Answers.Body: <p>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.)</p>
  
  <pre><code>[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+
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-09-27T09:54:37.570Z
Answers.Text: Nice position. Is it your own composition or did you find it somewhere?
Answers.UserDisplayName: Sir Cornflakes
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-09-27T13:27:21.823Z
Answers.Text: 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.
Answers.UserDisplayName: Rosie F
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-02-06T10:13:48.303Z
Answers.Text: @RosieF: To my best knowledge, it was done first by Noam Elkies and it's even somewhere here on StackExchange (maybe math, not chess).
Answers.UserDisplayName: Hauke Reddmann
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-09-27T07:55:28.097Z
Answers.LastActivityDate: 2016-09-27T07:55:28.097Z
Answers.UserDisplayName: Rosie F
Answers.Body: <p>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.</p>
  
  <p>Repetition is impossible.</p>
  
  <p>First, there obviously cannot be any pawn moves, castling or captures.</p>
  
  <p>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.</p>
  
  <p>Therefore, there cannot be any knight checks, or else the king would have to move or the knight would have to be captured.</p>
  
  <p>Therefore, all moves are moves by pieces, which means they must all block the previous checks.</p>
  
  <p>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.</p>
  
  <p>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)</p>
  
  <p>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.</p>
  
  <p>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.</p>
  
  <p>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.</p>
  
  <p>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.</p>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-09-25T17:40:43.077Z
Answers.LastActivityDate: 2016-09-25T17:58:50.413Z
Answers.LastEditDate: 2016-09-25T17:58:50.413Z
Answers.UserDisplayName: alphacapture
Answers.Body: <p>With Nightriders (NN) (a classical fairy piece) and Rooks, there are positions with mutual perpetual check. A position was given by <a href="https://pdb.dieschwalbe.de/search.jsp?expression=STIP%3D%27perpetual%27%0D%0A" rel="nofollow noreferrer">Arthur J. Roycroft</a> in 1976.</p>
  <p>(Formerly, I attributed it to <a href="https://www.chessvariants.com/index/listcomments.php?id=29540" rel="nofollow noreferrer">this comment on chessvariants.org</a> by H. G. Muller in 2012. The position is Black: Rb1, Rc1, Kb2; White NNa6, NNd6, Kb4; Black to move. Thanks to @RosieF for finding the reference to Roycroft, it was very recently added to the Schwalbe Problem Database.)</p>
  <p>It is also possible to construct a <a href="https://www.chessvariants.com/index/listcomments.php?id=29563" rel="nofollow noreferrer">mutual perpetual check with Nightriders and Bishops</a>: Black: Ba2 Bb1 Kb3 (two Bishops of the same colour); White NNf8, NNh6, Ke6; Black to move.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-02-05T09:14:49.653Z
Answers.Text: There has been a whole article on fairy perpetual check (grasshoppers work fine too, Circe also), but I don't remember where and when (I guess in "feenschach").
Answers.UserDisplayName: Hauke Reddmann
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-02-13T11:05:53.603Z
Answers.Text: Problem P1385725 at [PDB](https://pdb.dieschwalbe.de/search.jsp) shows this; its credit is Arthur John Roycroft, The Problemist, Jul 1976, p. 57, no. 3 after T. R. Dawson.
Answers.UserDisplayName: Rosie F
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-02-13T23:33:26.777Z
Answers.Text: @RosieF Thanks for the reference, added it to my answer.
Answers.UserDisplayName: Sir Cornflakes
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-02-13T23:42:31.857Z
Answers.Text: @jk-ReinstateMonica It so happens that I put it into the database. You're very welcome, ;). I also shared it here: https://chess.stackexchange.com/a/33883/15543
Answers.UserDisplayName: Rewan Demontay
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2016-09-26T10:29:39.653Z
Answers.LastActivityDate: 2021-02-14T01:25:40.320Z
Answers.LastEditDate: 2021-02-14T01:25:40.320Z
Answers.UserDisplayName: Sir Cornflakes
Answers.Body: <p>37 is the record so far <em>without promoted pieces.</em></p>
  <pre><code>[Title &quot;Giuseppe Ponzetto, Torre i Cavallo 1993, 37 Consecutive Checks&quot;] 
  [FEN &quot;4r1Q1/B2nr3/5b2/8/4p3/4KbNq/ppppppp1/RR3Nkn w - - 0 1&quot;]
  
  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+
  </code></pre>
  <p>A comment says that the record <em>with promoted pieces</em> is 53. However, according to Tim Krabbe's <a href="https://timkr.home.xs4all.nl/chess2/diary.htm" rel="nofollow noreferrer">Diary Entry #387</a>, the record is actually 54.</p>
  <pre><code>[Title &quot;Alexey Khanyan Sampsa Lahtonen, Tim Krabbe's Website Diary Entry #387 2013, 54 Consecutive Checks&quot;]
  [FEN &quot;5q2/2q1pn1B/R1Q1R3/1B2r1B1/1q1k1Knq/rr5Q/B2NPN1N/3Qb1QQ w - - 0 1&quot;]
  
  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+
  </code></pre>
  <p>I think that the theoretical hard limit is limited by which category you choose of promoted pieces allowed and not allowed. Additionally, the current records can be refined until only two kings and one piece giving check are left on the board.</p>
  <p>In the aforementioned Diary #387, Tim Krabbe even remarks that <strong>&quot;This is probably not the end to this task. The fact that in Lahtonen's 53 checks only 5 pieces remain in the final position and in the new 54-record there are 6, leads Khanyan to believe that if he could get this down to 5 pieces, too, he could add two checks and make it to 56. &quot;56 checks in a row do not seem to me as being higher than the clouds!&quot;</strong></p>
  <hr />
  <p>Additionally, just for fun, there are also records for <em>discovered checks.</em> Tim Krabbe's site is also the source.</p>
  <p>As Krabbe first shared in <a href="https://timkr.home.xs4all.nl/chess2/diary_7.htm" rel="nofollow noreferrer">Diary Entry #125,</a> and later in <a href="https://timkr.home.xs4all.nl/chess2/diary16txt.htm" rel="nofollow noreferrer">#308,</a>, the record without promoted pieces is 11 discovered checks.</p>
  <pre><code>[Title &quot;Ottavio Stocchi, L'Echiquier 1930, 11 Consecutive Discovered Checks&quot;]
  [FEN &quot;2R4B/3r2P1/4P2b/2N5/3k2BR/2N1n3/r1nK2bq/2Q5 w - - 0 1&quot;]
  
  1. g8=B+ Kc4+ 2. Nd3+ Bc6+ 3. Be2+ Nd4+ 4. Nxa2+ Nec2+ 5. Nf4+ Nxe2+ 6. exd7+
  </code></pre>
  <p>In Diary Entry <a href="https://timkr.home.xs4all.nl/chess2/diary19txt.htm" rel="nofollow noreferrer">#366</a> the record for with promoted piece is given. The author's full name and source date come from the <a href="https://pdb.dieschwalbe.de/P1183629" rel="nofollow noreferrer">Die Schwalbe Chess Problem Database.</a></p>
  <pre><code>[Title &quot;Werner Frangen, Feenschach 08/1974, 17 Consecutive Discovered Checks&quot;]
  [FEN &quot;B3R2B/1R1q3b/b4R2/RrNkPR2/4N3/1NbK2nr/B1rn3B/1b1r3Q w - - 0 1&quot;]
  
  1. e6+ Nxf5+ 2. Ng3+ Nf3+ 3. Nd2+ Rb3+ 4. Nxa6+ Bxa5+ 5. Rxb3+ Rc6+ 6. Rxb1+ Ke5+ 7. exd7+ Ne7+ 8. Ne4+ Nxh2+ 9. Rf3+
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-07-02T06:48:36.077Z
Answers.Text: According to Tim Krabbe's website, Alexey Khanyan first acheieved 53, and then improved it to 54. The comment you mentioned may be about the first one.
Answers.UserDisplayName: Cyriac Antony
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-02-05T15:11:51.670Z
Answers.Text: @HaukeReddmann Fixed! As for crosschecks, Khanyan has already done 9 discovered crosschecks, and 12 with all types, on SuperProblem.ru. Just scroll down a bit: http://superproblem.ru/archive/S_record-X.html
Answers.UserDisplayName: Rewan Demontay
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-02-05T17:29:24.190Z
Answers.Text: @RewanDemontay: THX for the info. I did 10 (w/o promoted): http://bambam42.de/problem/problem3.html Look under "feenschach" - or did I violate a condition? Or miscounted?
Answers.UserDisplayName: Hauke Reddmann
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-02-05T18:06:57.227Z
Answers.Text: Post-Edit Edit: Somehow the diagram was botched, Bg5 Rh5 must be Bg4 Rh4. Also note I optimized crosschecks, NOT discovered checks, two moves are non-battery crosschecks.
Answers.UserDisplayName: Hauke Reddmann
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-02-05T18:29:33.233Z
Answers.Text: @HaukeReddmann Without promoted pieces and all types, Khanyan has done 12 crosschecks, same link.
Answers.UserDisplayName: Rewan Demontay
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-02-06T08:31:57.807Z
Answers.Text: This is a terrific survey, and the new cross-check problems deserve a better consolidated home than just the comments. *However* with the greatest respect, all this valuable constructive data is only supporting evidence towards the actual question asked, is there a theoretical limit? I wish the questioner had stuck to their guns and not "accepted' this great answer, because we're really not there yet.
Answers.UserDisplayName: Laska
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Answers.CreationDate: 2019-04-07T01:02:37.367Z
Answers.LastActivityDate: 2023-09-08T06:36:32.470Z
Answers.LastEditDate: 2023-09-08T06:36:32.470Z
Answers.UserDisplayName: Rewan Demontay
Body: <p>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 <a href="http://timkr.home.xs4all.nl/chess/check.html" rel="noreferrer">this article</a> on Tim Krabbe's site entitled &quot;Check!&quot;</p>
  <p>Is there a theoretical limit for the length of such a sequence, or is a repetition possible, allowing checks forever?</p>
  
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2016-09-29T08:04:47.793Z
Comments.Text: http://www.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).
Comments.UserDisplayName: Eiko
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2019-04-08T18:01:13.687Z
Comments.Text: 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).
Comments.UserDisplayName: Peter
Tags: <problems><check>
Title: What is the theoretical limit for the number of consecutive checks?
ThreadText: # What is the theoretical limit for the number of consecutive checks?
  
  (Asked by: username_0 on 2015-04-01)
  
  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 [this article](http://timkr.home.xs4all.nl/chess/check.html) on Tim Krabbe's site entitled "Check!"
  
  Is there a theoretical limit for the length of such a sequence, or is a repetition possible, allowing checks forever?
  
  **Tags:** problems, check
  
  > **Comments:**
  > **username_1**: http://www.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).  
  > **username_0**: 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).  
  
  ---
  
  ## 4 answers
  
  ### Answer by username_6 on 2019-04-07. Score: 12 | Accepted answer
  
  37 is the record so far without promoted pieces.
  [Title "Giuseppe Ponzetto, Torre i Cavallo 1993, 37 Consecutive Checks"] 
  [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+
  
  A comment says that the record with promoted pieces is 53. However, according to Tim Krabbe's Diary Entry #387, the record is actually 54.
  [Title "Alexey Khanyan Sampsa Lahtonen, Tim Krabbe's Website Diary Entry #387 2013, 54 Consecutive Checks"]
  [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 by which category you choose of promoted pieces allowed and not allowed. Additionally, the current records can be refined until only two kings and one piece giving check are left on the board.
  In the aforementioned Diary #387, Tim Krabbe even remarks that "This is probably not the end to this task. The fact that in Lahtonen's 53 checks only 5 pieces remain in the final position and in the new 54-record there are 6, leads Khanyan to believe that if he could get this down to 5 pieces, too, he could add two checks and make it to 56. "56 checks in a row do not seem to me as being higher than the clouds!"
  
  Additionally, just for fun, there are also records for discovered checks. Tim Krabbe's site is also the source.
  As Krabbe first shared in Diary Entry #125, and later in #308,, the record without promoted pieces is 11 discovered checks.
  [Title "Ottavio Stocchi, L'Echiquier 1930, 11 Consecutive Discovered Checks"]
  [FEN "2R4B/3r2P1/4P2b/2N5/3k2BR/2N1n3/r1nK2bq/2Q5 w - - 0 1"]
  
  1. g8=B+ Kc4+ 2. Nd3+ Bc6+ 3. Be2+ Nd4+ 4. Nxa2+ Nec2+ 5. Nf4+ Nxe2+ 6. exd7+
  
  In Diary Entry #366 the record for with promoted piece is given. The author's full name and source date come from the Die Schwalbe Chess Problem Database.
  [Title "Werner Frangen, Feenschach 08/1974, 17 Consecutive Discovered Checks"]
  [FEN "B3R2B/1R1q3b/b4R2/RrNkPR2/4N3/1NbK2nr/B1rn3B/1b1r3Q w - - 0 1"]
  
  1. e6+ Nxf5+ 2. Ng3+ Nf3+ 3. Nd2+ Rb3+ 4. Nxa6+ Bxa5+ 5. Rxb3+ Rc6+ 6. Rxb1+ Ke5+ 7. exd7+ Ne7+ 8. Ne4+ Nxh2+ 9. Rf3+
  
  > **Comments:**
  > **username_7**: According to Tim Krabbe's website, Alexey Khanyan first acheieved 53, and then improved it to 54. The comment you mentioned may be about the first one.  
  > **username_6**: @HaukeReddmann Fixed! As for crosschecks, Khanyan has already done 9 discovered crosschecks, and 12 with all types, on SuperProblem.ru. Just scroll down a bit: http://superproblem.ru/archive/S_record-X.html  
  > **username_4**: @RewanDemontay: THX for the info. I did 10 (w/o promoted): http://bambam42.de/problem/problem3.html Look under "feenschach" - or did I violate a condition? Or miscounted?  
  > **username_4**: Post-Edit Edit: Somehow the diagram was botched, Bg5 Rh5 must be Bg4 Rh4. Also note I optimized crosschecks, NOT discovered checks, two moves are non-battery crosschecks.  
  > **username_6**: @HaukeReddmann Without promoted pieces and all types, Khanyan has done 12 crosschecks, same link.  
  > **username_8**: This is a terrific survey, and the new cross-check problems deserve a better consolidated home than just the comments. *However* with the greatest respect, all this valuable constructive data is only supporting evidence towards the actual question asked, is there a theoretical limit? I wish the questioner had stuck to their guns and not "accepted' this great answer, because we're really not there yet.  
  
  ---
  
  ### Answer by username_2 on 2016-09-27. Score: 10
  
  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+
  
  > **Comments:**
  > **username_3**: Nice position. Is it your own composition or did you find it somewhere?  
  > **username_2**: 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.  
  > **username_4**: @RosieF: To my best knowledge, it was done first by Noam Elkies and it's even somewhere here on StackExchange (maybe math, not chess).  
  
  ---
  
  ### Answer by username_5 on 2016-09-25. Score: 3
  
  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.
  
  
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  ### Answer by username_3 on 2016-09-26. Score: 1
  
  With Nightriders (NN) (a classical fairy piece) and Rooks, there are positions with mutual perpetual check. A position was given by [Arthur J. Roycroft](https://pdb.dieschwalbe.de/search.jsp?expression=STIP%3D%27perpetual%27%0D%0A) in 1976.
  
  (Formerly, I attributed it to [this comment on chessvariants.org](https://www.chessvariants.com/index/listcomments.php?id=29540) by H. G. Muller in 2012. The position is Black: Rb1, Rc1, Kb2; White NNa6, NNd6, Kb4; Black to move. Thanks to @RosieF for finding the reference to Roycroft, it was very recently added to the Schwalbe Problem Database.)
  
  It is also possible to construct a [mutual perpetual check with Nightriders and Bishops](https://www.chessvariants.com/index/listcomments.php?id=29563): Black: Ba2 Bb1 Kb3 (two Bishops of the same colour); White NNf8, NNh6, Ke6; Black to move.
  
  > **Comments:**
  > **username_4**: There has been a whole article on fairy perpetual check (grasshoppers work fine too, Circe also), but I don't remember where and when (I guess in "feenschach").  
  > **username_2**: Problem P1385725 at [PDB](https://pdb.dieschwalbe.de/search.jsp) shows this; its credit is Arthur John Roycroft, The Problemist, Jul 1976, p. 57, no. 3 after T. R. Dawson.  
  > **username_3**: @RosieF Thanks for the reference, added it to my answer.  
  > **username_6**: @jk-ReinstateMonica It so happens that I put it into the database. You're very welcome, ;). I also shared it here: https://chess.stackexchange.com/a/33883/15543  
  
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