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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.

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  • 2
    50 move rule in force?
    – Ian Bush
    Commented Jun 15, 2023 at 7:20
  • 1
    I am pretty sure x=5 is enough and x=3 is too weak. x=4 is most likely the border case...
    – Evargalo
    Commented Jun 15, 2023 at 12:31
  • @IanBush : certainly not, otherwise there is no solution...
    – Evargalo
    Commented Jun 15, 2023 at 12:32
  • @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.
    – Ian Bush
    Commented Jun 15, 2023 at 13:07
  • @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...
    – Evargalo
    Commented Jun 15, 2023 at 20:58

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