8

I should specify that I'm not asking about the longest forced checkmate (or forced checkmates in general); I am aware that the longest known forced mate is(was?) about 600 moves long, and this is not what I'm asking about.

My question is about the smallest number of legal moves a (valid) position can be from a checkmate, and specifically about how large that minimum can be (as a function of the position of the board). Think of it as 2 players trying to cooperatively end the game as soon as possible (without drawing).

For example, the starting position is only 2.5 moves/5 plies away from a checkmate via the Fool's Mate, and no checkmate can be made in fewer moves/plies. This shows that the greatest distance a game can be from a checkmate is at least 5 plies (though the true maximum will be larger).

I appreciate any responses, looking into this kept giving me info on longest forced checkmates so I decided to ask directly. Hopefully this has been looked into before!

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6
  • 1
    So to be clear: mate by either player counts?
    – Laska
    Commented Aug 3 at 9:56
  • 2
    @Laska possibly not, because by that standard the starting position is only 2.0 moves away (Qh4#).
    – Noam D. Elkies
    Commented Aug 3 at 19:20
  • 1
    Ok it seems undesirable that the record should always be for the player to move who delivers the mate (2.5 from game array). Most of the known long helpmates happen to be with black to move getting mated. Possibly best if you emphasize that it’s White who must deliver the mate
    – Laska
    Commented Aug 3 at 20:22
  • @Laska I have been thinking that the mate could be delivered by white or black, whichever happens to be longest. I've also meant to consider black/white to move as part of the data of the initial position, and to simply choose which option maximizes the distance to a mate (when that can be determined). I suppose you could specify that white must deliver mate, and then mirror any positions where black to mate would be longer, but that seems more like a technicality to limit the scope of the problem than it is enlightening.
    – Nylonco
    Commented Aug 4 at 20:11
  • Then there are really two versions of your challenge: (a) with white to mate and (b) with either to mate. The former is cleaner and would be calculated en route to the latter, so wouldn’t want to remove this part. Also I’ve already spent time answering it. On the other hand the player to move is usually a fixed part of the initial position definition. That should not vary. I've added what I think is shortest variant leading to mate by Black.
    – Laska
    Commented Aug 5 at 3:18

1 Answer 1

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4

TLDR: 97 ply

The following problem, by a famous composer, gives 96 ply, and I doubt that any better matrix is currently known.

However, one can cheekily modify the starting position by shifting wK to h2, and then White must play wKh2-h1 first, to reach the Fabel position. This is a common trick to squeeze the most out of a record problem, and I claim no credit.

   [Title "Karl Fabel - Schachmatt 1948 - h#48 C-"]
   [Startflipped "0"]
   [FEN "8/8/5p2/2p1pPp1/2PbPk1b/2pBp1pB/2P1P1P1/5BRK b - - 0 1"]

   1. g4 Bxg4 2. Bg5 Bf3 3. Bh6 Bg4 4. Bf8 Bf3 
   5. Bd6 Bg4 6. Bc7 Bf3 7. Ba5 Bg4 8. Bb4 Bf3
   9. Ba3 Bg4 10. Bc1 Bf3 11. Bd2 Bg4 12. Be1 Bf3
   13. Bf2 Bg4 14. Bxg1 Bf3 15. Kg5 Bg4 16. Kh6 Bf3
   17. Kg7 Bg4 18. Kf7 Bf3 19. Ke7 Bg4 20. Kd6 Bf3
   21. Kc6 Bg4 22. Kb6 Bf3 23. Ka5 Bg4 24. Ka4 Bf3
   25. Ka3 Bg4 26. Kb2 Bf3 27. Kc1 Bg4 28. Kd1 Bf3
   29. Ke1 Bg4 30. Kxf1 Bf3 31. Ke1 Kxg1 32. Kd1 Kf1
   33. Kc1 Ke1 34. Kb1 Kd1 35. Ka2 Kc1 36. Ka3 Kb1
   37. Ka4 Ka2 38. Ka5 Ka3 39. Ka6 Ka4 40. Ka7 Kb5
   41. Ka8 Kc6 42. Kb8 Kd5 43. Kc7 Ke6 44. Kd8 Kxf6 
   45. Ke8 Kg6 46. Kf8 f6 47. Kg8 f7+ 48. Kh8 f8=Q#

Question is whether a shorter helpmate exists in this position, but I think it's unlikely.

The diagram position is legal.

We mark this as "C-" (aka “cooked”) but uniqueness is not the composer's intention.

The first moves are forced: 1. g4 Bxg4. Now bBh4 must capture wRg1, and bK must capture wBf1, while the other wB wanders non-uniquely. Simple counting shows Kxf1 can be done at the earliest by move 30, after which can have 31. Ke1 Kxg1. Now need to promote some mating material for White so wK captures bPf6 and can be on a matable square by the time bPf6 promotes.

The only alternative I can see is that wK does not capture bBg1. This allows for mate without promotion: bKa8, bBb8, wKa6 & wBc6/d5. However this requires another 21 moves by bKf1 & bBg1, so will be slower.

EDIT: The questioner also asked what would happen if we are looking for the quickest helpmate by either player. I think that Black can checkmate wKa1 on the 14th move after the critical position at 30.0. This would give a total ply count of 88 (including the check in the diagram position).

   [Title "variation from 30.0, leading to wK mated"]
   [Startflipped "0"]
   [FEN "8/8/5p2/2p1pP2/2PbP3/1BpBp1p1/2P1P1P1/5kbK b - - 0 1"]

   1. Ke1 Ba4 2. Kd1 Bb3 3. Bf2 Ba4 4. Be1 Kg1
   5. Bf2+ Kf1 6. Bg1 Bb3 7. Kc1 Ke1 8. Kb2 Kd1
   9. Ka3 Kc1 10. Bf2 Kb1 11. Be1 Ka1 12. Bd2 Ba2
   13. Bc1 Bb1 14. Bb2#

If we mate wKa1, then bK is either on c1 or a3. If the former, then the kings must have passed one another but the corridor is too narrow so wK would have to go to the 8th rank. Therefore the shortest mate will be with bKa3. The bB must still pass wK, but this can be done in the corridor, just costing 4 Black moves. So I am confident that this is the shortest mate by Black. Of course it's not unique.

88 is still a big number. Since the other published problems of the same order are basically just Fabel's own variants of the one shown here, I think that the 88 record will stand as long as the 97 record stands.

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