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This is admittedly a bit of an obscure question, so I apologize for the awkward wording of the question. To begin, assume you are playing as White, and that a "forcing move" is any move that leaves Black with only one possible move.

Imagine a position where you have a mate in one. In addition, you also have a different move that gives black only one possible move that is also a mate in one. So, functionally, you have the option to play a mate in one or a mate in two. Extend this idea so you have a sequence of forcing moves that allows you to play a mate in one, mate in two, mate in three, and so on, all the way to mate in n. What is the largest value of n? For example, in the position below, White has a sequence of forcing moves that allow a mate in one, two, three, four, and five. Note that every move Black has in all variations is forced.

[FEN "6k1/8/8/4BQNB/8/8/8/6K1 w - - 0 50"]

1. Qf7# (1. Qh7+ Kf8 2. Qf7#) (1. Bf7+ Kf8 2. Qc8+ Ke7 3. Qe8#) (1. Bf7+ Kf8 2. Bd6+ Kg7 3. Qe5+ Kh6 4. Bf8#) (1. Bf7+ Kf8 2. Bd6+ Kg7 3. Qe5+ Kh6 4. Qd4 Kxg5 5. Qf4#)

For this position, there is a potential mate in six - 1. Bf7+ Kf8 2. Bd6+ Kg7 3.Qe5+ Kh6 4. Qd4 Nxg5 5. Qg7+ Kf5 6. Qg6# - but I am not counting this because the move 5... Kf5 is not, under this definition, forced. Black can also play 5... Kh5. To the best of my knowledge, there is not a series of forcing moves which allows for a mate in exactly six.

Of course, this position is not the end-all-be-all. I only created it for illustrative purposes. A position with mate in one through mate in fifty may exist, for all I know. Has anyone here heard of a similar problem before? If so, do you have any leads, or better yet, any ideas of your own?

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  • 1
    I'm tempted to call this sort of problem "toying with your opponent."
    – Cort Ammon
    Jan 11 at 18:09

1 Answer 1

6

Using a bit of wizardry to abuse the 50-move rule, I make n equal 3844. This is certainly toppable.

In this position, White has a mate in one at any point and all of Black's moves are forced. White has 45 pawns moves and 14 pieces to sacrifice. The Black knight capture adds on extra move. This allows for (45+14)*50+1=3842 moves to be added on.

[FEN "7k/8/8/6QN/5P2/5P2/PPPPPP2/RNBKnB1R w - - 0 1"]

1. Kxe1

To add on the last two moves, the limit due to threefold repetition, the White king must move up before the sacrifice of the second White knight.

[FEN "8/7k/8/6QN/8/8/8/4K3 w - - 0 1"]

1. Ke2 Kh8 2. Ke3 Kh7 3. Ke4 Kh8 4. Ke5 Kh7 5. Ke6 Kh8 6. Kf7 Kh7 7. Nf6+ Kh8 8. Nh7 Kxh7 9. Kf8 Kh8 10. Kf7 Kh7 11. Kf8 Kh8 12. Qg7#
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  • Why not go Kf8 Kh7 Nf6+ Kh8 Nh7 Kxh7 Kf7 Kh8 Kg6 and so forth? Or alternatively move the queen around all the while keeping h6 protected. E.g. in your line 9. Qg6+ Kh8 10.Qf6+ Kh7 11.Qe6 Kh8 and so forth.
    – koedem
    Jan 11 at 12:15
  • @koedem Good idea! Thanks for the improvement. Using [this question(chess.stackexchange.com/questions/22188/…) will help me. Jan 11 at 13:20
  • @Rewan: Obviously, an alternative challenge would be that the position is a #1 chess problem (only one White move to fulfil that!), if White doesn't make that move it is a #2 (ditto) etc., but you know my #1-#4 attempt from the MatPlusForum anyway :-) I wonder if a) it can be done with white all-promotion instead of the 4 moves of a white pawn on its initial field, b) #5. Jan 12 at 9:12

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