11

The known record for the longest sequence of consecutive checks (i.e. white checks, then black checks on the next move, white checks next, and so on) in a legal position without promoted pieces, is 37.

See http://timkr.home.xs4all.nl/chess/check.html

Is there a theoretical limit for the length of the sequence, or is a repetition possible, allowing checks forever?

  • 1
    hebdenbridgechessclub.co.uk/category/problems-and-compositions gives a legal position (with promotion, though) with a sequence of 53 moves. It's not a proof, but given the effort that has gone into it, I'd say there is some hard limit (i.e. no infinite cycle). – Eiko Sep 29 '16 at 8:04
  • What do you think makes my answer the one that should be accepted? – Rewan Demontay Apr 8 at 17:57
  • 54 half-moves is far more than I would have expected. It is unlikely that it can be broken. Moreover the other answers deal with pieces that do not exist in chess and one answered missed that neither the fifty-move rule plays a role nor it is not sufficient that ONE side gives checks (which would make the question utterly trivial). – Peter Apr 8 at 18:01
1

(If you are reading this, please fix the diagram for discovered checks, no promoted, pieces if you can as Nd4+ is not showing up for me, and delete this sentence when you are done.)

Foreword To Potential Downvoters: I have taken time to transcribe these games for you. This is for the benefit of all who come across this question.

I think that that 37 is the record so far WITHOUT promoted pieces. Here is the game for everyone’s convience.

[Title "G. Ponzetto Torre i Cavallo, 1993 "] 
[FEN "4r1Q1/B2nr3/5b2/8/4p3/4KbNq/ppppppp1/RR3Nkn w - - 0 1"]

1.Nh2+ f1=N+ 2.Rxf1+ gxf1=N+ 3.Ngxf1+ Bg5+ 4.Qxg5+ Bg2+ 5.Nf3+ exf3+ 6.Kd3+ Nc5+ 7.Qxc5+ Re3+ 8.Nxe3+ c1=N+ 9.Qxc1+ d1=Q+ 10.Qxd1+ e1=N+ 11.Qxe1+ Bf1+ 12.Nxf1+ f2+ 13.Ne3+ f1=Q+ 14.Qxf1+ Qxf1+ 15.Nxf1+ Re3+ 16.Nxe3+ b1=Q+ 17.Rxb1+ axb1=Q+ 18.Nc2+ Nf2+ 19.Bxf2+

One of the comments states that the record from promoted pieces is 53. However according to Tim Krabbe’s site (Journal Entry 387 https://timkr.home.xs4all.nl/chess2/diary.htm), this record has been broken since by 54. Here is that game as well, also for everyone’s convienence.

[Title "Alexey Khanyan, 2013"]
[FEN "5q2/2q1pn1B/R1Q1R3/1B2r1B1/1q1k1Knq/rr5Q/B2NPN1N/3Qb1QQ w - - 0 1"]

1. Qb6+ Rc5+ 2. Qd6+ Nxd6+ 3. Bf6+ Nxf6+ 4. Nfg4+ Qf2+ 5. Nhf3+ Kd5+ 6. e4+ Nfxe4+ 7. Nf6+ Nxf6+ 8. Be4+ Nfxe4+ 9. Qf5+ Nxf5+ 10. Rad6+ Nexd6+ 11. Bc4+ Nxc4+ 12. Re5+ Nxe5+ 13. Nc4+ Qfd2+ 14. Nxd2+ Rf3+ 15. Nxf3+ Qd2+ 16. Ncxd2+ Rc4+ 17. Bxc4+ Qxc4+ 18. Ne4+ Bd2+ 19. Qxd2+ Nd3+ 20. Qxd3+ Nd4+ 21. Nf6+ Qxf6+ 22. Qf5+ e5+ 23. Nxe5+ Rf3+ 24. Nxf3+ Qe5+ 25. Nxe5+ Nf3+ 26. Qd4+ Qxd4+ 27. Qe4+ Qxe4+

I think that the theoretical hard limit is limited to what category you choose-no promoted pieces and promoted allowed. Additionally, the current records can be refined until one pieceis left, as long as it gives check.


Slight Addition: Interestingly, to is possible to have mutual discovered checks. Here is the source, Journal Entry #366.

Here is the record without promoted pieces-11.

[Title "Alexey Khanyan, 2011"]
[FEN "q5B1/1p2rP1b/R1N1k1NR/5n2/4Kn1r/8/3B4/3Q4 w - - 0 1"]

1. f8=B+ Kd6+ 2. Nge5+ Ne6+ 3. Bf4+ Nd4+ 4. Ng6+ Nxf4+ 5. Nxe7+ bxa6+ 6. Nc6+

And with promoted pieces-17.

[Title "Alexey Khanyan, 2011"]
[FEN "Bb1R1r2/5B2/2RNN2q/RN1kn1n1/rb3K2/1R6/Q2n1nR1/2bR1r1B w - - 0 1"]

1. Nf5+ B4d6+ 2. Ned4+ Ngxf7+ 3. Rg5+ Nfe4+ 4. Rf3+ Nb3+ 5. Rcxc1+ Nc6+ 6. Nfxd6+ Ne5+ 7. N4f5+ Nd2+ 8. Nc4+ Nd7+ 9. Nd6+

I found a this brilliant example of mutual discovered checks elsewhere on Tim Krabe’s website (Journal Entry #265.)

He gives this series of 7 mutual discovered checks. What’s unique here is that all the moves, minus the first, are forced, which is what makes it unique.

[Title "V. Korolkov, 1940"]
[FEN "6B1/5Nb1/3p4/q2krP1R/Nn2p3/pPKnr3/Q1PB4/3R4 w - - 0 1"]

1. Nd8+ Re6+ 2. f6+ Ne5+ 3. Bxe3+ Nd3+ 4. b4#
5

Another way to get an infinite series of checks is by using a fairy piece. Consider this position, except that the black piece on e5 is not a knight but a camel (a (3,1)-leaper). Then the given sequence of four crosschecks restores the diagram position with White to move. (Unfortunately the PGN-viewer can't display it because of the fairy piece.)

[Title "Infinite series of checks"]
[fen "1k1b4/8/8/KN2n1r1/8/6B1/8/1R6 w - - 0 1"]
[StartFlipped "0"]

1. Nc7+ Cb6+ 2. Nb5+ Ce5+
  • Nice position. Is it your own composition or did you find it somewhere? – jknappen Sep 27 '16 at 9:54
  • 1
    It's my own composition, but it has been known before that such a construction is possible. I saw one on the web somewhere, but it is not on the Tim Krabbé web page cited above. Googling for it failed to locate anything suitable, so I constructed this one myself. – Rosie F Sep 27 '16 at 13:27
1

Edit: This doesn't work because I forgot about discovered checks. However, I think this progress is notable, so I'll leave the answer here.

Repetition is impossible.

First, there obviously cannot be any pawn moves, castling or captures.

Next, I claim that there cannot be any king moves. To prove this, note that a king move can give check only if it is a discovered check. So, in order for a king move to give check, the two kings must be in a line, whether vertical, horizontal, or diagonal. Given the position of one of the kings, the set of squares the other king can be on so that it can give check is the set of squares in the same line with the king and not the same square as the king or the squares next to that square. No two of these squares are adjacent, so the king cannot move from one such square to another in one move. Note that squares A and B are in a line if and only if squares B and A are in a line, so once one of the kings moves, they are no longer in a line, so no further king moves can give check. So, there is at most one king move in the cycle, but if it were a cycle the king would eventually have to move again, so there cannot be any king moves.

Therefore, there cannot be any knight checks, or else the king would have to move or the knight would have to be captured.

Therefore, all moves are moves by pieces, which means they must all block the previous checks.

For any metric on the set of squares of the chessboard, suppose it is true that, for any set of positions for the kings K1 and K2 and any square A which is in some line (vertical, horizontal, or diagonal) with the king, any blocking square B cannot increase the sum of the distances from the square to each of the kings (that is, d(A,K1)+d(A,K2)>=d(B,K1)+d(B,K2)). Then the sum of the distances to each of the squares of the kings must remain constant throughout the cycle.

It is easy to check that the following metrics satisfy that property: d(A, B)=|row(A)-row(B)| d(A, B)=|column(A)-column(B)| d(A, B)=|slope1diagonal(A)-slope1diagonal(B)| (By this I mean number the diagonals that are parallel to the A1H8 diagonal from 1-15) d(A, B)=|slope-1diagonal(A)-slope-1diagonal(B)| (Same as the previous, but parallel to the other diagonal)

In fact, it is easy to see that, for any of the above metrics, if the blocking square is not within the two parallel lines of those metrics (e.g. for the first metric, within the rectangle with sides made by the rows of each of the kings, and columns the sides of the board), then the sum of the distances will decrease with the next blocking square. Which would be a contradiction, so the blocking squares are restricted to be within each of the bounding parallel lines.

If the two kings are on the same row, column, or diagonal, using the argument from the paragraph above shows that all blocking squares must be in that row, column, or diagonal, clearly impossible.

Therefore, if we view the king positions as two opposite vertices of a rectangle with sides parallel to the sides of the board, by using the first two metrics, all blocking squares must be in or on the bounding rectangle. Using the other two metrics allows us to shrink this to a bounding parallelogram.

Note that the only possible blocking squares are those that are intersections of the rows, columns, and diagonals through each of the squares of the kings because they must give check to the other king and block a check. It is easy to see that there are always 2 possible blocking squares in the bounding parallelogram: the other two vertices of the parallelogram. But then, if we have one checking piece in each (which is necessary), then there are no squares from them to move to to give check, contradiction.

0

With Nightriders (NN) (a classical fairy piece) and Rooks, there are positions with mutual perpetual check. I attribute the discovery to this comment on chessvariants.org by H. G. Muller in 2012. The position is Black: Rb1, Rc1, Kb2; White NNa6, NNd6, Kb4; Black to move.

It is also possible to construct a mutual perpetual check with Nightriders and Bishops: Black: Ba2 Bb1 Kb3 (two Bishops of the same colour); White NNf8, NNh6, Ke6; Black to move.

-3

a player could be checked much more than 50 times in a row, the 50 moves rule winds back to zero if any pawn is moved or any piece captured. If white was checking black then a pawn move could be used to deliver a check every fiftieth move with the other 49 checks delivered by some other piece, since each of the 8 pawns can move 6 times, that's a potential 6 x 50 x 8 = 2400 checks in a row. Similarly black could escape checks by pawn moves leading to another 2400 potential checks.

30 pieces are capturable, you need one left to check, so maybe another 29x 50 = 1450 checks

so how about 6,250 checks in a row being possible - I think I could contrive a very boring game with that sort of number of checks in a row - as mentioned in a previous answer, you'd have to guard against 3 fold repetition, but I think that would be possible.

Infinity is definitely possible because of the fifty move rule which can only be wound back to zero by finite material leaving the board or finite pawn moves - chess itself has a longest possible game

  • 4
    It wasn't spelled out in the question itself, but from the linked example, it seems the OP is interested in sequences of moves where both sides are checking each time, and is wondering whether it is theoretically possible to construct a position allowing an infinite loop of such mutual checks back and forth. – ETD Apr 9 '15 at 22:55
  • @etd you got it! – Peter Apr 12 '15 at 22:46
  • Do not consider the 50-move-rule. The question is a theoretical one. – Peter Apr 12 '15 at 22:47
  • I see no reason for downvoting this answer. The question is vaguely worded that I couldn't get the OP's idea until after reading some of the answers. Anyone would understand "consecutive checks" as checks by the same side on every move. This has to be clearly stated in the question. – Wais Kamal Dec 18 '18 at 21:31
-3

Because of the fifty move rule, the limit is 50. If you ignore the 50 move rule, then there still is a limit because there is a finite number of chess positions. The fifty-move rule in chess states that a player can claim a draw if no capture has been made and no pawn has been moved in the last fifty moves (for this purpose a "move" consists of a player completing his turn followed by his opponent completing his turn).

Three fold repetition is when the position on the board repeats three times, a player can claim a draw.

  • 3
    The question is clearly of a more combinatorial nature. I don't think the practical rules of tournament chess are relevant to the answer. Or to put it differently: There is no player who would claim a draw. Only a chess problem composer trying to construct a position with no relevance for tournament chess whatsoever. – BlindKungFuMaster Apr 1 '15 at 14:59
  • well, then isn't the answer trivial because of repetition that we have allseen in games? I might be missing something. – CognisMantis Apr 1 '15 at 15:17
  • 1
    @CognisMantis I don't remember seeing a repetition where each move by both players is a check. – JiK Apr 1 '15 at 16:05
  • ok, i see where I went wrong. – CognisMantis Apr 1 '15 at 16:37
  • 1
    Easy mistake to make. I think I misunderstand every second of Peter's questions … at least initially. – BlindKungFuMaster Apr 2 '15 at 8:24

protected by Phonon Apr 7 at 3:35

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