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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).

I am pretty sure that this particular example can be done for any cardinal as well.

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.

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 by $7,8,9,\dots$ for an infinite board. For $1,2,\dots,6$ moves to mate, we do the same by only creating an $8\times 8$ board (which is sufficient as we know there are forced mate in $1,2,3,\dots,6$ in a normal chess board).

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).

I am pretty sure that this particular example can be done for any cardinal as well.

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) 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/911490181.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 by $7,8,9,\dots$ for an infinite board. For $1,2,\dots,6$ moves to mate, we do the same by only creating an $8\times 8$ board (which is sufficient as we know there are forced mate in $1,2,3,\dots,6$ in a normal chess board).

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 by $7,8,9,\dots$ for an infinite board. For $1,2,\dots,6$ moves to mate, we do the same by only creating an $8\times 8$ board (which is sufficient as we know there are forced mate in $1,2,3,\dots,6$ in a normal chess board).