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.

  • 1
    First you need sufficient mating material: a pawn, a queen, a rook, two bishops, or a bishop and knight. With a lone knight or bishop you can still mate if the opponent has other pieces or pawns to help smother their king. Exceptions are if the position is extremely closed with no way of opening it, or if the moves are forced to lead to stalemate, checkmate for the side that ran out of time, or capture of the only mating material left. The exceptions may be contrived, but I suspect they are usually easy to figure out. I look forward to counterexamples! – itub Sep 18 '17 at 20:33
  • @itub Those were my thoughts too, that exceptions should generally be easy to figure out. If there's a big material advantage, it's just a matter of giving up all/enough material to allow the other room to mate you, while in a closed position again it's either a matter of opening up the position and giving up all material, or realizing the position cannot be opened (which should be easy to assess?). But there may be some contrived helpmate-tricks or proof-game-tricks which I am not aware of to find exceptions. – TMM Sep 18 '17 at 20:39
  • PS: I forgot you can also helpmate with two knights! – itub Sep 18 '17 at 20:51
  • In practice, this does not usually seem to be difficult. If you think it might be, the onus is on you to demonstrate a difficult example Recollect that in helpmate problems the difficulty is often to find the order of the moves so that everything is legal. That does not apply here. – Philip Roe Sep 18 '17 at 22:17
  • This should not be difficult to figure out generally, because it does not need much material to mate. There are two potential scenarios to make it more difficult I could think of: (1) a very closed/blocked positions where it is not clear whether pieces could be "freed". or (2) some position where helpmate can only be achieved if say it is white's move. (thinking about positions with a knight, since you cannot lose/gain a tempo with a knight). – user1583209 Sep 19 '17 at 1:10
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"]
  • 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
  • Those are neat puzzles, and these are exactly the type of positions I'm looking for! In the second one I suppose white goes Be8-f7xg8, picks up only the g3-pawn via d4, picks up the f8-bishop (preventing stalemate due to black having g4-g3), and then mates easily. (Unless I'm miscounting that appears to be mate in 24 though.) – TMM Sep 23 '17 at 23:51
  • Ah, never mind, the black king is on g8 after move 19, so white needs to lose a tempo. – TMM Sep 23 '17 at 23:53
  • 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
  • Oh, I thought your edit was to include the position from your comment, but I see that's a different position now. So it's just a coincidence both are mate in 25. – TMM Sep 24 '17 at 0:26
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)

  • Nice! Without the restriction of the number of moves to helpmate, only the first few moves are forced though - once the threat of white getting mated has been averted, there are many ways to mate black. (Still the first few moves take some effort to find.) – TMM Sep 22 '17 at 13:09
  • 1
    Note that the question "Can White win this on time if Black (on the move) couldn't find a Queen or Rook before their flag fell?" is not precise enough: if Black can't find a Queen, looks at her clock, panics, grabs a Bishop, plays h1B and get flagged, she will still save half a point since no mate for White is possible anymore after 1...h1B?! – Evargalo Sep 22 '17 at 15:09
  • 4
    The solution can also be found in this nice paper: plouffe.fr/simon/OEIS/citations/knight.pdf but its author must be too modest... – Evargalo Sep 22 '17 at 15:50
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

  • Hm, I think the answer is Kxe8. – Deduplicator Oct 14 '17 at 10:11
  • @Deduplicator: hi again! You need to ask yourself if in the prior position there is any way that the game can end in a checkmate. This is quite a hard problem I think - but that's what OP asked for :D:D – Laska Oct 14 '17 at 10:28
  • I'm with Deduplicator here. Are we missing something? Or are you somehow counting "incomplete" moves like black playing axb3 and already taking white's piece on b3, but not yet moving the pawn? (I don't see any opportunities for en passant, castling, etc. and otherwise this seems like a dead draw. White can maneuver the bishop all the way to h7, but that doesn't change anything it seems.) – TMM Oct 15 '17 at 17:43
  • @TMM: you asked "whether positions exist for which it is hard to decide if a mating sequence exists". I hope you will shortly believe that the answer is "yes" :D. The current position is indeed a dead draw for a subtle reason. However before the last move, the game could have ended in helpmate. It follows that the last move was not Kxe8 since that would have been forced. By elimination there's only one possible last move. There are no tricks: there is something you haven't spotted. – Laska Oct 16 '17 at 4:27
  • @TMM: I will add a hint to the answer – Laska Oct 18 '17 at 4:17
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.

  • The 75 move rule is irrelevant in this case (unless the flag falls after the 75th move with no pawn moves or captures has been made). Article 6.9 says only that "the game is drawn if the position is such that the opponent cannot checkmate the player’s king by any possible series of legal moves." Note "any possible series of legal moves" with no limit as to number. – Brian Towers Sep 20 '17 at 22:42
  • (+1) It's still not quite a conclusive answer, but it shows some interesting scenarios I hadn't thought of before. (And that again highlights why it's risky to go from "we don't think it's possible" to "it's certainly not possible" - there may always be some creative position we hadn't thought of before.) – TMM Sep 21 '17 at 20:18
  • @TMM: BTW, it's possible to have positions where it would be impossible to achieve a helpmate without castling. – supercat Sep 21 '17 at 20:59
  • @TMM: I just posted an example of such a situation. Cute, eh? – supercat Sep 21 '17 at 23:53
  • So that's an even stronger example than the one I posted: where it is not even sufficient to know the immediate history to determine if a helpmate is possible, but the entire history of the game must be known. Thanks for sharing! (Still it would be even nicer to have a situation where, even with knowledge of all previous moves, it would take someone a considerable amount of time to see if mate is possible.) – TMM Sep 21 '17 at 23:58
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.

  • 3
    You're missing the point of my question, and your answer provides no answer to the question I asked, which is whether positions exist for which it is hard to decide whether a helpmate exists. For those en passant and castling are not necessarily irrelevant. And you don't need to write moves in a rapid game, so it's not an as nonsensical scenario as you make it out to be. – TMM Sep 18 '17 at 22:40
  • Helpmate when there is no limit on the number of moves is always trivial. That and your suggestion of deliberately breaking the rules to somehow "thwart" the arbiter makes your question a particularly silly one. – Brian Towers Sep 19 '17 at 10:11
  • Do you even know how to read? In a rapid game you are not required to write down moves, so it's not a matter of "thwarting the arbiter" but simply adhering to the rules. And again, I am looking either for a convincing argument/proof that this is always trivial, or a counterexample. Saying "it is trivial" is not a convincing argument. – TMM Sep 19 '17 at 10:15
  • 1
    @BrianTowers: remember there exist joke problems where white doesn't have a choice but mate black in x moves, probably also where stalemate is forced in x moves. You'd at least have to make sure that one of those things isn't happening. And anyway it's not a question about practical games, but rather about the theoretical possibility of hard to decide positions. – RemcoGerlich Sep 19 '17 at 11:45
  • @RemcoGerlich By definition "problems where white doesn't have a choice but mate black in x moves" are not positions which are hard to decide. It is important only to look for such forced moves. In a FIDE rated tournament where the arbiter is likely to send the games to FIDE (varies from federation to federation, I know) the arbiter will see this when he enters the game into Chessbase with the engine running (which I normally do) when the result can be corrected. The only possible hard positions are blocked ones and they are not hard given the license to play silly moves to open them. – Brian Towers Sep 19 '17 at 15:45
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...

  • There's a difference between the world of chess compositions and the world of chess games. You will save a lot of petty arguments if you park your problem firmly in the well-policed domain of compositions, rather than worrying about arbiters, score-sheets, etc. Only the basic rules apply by default to compositions. There are conventions to say whether you can assume an apparent castling or e.p. is viable. You can mention in the stipulation that e.g. White has just timed out. E.p. is by default not allowed unless you can prove using retro logic that the last move was double pawn hop. – Laska Oct 14 '17 at 4:57
  • Nit: Black can't time out when White's on move. – Laska Oct 14 '17 at 5:02
  • @Laska What if Black moves his piece, and just before pushing the button time runs out? How is that counted? – Deduplicator Oct 14 '17 at 10:04
  • @Deduplicator the turn is completed when the button is pushed. Until that point the move has not been made. – Laska Oct 14 '17 at 10:27
  • 1
    @Laska When I posted this answer, we were still exploring better examples. By now people have come up with better examples for sure. And re: nit: black could make a move, press the clock and then white could notice black's flag fell. Although I guess that would officially still count as "black's move" since he did not complete his move in a legal manner. (But I wrote "either side times out" - if white times out, that would not change the situation of not being able to determine whether a winning sequence of moves exists.) – TMM Oct 14 '17 at 15:24
2

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

From one of C.J. Morse's chess records book, a preview of which can be seen on Google Books, a dual free problem of a 28 move helpmate is given.

[Title "A. Hegermann, Black To Move, White To Mate 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#

If you don't mind an illegal position, I've found this helpmate in 45 by Ottó Titusz Bláthy of course The original be problem was 47, but it was cooked down to 45, so that it what . The problem can be found in the Schwalbe PBD with the ID number of P0569739.

However, the given cook solution has notations that are are all wrong, and as such they are unreadable as frick to me. Until I can figure it out, I will be posting a solution of 46 moves, which is a solution that I could actually fugure out. If you can dig out that 45, please do let me know.

[Title "Ottó Titusz Bláthy, 1924, Black To Move, White To Mate In 45 Moves"]
[FEN "7k/2p5/2p5/2p1p3/2PbPp2/2pBpPp1/2P1P1P1/4N1K1 b - - 0 1"]

1... Kg8 2. Kh1 Kf8 3. Kg1 Ke8 4. Kh1 Kd8 5. Kg1 Kc8 6. Kh1 Kb7 7. Kg1 Kb6 8. Kh1 Ka5 9. Kg1 Kb4 10. Kh1 Ka3 11. Kg1 Kb2 12. Kh1 Kc1 13. Kg1 Kd1 14. Kh1 Kxe1 15. Kg1 Kd1 16. Kf1 Kc1 17. Ke1 Kb1 18. Kd1 Ka2 19. Kc1 Ka3 20. Kb1 Kb4 21. Ka2 Ka5 22. Ka3 Kb6 23. Ka4 Kb7 24. Ka5 Kc8 25. Ka6 Kd7 26. Kb7 Kd8 27.  Kxc6 Kc8 28. Kd5 Kb7 29. Ke6 Ka6 30. Kf5 Ka5 31. Kg4 Kb4 32. Kh3 Ka3 33. Kg4 Kb2 34. Kh4 Kc1 35. Kh5 Kd1 36. Kg5 Ke1 37. Kg4 Kf1 38. Kh5 Kxg2 39. Kh4 Kh1 40. Kh3 g2 41. Kg4 g1=N 42. Kh4 Nxe2 43. Bxe2 c6 44. Bf1 e2 45. Kh3 Bg1 46. Bg2#

I made my own personal extension of Otto Blathy’s selfmate that uses a ton of extra material.

[Title "White To Move, Helpmate In 59 Moves"]
[FEN "8/2p1p3/2PbP3/2pBp3/2PbPp1k/2pBpPp1/2P1P1P1/4NKq1 w - - 0 1"]

1. Kxg1 Kg5 2. Kh1 Kf6 3. Kg1 Kg7 4. Kh1 Kf8 5. Kg1 Ke8 6. Kh1 Kd8 7. Kg1 Kc8 8. Kh1 Kb8 9. Kg1 Ka7 10. Kh1 Kb6 11. Kg1 Ka5 12. Kh1 Kb4 13. Kg1 Ka3 14. Kh1 Kb2 15. Kg1 Kc1 16. Kh1 Kd1 17. Kg1 Kxe1 18. Kh1 Kd1 19. Kg1 Kc1 20. Kf1 Kb2 21. Ke1 Ka3 22. Kd1 Kb4 23. Kc1 Ka5 24. Kb1 Kb6 25. Ka2 Ka7 26. Kb3 Kb8 27. Ka4 Kc8 28. Kb5 Kd8 29. Ka6 Ke8 30. Kb7 Kf8 31. Kc8 Kg7 32. Kd8 Kh6 33. Ke8 Kh7 34. Kf7 Kh8 35. Kg6 Kg8 36. Kh5 Kf8 37. Kh4 Ke8 38. Kg4 Kd8 39. Kh4 Kc8 40. Kg4 Kb8 41. Kh4 Ka7 42. Kg4 Kb6 43. Kh4 Ka5 44. Kg4 Kb4 45. Kh4 Ka3 46. Kg4 Kb2 47. Kh4 Kc1 48. Kg4 Kd1 49. Kh4 Ke1 50. Kg4 Kf1 51. Kh4 Kxg2 52. Kg4 Kh2 53. Kh4 g2 54. Kg4 g1=N 55. Kh4 Nxe2 56. Bxe2 Kh1 57. Bf1 e2 58. Kh3 Bg1 59. Bg2#

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#

There's also this problem in of arbitrary length by someone else that contains no promoted pieces, in which castling must be the first move that is 7 moves long.

[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 bethe 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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