I think it should be possible to get $3\omega$, $4\omega$, etc. by adding more Black pawns on the same file or on multiple columns, though some care may be needed to get the counts right because the White Rook could eliminate more than one pawn at each iteration.

Answer 8

The white queen moves anywhere to the east, then the black rooks force the king east, back rank mate-style, until they've either skewered, pinned, or forked the queen and king. Worst case scenario, black loses 2 rooks, and can still mate. If black ever doesn't check, white will have a perpetual. Note that black can't force mate, as white's strategy can always be "go in a northerly direction to escape check."

http://www.freeimagehosting.net/t/56042.jpg

P.S. Sorry, I switched the colors.

Answer 9

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.

Answer 10

See On numbers and endgames: Combinatorial game theory in chess endgames by Elkies for some chess positions with non-integer values.

Answer 11

Dropping the assumption of finitely many pieces as in this answer, we construct for any countable ordinal $\alpha$ a position having mate in $\beta > \alpha$, so $\gamma = \omega_1$ in the context of positions with infinitely many pieces.

Our strategy to prove this is the same as the one used for 3-dimensional chess by Joel David Hamkins and C. D. A. Evans in their paper Transfinite Game Values in Infinite Chess. As in their paper, for a tree $T$ on $\omega$ (Meaning $T \subseteq \omega^{<\omega}$ and any initial segment of an element of $T$ is also in $T$) define the Climbing-through-$T$ game as follows. White's only role is to watch. Black begins on the root node of the tree. At any stage of the game, black is on a node of $T$, and selects an immediate successor to move to. Black loses if and only if they move to a leaf node. It is clear White wins Climbing-through-$T$ if and only if $T$ is well-founded, and that in this case the game value of Climbing-through-$T$ is the same as the ordinal rank of $T$. As such trees of arbitrarily large countable ordinal rank exist, if for all trees $T$ on $\omega$ we can embed a game equivalent to Climbing-through-$T$ into infinite chess, we will have proven $\gamma = \omega_1$.

In the positions we construct, Black's king is imprisoned in the White court

Black is in local zugzwang - currently white cannot give mate, but if black is ever reduced to no moves outside the court, then they will be forced to take on d8 and will be promptly mated after 1... Kxd8 2. Rc8+ Kd9 3. Rd8+ Kxd8 4. Qc9#. This will determine both player's play as black desperately tries to achieve an infinite source of moves, while white's only winning plan is to prevent this.

We now begin the construction of the Climbing-through-$T$ game, beginning with analysis of a local position which will act like a node of our tree. The box from b2-f6 is simply to initialize the position, and will be empty for generic nodes.

In the above position, we assume the court is somewhere far below and that otherwise Black's only legal moves are with the bishop on e5. Thus, if Black moves this bishop to somewhere unprotected, White will simply capture it, leaving black with no option but to checkmate themself within the court. Therefore, Black's only reasonable option shown is the indicated Bn14 (In the final position, there will be more protected squares further along the infinite a1-w23 diagonal). Black now has the potent threat of Bo13, escaping the eyes of the d4 bishop in order to eventually leave along the a5-s23 or a7-q23 diagonals, after which white can never hunt down the bishop on an open board. Thus white is forced to play Bxn14, allowing mxn14.

At this point, black plans to move the n14 pawn down to take o6, then push the o pawns, eventually freeing the p10 bishop to escape via o9 (Other options for black will result in play ending in finite time if white just plays waiting moves). White therefore must play to free their v2 bishop by pushing the u pawns, as the only other local options simply speed up blacks plan and allows the p10 bishop to escape to the wilderness.

We thus arrive at the following position with the last move having been u4 and black in a situation very similar to that which we began in. The local position leaves no pawn moves for black, and Black's only hope to prolong play more than a few moves is to move the o9 bishop to some space along the infinite w1-a23 diagonal. White's v2 bishop will then snatch it, potentially starting this process again.

We now indicate how to build Climbing-through-$T$ . Easiest is to first build a tree $\mathcal T$ where at every node black is faced with countably many choices. Above is the start of our embedding of $\mathcal T$ with two full nodes in view, together with white's court. Note the crucial fact that the court has been placed such that no white piece outside it (All pawns and bishops of a single color) can take the black pawns holding it in place, so whites only plan is indeed the zugzwang idea. Below is a more zoomed out view, in order to give a better idea of the structure. To describe the specific positions of all the nodes, label the green diagonal $R_0$ and the yellow diagonals parallel to it $R_n$, and the yellow diagonals in the other direction $L_n, n \in \mathbb Z^+$. Then we build nodes on the intersections of certain diagonals, so that $R_n$ exits onto $\{L_m | m \equiv 2^n \pmod {2^{n+1}}\}$ and $L_n$ exits onto $\{R_m | m \equiv 2^{n-1} \pmod {2^{n}}\}$. One checks that this indeed is a tree, and is isomorphic to $\mathcal T$ as every node has countably infinite many children.

Play now proceeds as in the Climbing-through-$\mathcal T$ game. In the main line, black begins by moving their single mobile bishop along the green diagonal to one of the infinitely many protected squares along the diagonal indicated in red, choosing the next diagonal play will occur on. Our earlier analysis then applies (We confirm that blacks bishops can still easily escape outside the tree if white ever deviates), and will result in black eventually moving a bishop along the yellow diagonal emerging from black's chosen node, selecting a new diagonal from infinitely many choices and beginning the process again.

As every tree $T$ on $\omega$ is isomorphic to a subtree of $\mathcal T$, it is clear that in order to build Climbing-through-$T$ , all we must do is not build those nodes in $\mathcal T$ but not $T$, so any countable ordinal can indeed be achieved.

Answer 12

Yes there is if the finite number of pieces are large enough. But to answer your question I want black to win (because I hate it when problems most often require white to win). Let's consider the case for $N\times N$ boards and extrapolate it and then I will show how this is done for infinite boards. For a usual $8\times 8$ board there is no argument on the fact that we have mates in $1,2,\dots,7$ (and Im sure even more). I claim that for an $N\times N$ board we can have mates in $1,2,\dots,N-1$, and then I show how this is done for an infinite board.

Consider this classical mate in 7 (white to move, black mates in 6 plies, white moves maximum 7 plies) position (actually its mate in 8 if a queenside castling was allowed and we had black rook in a8 and king in e8, its funny position I always show people who never considered castling when solving such problems):

alt text http://i.harepix.com/i/389837748.jpg

Now you can reproduce the same position for an $N\times N$ board and get a mate by $N-1$ for any $N>8$. To make this work for an infinite board just surround an infinite board by black's pawn to "create" a bounded $N\times N$ board. So to make an $8\times 8$ board like the one in the diagram below. Just surround the $10\times 10$ area by Black's pawn. White's king cannot move away from the "boundary" because of the pawns. So in this way we see that we get mate in $7,8,9,\dots$ for an infinite board. For $1,2,\dots,6$ moves to mate, we do the same by only creating an $8\times 8$ board but hopefully in such a way that the "boundary" cannot be taken or moved by white (I don't think its difficult to provide the particular examples here).

My instincts tell me that this particular example can be done for any ordinal as well.