Answers.Body: <p>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 &quot;best defense.&quot; This involves one side leaving the other with only a single possible continuation in a vain hope.</p>
  <pre><code>[FEN &quot;&quot;]
  
  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#
  </code></pre>
  <p>A #5 can be done in 6 plies.</p>
  <pre><code>[FEN &quot;&quot;]
  
  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#
  </code></pre>
  <p>A #6 can be done in 5 plies; see Noam Elkies's answer.</p>
  <p>A #7 can be done in 8 plies.</p>
  <pre><code>[FEN &quot;&quot;]
  
  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#
  </code></pre>
  <p>A #8 can be done in 10 plies.</p>
  <pre><code>[FEN &quot;&quot;]
  
  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#
  </code></pre>
  <p>A #9 can be done in 6 plies.</p>
  <pre><code>[FEN &quot;&quot;]
  
  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#
  </code></pre>
  <p>A #10 can be done in 5 plies.</p>
  <pre><code>[FEN &quot;&quot;]
  
  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#
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-04-22T12:40:48.873Z
Answers.LastActivityDate: 2021-07-14T13:49:11.733Z
Answers.LastEditDate: 2021-07-14T13:49:11.733Z
Answers.UserDisplayName: Rewan Demontay
Answers.Body: <p>Black can have a forced mate in 4 after only <strong>2.5 moves</strong>,
  in this line from the Queen's Bongcloud Attack:</p>
  <pre><code>[FEN &quot;&quot;]
  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#
  </code></pre>
  <p><strong>Mate in 6</strong> is also possible after <strong>2.5 moves</strong> in this
  variation of the Fred Defense reversed, with one unique variation:</p>
  <pre><code>[FEN &quot;&quot;]
   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#)
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-08-07T08:23:14.923Z
Answers.Text: Although good and perfectly valid answer, this is the first time I see From's gambit referred to as Fred Defense reversed:-)
Answers.UserDisplayName: comodoro
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-08-12T08:25:59.263Z
Answers.Text: @comodoro Probably because 2.Kf2 is not usual in the From gambit, but 2...Kf7 is the main idea of the Fred defense.
Answers.UserDisplayName: Evargalo
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-04-23T19:45:33.500Z
Answers.LastActivityDate: 2021-04-23T19:45:33.500Z
Answers.UserDisplayName: Noam D. Elkies
Answers.Body: <p>We can reach <strong>#9 within only 8 plies</strong>:</p>
  <pre><code>[Title &quot;Reaching #9 afte 8 plies&quot;]
  [FEN &quot;&quot;]
  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#
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-05-25T06:51:26.270Z
Answers.LastActivityDate: 2021-07-13T19:17:29.993Z
Answers.LastEditDate: 2021-07-13T19:17:29.993Z
Answers.UserDisplayName: user21820
Body: <p>Obviously inspired by the analogous <a href="https://chess.stackexchange.com/questions/33168/the-quickest-selfmate-in-1">selfmate</a>  question. (Helpmate would also be possible in principle, but looks a bit silly.)</p>
  <p>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.</p>
  <p>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.</p>
  
Tags: <problems><checkmate><records>
Title: The quickest mate in n
ThreadText: # The quickest mate in n
  
  (Asked by: username_0 on 2021-04-22)
  
  Obviously inspired by the analogous [selfmate](https://chess.stackexchange.com/questions/33168/the-quickest-selfmate-in-1) 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.
  
  **Tags:** problems, checkmate, records
  
  
  ---
  
  ## 3 answers
  
  ### Answer by username_1 on 2021-04-22. Score: 6 | Accepted answer
  
  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 username_2 username_2'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#
  
  ```
  
  
  ---
  
  ### Answer by username_2 on 2021-04-23. Score: 4
  
  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#)
  
  ```
  
  > **Comments:**
  > **username_3**: Although good and perfectly valid answer, this is the first time I see From's gambit referred to as Fred Defense reversed:-)  
  > **username_4**: @username_3 Probably because 2.Kf2 is not usual in the From gambit, but 2...Kf7 is the main idea of the Fred defense.  
  
  ---
  
  ### Answer by username_5 on 2021-05-25. Score: 1
  
  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#
  
  ```
  
  
  ---
  
  