Consider a board with only two edges that are perpendicular.
.------------------------------------------------------ | ^ | |m |<------------n---------------->bK wK v | wR | | |
The edge on the right-side doesn't exist (or is very far away), so the task of white is to first drive the black king to the left, into the corner (or is it?).
Assume that both kings are on the edge and opposing each other as shown. Further more assume that there are n empty squares between the left-edge and the black king.
Lets use a 8x8 board to illustrate things below.
n = 4 and the rook is placed such that the black king has only 1 row (
m = 1).
The optimal way to mate here is (with white to play):
[FEN "4k1K1/7R/8/8/8/8/8/8 w - - 0 1"] 1. Rg7 Kd8 2. Kf8 Kc8 3. Ke8 Kb8 4. Kd8 Ka8 5. Kc7 Ka7 6. Rg6 Ka8 7. Ra6#
Thus, we can conclude that it in the general case (where the rook is far to the right) it will take
n+3 moves for white to give mate. But note how white began with losing a tempo by playing a rook move.
Now consider the case where the rook is on the third row from the edge (
m = 2).
If black uses the strategy to walk away straight from the
wK (ie, remains on the second rank), he loses as fast as in the previous case:
[Variant "From Position"] [FEN "8/4k1K1/7R/8/8/8/8/8 w - - 0 1"] 1. Rf6 Kd7 2. Kf7 Kc7 3. Ke7 Kb7 4. Kd7 Ka7 5. Kc7 Ka8 6. Ra6#
n + constant. A better strategy is to move to the edge at the moment the white king is on the second rank (so the rook can't be played to the second rank):
[Variant "From Position"] [FEN "8/4k1K1/7R/8/8/8/8/8 w - - 0 1"] 1. Rf6 Kd7 2. Kf7 Kc8 3. Ke7 Kc7 4. Rh6 Kb8 5. Kd7 Kb7 6. Rg6 Ka8 7. Kc7 Ka7 8. Rh6 Ka8 9. Ra6#
The pattern starts different, with
1... Kd7 as opposed to
1... Kd8 because after
1... Kd8? 2. Rf7! resulting in a much faster win for large
In fact, an 8x8 board is already too small to properly investigate the best algorithm for large
m=2. At first sight the number of moves is going to be
2n + constant, but of course, white has the trick to play the rook to the left of its king!
[Variant "From Position"] [FEN "8/5k1K/7R/8/8/8/8/8 w - - 17 9"] 1. Ra6 Ke7 2. Kg7 Kd7 3. Kf7 Kc7 4. Ke7 Kb7 5. Rh6 Kc7 6. Rg6 Kc8 7. Kd6 Kb7 8. Kd7 Kb8 9. Kc6 Ka7 10. Kc7 Ka8 11. Ra6#
Effectively still shifting one position to the left per move (or else the rook goes to the second rank), and it is in fact mate in
n + 6.
Now, given the relative positions of the three pieces from the last starting position (
wK opposing and
wR on the same file as the
wK; white to move), with
n empty squares on the left of the
bK and the
m rows from the edge (
m=2 in the last position).
Then in how many moves can white give mate on said infinitely large board, for large
n and any
m > 0 ?
One more remark though - notice this interesting zigzag strategy of white where the king goes in front of its own rook in order to achieve a
2n + c with
m = 2:
[Variant "From Position"] [FEN "4k3/6K1/7R/8/8/8/8/8 w - - 0 1"] 1. Kf6 Kd7 2. Kf7 Kd8 3. Ke6 Kc7 4. Ke7 Kc8 5. Kd6 Kb7 6. Kd7 Kb8 7. Kc6 Ka7 8. Kc7 Ka8 9. Ra6#
Which thus can be improved for large
n (the above is an optimal mate in 9 though), but which returns shortly with
m = 3:
[Variant "From Position"] [FEN "8/4k3/6K1/7R/8/8/8/8 w - - 0 1"] 1. Kf5 Kd6 2. Kf6 Kd7 3. Ke5 Kc6 4. Ke6 Kc7 5. Kd5 Kd7 6. Rh7+ Ke8 7. Ke6 Kd8 8. Rf7 Kc8 9. Kd6 Kb8 10. Kc6 Ka8 11. Kb6 Kb8 12. Rf8#
where mate in 12 is indeed the fastest possible (on 8x8).