3

What are the longest checkmates from initial positions which are (almost) fully covered in pieces? Down below I'll give examples of what I mean. An almost fully covered position can't be legal, obviously.

I don't know the source of the following puzzle, I found it on Reddit (also here):

[White ""]
[Black ""]
[FEN "rrnrkbnr/rnpnbrbp/brnnbnbn/rbbrbnrn/nnrrrpnr/brrnbbbb/nnrbrbrb/KnbnbnNn w kq - 0 1"]
[Variant "From Position"]

1. Nxf3 Rg1 2. Nxg5 Bg2 3. Nxe6 Bh3 4. Nxc7# 

Next puzzle is by MonkeyBusiness from SE:

[White ""]
[Black ""]
[FEN "QQQQQRBK/QQQQRBpB/QQQQBRBR/QQQQRBRB/QQQQBkBR/QQQQRBRB/QQQQQRBR/QQQQQBRB b - - 0 1"]
[Variant "From Position"]

1... gxf6 2. Kg7 fxg5 3. Kf6 gxh4 4. Kg7 hxg3 5. Kf6 gxh2 6. Kg7 hxg1=B 7. Kf6 Bh2 8. Kg7 Bg3 9. Kh8 Bh4 10. Kg7 Bg5 11. Kh8 Bf6#

Next three puzzles are by me:

[White ""]
[Black ""]
[FEN "KRbqbbNb/BpBppppk/pppppppn/qqqqqqrb/qqqqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq w - - 0 1"]
[Variant "From Position"]

1. Bxd8 Kxg8 2. Bc7 Kh7 3. Rxc8 Ng8 4. Rxe8 Kh6 5. Rxf8 Kh7 6. Rxf7 Nh6 7. Rf8 Ng8 8. Kxb7 Nh6 9. Bab8 Ng8 10. Bd8 Nh6 11. Kc7 Ng8 12. Re8 Nh6 13. Kxd7 Ng8 14. Bbc7 Nh6 15. Bxe7 Nf7 16. Bf8 Nh6 17. Bd8 Nf7 18. Bde7 Nh6 19. Bxg7 Bxg7 20. Ra8 Nf7 21. Bf8 Bxf8 22. Rxf8 Nh6 23. Ra8 Nf7 24. Ke8 Kg8 25. Ra7 Nh8 26. Rb7 Nf7 27. Rxf7 Kh8 28. Rf8+ Kg7 29. Ke7 Kh6 30. Kf7 Kh7 31. Rd8 Kh6 32. Rh8#

Here Black gives a checkmate in 32 moves or less:

[White ""]
[Black ""]
[FEN "RNRQRNRQ/NRNPPNPB/RNPRPPPR/KRRPPPRP/BRRPPRPP/QPPPRPPR/nPPPPPPP/BBBnnnkn b - - 0 1"]
[Variant "From Position"]

1... Nxc1 2. Qa2 Nhxf2 3. Qa3 Kxg2 4. Ba2 Nxh3 5. Bb1 Kf2 6. Ba2 Ng2 7. Bb1 Ke1 8. Ba2 Ndf2 9. Bb1 Kxd2 10. Qa2 Kxe3 11. Qa3 Nxe2 12. Ba2 Nxh2 13. Bb1 Nxg3 14. Qa2 Kxf4 15. Qa3 Nxf3 16. Ba2 Ngxh4 17. Bb1 Nfxg5 18. Qa2 Nxg4 19. Qa3 N3xe4 20. dxe4 Nhf2! 21. Qa2 Nfxe4! 22. Qa3 Ne3! 23. Ba2 Nhxf5! 24. Bb1 Nfxd6! 25. Ba2 Nef5! 26. Bb1 Ke3! 27. exd6 Nexd6 28. Nxd6 Nxd6 29. Ba2 Nge4 30. f7 Nxb7+ 31. cxb7 Nd6 32. Nbc6 Nxb7# 

The last position is unique. It's not 100% covered in pieces, but only 2 squares out of 64 are empty. And despite the absurdity of the position it plays out like some immortal game:

[White ""]
[Black ""]
[FEN "qrqqqqqq/qqqqqBqN/qrrPnpqq/Nnr1nppr/qrBNPqpb/rPNBnN1k/pPQPQQRP/bNBqKRRb w - - 1 1"]
[Variant "From Position"]

1. Nxd1!! Nxg2+ 2. Rxg2 gxf3 3. Qxh4+ Qxh4+ 4. Nf2+ Kxg2 5. Nxf3!! Nxf3+ 6. Qxf3+ Kxf3 7. Be2+ Kg2 8. Bf3+ Kxf3 9. Qd1+ Kg2 10. Rg1+ Kxh2 11. Rxh1+ Kg2 12. Rg1+ Kh2 13. Kf1!! Qh3+ 14. Nxh3 Rbxb3 15. Nc3!! Rxc3 16. Qe2+ Kxh3 17. dxc3 Qxc4 18. Nxc4!! Qexd6 19. Qg2+ Kh4 20. Rh1+ Qh2 21. Rxh2+ Qxh2 22. Qxh2+ Kg4 23. Ne3+ Kf3 24. Qg2+ Kf4 25. Nc4+ Qd2 26. Bxd2+ Qxd2 27. Qxd2+ Kg4 28. Qg2+ Kh4 29. Qf2+ Kh3 30. Qf3+ Kh4 31. Kg2!! Nf4+ 32. Kh2 Rxc3 33. bxc3 Rxc4 34. Qg3# 
2
  • Is my answer sufficient to be accepted? Also, new question: Since we're dealing with illegal positions anyway, must we have a White king? You do not specify, so I want to be sure. Commented Jun 26 at 22:16
  • @Rewan, see my response. I do think your answer is the best (already planned to accept it) and that there's probably no new mechanism which can lead to even longer checkmates. Don't know how to answer your King question. Usually illegal positions don't imply "there may be no King". Commented Jun 27 at 9:59

2 Answers 2

4

I offer 85 moves long, utilizing the holes matrix idea presented by Hauke. It is much slower to force a rook through, as a knight is too slippery and allows for much shorter mates. I do not foresee much more potential therein; of course, I may very well be wrong.

[FEN "bBbBbBbB/RpBpBpBp/pBpBpBpB/BpBpBpBp/pBpBpBpB/BpBpBpBb/NPpPpBkB/KbBbBbBN w - - 0 1"]

1. Rxa8 Kxh1 2. B8a7 Kg2 3. Rxc8 Bxa2 4. Bcb8 Bb1 5. Bd8c7 Kh1 6. Rxe8 Kg2 7. Bed8 Kh1 8. Bf8e7 Kg2 9. Rxg8 Kh1 10. Bgf8 Kg2 11. Rg7 Kh1 12. Rxh7 Kg2 13. Rxf7 Kh1 14. Bf6g7 Kg2 15. Be7f6 Kh1 16. Rxd7 Kg2 17. Bd6e7 Kh1 18. Bc7d6 Kg2 19. Rxb7 Kh1 20. Bb6c7 Kg2 21. Rb6 Kh1 22. Rxa6 Kg2 23. Rxc6 Kh1 24. Bc5b6 Kg2 25. Bd6c5 Kh1 26. Rxe6 Kg2 27. Be5d6 Kh1 28. Bf6e5 Kg2 29. Rxg6 Kh1 30. Rf6 Kg2 31. Rxf5 Kh1 32. Be5f6 Kg2 33. Rxd5 Kh1 34. Bd4e5 Kg2 35. Bc5d4 Kh1 36. Rxb5 Kg2 37. Rc5 Kh1 38. Rxc4 Kg2 39. Bd4c5 Kh1 40. Rxe4 Kg2 41. Rd4 Kh1 42. Rxd3 Kg2 43. Rd4 Kh1 44. Rc4 Kg2 45. d4 Kh1 46. d5 Kg2 47. Bc5d4 Kh1 48. Bd6c5 Kg2 49. d6 Kh1 50. d7 Kg2 51. Bc7d6 Kh1 52. Bd8c7 Kg2 53. d8=Q Kh1 54. Bc3d2 Kg2 55. Bbc3 Kh1 56. Rxa4 Kg2 57. Ba3b4 Kh1 58. Qe8 Kg2 59. Qf7 Kh1 60. Qxb3 Kg2 61. Qf7 Kh1 62. Qxh5 Ba2 63. Kxa2 Bfg2 64. Ba3 Bf1 65. Bcd8 Bhg2 66. b4 Bh3 67. Bb6c7 Kg2 68. b5 Kh1 69. b6 Bhg2 70. b7 Bh3 71. Bc7b6 Bfg2 72. Bb8c7 Bf1 73. b8=Q Bhg2 74. Ba1 Bh3 75. Bdb2 Bhg2 76. Bec3 Bh3 77. B4e5 Kg2 78. Rxg4 Kh1 79. Qa8 Bfg2 80. Bg3f4 Bf1 81. Qxf3+ Bfg2 82. Qxg2+ Bxg2 83. Rxg2 Kxg2 84. Qg4+ Kf1 85. Qh3#

The longest holes matrix I've found with a knight in play.

[FEN "pBpBbBbB/NpBpBpBp/pBpBpBpB/BpBpBpBp/pBpBpBpB/BpBpBpBb/bPpPpBkB/KbBbBbBN w - - 0 1"]

1. Nxc8 Kxh1 2. B6a7 Kg2 3. Nb6 Kh1 4. Nxa8 Kg2 5. Bc7b6 Kh1 6. Nc7 Kg2 7. Nxe8 Kh1 8. Bd6c7 Kg2 9. Nd6 Kh1 10. Nxb7 Kg2 11. Nd6 Kh1 12. Nc8 Kg2 13. Be7d6 Kh1 14. Ne7 Kg2 15. Nxg8 Kh1 16. Bf6e7 Kg2 17. Nf6 Kh1 18. Nxh7 Kg2 19. Nf6 Kh1 20. Nxd7 Kg2 21. Be5f6 Kh1 22. Ne5 Kg2 23. Nxf7 Kh1 24. Ne5 Kg2 25. Nxg6 Kh1 26. Ne5 Bfg2 27. Nxc6 Bf1 28. Bd4e5 Bfg2 29. Nd4 Bf1 30. Nxe6 Bfg2 31. Bc5d4 Bf1 32. Bd6c5 Bfg2 33. Bc7d6 Bf1 34. Nc7 Kg2 35. Nxa6 Kh1 36. Nc7 Kg2 37. Nxd5 Kh1 38. Nc7 Bhg2 39. Nxb5 Bh3 40. Bd6c7 Bhg2 41. Nd6 Bh3 42. Nxf5 Bfg2 43. Nd6 Bf1 44. Nxc4 Bhg2 45. Nd6 Bh3 46. Nxe4 Bfg2 47. Bc5d6 Bf1 48. Nc5 Bfg2 49. Nxd3 Bf1 50. Nc5 Bfg2 51. Ne4 Bf1 52. Bd4c5 Bfg2 53. d4 Bf1 54. d5 Bfg2 55. Be5d4 Bf1 56. Bd6e5 Kg2 57. d6 Kh1 58. d7 Bfg2 59. Be3d2 Bf1 60. Be7d6 Kg2 61. Bd8e7 Kh1 62. d8=N Bfg2 63. Ne6 Bf1 64. Bf4e3 Bfg2 65. Bed8 Bf1 66. Bd6e7 Bhg2 67. Be5d6 Bh3 68. Bge5 g3 69. Nxg3+ Kg2 70. Nf4#

Without holes, my initial presentation was 76 moves long; I can see no further moves to be squeezed from this matrix.

[FEN "qqqqqqBN/PqqqqqpN/qBqqqqpk/qqBqqqNP/qqqBqPRN/qqqrBRPR/qqqqqBPK/qqqqqqNn w - - 0 1"]

1. hxg6 Qxg1+ 2. Bxg1 Qxg1+ 3. Bxg1 Qxg1+ 4. Bxg1 Qxg1+ 5. Bxg1 Qxg1+ 6. Bxg1 Qxg1+ 7. Kxg1 Qaa1+ 8. Kh2 Qg1+ 9. Kxg1 Qaa1+ 10. Kh2 Qg1+ 11. Kxg1 Qaa1+ 12. Kh2 Qg1+ 13. Kxg1 Qaa1+ 14. Kh2 Qg1+ 15. Kxg1 Qaa1+ 16. Kh2 Qg1+ 17. Kxg1 Qaxa7+ 18. Kh2 Qg1+ 19. Kxg1 Qbb1+ 20. Kh2 Qg1+ 21. Kxg1 Qbb1+ 22. Kh2 Qg1+ 23. Kxg1 Qbb1+ 24. Kh2 Qg1+ 25. Kxg1 Qbb1+ 26. Kh2 Qg1+ 27. Kxg1 Qbb1+ 28. Kh2 Qg1+ 29. Kxg1 Qbb1+ 30. Kh2 Qg1+ 31. Kxg1 Qcc1+ 32. Kh2 Qg1+ 33. Kxg1 Qcc1+ 34. Kh2 Qg1+ 35. Kxg1 Qcc1+ 36. Kh2 Qg1+ 37. Kxg1 Qcc1+ 38. Kh2 Qg1+ 39. Kxg1 Qcc1+ 40. Kh2 Qg1+ 41. Kxg1 Qcc1+ 42. Kh2 Qg1+ 43. Kxg1 Qdd1+ 44. Kh2 Qg1+ 45. Kxg1 Qdd4+ 46. Kh2 Qg1+ 47. Kxg1 Qdd4+ 48. Kh2 Qg1+ 49. Kxg1 Qdd4+ 50. Kh2 Qg1+ 51. Kxg1 Qdd4+ 52. Kh2 Qg1+ 53. Kxg1 Qe1+ 54. Kh2 Qg1+ 55. Kxg1 Qe1+ 56. Kh2 Qg1+ 57. Kxg1 Qe1+ 58. Kh2 Qg1+ 59. Kxg1 Qe1+ 60. Kh2 Qg1+ 61. Kxg1 Qe1+ 62. Kh2 Qg1+ 63. Kxg1 Qe1+ 64. Kh2 Qg1+ 65. Kxg1 Q8c5+ 66. Kh2 Qg1+ 67. Kxg1 Qa7+ 68. Kh2 Qg1+ 69. Kxg1 Qd4+ 70. Kh2 Qg1+ 71. Kxg1 Qc5+ 72. Kh2 Qg1+ 73. Kxg1 Rd1+ 74. Kh2 Kh5 75. Nf5+ Kxg4 76. Ne3
2
  • As I understand, the original idea was more complicated: pdb.dieschwalbe.de/P1003966 . You use the knight to block the protection of a bishop and force the king to capture one of the bishops, creating more open space... leading to a checkmate. Is it possible? Does it lead to a longer checkmate? At least if we leave some squares unoccupied. Maybe something like this, maybe with a Knight instead of a Rook: rBrBrBkB/BnBpBnBp/nBnBpBPB/BnBpBpBp/rBrBpBpB/BrBpBpBp/K2BpBpB/1RBrBrBr w - - 0 1 Commented Jun 27 at 9:52
  • Your 76 move puzzle also works like this (maybe): qqqqqqbN/PqqqqqrN/qBqqqqpk/qqBqqqNP/qqqBqrRN/qqqrBRPR/qqqqqBPK/qqqqqqNn w - - 0 1 (bRf4, wNh7, wNh8, bRg7, bBg8). Don't claim any credit. Commented Jun 29 at 13:30
4

I offer a board that is only half full - sorry! Plaster the board with white bishops on black, add a white knight and both kings, and then remove two white bishops. (Published as Schwalbe (194/11467), 2002-04) For example:

[FEN "1B1B1BkB/B1B1B1B1/1B1B1B1B/B1B1B1B1/1B1B1B1B/B1B1B1B1/K2B1B1B/1NB1B1B1 w - - 0 1"]

In the publication, I gave a valid algorithm for mating, which (since you must transport the knight from "hole to hole" a few times) takes quite long, I guess 50 moves at least. (Someone should make a tablebase...) R instead of N also works.

2
  • I've found the Shwable link and understood the idea (placing the knight on g7), but is it really forced, is there any kind of proof? A position from where a computer can calculate the checkmate? Commented Jul 15 at 4:42
  • 1
    @FlowyPoosh: Ye-e-e-s. I think the algorithm is watertight, but a formal proof of its correctness was not given. Naturally I didn't try what happened with a mate search either in LiChess (hey, it does NOT complain!) or Gustav (most promising, alas, I don't have it) or Popeye (takes too long, I bet). Commented Jul 15 at 8:49

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