4

Obviously inspired by the analogous selfmate question. (Helpmate would also be possible in principle, but looks a bit silly.)

Thus, 1.f3 e5 2.g4 allows Black a #1, and 1.f3 e5 2.h3 a #2. 1.f4 e5 2.Nf3 Be7 3.h3 is a #3. This should be the quickest games.

Find a shortest helpplay to a position where one side has a forced #4 (and of course it must be unique and not be shortenable), for starters.

5

A mere 3 moves are needed to reach a position in which White has a forced #4.

[FEN ""]

1. e3 b6 2. Bc4 Bb7 3. Qf3 d6 {Now the helping play is over.}  4. Qxf7+ Kd7 5. Qf5+ e6 6. Bxe6+ Kc6 (6... Ke7 7. Qf7#) 7. Qd5#

However, longer mates are also possible to produce in just three moves.

Here is a #5.

[FEN ""]

1. e4 e5 2. Bc4 f5 3. Qe2 h6 {Now the helping play is over.} 4. Qh5+ g6 5. Qxg6+ Ke7 6. Qf7+ Kd6 7. Qd5+ Ke7 8. Qxe5#

And here is a #6.

[FEN ""]

1. e3 d6 2. Bc4 b6 3. Qf3 b5 {Now the helping play is over.} 4. Qxf7+ Kd7 5. Qf5+ e6 6. Qxe6+ Kc6 7. Qd5+ Kd7 8. Qf5+ Kc6 9. Qxb5#

I have discovered a #7, but it's partially non-unique, so it doesn't fully count.

[FEN ""]

1. e3 e5 2. Bc4 Ke7 3. Qh5 Kf6 {Now the helping play is over.} 4. Qxf7+ Kg5 5. g4 {This is where the minor dual occurs.} (5. h4+ Kh6 6. g4 g6 7. g5+ Qxg5 8. hxg5+ Kxg5 9. Nf3+ Kg4 10. Rh4#) g6 6. h4+ Kh6 7. g5+ Qxg5 8. hxg5+ Kxg5 9. Nf3+ Kg4 10. Rh4#
5
  • A really clever one, especially as the Qf5+ was not obvious. – Hauke Reddmann Apr 22 at 17:22
  • Addendum - As I immediately suspected, Black can drop the Bb7 move. After 3...h6, creating another hole on white, it is still a forced mate (Stockfish declares it 6#)! After a pass (say, h5 instead) it's still a #7. Maybe someone can find the missing 5#? – Hauke Reddmann Apr 22 at 17:29
  • @HaukeReddmann Firstly, that is not a #7; it's still #6. There are plenty of #6's, but you asked specifically for a #4, so I did not put one in. Also, I found a #5! – Rewan Demontay Apr 22 at 21:02
  • Stockfish said #7 (only at Bd5+ he - and I - realizes it's #6) and I believed it as my solving program is far too slow (and I'm too old :-) OK, so the longest unsolved is now #7. P.S. Greatest LOL - LiChess has an actual game: lichess.org/GRKeuiPj – Hauke Reddmann Apr 22 at 21:36
  • @HaukeReddmann Actually, I've already found #7, with a minor dual, so it's mostly solved. And lmao, that game. – Rewan Demontay Apr 22 at 21:44
1

Black can have a forced mate in 4 after only 2.5 moves, in this line from the Queen's Bongcloud Attack:

[FEN ""]
1.d3 e6 2.Kd2? Bd6 3.Ke3?? Qg5+ 4.Ke4 (4.Kf3? Qf5+) (4.f4?? Qxf4#)
(4.Kd4 Qe5+ (Nc6+) 5.Kc4 Qc5+ 6.Kb3 Qb4#)
Qe5+ 5.Kf3 Qf5+ 6.Ke3 (6.Bf4 Qxf4#) Qf4#

Mate in 6 is also possible after 2.5 moves in this variation of the Fred Defense reversed, with one unique variation:

[FEN ""]
 1.f4 e5 2.Kf2? exf4 3.h3?? Qh4+ 4.Kf3 (4.g3?? Qxg3#) Qg3+ 5.Ke4 d5+ 6.Kxd5
(6.Ke5? Nf6!) (6.Kd4 Nc6+ 7.Kxd5 Be6+ 8.Ke4 Nf6# (8...f5#))
Be6+ 7.Ke4 Nf6+ 8.Ke5 Nc6# (8.Kd4 Nc6#)

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