Answers.Body: <p>I think there can be no slower one than mate in 3. My standard checkmating attempt would be like the starting positions 1.e4 e6 2.Ke2 Qh4 3. pass Qxe4#. What can go wrong there? 1. Ke2 could be illegal in a position like Kb1 Qh8. With the king on b1 the a1 piece has to be a rook per rules of piece set up. In this case 1.a4 g5 2.Ka2 Qd4 3.Rb1 Qxa4#.</p>
  <p>So finally we are left with a knight being a blocker. This can only be the case if there is a knight next to the king. In that case there is a knight on the opposite side that can go for a suffocated checkmate. So if we have e.g. Ke1 Nd1 then the Nd8-e6 can go to f4xg2 or d4xc2 for checkmate. If either is only defended once, white can put the defending piece on the respective square. So the only case left is the one where both these squares are defended twice. Furthermore there can't be a heavy piece on either side of the king as otherwise the original mate would work, with a pawn advance diagonal to the king, king advance and the heavy piece blocking the retreat path. The position that's left over is something like this</p>
  <pre><code>[FEN &quot;brnknqrb/pppppppp/8/8/8/8/PPPPPPPP/BRNKNQRB w - - 0 1&quot;]
  </code></pre>
  <p>In that case the defending queen can only be on one side of the board, the other side is defended by bishop, rook or knight in some combination. In that case if it is a rook that rook can move out of the way a la b3 Bb2 Ra1. A defending knight can move away to no longer defend, with two moves left over to fianchetto either rook or bishop.
  This can only go wrong if the non queen side is defended by three pieces, however this does not work as the bishop not defending that pawn has no room. (both the defending knight and rook as well as the rook on the other side plus the king have to be on the same colour leaving no room for the bishop)</p>
  <p>Sorry for the many edits, I hope I cleared up my not so clean initial explanation.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-28T23:36:41.387Z
Answers.Text: @RewanDemontay yes, I just gave this as an example of roughly how a counter example would have to look like to then show that there can't be one. It should be straightforward to compose one needing exactly 6 plies though. (by that idea of defending both knight entry points twice)
Answers.UserDisplayName: koedem
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-29T09:18:42.730Z
Answers.Text: Those were exactly my thoughts (in short: either K*7 Q*5 or a smothered mate are possible; even Ka*, impossible in Chess960, doesn't work, as it seems). I hoped, though, that someone wrote a script. :-)
Answers.UserDisplayName: Hauke Reddmann
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-29T13:56:18.333Z
Answers.Text: @HaukeReddmann why would you write a script when you can just prove it to be impossible? :D
  For the mathematician in me, a result like this is much more satisfying. (even though I suppose the write up could be a bit prettier)
Answers.UserDisplayName: koedem
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-30T10:17:03.500Z
Answers.Text: @koedem: Because the script gives also the exact shortest length (which varies a bit) and the number of such variations ;-) I wouldn't exclude that a certain setup of the major pieces (ignoring 960 rules) gives a shortest proof game. That would be a quite interesting result hardly to come by handish!
Answers.UserDisplayName: Hauke Reddmann
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-28T23:23:10.590Z
Answers.LastActivityDate: 2021-01-28T23:32:17.957Z
Answers.LastEditDate: 2021-01-28T23:32:17.957Z
Answers.UserDisplayName: koedem
Answers.Body: <p>Hi Hauke: you didn't clarify on the n.0 versus n.5 ambiguity.</p>
  <p>There are several positions where h#3.5 is the shortest mate with White to move and White to mate:</p>
  <pre><code>[Title &quot;shortest helpmate is 3.5 (numerous)&quot;]
  [FEN &quot;brnqknrb/pppppppp/8/8/8/8/PPPPPPPP/BRNQKNRB w - - 0 0&quot;]
  </code></pre>
  <p>If you set this as Black to move, White to mate, then there are solutions in 3.0, but Black to move isn't the starting position. Equivalently, you can set White to move but Black to mate (so-called &quot;HalfDuplex&quot;). This isn't a normal helpmate, although you maybe argue that Fool's Mate is a HalfDuplex too, so this is legitimately included in your scope for this task. But it would need to be stipulated, as absent duplex, White does the mating in a helpmate, whoever begins.</p>
  <p>So while both my position &amp; koedem’s are 3.0 as halfduplex, mine is 3.5 as regular helpmate, compared to his is as 2.5.</p>
  <p>Btw the wiki page you link to is crazy. It says White to move problems are unorthodox: in the same paragraph as fairy problems!?!</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-30T10:20:58.357Z
Answers.Text: Complaints please to Rewan(?!) who linked it and the problem community who didn't edit in a more precise wording, given Wiki is public :-)
Answers.UserDisplayName: Hauke Reddmann
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-30T10:24:21.203Z
Answers.Text: @Hauke yes but ignoring these others, how do you want to define your own cute task? :-)
Answers.UserDisplayName: Laska
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Answers.CreationDate: 2021-01-30T10:42:46.463Z
Answers.Text: I'm a scientist - working hypotheses are defined as soon as you have some preliminary results :-) (If any 960 is at most h#3 by dozens of solutions, the problem isn't that interesting anymore and one should consider to alter the rules. E.g. "starting position, twin by 1 change such that the position is an unique h#2 with White mated, ignoring legality". (I found only one.) I post a new problem as new question!
Answers.UserDisplayName: Hauke Reddmann
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-29T16:51:38.837Z
Answers.LastActivityDate: 2021-01-30T04:47:06.083Z
Answers.LastEditDate: 2021-01-30T04:47:06.083Z
Answers.UserDisplayName: Laska
Body: <p>The fastest Fool's Mate in Chess960 is obviously 1. d4 g5 2. Qxg5# with any ..Q.BRKR setup, which even beats standard chess. But which is the starting position with the slowest mate? I checked a few positions with Olive and all were <a href="https://en.wikipedia.org/wiki/Helpmate" rel="nofollow noreferrer">helpmates</a> in 3 (assuming White starts, which is not the default for h#). I daresay this is the maximum (Googling gave no hits).</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-01-29T05:59:06.663Z
Comments.Text: Helpmates in 3.0 with black mating or 2.5 with white mating? There are game arrays in chess960 in which parity might have been switched, even without losing castling rights.
Comments.UserDisplayName: Laska
Tags: <problems><chess-variants><chess960><fools-mate>
Title: The slowest Fool's Mate in Chess960
ThreadText: # The slowest Fool's Mate in Chess960
  
  (Asked by: username_0 on 2021-01-28)
  
  The fastest Fool's Mate in Chess960 is obviously 1. d4 g5 2. Qxg5# with any ..Q.BRKR setup, which even beats standard chess. But which is the starting position with the slowest mate? I checked a few positions with Olive and all were [helpmates](https://en.wikipedia.org/wiki/Helpmate) in 3 (assuming White starts, which is not the default for h#). I daresay this is the maximum (Googling gave no hits).
  
  **Tags:** problems, chess-variants, chess960, fools-mate
  
  > **Comments:**
  > **username_1**: Helpmates in 3.0 with black mating or 2.5 with white mating? There are game arrays in chess960 in which parity might have been switched, even without losing castling rights.  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_2 on 2021-01-28. Score: 3 | Accepted answer
  
  I think there can be no slower one than mate in 3. My standard checkmating attempt would be like the starting positions 1.e4 e6 2.Ke2 Qh4 3. pass Qxe4#. What can go wrong there? 1. Ke2 could be illegal in a position like Kb1 Qh8. With the king on b1 the a1 piece has to be a rook per rules of piece set up. In this case 1.a4 g5 2.Ka2 Qd4 3.Rb1 Qxa4#.
  So finally we are left with a knight being a blocker. This can only be the case if there is a knight next to the king. In that case there is a knight on the opposite side that can go for a suffocated checkmate. So if we have e.g. Ke1 Nd1 then the Nd8-e6 can go to f4xg2 or d4xc2 for checkmate. If either is only defended once, white can put the defending piece on the respective square. So the only case left is the one where both these squares are defended twice. Furthermore there can't be a heavy piece on either side of the king as otherwise the original mate would work, with a pawn advance diagonal to the king, king advance and the heavy piece blocking the retreat path. The position that's left over is something like this
  [FEN "brnknqrb/pppppppp/8/8/8/8/PPPPPPPP/BRNKNQRB w - - 0 1"]
  
  In that case the defending queen can only be on one side of the board, the other side is defended by bishop, rook or knight in some combination. In that case if it is a rook that rook can move out of the way a la b3 Bb2 Ra1. A defending knight can move away to no longer defend, with two moves left over to fianchetto either rook or bishop.
  This can only go wrong if the non queen side is defended by three pieces, however this does not work as the bishop not defending that pawn has no room. (both the defending knight and rook as well as the rook on the other side plus the king have to be on the same colour leaving no room for the bishop)
  Sorry for the many edits, I hope I cleared up my not so clean initial explanation.
  
  > **Comments:**
  > **username_2**: @RewanDemontay yes, I just gave this as an example of roughly how a counter example would have to look like to then show that there can't be one. It should be straightforward to compose one needing exactly 6 plies though. (by that idea of defending both knight entry points twice)  
  > **username_0**: Those were exactly my thoughts (in short: either K*7 Q*5 or a smothered mate are possible; even Ka*, impossible in Chess960, doesn't work, as it seems). I hoped, though, that someone wrote a script. :-)  
  > **username_2**: @HaukeReddmann why would you write a script when you can just prove it to be impossible? :D For the mathematician in me, a result like this is much more satisfying. (even though I suppose the write up could be a bit prettier)  
  > **username_0**: @username_2: Because the script gives also the exact shortest length (which varies a bit) and the number of such variations ;-) I wouldn't exclude that a certain setup of the major pieces (ignoring 960 rules) gives a shortest proof game. That would be a quite interesting result hardly to come by handish!  
  
  ---
  
  ### Answer by username_1 on 2021-01-29. Score: 2
  
  Hi Hauke: you didn't clarify on the n.0 versus n.5 ambiguity.
  There are several positions where h#3.5 is the shortest mate with White to move and White to mate:
  [Title "shortest helpmate is 3.5 (numerous)"]
  [FEN "brnqknrb/pppppppp/8/8/8/8/PPPPPPPP/BRNQKNRB w - - 0 0"]
  
  If you set this as Black to move, White to mate, then there are solutions in 3.0, but Black to move isn't the starting position. Equivalently, you can set White to move but Black to mate (so-called "HalfDuplex"). This isn't a normal helpmate, although you maybe argue that Fool's Mate is a HalfDuplex too, so this is legitimately included in your scope for this task. But it would need to be stipulated, as absent duplex, White does the mating in a helpmate, whoever begins.
  So while both my position & username_2’s are 3.0 as halfduplex, mine is 3.5 as regular helpmate, compared to his is as 2.5.
  Btw the wiki page you link to is crazy. It says White to move problems are unorthodox: in the same paragraph as fairy problems!?!
  
  > **Comments:**
  > **username_0**: Complaints please to Rewan(?!) who linked it and the problem community who didn't edit in a more precise wording, given Wiki is public :-)  
  > **username_1**: @Hauke yes but ignoring these others, how do you want to define your own cute task? :-)  
  > **username_0**: I'm a scientist - working hypotheses are defined as soon as you have some preliminary results :-) (If any 960 is at most h#3 by dozens of solutions, the problem isn't that interesting anymore and one should consider to alter the rules. E.g. "starting position, twin by 1 change such that the position is an unique h#2 with White mated, ignoring legality". (I found only one.) I post a new problem as new question!  
  
  ---
  
  