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.


A mere 6 plies are needed to reach a position in which White has a dual free #4. For any length record, as long as there is at least one unique mating line, it qualifies. This is in spirit as some checkmating problems are created with a line of "best defense." This involves one side leaving the other with only a single possible continuation in a vain hope.

[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#

A #5 can be done in 6 plies.

[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#

A #6 can be done in 5 plies; see Noam Elkies's answer.

A #7 can be done in 8 plies.

[FEN ""]

1. e4 e5 2. Bc4 d5 3. Qh5 Ke7 4. f4 Kf6 {Now the helping play is over.} 5. Qg5+ Ke6 6. Qxe5+ Kd7 7. Qxd5+ Ke7 8. Qxf7+ Kd6 9. Qd5+ Ke7 10. Qe5+ Be6 (10... Kd7 11. Qe6#) 11. Qxe6#

A #8 can be done in 10 plies.

[FEN ""]

1. e3 e5 2. Qh5 Ke7 3. Bc4 Kf6 4. c3 g6 5. d3 a6 {Now the helping play is over.} 6. Qf3+ Kg7 7. Qxf7+ Kh6 8. e4+ Qg5 9. Bxg5+ (9... Kxg5 10. h4+ Kh6 11. Qxf8+ Kh5 12. g4+ Kxg4 13. Qf3#) Kh5 10. g4+ Kxg4 11. h4 Nf6 12. Qxf6 Kh5 13. Qf3#

A #9 can be done in 6 plies.

[FEN ""]

1. f4 e5 2. Kf2 Qh4+ 3. Ke3 Bc5+ {Now the helping play is over.} 4. Kd3 e4+ 5. Kc3 Qf6+ 6. d4 exd3+ 7. Kd2 Qxf4+ 8. e3 Qxe3+ 9. Kc3 d2+ 10. Bd3 Qd4+ 11. Kxd2 Qf4+ 12. Kc3 (12. Ke2 Qf2#) (12. Ke1 Qf2#) 12... Qb4#

A #10 can be done in 5 plies.

[FEN ""]

1. d4 e5 2. Kd2 Bb4+ 3. Ke3 {Now the helping play is over.} Qg5+ 4. Kd3 e4+ 5. Kc4 b5+ 6. Kxb4 a5+ 7. Kc3 b4+ 8. Kb3 a4+ 9. Kc4 Ba6+ 10. Kxb4 Nc6+ 11. Kc3 Qa5+ 12. b4 Qxb4#
  • BTW, the 8# play reminds a bit of the Greco analysis of the Damiano Accepted (1.e4 e5 2.Nf3 f6 3.Nxe5 fxe5). May 25 at 8:28

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#)
  • 2
    Although good and perfectly valid answer, this is the first time I see From's gambit referred to as Fred Defense reversed:-)
    – comodoro
    Aug 7 at 8:23
  • 1
    @comodoro Probably because 2.Kf2 is not usual in the From gambit, but 2...Kf7 is the main idea of the Fred defense.
    – Evargalo
    Aug 12 at 8:25

We can reach #9 within only 8 plies:

[Title "Reaching #9 afte 8 plies"]
[FEN ""]
1. e4 d5 2. Qf3 Kd7 3. Bc4 Kc6 4. Nc3 d4!
5. e5+ Kd7 6. Qd5+ Ke8 7. Qxf7+ Kd7 8. e6+ Kc6 9. Qf3+ Kb6 
10. Na4+ Ka5 11. b4+ Kxb4 12. Qb3+ Ka5 13. Qb5#

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