show/hide this revision's text 4 See below for an explicit and improved construction of Mate in $N\omega$

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
show/hide this revision's text 3 added 218 characters in body

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.

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
show/hide this revision's text 2 Changed "K+B vs. K+R" twice to "K+R vs. K+B" which is clearer

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+B K+R vs. K+RK+B. I claim that this is sufficient!

In the orthodox game K+B K+R vs. K+R 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.

(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.

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
show/hide this revision's text 1