A "reverse checkers" position is a position where every piece for one player is on one colour and every piece for the other player is on the other colour. Consider this position, for example:

example position

It is one piece away from being a full "reverse checkers" position - the black king needs to be either on h7 or g8.

What's the smallest number of moves needed to reach such a position?

  • $\begingroup$ ask to chess, not puzzling? $\endgroup$
    – Fmbalbuena
    Commented Dec 29, 2021 at 20:09

1 Answer 1


It can be done in

8 moves

1.e3 e6 2.Qg4 c6 3.Qxg7 Qg5 4.Qxh8 Qxg2 5.c3 Qxh1 6.a3 a6 7.Bxa6 Bxa3 8.Nxa3 Nxa6 *

final position


In the initial position exactly half the pieces are on "wrong" squares. Those must be either captured or moved to good squares. To do so in fewer than 16 plies, there would have to be at least one piece that captured an unmoved opponent's piece on a good (wrong from opponent's perspective) square with every ply it makes. Also, each such piece would get rid of just one bad piece more than plies. From the initial position this is possible only for the Qs and Rs and only after the two pawns in their file have been removed. As these pawns are both on good squares (from their owner's perspective) this cannot be achieved without losing a ply somewhere.

Therefore the given solution is optimal.

Let us annotate the moves from this angle:

1.e3 e6 (0) both players move a piece from a bad to a good square, thus trading even 2.Qg4 (-1) white moves the Q without immediate gain 2 ... c6 (0) 3.Qxg7 (+1) white captures a bad piece and moves to a good square 3 ... Qg5 (-1) 4.Qxh8 (0) 4 ... Qxg2 (+1) 5.c3 Qxh1 (0) 6.a3 a6 (0) 7.Bxa6 Bxa3 (-1,-1) both black and white capture pieces that are already on good (for them) squares 8.Nxa3 Nxa6 * (+1,+1) both black and white move their Ns from bad to good squares while capturing pieces on bad squares.

  • 1
    $\begingroup$ @double-beep In the initial position exactly half the pieces are on "wrong" squares. Each ply can rectify at most two of them by either moving a wrong placed piece to a right colour square or by capturing a wrong placed opponent's piece or by doing both at the same time. So a heuristic strategy is trying to only move and capture wrong placed pieces. The rest is just trial and error. $\endgroup$
    – loopy walt
    Commented Dec 29, 2021 at 20:32
  • 1
    $\begingroup$ @double-beep a matter of taste I suppose. I personally find it a bit to vague to put it in the "official" answer. $\endgroup$
    – loopy walt
    Commented Dec 29, 2021 at 21:52
  • $\begingroup$ This answer does not provide proof or reason why it shouldn't be possible in fewer moves. $\endgroup$
    – Magma
    Commented Jan 1, 2022 at 21:09
  • 1
    $\begingroup$ @Magma feel free to add an improved answer. $\endgroup$
    – loopy walt
    Commented Jan 1, 2022 at 22:30

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