5
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The white pieces in the above position should maneuver themselves into the flipped starting position (i.e. king in e3). All pieces makes regular moves within the 4x4 center squares. By removing a pawn from the board to allow a moving space, find a way with least moves.

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  • $\begingroup$ The starting position has a removed pawn thus its symmetry. removing not considered a move. $\endgroup$
    – TSLF
    Sep 3, 2020 at 21:08
  • $\begingroup$ What I do first is to calculate the number of status of the board, which (with one pawn removed) is at most $16! / (7! 2!2!2!) = 518918400$. It should be even less, because pawns only move forward. Thus it is perfectly bruteforceable (: $\endgroup$
    – WhatsUp
    Sep 3, 2020 at 23:50
  • $\begingroup$ @WhatsUp- It is also perfectly learnable. 1) side pawns are not removable 2) pawns makes 14 or 15 moves only. $\endgroup$
    – TSLF
    Sep 4, 2020 at 5:27
  • $\begingroup$ Just verifying - e6 goes to e3? $\endgroup$
    – Moti
    Sep 5, 2020 at 18:49
  • $\begingroup$ @Moti- yes, the only possible flip $\endgroup$
    – TSLF
    Sep 6, 2020 at 16:08

1 Answer 1

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No , and since this question was arguably dead, I don't even feel bad :)

The optimum is 43 (so close...) if you remove d4 or e4. For d3 and e3, the optimum is 49. Side pawns are obviously impossible.

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