11

One of the FIDE rules states that if someone's flag falls, and there exists a legal sequence of moves such that the other player mates the first player, then the position is a win for the second player. This got me thinking whether this rule can theoretically be hard to enforce for arbiters, i.e. whether it is possible that the arbiter cannot (easily) decide whether a game is winnable for one side or not:

Do there exist "hard" chess puzzles where the objective is to find a sequence of moves of any length, with both sides helping, so that one side wins? So essentially helpmate puzzles, but without specifying the number of moves until mate? Or is it always rather straightforward to determine whether, from a given position, there is a sequence of moves leading to mate?

Perhaps one way to make the job for the arbiter difficult is to not keep track of previous moves, and then present a position to the arbiter (when the flag falls) where it is difficult to prove whether one side can legally castle or not, or take en passant or not etc. - if such positions can only be won when say en passant is available, the arbiter (or the second player claiming a win on time, rather than a draw) would have to construct a proof game to show that the game can indeed be won.

In any case: I cannot come up with examples which are hard helpmates of any length, but I am in no way an expert when it comes to helpmates. Any thoughts or comments are appreciated!

Edit: This question is not about what is sufficient mating material, but about whether positions exist for which it is hard to decide if a mating sequence exists. This is more in the territory of artificially constructed helpmates/proof game problems than about simple, realistic adjournment situations for arbiters.

9

I don't remember where I first saw this kind of trick:

[Title "Could White (or Black) possibly win this?"]
[FEN "8/1p4p1/1Pp3p1/k1P3p1/1pP3Pb/1P4p1/6P1/7K w - - 0 0"]

Yes, the Kingside is necessary!

[Title "White mates in 19 (with Black's help)"]
[FEN "8/1p4p1/1Pp3p1/k1P3p1/1pP3Pb/1P4p1/6P1/7K w - - 0 0"]

1.Kg1 Ka6 2.Kf1 Ka5 3.Ke2 Ka6 4.Ke3 Ka5 5.Ke4 Ka6 6.Ke5 Ka5 7.Ke6 Ka6 8.Kf7! Ka5 9.Kxg6! Ka6 10.Kf7 Ka5 11.Ke7 Ka6 12.Kd7 Ka5 13.Kc7 Ka6 14.Kc8 Ka5 15.Kxb7 g6! 16.Ka8 Ka6 17.b7 Ka5 18.b8=Q Ka6 19.Qa7#

Black could mate a few moves later after (say) 19.Qf4 Ka5 20.Qf2 gxf2 etc.

P.S. It turns out the I first noticed this trick in a help-stalemate problem by Tivadar Kardos (see my comment below). Looking at that problem and some of his others suggests this improved setting of the same idea:

[Title "White's turn.  Could either side possibly win this? (NDE after T.Kardos)"]
[FEN "5brk/4p1p1/3pP1P1/1B1P2p1/3p2p1/3P4/4K1P1/8 w - - 0 0"]
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  • Found my source for this tempo trick: Tivadar Kardos, The Problemist 1997 #F1705, helpmate in 25 8/2p1p1p1/2P5/2P3p1/2P4k/2P1p1p1/2P1P1P1/5B1K (with a minor dual at White's 22nd move). – Noam D. Elkies Sep 23 '17 at 5:29
  • Thanks. In the second position (starting 1 Be8), I count 25 moves too, but it's just an accident that this is the same as the length of the Kardos help-stalemate. After 21 Kxf8 g3 22 Kxe7 Kg8 White and Black can also cooperate to let Black checkmate, but that will take more than 30 moves in total. – Noam D. Elkies Sep 24 '17 at 0:09
6

Here's a famous helpmate problem where at first it is not clear that White can win at all:

[Title "Helpmate in 10 (Gabor Cseh, StrateGems 2001)"]
[FEN "8/5NR1/6PB/1P2P1P1/4P3/2p2p1p/pprp1p1p/kbqrbK2 w - - 0 0"]

Can White win this on time if Black (on the move) couldn't find a Queen or Rook before their flag fell?

(solution at https://www.chess.com/forum/view/more-puzzles/best-help-mate-puzzles)

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5

Here's an old problem built round the idea of finding a tricky helpmate:

[TITLE "White to move. Last move? (A.Buchanan, StrateGems 2002)"]  
[FEN "Bb1k1b2/bKp1p1p1/1pP1P1P1/1P6/p5P1/P7/8/8 w - - 0 1"]

Hint: by FIDE Law A1.3, if a future mate is not possible, even cooperatively, the game is over, drawn "in dead position". If the diagram position here is dead then the game is ended. But the prior position before the last move must have been alive though or the game would have ended then.

A1.3 applies "symmetrically", asking whether either player could deliver mate as a helpmate, and all the compositions so far in this thread focus on that. However off-hand I don't know of any composition which relies on the asymmetric rule (FIDE Law 6.9) associated with flag fall, where we only ask about one player's ability to helpmate. So there's a creative opportunity here... :D

Hint: bK is mated on:

e8

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5

Positions where a player would have adequate mating material but unable to achieve a helpmate are very rare. In almost call such cases, both sides will have a very limited number of possible moves until either the game was over (due to stalemate, checkmate, or draw by repetition) or the position opened up enough that a helpmate could easily be achieved.

While one could have a position where both sides have a substantial number of moves and yet no mate would be possible, e.g.

[FEN "k1bB/8/2p1p1p1/1pPpPpP1/1P1P1P2/8/8/K1Bb w - - 0 1"]

[each side has promoted at least one pawn for a bishop]

One may have to study the board for a little while to determine whether there is any means of either breaking the pawn barrier or producing a checkmate with pieces that are stuck on one side of it, but I wouldn't expect that any deep analysis should be required. Simply figure the range of squares that each piece could occupy without breaking the pawn barrier, and see if any piece could break the pawn barrier or if pieces could be arranged to create a checkmate. In the situation above, checkmate can be seen to be impossible because none of White's pieces except the king can ever control any light squares, the two kings can't get close, there's no dark square the Black king could occupy which doesn't have at least two light escape squares, and there's only one Black piece that could block a light escape square. Similar issues would prevent Black from mating White.

There may be some situations which would require a little care to determine whether a mate was possible, but analysis should be straightforward.

BTW, it's possible to achieve a position where White could have all his starting pieces, and be able to play hundreds of moves without a stalemate or threefold repetition, and yet be unable to achieve a helpmate.

[FEN "nb2kBBN/2p1PRRN/1pP2PPQ/1P4PP/8/8/4P3/6K1 w - - 0 1"]

Black can move his bishop back and forth and there's no danger of his becoming unable to do so, White can move his king all around the lower part of the board, and White may also push his e pawn four times. White's king can never get in a position to capture anything, the pawn can neither capture nor give check, and no other pieces can do anything at all.

On the other hand, there is one scenario where things might be complicated for an arbiter: if a position arises where the game will be drawn if it reaches a certain number of moves, the question of whether a helpmate can be executed before the game is drawn could be tricky to resolve since it would basically turn into a normal "helpmate in N" problem.

As a further observation, it is possible, at least in contrived situations, for one side to have a helpmate available only if castling is possible.

[FEN "6Bk/5P1P/7P/8/8/1p1p1p2/pp1P1P2/brn1K2R w - - 0 1"]

White can only helpmate black by castling. If White does anything else, Black's next move must be ...Ne2++. After O-O Ne2+, however, White can move Kh2, escaping the check, and helpmate will become trivial.

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2

One pawn is sufficient for helpmate.

A bishop plus a knight is sufficient for helpmate.

A king plus bishop can deliver helpmate if the other side has at least one piece which can limit the king's mobility without stopping the bishop check and without checking the other king, so knight or bishop of the other colour.

Similarly a king plus knight can deliver helpmate if the other side has a bishop or a knight.

A lone king cannot deliver mate.

Any major piece is sufficient for helpmate.

Castling is irrelevant for helpmate. En passant is only relevant in blocked positions and players are obliged by 11.11 to help the arbiter in such situations.

In any case: I cannot come up with examples which are hard helpmates of any length, but I am in no way an expert when it comes to helpmates.

Helpmates of arbitrary length are always trivial because it is either just a case of the mating side queening a pawn and mating or it is just a matter of one side getting rid of the right material and moving into position to be mated. What is difficult is helpmate in a set number of moves.

Perhaps one way to make the job for the arbiter difficult is to not keep track of previous moves

This is nonsense!

Article 8.1.1

In the course of play each player is required to record his own moves and those of his opponent in the correct manner, move after move, as clearly and legibly as possible, in the algebraic notation (Appendix C), on the ‘scoresheet’ prescribed for the competition.

As an arbiter when I see somebody not recording the moves I first ask them to bring their scoresheet up to date with all the moves. If they refuse I impose a time penalty. Repeated refusal results in loss of the game.

Note also article 11.11 of the FIDE Laws of Chess:

Both players must assist the arbiter in any situation requiring reconstruction of the game, including draw claims.

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2

A simple but partial counterexample, showing that if the history of the past moves is unknown, it can theoretically be impossible to decide whether one side can win or not.

[FEN "8/k7/2p5/KpP5/pP6/P7/8/8 w - - 0 1"]

If either side loses on time in this position with white to move, the other side would legally win if there is a legal sequence of moves to mate the opponent. However, it seems that white is in a stalemate position, and thus it's a draw. Unless black's last move was b7-b5, in which case cxb6 (en passant) would lift the stalemate and easily allow either side to win with the opponent's assistance.

In this situation, the immediate history of the position resolves the issue though - the arbiter could ask either player what was the last move, and use this information to draw the right conclusion. So perhaps there exist better counterexamples where drawing conclusions is even harder...

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2

If you are looking for helpmate problems of arbitrary length, than I have a few for you.

The longest legal helpmate with a unique solution is 28 moves long, and this record has stood for 76 years!

[Title "A. Hegermann, Fairy Chess Review 1934, Helpmate In 28 Moves"]
[FEN "6kR/4p1p1/1p2P1P1/1P2p3/1P2P3/1P2p1p1/4P1P1/5BK1 b - - 0 1"]

[FEN "6kR/4p1p1/1p2P1P1/1P2p3/1P2P3/1P2p1p1/4P1P1/5BK1 b - -"]

1... Kxh8 2. Kh1 Kg8 3. Kg1 Kf8 4. Kh1 Ke8 5. Kg1 Kd8 6. Kh1 Kc7 7. Kg1 Kd6 8. Kh1 Kxe6 9. Kg1 Kf6 10. Kh1 Kg5 11. Kg1 Kf4 12. Kh1 Kxe4 13. Kg1 Kd4 14. Kh1 Kc3 15. Kg1 Kxb3 16. Kh1 Kxb4 17. Kg1 Kxb5 18. Kh1 Kc4 19. Kg1 b5 20. Kh1 b4 21. Kg1 b3 22. Kh1 b2 23. Kg1 b1=Q 24. Kh1 Qf5 25. Kg1 Qf7 26. gxf7 Kc3 27. f8=Q Kd2 28. Qc8 Ke1 29. Qc1#

The longest legal helpmate without a unique solution, as far as I know, is 48 moves long.

[Title "Karl Fabel, Schachmatt  1948, Helpmate In 48 Moves "]
[FEN "8/8/5p2/2p1pPp1/2PbPk1b/2pBp1pB/2P1P1P1/5BRK b - - 0 1"]

1... g4 2. Bxg4 Kg5 3. Bf3 Kh6 4. Bg4 Kg7 5. Bf3 Kf7 6. Bg4 Ke7 7. Bf3 Kd7 8. Bg4 Kc7 9. Bf3 Kb6 10. Bg4 Ka5 11. Bf3 Ka4 12. Bg4 Ka3 13. Bf3 Kb2 14. Bg4 Kc1 15. Bf3 Kd1 16. Bg4 Bg5 17. Bf3 Bh6 18. Bg4 Bf8 19. Bf3 Bd6 20. Bg4 Bc7 21. Bf3 Ba5 22. Bg4 Bb4 23. Bf3 Ba3 24. Bg4 Bc1 25. Bf3 Bd2 26. Bg4 Be1 27. Bf3 Bf2 28. Bg4 Bxg1 29. Kxg1 Ke1 30. Kh1 Kxf1 31. Bf3 Ke1 32. Kg1 Kd1 33. Kf1 Kc1 34. Ke1 Kb2 35. Kd1 Ka3 36. Kc1 Ka4 37. Kb1 Ka5 38. Ka2 Kb6 39. Ka3 Kc6 40. Ka4 Kd6 41. Kb5 Ke7 42. Kc6 Kf7 43. Kd6 Kg7 44. Ke6 Kh8 45. Kxf6 Kg8 46. Kg6 Kh8 47. f6 Kg8 48. f7+ Kh8 49. f8=Q#

Lastly, the longest helpmate in an illegal position is 111 moves long! It is hard to imagine that there exists checkmating sequence in this position at all!

[Title "James Malcom & Arno Tüngler, Matplua.net Forum 1/30/2020, Helpmate In 111 Moves"]
[FEN "5b1b/2p1p1b1/1bPbPbpp/b1pPp1b1/1bPpPpkp/b1pBpPpP/1bPbPbPb/bBbBb1bK b - - 0 1"]

1... Kh5 2. Ba2 Ba7 3. Bb1 B5b6 4. Ba2 B4a5 5. Bb1 B3b4 6. Ba2 B2a3 7. Bb1 Bb8 8. Ba2 B6a7 9. Bb1 B5b6 10. Ba2 B4a5 11. Bb1 B3b4 12. Ba2 Bca3 13. Bb1 Bdc1 14. Ba2 Bcb2 15. Bb1 Bd2 16. Ba2 Bdc1 17. Bb1 Be1 18. Ba2 Bed2 19. Bb1 Bf2 20. Ba2 Bfe1 21. Bb1 Bg1 22. Kxg1 Bf2+ 23. Kf1 Bde1 24. Ba2 Bg1 25. Kxe1 Bd2+ 26. Kf1 Bbc1 27. Bb1 Be1 28. Kxe1 Bd2+ 29. Kf1 Be1 30. Kxe1 Bc1 31. Kf1 Bd2 32. Ba2 Be1 33. Bb3 Ba3 34. Ba4 Bc1 35. Bb5 Bb4 36. Ba6 Bba3 37. Bc8 Ba5 38. Bd7 B5b4 39. Be8 Bb6 40. Bf7 B8a7 41. Kxe1 Bd2+ 42. Kf1 Be1 43. Bg8 Bc1 44. Bh7 Bcd2 45. Bxg6+ Kxg6 46. Kxg1 Kh7 47. Kh1 Kg8 48. Kg1 h5 49. Kh1 B7h6 50. Kg1 Bf8g7 51. Kh1 Kf8 52. Kg1 Ke8 53. Kh1 Kd8 54. Kg1 Kc8 55. Kh1 Kb8 56. Kg1 Ka8 57. Kh1 Bb8 58. Kg1 Ka7 59. Kh1 Ka6 60. Kg1 Ka5 61. Kh1 Ka4 62. Kg1 Ka3 63. Kh1 Kb2 64. Kg1 Kc1 65. Kh1 Kxd1 66. Kg1 Kc1 67. Kh1 Kb2 68. Kg1 Ka3 69. Kh1 Ka4 70. Kg1 Ka5 71. Kf1 Bc1 72. Kxe1 Ka6 73. Kd1 Ka7 74. Kxc1 Ka8 75. Kb1 B6a5 76. Ka2 Ba7 77. Kb3 Kb8 78. Ka4 Kc8 79. Kb5 Kd8 80. Ka6 Ke8 81. Kxa7 Kf8 82. Kb7 Kg8 83. Kc8 Bf8 84.  Kd7 Bh8g7 85. Ke8 Kh8 86. Kf7 Bb6 87. Kg6 Kg8 88. Kf5 Bh8 89. Kg6 Bf8g7 90. Kf5 Kf8 91. Kg6 Ke8 92. Kf5 Kd8 93. Kg6 Kc8 94. Kf5 Kb8 95. Kg6 Ka7 96. Kf7 Ka6 97. Ke8 Ka5 98. Kd7 Ka4 99. Kc8 Ka3 100. Kb7 Kb2 101. Ka6 Kc1 102. Kb5 Kd2 103. Ka4 Ke1 104. Kb3 Kf2 105. Ka2 Kxg2 106. Kb1 Kxh3 107. Kc1 g2 108. Kd1 g1=N 109. Ke1 Nxf3+ 110. Kf1 Kg4 111. Kg2 Bb2 112. exf3# 

Also, I would like to contribute to @supercat's idea of castling being the only move that works in a helpmate. I offer a helpmate in three setting that is much simpler than @supercat's problem.

[FEN "k7/Pp6/1P6/8/8/3p1p2/3P1P1p/R3K1nr w KQ - 0 1"]

1. O-O-O Nh3 2. Re1 Ng1 3. Re8#

Here is the longest problem in which White must castle on the first move:

[Title "Frank Richter, 10945 feenschach (207) 05-06/2014"]
[FEN "4k3/8/4p3/4P3/p1p5/PpPpPp1p/1P1PbP1P/R3K3 b Q - 0 1"]

1... Bf1 2. O-O-O Bg2 3. Rg1 Bh1 4. Rg3 Kf7 5. Rxf3+ Kg6 6. Rg3+ Kf5 7. f4 Be4 8. Rg5#

And when it is to be the last move:

[Title "Mirko Degenkolbe & Steven B. Dowd, 2010"]
[FEN "8/8/8/2p5/p1PNp3/P1Pp1n2/Pn1P1Ppp/RNB1Kbkr w Q - 0 1"]

1. Nxf3+ exf3 2. Bxb2 Be2 3. Bc1 Bd1 4. Bb2 Bb3 5. Bc1 Bxc4 6. Bb2 Bxa2 7. c4 Bxb1 8. Bd4 Ba2 9. O-O-O#
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