Retrograde puzzle

Is this position, with Black to move, legal?

This position is part of my random sample of chess positions at

https://github.com/tromp/ChessPositionRanking/blob/main/sortedRnd1kFENs and determining its legality will help to estimate the number of legal chess positions. See also the discussion at https://www.chess.com/forum/view/general/on-the-number-of-chess-positions

Wow this puzzles sooo confusing rn

Yes black has a couple legal moves like ka8 or kc8

It's not about what moves can be made in this position, but whether the position could have been reached from the starting one by legal moves. That is what it means for a position to be legal.

Here's another one:

Legal or not?

I believe the answer to this one is yes.  Each side has 2 pawns.  That leaves 6 pawns each for promotion or being captured.  White's LSB has also been captured.

Each side has a minimum of 9 promotions - Bishop, Knight, 2 Rooks, and a Queen for White, 3 Bishops and a rook for Black.  That allows for 5 captures.  Black Queen, 2 Black pawns, White Bishop, and White pawn.  All Bishops can get out as there is no case of B and d or e and g pawns of one color on their original squares (i.e. Black pawns on b7 and d7) and so no extra promotion is necessary.

So you need to be able to make 9 promotions with 5 captures.  Two Black promotions must be on light, one on dark.  One of White's promotions must be on dark.

Black throws his a-pawn away to a White piece and White promotes the a-pawn.  That leaves 4 promotions each with BP, BQ, WP, and W LSB to boot.

Next the Black Queen is given away on the d-file behind his d-pawn to the White c-pawn.  Black gets a promotion on c1 and White gets a d8 promotion.  That leaves 3 more promotions for each play, 2 on light squares for Black, with WP, BP, and WB to boot.

Black surrenders his e-pawn to White's d-pawn and White gets two promotions on e8.  This leaves 3 promotions for Black, one for White with a Bishop and Pawn, both White, to boot.

White rides the h-pawn to h6 for Black to capture.  This gives Black two h1 promotions (there are his 2 light square Bishop promotions) and White's g-pawn promotes.

Last, White surrenders the Bishop to the Black f-pawn for a g-file Black promotion.

Therefore, in summary.  The promotions are:

White - a8, d8 (his dark square Bishop promotion), twice on e8, and g8.

Black - c1 (his dark square Bishop promotion), g1, and twice on h1 (his two light square Bishop promotions).  Black maintains b and d pawns, White b and f.

Also, last move is Nc7-b5 for discovered check.

I suppose the position could have been reached  

See posts 6 and 7.  That is how you have to figure these out.  Knights, Rooks, Kings and Queen's can get to any square.  It is all about the following:

Number of captures

What pieces and pawns are captured

Bishops - You have to make sure that they could get out from their start square, promotion to Bishop must be on the correct color promotion square, etc.

Pawns - How to get enough promotions given the number of captures you have, and making sure you have enough promotions on the correct color squares for Bishop promotions.

Lastly, what was the last move if a King is in check.  If no check, last move is irrelevant.

Also, if a Bishop is hemmed in on the start square, then the Rook can only get out with the Rook pawn pushed at least 2 squares.  If you have say, BPe7, BPg7, BBf8 (Meaning it has not been captured), BPh6, and a rook outside of h8, g8, or h7, then it means one of your captures must be of the Black Rook on h8 and a promotion of Black's must be to an extra Rook.  This wasted capture and promotion can also make it impossible, and this is why I clarified in the solution that no Bishop is hemmed in its original square and so not extra capture and promotion to Bishop or Rook was necessary.

Proving a position illegal is done with reasoning of the kind Thriller Fan has done.

Proving a position legal is done with a proof game.

I have a game of 70 moves leading to the first position which you have posted: https://www.chess.com/a/ojscGi4W2KhQa

For sure it can be done with a few moves less, but the question was whether it is legal, and yes it is!

Good job, rjehn. Here is your proof game without the side variations.

Good job on the first one. I found 52 legal positions in the random sample of 1000 positions. But rather than give a full proof game for each, I'll just provide a possible sequence of pawn captures and pawn promotions, which seem to be the critical elements.

What do you think about the second position?

I think right now that it's illegal. I don't think the four available captures are enough to create the open files needed for the seven required promotions, but I don't have a solid proof. I showed the position in the Illegal Positions thread to see if anyone there can give a valid illegality proof or compose a proof game.

The second position does seem to be illegal. It's possible to get the 3 white and 4 black promotions, but Black cannot promote to both bishops on light squares, just one. 

#5
I also think it is illegal. What possible sequence of pawn captures and pawn promotions did you produce to conclude it would be legal? If the method is wrong for this one, it may be wrong for other as well and the real fraction of legal positions may be lower.

I didn't say the 2nd position is from the set of 52 I found to be legal...

I come to the same conclusion when trying to find the matching promotions. There are just enough captures available to get 4 black pawns to promotion fields, but only the b-pawn can make it to a light square. d1 and f1 are blocked and the h-file cannot be exploited because the white h-pawn would need to be captured (if it captures a piece itself to get to the g-file, that would not work, because the pawn on g7 is still sitting there). But we need the capture of the only white missing piece for creating the separation of the pawns in the other rows. So this position is illegal

In my analysis (trying to limit myself to one line) I wrote about the 2nd position:

Illegal since black pawns promoted on b1 c1 e1 g1, only 1 light colored