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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, as is the number of moves since last capture/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 summarise 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 count all of them. Someone touched on castling, which is very important. Can you see how to handle castling systematically?

En passant is less important but amazingly there are vampire en passant positions e.g.: r1bqkb1r/2pppppp/8/1pP5/8/8/PPPPP1PP/R1BQKB1R where White retains at least one castling right, Black has at least queenside castling rights and White can capture e.p.

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    @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. – Glorfindel Mar 30 at 7:38
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    @Glorfindel: Good point. Is this (and similar positions where queen/king could not have moved) the only exception? – user1583209 Mar 30 at 7:42
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    @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. – D Krueger Mar 30 at 11:18
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    @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. – Remellion Mar 31 at 14:53
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    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. – Remellion Mar 31 at 16:19
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Odd if you count the initial position and Even if not.

Mirrored positions

Almost any position, including checkmates and stalemates can be flipped by switching colors.

Mirroring the chess board horizontally between 4 and 5 rows, you cannot find a position that colors can be switched and produce an illegal position. So for each position there exists a mirrored position, except the start position.

The challenge would be to find a position that cannot be mirrored.

Non Mirrored positions / Corresponding positions

There can exist positions where it cannot be mirrored due to the "turn that cannot be switched". For its this position, there would be a "corresponding position" that also cannot be mirrored.

e.g. 1. Nf3 Nf6 2. Ng1 Nd5 3. Nf3 Nf4 4. Ng1 Nd3+

its corresponding condition is:

  1. Nf3 Nf6 2. Nd4 Ng8 3. Nf5 Nf6 4. Nd6+

Both cannot be mirrored.

Triangulation

Due to Triangulation, all positions after a pawn is pushed can be mirrored. Non mirrored positions can be constructed only with knight moves, moving only knights cannot produce Triangulation.

Start position

The start position, before any moves, the turn cannot be switched so that black makes the first move.

Calculating positions from knight moves

To solve this for sure, we need to calculate the total positions that arise from only knight moves.

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    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. – D M Mar 30 at 20:57
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    Also just 1. Nf3 Nf6. Like the start position, it's its own mirror image. – academic Mar 30 at 22:27
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    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? – user1583209 Mar 31 at 8:01
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    @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.... – user1583209 Mar 31 at 8:11
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    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. – Especially Lime Mar 31 at 9:29

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