Answers.Body: <p><strong>Even</strong></p>
  <p>For each position, we can <em><strong>produce 1 &quot;pair&quot; position using the following cases and vice versa</strong></em>.</p>
  <p>Lets break down the game to the following cases:</p>
  <p><strong>Case 1</strong>) Whenever we can use <strong>triangulation</strong> to produce a <strong>pair position</strong> with the <strong>same pieces in the mirrored positions</strong> between the rows 4 and 5 <strong>with the opponent to move</strong>.</p>
  <p>For example:</p>
  <p><a href="https://i.sstatic.net/1bE6U.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1bE6U.png" alt="enter image description here" /></a></p>
  <pre><code>1. c3 (1. d3) (1. b3 Nc6 2. Ba3)
  
  1. Nc3 b6 2. Nb1 Bb7 3. Nc3 Ba6
  </code></pre>
  <p><strong>Case 2</strong>) Every position where triangulation cannot be done.</p>
  <p>Every position can have a <strong>pair position with or without castling done for white</strong>.</p>
  <p>By moving the white knight on g1 and then the white rook to g1 and then back to h1, the castling rights are lost.</p>
  <p>eg. Starting with the following set of moves</p>
  <pre><code>[FEN &quot;&quot;]
  1. Nf3 Nf6 2. Rg1 Ng8 3. Rh1 Nf6 4. Ng1 Ng8
  </code></pre>
  <p>we produce the start position without castling rights for white, all positions after that until we can use Case (1) can be with and without castling rights for white.</p>
  <p>The initial position falls to Case (2).</p>
  <p>References</p>
  <ul>
  <li><a href="https://en.wikipedia.org/wiki/Triangulation_(chess)" rel="nofollow noreferrer">Chess triangulation</a></li>
  </ul>
  <p>PS: Updated the answer, so the comments till Jan 7th 2024 are for the old answer</p>
  
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Answers.CreationDate: 2020-03-30T20:57:48.890Z
Answers.Text: 1. Nf3 Nf6 2. Ng1 Nd5 3. Nf3 Nf4 4. Ng1 Nd3+ I don't think you can mirror this with Black to move, as there's no way to lose the tempo.
Answers.UserDisplayName: D M
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Answers.Text: Also just 1. Nf3 Nf6. Like the start position, it's its own mirror image.
Answers.UserDisplayName: academic
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Answers.CreationDate: 2020-03-31T08:01:17.310Z
Answers.Text: I am confused you have "switching colors", "mirrored horizontally" and "corresponding positions". Could you make the difference more clear, specifically when counting, do you count the color-switched or the mirrored positions?
Answers.UserDisplayName: user1583209
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Answers.CreationDate: 2020-03-31T08:07:13.383Z
Answers.Text: Mirrored positions have exactly the same piece positions on the board switched both piece colors and turn. You must have both, due to checks you cannot only switch turn or colors. Corresponding have the turn switched, but different piece layout on the board, actually ALL positions with only knights moved cannot be mirrored, but have a corresponding position that also cannot be mirrored.
Answers.UserDisplayName: Jannes Botis
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Answers.CreationDate: 2020-03-31T08:11:37.440Z
Answers.Text: @JannesBotis: Still confused. If I just take your sentence: **"you cannot find a position that colors can be switched and produce an illegal position"** and apply this to the initial position, I end up with all white pawns on the 7th rank and all black pawns on the 2nd. This is clearly illegal because the pawns could not pass each other. Perhaps I misunderstand something....
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Answers.Text: You have also to "flip/ mirror" the chess board horizontally between 4 and 5 rows. The goal is to produce pairs of positions, so to prove the total is even.
Answers.UserDisplayName: Jannes Botis
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Answers.CreationDate: 2020-03-31T09:29:13.643Z
Answers.Text: It's not true that non-mirrored positions can only involve knight moves. You can also push rook pawns a single square, and move the rooks a single square forward or sideways. Since this does not give a rook enough freedom to move more than a single square, no triangulation can happen.
Answers.UserDisplayName: Especially Lime
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Answers.CreationDate: 2020-03-31T09:30:58.737Z
Answers.Text: If it really comes down to the knight-move only group, that should be even. The special case is check. There are two checking squares and for both, the non-checking knights have equivalent possibilities. So those are even. All the non-checking positions can be mirrored, with the caveat that that may not change the move. But there are an even number of those as well.
Answers.UserDisplayName: D Krueger
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Answers.CreationDate: 2020-03-31T14:41:44.857Z
Answers.Text: Triangulation section is wrong. The moves a3, h3, ...a6, and ...h6 also preserve the parity of the position.
Answers.UserDisplayName: Remellion
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Answers.CreationDate: 2022-04-19T10:04:23.987Z
Answers.Text: I see no reason why there would be one and exactly one "corresponding position" for each "non-mirrorable position".
Answers.UserDisplayName: Evargalo
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Answers.CreationDate: 2023-05-02T07:08:03.447Z
Answers.Text: If anything, you haven't shown how to build such a "corresponding position"...
Answers.UserDisplayName: Evargalo
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Answers.CreationDate: 2024-01-08T10:40:50.047Z
Answers.Text: I updated my answer anew using a new approach
Answers.UserDisplayName: Jannes Botis
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Answers.CreationDate: 2024-11-14T23:39:19.003Z
Answers.Text: @JannesBotis What about the initial position with the queens removed, and the kings moved to d1 and d8? There is no castling possible, I don't think there is any triangulation possible (the only moves have been knight moves, rook moves between a1/b1, a8/b8, g1/h1, g8/h8, and king moves between d1/e1, d8/e8). The position is its own mirror. So it doesn't seem to have a partner under your proposal. There are further positions like this, with the knights in different places, and with either the a pawns, or the h pawns, or both, advanced by a single square.
Answers.UserDisplayName: James Martin
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Answers.Text: A similar effect can be achieved eg by swapping all the rooks with their neighbouring knights in the initial position
Answers.UserDisplayName: James Martin
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Answers.LastEditDate: 2024-01-08T10:31:31.847Z
Answers.UserDisplayName: Jannes Botis
Answers.Body: <p>Happy Halloween!</p>
  <p>Basically the only possible approach to determining the parity of the number of positions is to pair off positions with one another, and then consider the remainder. The position after 1.e4 can be paired with the position after 1.Nc3 e6 2.Nf6 e5. These are mirror images of one another, with the board and colours reversed, and the other player to move. Most positions in the game are of this kind. To compute the parity of the number of positions overall, we need only consider the relatively small but still uncounted positions with no mirror image. With an attempt at whimsy, we call these vampires. The game array itself is an example: it can only be seen with White to move.</p>
  <p>It might be possible to pair off the vampires with one another e.g. ones where wQd1 versus ones where wQ has been captured. But we think this exercise is tricky and probably prone to error. So we are engaged in computer analysis of the vampires as a whole. This will be less error-prone and also give us more information: the number of vampires with White to move and the number of vampires with Black to move.</p>
  <p>Considerable progress has been made on this challenge by Miguel Ambrona. He has also developed an app that allows you to play <a href="https://chasolver.org/vampires/" rel="nofollow noreferrer">with chess vampires</a>. I love the way the figurines revert to the orthodox ones, when the engine determines that the position is no longer a vampire. (E.g. compare 1. h3 with 1. g3.) Left and right arrows allow you to go back and forth. It is explained in the &quot;What is this&quot; button.</p>
  <p>For now we are focusing just on one of the 16 vampire clans. This is the one in which all castling rights remain. No king or rook can move or rook be captured. Not even castling can happen because that costs the other castling right for that player too. The post-castling position if it were still a vampire would fall into another clan. The general shape of results will be the same for all clans.</p>
  <p>Miguel has discovered that a vast cavern of simple vampires starting with Ghe game array, depending upon parity, lead to a number of long narrower passages which remain vampiric due to one player being on the clock, and only having a limited number of retractions before being in retro-stalemate. He has published a number of puzzles based on this idea, e.g. <a href="https://www.thehoppermagazine.com/OP033" rel="nofollow noreferrer">this one</a>.</p>
  <p>As he is still exploring the passages, he cautions that in his app: &quot;Do not trust the output after 8 plies on an 'on-the-clock' position, as that precomputation is not available yet.&quot;</p>
  <p><a href="https://i.sstatic.net/UmKkCAzE.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UmKkCAzE.png" alt="Chess vampires" /></a></p>
  
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Answers.CreationDate: 2024-11-11T11:50:33.883Z
Answers.Text: Could you please clarify how this answers the question?
Answers.UserDisplayName: Cyriac Antony
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Answers.CreationDate: 2024-11-11T13:22:48.423Z
Answers.Text: Have expanded the answer @CyriacAntony - hope this helps
Answers.UserDisplayName: Laska
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Answers.CreationDate: 2024-11-12T16:30:41.230Z
Answers.Text: Great. Thank you.
Answers.UserDisplayName: Cyriac Antony
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Answers.LastActivityDate: 2024-11-11T15:03:42.623Z
Answers.LastEditDate: 2024-11-11T15:03:42.623Z
Answers.UserDisplayName: Laska
Body: <p>Is the number of legal chess positions odd or even? Two positions are not the same if they differ in castling rights (i.e. whether K or R have actually moved) or en passant capability (i.e. whether the move can actually be made) or who has the move.</p>
  <p>The number of times position has been repeated is not part of the definition of position for obvious reasons; nor is the number of moves since last capture or pawn move.</p>
  <p>I don’t have the answer myself. This is hard but doable with a bit of programming I think.</p>
  <p>EDIT: The answer so far and the comments are clearly heading along the right lines, although not always correct yet: terrific work. I want to give multiple +1s! I will summarize the key points to help focus the work. Pairing mirrored positions is essential i.e. those which allow triangulation. I term positions which have no mirror image “vampires” :-) Any vampire (except for the starting position) must be immediate offspring of a vampire, so it’s maybe easiest to just focus on them.</p>
  <p>Someone touched on castling, which is very important. We can partition the vampires into 16 different clans. The head of each clan is an &quot;initial game array&quot; position (i.e. with 32 pieces on apparently original squares) but with specific castling rights disabled (by some initial dance of rooks and knight taking 2n.0 moves).</p>
  <p>In a clan, if a castling right remains, then that implies that the rook's pawn can safely capture off the file, the f-pawn can move forward a single square or capture or (on the kingside) the bishop can be captured, without allowing triangulation. If the right is lost then the rook's pawn can only move forward a single square, and the bishops cannot be captured.</p>
  <p>En passant is less important but amazingly there are vampire en passant positions. For example:</p>
  <pre><code>r1bqkb1r/2pppppp/8/1pP5/8/8/PPPPP1PP/R1BQKB1R w KQkq b6 0 12
  </code></pre>
  <p>where White retains at least one castling right, Black has at least queenside castling rights and White can capture e.p.</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-30T07:31:19.083Z
Comments.Text: Shouldn't it be even because to every position there is a mirrored position, mirrored on the axis between the e and d files. (Note that a position cannot be mirror symmetric because of the kings.
Comments.UserDisplayName: user1583209
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Comments.CreationDate: 2020-03-30T07:38:19.847Z
Comments.Text: @user1583209 the mirror image of the initial position isn't exactly legal (or at least the OP should clarify whether it is) - it cannot be reached from the starting position.
Comments.UserDisplayName: Glorfindel
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Comments.CreationDate: 2020-03-30T07:42:17.067Z
Comments.Text: @Glorfindel: Good point. Is this (and similar positions where queen/king could not have moved) the only exception?
Comments.UserDisplayName: user1583209
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Comments.CreationDate: 2020-03-30T09:26:00.827Z
Comments.Text: Clearly there has to be some matching strategy based on symmetry, but flipping left/right isn’t viable I think, because it’s hard to marshal the exceptional positions.
Comments.UserDisplayName: Laska
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-30T11:18:09.623Z
Comments.Text: @bot I think the approach would be to count the positions that cannot be reached legally with the opposite color on the move. For example, you can reach the equivalent of `1. e3` with `1. Nf3 e6 2. Ng1 Bd6 3. Nf3 Be7 4. Ng1 Bf8`, so there are an even number of legal positions from that point. Of white's 20 opening options, only the four knight moves, a3, f3, and h3 have unreachable equivalents with black. I think it could still be very difficult, though.
Comments.UserDisplayName: D Krueger
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Comments.CreationDate: 2020-03-30T11:23:07.003Z
Comments.Text: @DKrueger Even then, the number will be so high, it probably cannot be counted to in billions of lifetimes. I think that the only way is some kind of mathematical proof, like on a much lower level, why a negative times a negative is a positive, for example.
Comments.UserDisplayName: PhishMaster
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Comments.CreationDate: 2020-03-30T11:34:43.417Z
Comments.Text: @DKrueger Unfortunately, that includes not just opening positions, but every position which includes a check.
Comments.UserDisplayName: D M
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Comments.CreationDate: 2020-03-31T12:15:18.747Z
Comments.Text: @bof it’s a good sub-puzzle so I won’t give the answer - it’s only a corner case. Dealing with castling properly is a necessary first step.
Comments.UserDisplayName: Laska
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Comments.CreationDate: 2020-03-31T12:49:07.027Z
Comments.Text: @bof This is a question I want the answer to because I don’t know it. But I have some partial results and I want you guys to reach the frontier of my thinking. One cute thing I have found is this e.p. idea: someone else might enjoy looking for it. You need to think about the different sets of castling rights which are possible and the implications so that you don’t double count. I am only doing matching based on non-vampire mirroring. I don’t see any other way to be sure that the matching is 1-1. Hope this helps
Comments.UserDisplayName: Laska
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Comments.CreationDate: 2020-03-31T13:59:06.987Z
Comments.Text: @bof attitude? What attitude? I’m baffled
Comments.UserDisplayName: Laska
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Comments.CreationDate: 2020-03-31T14:53:23.833Z
Comments.Text: @Laska There is some strange interaction between castling and parity which I can't quite see how to count effectively yet. Take for instance the position after 1. Nf3 Nf6 2. Ne5 Ng8 3. Ng6 Nf6 4. Nxf8 Ng8. This position cannot be "mirrored" (inversion about horizontal axis + colour inversion) because of the side to move, despite a piece being captured -- you can't reach the "mirror" position while preserving white kingside castling rights.
Comments.UserDisplayName: Remellion
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Comments.CreationDate: 2020-03-31T15:06:56.580Z
Comments.Text: @Remellion: I think there are 16 independent vampire clans, each for a different combination of castling rights. The head of each clan has the pieces all in game array squares. If a player has one castling right then the only way they can move K and R is by castling. Any other move would duplicate a position in one of the other less entitled clans
Comments.UserDisplayName: Laska
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-31T15:08:22.587Z
Comments.Text: @bof so is the number of en passant vampires odd or even? :)
Comments.UserDisplayName: Laska
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-31T15:45:25.350Z
Comments.Text: @bof Yes I think so too. Two possible double pawn moves in every case. If as was suggested we count all the vampires, then we should count all of these little guys separately
Comments.UserDisplayName: Laska
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-31T16:19:54.860Z
Comments.Text: If I have to rely on humanly identifying matrices like the one I posted to do a census of the clans, then I wouldn't call that "effective". :P Too easy to miss an idea or wrongly identify a corner case.
Comments.UserDisplayName: Remellion
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-03-31T16:25:56.350Z
Comments.Text: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/106162/discussion-on-question-by-laska-is-the-number-of-legal-chess-positions-odd-or-ev).
Comments.UserDisplayName: Brian Towers
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-04-02T10:24:54.840Z
Comments.Text: No warning? And this is intended as communal solving of a difficult unsolved problem. Does that have no space here? Thanks for spoiling
Comments.UserDisplayName: Laska
Tags: <rules><mathematics>
Title: Is the number of legal chess positions odd or even?
ThreadText: # Is the number of legal chess positions odd or even?
  
  (Asked by: username_0 on 2020-03-30)
  
  Is the number of legal chess positions odd or even? Two positions are not the same if they differ in castling rights (i.e. whether K or R have actually moved) or en passant capability (i.e. whether the move can actually be made) or who has the move.
  The number of times position has been repeated is not part of the definition of position for obvious reasons; nor is the number of moves since last capture or pawn move.
  I don’t have the answer myself. This is hard but doable with a bit of programming I think.
  EDIT: The answer so far and the comments are clearly heading along the right lines, although not always correct yet: terrific work. I want to give multiple +1s! I will summarize the key points to help focus the work. Pairing mirrored positions is essential i.e. those which allow triangulation. I term positions which have no mirror image “vampires” :-) Any vampire (except for the starting position) must be immediate offspring of a vampire, so it’s maybe easiest to just focus on them.
  Someone touched on castling, which is very important. We can partition the vampires into 16 different clans. The head of each clan is an "initial game array" position (i.e. with 32 pieces on apparently original squares) but with specific castling rights disabled (by some initial dance of rooks and knight taking 2n.0 moves).
  In a clan, if a castling right remains, then that implies that the rook's pawn can safely capture off the file, the f-pawn can move forward a single square or capture or (on the kingside) the bishop can be captured, without allowing triangulation. If the right is lost then the rook's pawn can only move forward a single square, and the bishops cannot be captured.
  En passant is less important but amazingly there are vampire en passant positions. For example:
  r1bqkb1r/2pppppp/8/1pP5/8/8/PPPPP1PP/R1BQKB1R w KQkq b6 0 12
  
  where White retains at least one castling right, Black has at least queenside castling rights and White can capture e.p.
  
  **Tags:** rules, mathematics
  
  > **Comments:**
  > **username_1**: Shouldn't it be even because to every position there is a mirrored position, mirrored on the axis between the e and d files. (Note that a position cannot be mirror symmetric because of the kings.  
  > **username_2**: @username_1 the mirror image of the initial position isn't exactly legal (or at least the OP should clarify whether it is) - it cannot be reached from the starting position.  
  > **username_1**: @username_2: Good point. Is this (and similar positions where queen/king could not have moved) the only exception?  
  > **username_0**: Clearly there has to be some matching strategy based on symmetry, but flipping left/right isn’t viable I think, because it’s hard to marshal the exceptional positions.  
  > **username_3**: @bot I think the approach would be to count the positions that cannot be reached legally with the opposite color on the move. For example, you can reach the equivalent of `1. e3` with `1. Nf3 e6 2. Ng1 Bd6 3. Nf3 Be7 4. Ng1 Bf8`, so there are an even number of legal positions from that point. Of white's 20 opening options, only the four knight moves, a3, f3, and h3 have unreachable equivalents with black. I think it could still be very difficult, though.  
  > **username_4**: @DKrueger Even then, the number will be so high, it probably cannot be counted to in billions of lifetimes. I think that the only way is some kind of mathematical proof, like on a much lower level, why a negative times a negative is a positive, for example.  
  > **username_5**: @DKrueger Unfortunately, that includes not just opening positions, but every position which includes a check.  
  > **username_0**: @bof it’s a good sub-puzzle so I won’t give the answer - it’s only a corner case. Dealing with castling properly is a necessary first step.  
  > **username_0**: @bof This is a question I want the answer to because I don’t know it. But I have some partial results and I want you guys to reach the frontier of my thinking. One cute thing I have found is this e.p. idea: someone else might enjoy looking for it. You need to think about the different sets of castling rights which are possible and the implications so that you don’t double count. I am only doing matching based on non-vampire mirroring. I don’t see any other way to be sure that the matching is 1-1. Hope this helps  
  > **username_0**: @bof attitude? What attitude? I’m baffled  
  > **username_6**: @username_0 There is some strange interaction between castling and parity which I can't quite see how to count effectively yet. Take for instance the position after 1. Nf3 Nf6 2. Ne5 Ng8 3. Ng6 Nf6 4. Nxf8 Ng8. This position cannot be "mirrored" (inversion about horizontal axis + colour inversion) because of the side to move, despite a piece being captured -- you can't reach the "mirror" position while preserving white kingside castling rights.  
  > **username_0**: @username_6: I think there are 16 independent vampire clans, each for a different combination of castling rights. The head of each clan has the pieces all in game array squares. If a player has one castling right then the only way they can move K and R is by castling. Any other move would duplicate a position in one of the other less entitled clans  
  > **username_0**: @bof so is the number of en passant vampires odd or even? :)  
  > **username_0**: @bof Yes I think so too. Two possible double pawn moves in every case. If as was suggested we count all the vampires, then we should count all of these little guys separately  
  > **username_6**: If I have to rely on humanly identifying matrices like the one I posted to do a census of the clans, then I wouldn't call that "effective". :P Too easy to miss an idea or wrongly identify a corner case.  
  > **username_7**: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/106162/discussion-on-question-by-laska-is-the-number-of-legal-chess-positions-odd-or-ev).  
  > **username_0**: No warning? And this is intended as communal solving of a difficult unsolved problem. Does that have no space here? Thanks for spoiling  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_8 on 2020-03-30. Score: 3
  
  Even
  For each position, we can produce 1 "pair" position using the following cases and vice versa.
  Lets break down the game to the following cases:
  Case 1) Whenever we can use triangulation to produce a pair position with the same pieces in the mirrored positions between the rows 4 and 5 with the opponent to move.
  For example:
  
  1. c3 (1. d3) (1. b3 Nc6 2. Ba3)
  
  1. Nc3 b6 2. Nb1 Bb7 3. Nc3 Ba6
  
  Case 2) Every position where triangulation cannot be done.
  Every position can have a pair position with or without castling done for white.
  By moving the white knight on g1 and then the white rook to g1 and then back to h1, the castling rights are lost.
  eg. Starting with the following set of moves
  [FEN ""]
  1. Nf3 Nf6 2. Rg1 Ng8 3. Rh1 Nf6 4. Ng1 Ng8
  
  we produce the start position without castling rights for white, all positions after that until we can use Case (1) can be with and without castling rights for white.
  The initial position falls to Case (2).
  References
  
  Chess triangulation
  
  PS: Updated the answer, so the comments till Jan 7th 2024 are for the old answer
  
  > **Comments:**
  > **username_5**: 1. Nf3 Nf6 2. Ng1 Nd5 3. Nf3 Nf4 4. Ng1 Nd3+ I don't think you can mirror this with Black to move, as there's no way to lose the tempo.  
  > **username_9**: Also just 1. Nf3 Nf6. Like the start position, it's its own mirror image.  
  > **username_1**: I am confused you have "switching colors", "mirrored horizontally" and "corresponding positions". Could you make the difference more clear, specifically when counting, do you count the color-switched or the mirrored positions?  
  > **username_8**: Mirrored positions have exactly the same piece positions on the board switched both piece colors and turn. You must have both, due to checks you cannot only switch turn or colors. Corresponding have the turn switched, but different piece layout on the board, actually ALL positions with only knights moved cannot be mirrored, but have a corresponding position that also cannot be mirrored.  
  > **username_1**: @JannesBotis: Still confused. If I just take your sentence: **"you cannot find a position that colors can be switched and produce an illegal position"** and apply this to the initial position, I end up with all white pawns on the 7th rank and all black pawns on the 2nd. This is clearly illegal because the pawns could not pass each other. Perhaps I misunderstand something....  
  > **username_8**: You have also to "flip/ mirror" the chess board horizontally between 4 and 5 rows. The goal is to produce pairs of positions, so to prove the total is even.  
  > **username_10**: It's not true that non-mirrored positions can only involve knight moves. You can also push rook pawns a single square, and move the rooks a single square forward or sideways. Since this does not give a rook enough freedom to move more than a single square, no triangulation can happen.  
  > **username_3**: If it really comes down to the knight-move only group, that should be even. The special case is check. There are two checking squares and for both, the non-checking knights have equivalent possibilities. So those are even. All the non-checking positions can be mirrored, with the caveat that that may not change the move. But there are an even number of those as well.  
  > **username_6**: Triangulation section is wrong. The moves a3, h3, ...a6, and ...h6 also preserve the parity of the position.  
  > **username_11**: I see no reason why there would be one and exactly one "corresponding position" for each "non-mirrorable position".  
  > **username_11**: If anything, you haven't shown how to build such a "corresponding position"...  
  > **username_8**: I updated my answer anew using a new approach  
  > **username_12**: @JannesBotis What about the initial position with the queens removed, and the kings moved to d1 and d8? There is no castling possible, I don't think there is any triangulation possible (the only moves have been knight moves, rook moves between a1/b1, a8/b8, g1/h1, g8/h8, and king moves between d1/e1, d8/e8). The position is its own mirror. So it doesn't seem to have a partner under your proposal. There are further positions like this, with the knights in different places, and with either the a pawns, or the h pawns, or both, advanced by a single square.  
  > **username_12**: A similar effect can be achieved eg by swapping all the rooks with their neighbouring knights in the initial position  
  
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  ### Answer by username_0 on 2024-11-01. Score: 2
  
  Happy Halloween!
  
  Basically the only possible approach to determining the parity of the number of positions is to pair off positions with one another, and then consider the remainder. The position after 1.e4 can be paired with the position after 1.Nc3 e6 2.Nf6 e5. These are mirror images of one another, with the board and colours reversed, and the other player to move. Most positions in the game are of this kind. To compute the parity of the number of positions overall, we need only consider the relatively small but still uncounted positions with no mirror image. With an attempt at whimsy, we call these vampires. The game array itself is an example: it can only be seen with White to move.
  
  It might be possible to pair off the vampires with one another e.g. ones where wQd1 versus ones where wQ has been captured. But we think this exercise is tricky and probably prone to error. So we are engaged in computer analysis of the vampires as a whole. This will be less error-prone and also give us more information: the number of vampires with White to move and the number of vampires with Black to move.
  
  Considerable progress has been made on this challenge by Miguel Ambrona. He has also developed an app that allows you to play [with chess vampires](https://chasolver.org/vampires/). I love the way the figurines revert to the orthodox ones, when the engine determines that the position is no longer a vampire. (E.g. compare 1. h3 with 1. g3.) Left and right arrows allow you to go back and forth. It is explained in the "What is this" button.
  
  For now we are focusing just on one of the 16 vampire clans. This is the one in which all castling rights remain. No king or rook can move or rook be captured. Not even castling can happen because that costs the other castling right for that player too. The post-castling position if it were still a vampire would fall into another clan. The general shape of results will be the same for all clans.
  
  Miguel has discovered that a vast cavern of simple vampires starting with Ghe game array, depending upon parity, lead to a number of long narrower passages which remain vampiric due to one player being on the clock, and only having a limited number of retractions before being in retro-stalemate. He has published a number of puzzles based on this idea, e.g. [this one](https://www.thehoppermagazine.com/OP033).
  
  As he is still exploring the passages, he cautions that in his app: "Do not trust the output after 8 plies on an 'on-the-clock' position, as that precomputation is not available yet."
  
  [![Chess vampires](https://i.sstatic.net/UmKkCAzE.png)](https://i.sstatic.net/UmKkCAzE.png)
  
  > **Comments:**
  > **username_13**: Could you please clarify how this answers the question?  
  > **username_0**: Have expanded the answer @CyriacAntony - hope this helps  
  > **username_13**: Great. Thank you.  
  
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