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The "Fireside Book of Chess" by I. Chernev and F. Reinfeld includes the following diagram

[fen "3nk3/3NN3/3PP3/3BB3/3PP3/3PP3/3PP3/2RQKR2 w - - 0 1"]

Composed by J.N. Babson for Brentano's Chess Monthly in 1882. Mate on the 1220th move, after compelling Black to make three successive and complete Knight's tours.

(Note that there is a mate in one. The problem asks for something more specific.)

Facts:

  • Babson is a real composer, famous for long mate problems.
  • Brentano's Chess Monthly was a real publication from 1880 to 1882, and Babson did have problems published there.
  • A FIDE page that lists people with FIDE titles has a biographical sketch of Babson citing a 1220 move problem, and a 1990 move problem on a 10x10 board.

Issues:

  • I can't find any reference to this problem on the web, other than the above mentioned book.
  • Nobody ever mentions any problem with this many moves... anywhere!
  • A knight tour is the motion of a knight through all the board squares. Three of these would mean only 192 moves.

So here are my questions:

  1. Is the problem for real?
  2. How should one interpret the knight tour condition? Probably the knight will capture most of the central pieces, but do the other pieces have to move aside to let the knight through?
  3. What is the solution?
  4. Why is this not more widely known?
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  • 8
    Interesting question, +1. But I'm not sure why you compare this to the 517-mover of Konoval and Bourzutschky (chess.stackexchange.com/a/674/167). For that position, it is optimal play by both sides that results in conversion to a won 6-piece endgame after 517 moves. The purported 1220 moves for this position has nothing to do with optimal play, of course, as it's noted in the question that White's best play is simply mate in one: 1.Rf8#.
    – ETD
    Sep 8, 2013 at 14:27
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    @Ed Dean: The instructions of a problem tell you what your goal is. In the 517-mover the goal is to reach a theoretically won position through optimal play n both sides. In this other problem the instructions say nothing about optimal play; the mate in 1 is irrelevant because it does not fulfill the goal. I compare them both because they both need a huge number of plays to reach the requested goal.
    – yrodro
    Sep 8, 2013 at 16:29
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    Oh, I understand that these problems have very different goals. Indeed, that was exactly my point: it just struck me as uninformative to compare the numbers involved, since it's 517 apples and 1220 oranges, precisely because the stated goals are different. (Just to emphasize: I consider this only a minor quibble, and won't harp on it any further. I very much like the question, and hope to see an answer that touches on the intended solution.)
    – ETD
    Sep 8, 2013 at 16:56
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    Sidenote: This should be termed a "fairy chess problem", due to the side conditions. And when it comes to fairy chess, ~1000 moves are nothing, if my memory serves me well. (I dimly remember one with ~10000.)
    – Hauke Reddmann
    Aug 28, 2016 at 20:45
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    Mike Fox & Richard James copied it in their The Complete Chess Addict (pub.Faber 1987), p.174, but don't give any clue as to a solution.
    – Rosie F
    Oct 27, 2018 at 17:42

3 Answers 3

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After some searching, it turns out that this monster is very much a real chess problem. The official name for it is "The Obelisk."

The earliest known reference I have found is from February of 1882, mere weeks after its publication. Bizarrely, it comes from an Australian magazine. The site I found it on is trove.nla.gov.au.

It appears in the February 25th, 1882 edition of the South Australian Weekly Chronicle, published in Adelaide, on page 25. It is part of an overall review of Brentano's January 1882 issue.

Here is a clipping of the relevant section to read for yourself.

enter image description here

I also found mention of it in the "American Chess Review, Volume 1, Issues 1-6"readable on Google books. It is dated to 1886, four years later.

The information is given in the middle of the page. The position itself is given near the end of the page in descriptive notation. More about what that is on Wikipedia. For your information, the "S" is the German notation for the knight; it was semi-standard back then.

enter image description here

Here are later sources.

Chess Review 1936, issue 37 page 17; viewable online in the US Chess Federation's online archive

Curious Chess Facts, by Irving Chernev, 1937

Wonders and Curiosities of Chess, page 205, by Irving Chernev, 1974

The Complete Chess Addict, by Mike Fox & Richard James, 1987

The Even More Complete Chess Addict, by Mike Fox & Richard James, 1994

Unsurprisingly, there is no solution given anywhere at all. The reason for this is that the magazine ceased publication soon after the mate in 1220 was printed. Thus, another edition with the solution was never made.

As for how to interpret the stipulation, I believe it is very simple. White forces the Black knight to visit each and every square three times. Then White gives checkmates. Most of the moves come from White's pieces dancing around to do said forcing. If anyone is crazy enough to try and solve it, they will surely find a much shorter solution than whatever Babson had in mind.

However, when it comes to solving the problem, I have found a sequence that allows for all of White's pawns to be promoted. This is surely more than enough force for the whole solution for whosoever seeks to truly solve it.

[FEN "3nk3/3NN3/3PP3/3BB3/3PP3/3PP3/3PP3/2RQKR2 w - - 0 1"]

1. Nf6+ Kf8 2. Nh7+ Ke8 3. Qa4+ Nc6 4. Qb5 Kd8 5. Qb6+ Ke8 6. d7+ Kxe7 7. Bf6+ Kd6 8. Qa6 Kc7 9. e5 Kb8 10. Qb6+ Ka8 11. Qb5 Ka7 12. Ra1+ Na5 13. Bc4 Ka8 14. e7 Ka7 15. d5 Ka8 16. e8=Q+ Ka7 17. Qh5 Ka8 18. d6 Ka7 19. Be6 Ka8 20. d8=Q+ Ka7 21. Bc8 Ka8 22. e6 Ka7 23. d7 Ka8 24. e7 Ka7 25. d4 Ka8 26. d5 Ka7 27. d6 Ka8 28. e4 Ka7 29. e5 Ka8 30. e6 Ka7 31. d4 Ka8 32. e4 Ka7 33. d5 Ka8 34. e5 Ka7 35. Qh8 Ka8 36. e8=Q Ka7 37. Qeg8 Ka8 38. d8=Q Ka7 39. Qdf8 Ka8 40. e7 Ka7 41. e8=Q Ka8 42. Qeg6 Ka7 43. e6 Ka8 44. e7 Ka7 45. e8=Q Ka8 46. Qee6 Ka7 47. d7 Ka8 48. d8=Q Ka7 49. d6 Ka8 50. d7 Ka7 51. Qde8 Ka8 52. d8=Q Ka7
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  • Interestingly, in the Australian newspaper's clip, the piece on e1 seemes to be a White Queen and not a White King. That's certainly a misprint, but another source from the same epoch is needed to prove that Babson's original problem didn't have 2 wQs and no wK, for whatever feeric reason.
    – Evargalo
    Jun 14, 2022 at 13:11
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    @Evargalo I have no doubt that it is a king, given the century-and-a-half-plus of printed diagrams, despite that one misprint.
    – Rewan Demontay
    Jun 14, 2022 at 21:58
  • Ref a much shorter solution than whatever Otto Blathy had in mind : although O.Blathy is also famous for his long problems, I think you mean J.Babson here...
    – Evargalo
    Feb 23 at 12:59
  • If anyone tries to solve the Obelisk, it might be worth to first try to solve the much easier "Cross" problem given in the clip from the American Chess Review. Interestingly, it has no wP, so Babson seem to infer that one can force the bN to make a full knight tour without multiple wQs...
    – Evargalo
    Feb 23 at 13:03
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    @Evargalo Thanks, fixed. And no doubt that is interesting...
    – Rewan Demontay
    Feb 23 at 22:24
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To force the knight into a corner a piece must be sacrificed. 3 knight tours, 4 corners per tour, so that's 12 White pieces that must be lost. And none of the lost pieces can be knights (from where did it get there to give check?).

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    So White starts Nf6+ (else Nxe6 is annoying, making it KNN vs KN after the tours), Black is forced to play Kf8. Then I think Rc8 forcing Kg7.
    – Post-It-Note
    Aug 5, 2016 at 23:23
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    Can you explain why a piece has to be sacrificed to force a knight into the corner? Not so obvious to me.
    – Steve Bennett
    Jun 15, 2021 at 3:57
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    @SteveBennett: In order for a (Black) knight move to be forced, White must give check, because a pinned knight has no legal moves, and an unpinned knight in the absence of check would have more than one legal move. So the knight move must stop the check. But if the knight moves into the corner, it cannot block a check, because the checking piece would need to be on "the other side" of the knight. Therefore, by process of elimination, the knight move must capture the checking piece.
    – Kevin
    Apr 17, 2022 at 16:29
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Looking at the position and the phrasing of the question, I would have to say that the problem is flawed. The directions are far too restrictive - how can white compel black to make three knight tours? If this was defined as a helpmate that's solvable in 1220 moves, then the problem would make more sense. However, even in this case, I cannot fathom how that would be possible as the number of moves that a three knight tours sum to is far less than 1220.

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  • 10
    I think the idea is probably to force and trap the black king somewhere that a check can only be escaped by moving the knight to the next square on the tour (either interposing or capturing the piece there). The knight would thus move exactly 192 times, but there would be a lot of moves by the black king.
    – supercat
    Mar 18, 2014 at 1:59

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