Is it possible to set up a legal chess position with all 32 pieces on the board, where there can be a string of unbroken captures such that only the kings are on the board when it is over?

You can see an attempt I made here. In the resulting position (after 13. .. Qxg4), the puzzle would be solved if not for the fact that white's capture of the rook with the queen was check, which unfortunately forces black to recapture. This means that if we could switch the places of the white (black) queen with the white (black) rook in the starting position, the puzzle would be solved. The problem is that that position is not legal, as the rook has no way of jumping over the pawn chain.

Perhaps this can be done easier with knights capturing everything?

PS. I hope this is an appropriate question here. If not, please let me know what to do to improve it. Also, if you have tag suggestions, please do share.

  • $\begingroup$ What makes swapping the rooks and queens an invalid starting position? Couldn't they just go around the pawns when one has moved 2 spaces forward and an adjacent one is unmoved? $\endgroup$ – Apep Aug 26 '17 at 11:41
  • $\begingroup$ @Apep Hm, you probably right, dunno why I overlooked that. I'd accept that answer. Thanks. $\endgroup$ – Bobson Dugnutt Aug 26 '17 at 12:04
  • $\begingroup$ Flagging as too broad, many answers could be right, as far as i can see $\endgroup$ – TrojanByAccident Aug 26 '17 at 22:17
  • $\begingroup$ @TrojanByAccident Is it a requirement that there is only one answer? $\endgroup$ – Bobson Dugnutt Aug 27 '17 at 8:49
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    $\begingroup$ @TrojanByAccident There is actually one answer, yes or no. There may be many examples that prove yes, but there needs be only one provided. $\endgroup$ – micsthepick Sep 5 '17 at 1:22

If you want the fastest string of unbroken captures from the starting position, it can be done it just 17 moves starting on move 3 as shown by the below version of this chess problem. Enjoy the elegant solution!

Gerd Wilts & Norbert Geissler, PDB CD-ROM 1998-Version

enter image description here

PGN: 1. d3 d6 2. Bg5 Bg4 3. Bxe7 Bxe2 4. Bxd6 Bxf1 5. Bxc7 Qxd3 6. Bxb8 Qxc2 7. Bxa7 Rxa7 8. Qxc2 Rxa2 9. Qxh7 Rxb2 10. Qxg8 Rxb1+ 11. Rxb1 Rxh2 12. Rxb7 Rxh113. Rxf7 Rxg1 14. Rxg7 Rxg2 15. Qxf8+ Kxf8 16. Kxf1 Rxf2+ 17. Kxf2 Kxg7

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    $\begingroup$ Is there any reason the last two moves play Kxg7 and Kxg2 instead of Rxf2 Kxf2 Kxg7 and then you don't have to leave the pawn behind? $\endgroup$ – Michael Moschella Jun 16 '20 at 16:55

Just to close the books on this one, here is a solution, which was created by swapping the queens and rooks (as suggested by @Apep). Since the rook doesn't move as freely as the queen, some minor asymmetries had to be introduced.

I checked that the initial position is indeed reachable within the rules of chess.

By the way, if anyone has a more elegant solution, please do share.


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