Answers.Body: <p>Using wizardry to abuse the 50-move rule, I can make <strong>n equal 3053.</strong> White has a mate in one at any point and all of Black's moves are forced.</p>
  <p>Here, White has 47 pawn moves and 14 pieces to sacrifice. The Black knight capture adds an extra move. This allows for (47+14)*50+1=3051 moves.</p>
  <pre><code>[FEN &quot;7k/8/8/6QN/5P2/5P2/PPPPPP2/RNBKnB1R w - - 0 1&quot;]
  
  1. Kxe1
  </code></pre>
  <p>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.</p>
  <pre><code>[FEN &quot;8/7k/8/6QN/8/8/8/4K3 w - - 0 1&quot;]
  
  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#
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2022-01-12T09:12:59.233Z
Answers.Text: @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.
Answers.UserDisplayName: Hauke Reddmann
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2022-01-11T02:59:34.423Z
Answers.LastActivityDate: 2024-06-23T15:25:19.347Z
Answers.LastEditDate: 2024-06-23T15:25:19.347Z
Answers.UserDisplayName: Rewan Demontay
Body: <p>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 &quot;forcing move&quot; is any move that leaves Black with only one possible move.</p>
  <p>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 <em>also</em> 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 <em>n</em>. What is the largest value of <em>n</em>? 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.</p>
  <pre><code>[FEN &quot;6k1/8/8/4BQNB/8/8/8/6K1 w - - 0 50&quot;]
  
  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#)
  </code></pre>
  <p>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 <em>not</em> 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.</p>
  <p>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?</p>
  
Tags: <problems><checkmate><forced-move>
Title: What is the largest n such that discrete forced mate in one through mate in "n" exist?
ThreadText: # What is the largest n such that discrete forced mate in one through mate in "n" exist?
  
  (Asked by: username_0 on 2022-01-11)
  
  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?
  
  **Tags:** problems, checkmate, forced-move
  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_1 on 2022-01-11. Score: 6
  
  Using wizardry to abuse the 50-move rule, I can make **n equal 3053.** White has a mate in one at any point and all of Black's moves are forced.
  
  Here, White has 47 pawn moves and 14 pieces to sacrifice. The Black knight capture adds an extra move. This allows for (47+14)\*50+1=3051 moves.
  
  ```
  [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#
  
  ```
  
  > **Comments:**
  > **username_2**: @username_1: 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.  
  
  ---
  
  