Body: <p>Suppose White has a king and a range-x rook and black has a bare king on a chessboard with dimensions y times y.
  What is the minimum value of x such that for all sufficiently large y, White, moving first, can force a win no matter where the pieces are initially placed? Passing is not allowed and stalemate is a draw.</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2023-06-15T07:20:59.197Z
Comments.Text: 50 move rule in force?
Comments.UserDisplayName: Ian Bush
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2023-06-15T12:31:02.893Z
Comments.Text: I am pretty sure x=5 is enough and x=3 is too weak. x=4 is most likely the border case...
Comments.UserDisplayName: Evargalo
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2023-06-15T12:32:09.093Z
Comments.Text: @IanBush : certainly not, otherwise there is no solution...
Comments.UserDisplayName: Evargalo
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2023-06-15T13:07:35.623Z
Comments.Text: @Evargalo Well x=8;y=8 certainly has a solution with the 50 move rule in force - if it does apply it "just" make the problem more complex.
Comments.UserDisplayName: Ian Bush
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2023-06-15T20:58:21.847Z
Comments.Text: @IanBush : sure, but there is no x *such that for all sufficiently large y, White, moving first, can force a win no matter where the pieces are initially placed* if the 50 moves rules applies. (just place the bK 50*x files away from the rook and there won't even be a *check*). Anyway, the 50 moves never applies in fairy chess problems...
Comments.UserDisplayName: Evargalo
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2023-06-15T22:20:35.200Z
Comments.Text: To clarify, 50-move rule not used, threefold repetition used.
  Also, I agree that x=5 is enough.
Comments.UserDisplayName: Lucenaposition
Tags: <checkmate><chess-variants><rook-endgame><unorthodox-pieces>
Title: King and short rook checkmate
ThreadText: # King and short rook checkmate
  
  (Asked by: username_0 on 2023-06-15)
  
  Suppose White has a king and a range-x rook and black has a bare king on a chessboard with dimensions y times y.
  What is the minimum value of x such that for all sufficiently large y, White, moving first, can force a win no matter where the pieces are initially placed? Passing is not allowed and stalemate is a draw.
  
  **Tags:** checkmate, chess-variants, rook-endgame, unorthodox-pieces
  
  > **Comments:**
  > **username_1**: 50 move rule in force?  
  > **username_2**: I am pretty sure x=5 is enough and x=3 is too weak. x=4 is most likely the border case...  
  > **username_2**: @username_1 : certainly not, otherwise there is no solution...  
  > **username_1**: @username_2 Well x=8;y=8 certainly has a solution with the 50 move rule in force - if it does apply it "just" make the problem more complex.  
  > **username_2**: @username_1 : sure, but there is no x *such that for all sufficiently large y, White, moving first, can force a win no matter where the pieces are initially placed* if the 50 moves rules applies. (just place the bK 50*x files away from the rook and there won't even be a *check*). Anyway, the 50 moves never applies in fairy chess problems...  
  > **username_0**: To clarify, 50-move rule not used, threefold repetition used. Also, I agree that x=5 is enough.  
  
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  ## 0 answers
  
  