7

Found this construction problem from '83 and liked it. I don't know the best solution (yet), and maybe you can top it anyway.

Consider the (help) "game" 1.e4 a6 2.Bc4 a5 3.Qh5 a4. Now White can mate in two ways (and I doubt this value can be increased having 6 preceding half-moves). Maximize the number of possible White mates after 12 half-moves (which you choose).

Hauke Reddmann
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    The count for 6 preceding half-moves can be increased! 1.e4 f5 2.Bc4 fxe4 3.Qf3 g5, followed by 4.Qh5#, 4.Qf7# or 4.Bf7# – James Martin Jan 23 '24 at 00:33
  • THX. If I had had formulated as a question, the post might have been closed due to one question too many :-))) – Hauke Reddmann Jan 23 '24 at 07:29
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    I can confirm by brute force that 1. e4 f5 2. Bc4 fxe4 3. Qf3 g5 is the only position with 3 different mates after 6 half-moves. There are also 5 other transpositions of those moves that yield the same position. – LegionMammal978 Jan 23 '24 at 18:04

4 Answers4

10

11 without discovered checks:

[FEN ""]
[startply "12"]
  1. e3 e6 2. d4 Ke7 3. Bc4 Kf6 4. Nh3 Kf5 5. Bxe6+ Ke4 6. Qh5 Qe8

Noam D. Elkies
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8

5.5 moves to 14 fool's mates:

[FEN ""]
[startply "11"]
  1. b4 e6 2. Bb2 Qh4 3. Bxg7 Ne7 4. Bb2 Rg8 5. g3 Rxg3 6. f4

For 6.0 moves, start with something like 1. a3 and then reverse the colors (1. a3 b5 2. e3 Bb7 3. Qh5 Bxg2 etc.).

Noam D. Elkies
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3

I bid 7.

[FEN ""]
[startply "12"]
  1. e4 f6 2. Qf3 Kf7 3. Be2 Kg6 4. e5 Kg5 5. d4+ Kh4 6. d5 e6

Rewan Demontay
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2

15:

[FEN ""]
1. e3 f6 2. a4 Kf7 3. a5 Kg6 4. Ra4 Kf5 5. Bc4 Ke4 6. Qh5 Qe8

There are 11. discovered check mates plus 7. Qd5# 7. d3# 7. f3# and 7. Nc3#

Albert.Lang
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