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You sit down to play a game of chess. Your side has the white pieces, arranged in their regular starting positions. One problem: your opponent has left, and has taken all the black pieces!

Can you, by making regular chess moves with only the white pieces, maneuver your pieces in such a way that you eliminate all possible moves for yourself?

Rules:

  1. No black pieces are on the board.

  2. You are allowed to remove some, or none, of the white pieces from the board originally, before play begins. Once play begins, however, white is only allowed to make a sequence of regular chess moves.

  3. There is one exception to rule #2: you must retain a King. You are not allowed to remove the white King before play begins.

  4. The pieces that you originally choose to keep must begin in their normal beginning locations. So, for example, whatever pawns you choose to start with must begin on your second rank.

  5. As a consequence of rule #4, you are not allowed to start with more than the usual amount of white pieces.

  6. The pieces move normally. In particular, you will not be able to capture any pieces (since white pieces cannot capture each other, and there are no black pieces). In particular, capturing en passant is clearly impossible. However, you will be allowed to castle (assuming the King and Rook involved have not yet moved), and pawn promotion works as usual.

    (In particular, assume that there are plenty of white pieces lying around. So, if you want to promote all 8 of your pawns to Queens, you can find enough white Queens to continue playing. It's just the black pieces that are gone!)

  7. The point is to achieve a stalemate, not a draw. Thus, draws by threefold repetition of a position, or by the fifty-move rule, etc., do not count.

    https://en.wikipedia.org/wiki/Stalemate

    https://en.wikipedia.org/wiki/Draw_(chess)

  8. In case it matters, those other draw rules will not be enforced at all. So, if you need to make fifty-one consecutive moves without moving a pawn, that's fine. (I don't think there's any advantage in repeating a position three times, though...)

In short: in order to succeed, you need to achieve a board position where no "regular" moves remain for white.

Further notes:

According to rules 2 and 3, you are allowed to begin play with only a King. However, this choice will not succeed, since a King on an otherwise empty board will always have moves available.

Posting an entire list of moves would likely be tedious. As such, you can just give the ending position, if it's obvious that that position is achievable through regular chess moves. However, if it's not obvious that your position is achievable, then you should probably give some kind of explanation, like a move list or at least a description of the tricky parts.

This puzzle was inspired by @Ian McDonald's comments at: Mate in one with NO PIECES?

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Already posted this as an answer to another puzzle, but here we go :)

Remove both knights and the queen. Then promote three pawns into a bishop and two rooks.

enter image description here

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    $\begingroup$ Hey, thanks for pointing out the other puzzle--somehow I didn't find it when looking for duplicates. Anyway, nice job! $\endgroup$ – mathmandan Oct 24 '18 at 22:14
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My original solution looked similar to Jafe's answer:

One-sided stalemate 2 FEN:KQRBRBRB/PPP1P1P1/8/8/8/8/8/8 w - -

There are plenty of variants of this idea. For example:

One-sided stalemate 1 FEN:BKQBRBRB/PPPPP1P1/8/8/8/8/8/8 w - -

This latter solution uses

14 pieces; we start by removing both Knights.

I am not sure if

13, as shown in Jafe's answer, is the minimum number of pieces needed for a solution. (A trivial lower bound on the number of pieces required is 4, and it doesn't take too much work to improve that to 5.)

Similarly, I am not sure if

14 is the maximum. (I guess 16 is an upper bound!)

Finally, I am not sure if there is any solution which

uses one or more Knights. Note that any solution which uses more than 14 pieces must use at least one Knight, so if 14 is not the maximum then there is a solution with a Knight. Equivalently, if it's impossible to use Knights, then 14 is the maximum.

Anyway, to achieve the latter solution,

immediately advance your f- and h-file pawns and promote them to Bishops. Next, play g4 followed by Bg2 and Ba1. Place your other original Bishop similarly, with d4, Bg5, Bd8. Now maneuver your King, Queen, and both Rooks into position. Finally, advance all remaining pawns.

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I believe this works, in response to two conjectures by @mathmandan:

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No spoilers because this is an old question. Mainly exploring minimum and maximum bound.

Some trivial bounds: at least 8, for a start. If a column is empty, it allows movement of any piece from a neighboring column (except a pawn, but you can't have just a pawn in a column). Because of king, you easily increase that bound to 10 - two columns will require 2 pieces each (eg KB,PP) to prevent king's movement.

Now, how many of these columns can remain with a single piece? The only way to have a single is if the "bottom" of that column contains a bishop*. Neighboring columns then require 2. *it might contain knight by the first glance, but knights are tricky and require columns with 3 pieces anyway so they can be ignored.

Adding both requirements (king packed in + those empty columns require neighbors with 2), you end up with 13 pieces at minimum. Some solutions have been found already and you can't find anything relevant new because there are so few options for 3 single piece columns.

Now, for the maximum, we need to throw in some knights. Those are tricky beasts, because each knight requires at least one column with 3 pieces. Suppose knight is on A8 - it can move to B6, so that tile needs to have a pawn (any figure requires neighbors with even more pieces). Pawn needs to be blocked from going forward. If you put a rook there, you need B8 with something = 3 pieces in column. If you put a bishop, you need C6 and C8 with something to block bishop + C7 to block knight = a column with 3 pieces again.

Second knight requires another column with 3. So, we need to squeeze in 2 columns with a single bishop.

You can almost do it:

FEN: BRBRKNQN/1P1PRPRP/4P1P1/8/8/8/8/8 w - -

The only problem here is that we have no dark bishop and 2 light instead.

I haven't found an elegant proof of impossibility here, but you can see you can't position knights by trying all the options on 8th row (so, AB, AC, AD... BC ... DH - the other half is just mirrored). AB fails because columns ABC have 3 pieces and D has 2, so you lack one piece. AH fails because only DE can have a single piece so again you lack one (other stuff is obviously impossible)

The maximum is therefore 15 pieces. Again, it was found already.

Note that one knight solution requires 15 pieces. One column has 3. To get solution with 14, there need to be three columns that have single bishop. This is possible only if you permit all bishops are light or dark, similar to 16 piece solution above.

Summary of possible options is therefore:

  • 13 pieces without knights. (in addition to knights you need to remove either queen or bishop, same reasoning as above)
  • 14 pieces without knights.
  • 15 pieces with one knight.
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  • $\begingroup$ Starting from it might contain knight by the first glance ..., the arguments are no longer as rigorous as previously. $\endgroup$ – WhatsUp Jul 21 at 12:39
  • $\begingroup$ @WhatsUp this knight is focused on later. It is trivial to see that knight requires at least 2 pieces in columns next to it, same as bishop: Put a knight on say D8. E6 needs pawn (any other piece can move to column D, invalidating "knight is the only piece in column"). Then you need another piece on E7 to block it. So you are at 2 pieces. A tiny step forward is observation that E7 cannot be a pawn and the piece can move to column D, invalidating "knight is the only piece in column", so I put it the way I did - but it has no effect on minimum bound so I didn't write all the reasoning. $\endgroup$ – Zizy Archer Jul 21 at 13:05
  • $\begingroup$ @WhatsUp The shaky bits do start - but I believe they start only after the FEN. I have no idea how to elegantly prove these things. Some options are eliminated by need to have just light square bishop or just dark square bishop which is harder to prove than just manually checking all options and seeing nothing fits. $\endgroup$ – Zizy Archer Jul 21 at 13:14

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