In this position, White’s sole aim is to castle. But in order to do so, they must promote their pawn to a queen to force a Black bishop move. The 10 black kings represent where Black’s king can safely be put so that it can’t prevent the promotion.

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FEN: 8/8/6p1/6pb/5prp/5PpN/P5Pb/k3K2R w K - 0 1

On which square/s should the Black king be placed as to gurantee the longest string of moves before White can castle?

Computers are allowed.

  • $\begingroup$ The final sequence of moves, I believe, will be 1... Bg1 2. Nxg1 h3 3. Nxh3, after which Black is stalemated and White cannot castle. $\endgroup$
    – Cloudy7
    Commented Apr 3, 2020 at 16:59
  • $\begingroup$ Oops, that is correct. After 2... h3, the rook can move. $\endgroup$
    – Cloudy7
    Commented Apr 3, 2020 at 17:02

3 Answers 3


Here's my answer:

The initial position of the king does not matter.

Reasoning :

First of all, let's reformulate the problem.
Promoting the pawn takes 5 moves, and it will always be promoted on the same square.
So we want to find how many moves the queen will take to stalemate the king for every possible black king position in this position : Q7/8/6p1/6pb/5prp/5PpN/6Pb/4K2R w K - 0 1

Moreover, there are 11 of these squares that the king can access in 5 moves, regardless of its original position : c3, d3, e3, c4, d4, c5, e5, f5, d6, e6 and f6.
Therefore if one of these square have a 'stalemate time' higher (or equal) than every single square in this list : h6, h7, g7, f7, e7, d7, c1, c2, c7,b1, b2, b3, b4, b5, b6 (squares that you can only access in 5 moves from certain initial king positions), then the initial king position doesn't matter.

This is exactly what happens ; the b6 square takes 8 moves to stalemate and it's the maximum of list 2. But the d4 square, that we can acces in 5 moves from all the 10 king positions, also takes 8 moves to stalemate.
We end up with the following string of moves, supposing the king started of b2 (but we can also make it start on a1, b1, c1, c2, h6, h7, h8, g7 or g8) :
1. a4 Kb3 2. a5 Kc3 3. a6 Kc4 4. a7 Kc3 5. a8=Q Kd4 6. Qa5 Kc4 7. Qe5 Kb4 8. Qd5 Ka4 9. Qb7 Ka5 10. Qb3 Ka6 11. Qb4 Ka7 12. Qb5 Ka8 13. Qb6 Bg1 14. Nxg1 h3 15. Nxh3 Rh4 16. O-O

But I might be wrong.

  • $\begingroup$ Is it the correct answer ? I could not find any square in the second list that was longer to stalemate than the square I used in my string of moves. $\endgroup$ Commented Apr 5, 2020 at 14:00

Maybe I'm missing something, but it seems that

regardless of where the black king is placed, the white queen will be able to force them into a position where they cannot move, resulting in the sequence of moves: 1. ...Bg1 2. Nxg1 h3 3. Nxh3 Rh4 4.O-O So from this the only question is what position of the black king will require the longest stalemate sequence. I'm guessing this is c5? Which the king could reach from any initial position before the pawn promotes.

But I'm not certain.

  • 1
    $\begingroup$ Nice! To be honest I'm unsure of where to even start when it comes to rigorously figure out the last part, so I'll probably leave the full solve for someone else. Interesting puzzle though! $\endgroup$
    – H Rogers
    Commented Apr 3, 2020 at 16:20

I believe the answer is

g7 or g8.

The reason being

If White plays 1. a3, then Black plays 1... Kh7! and there follows
2. a4 Kh8
3. a5 Kh7
4. a6 Kh8
5. a7 Kh7
6. a8Q Kg7
and White is forced to stalemate in one of the other corners, as the Queen cannot move any closer than the e-file and Black will always have at least two squares to maneuver.

Meanwhile, if White plays 1. a4, then Black plays 1... Kh8! and there follows the same sequence as described above.

Maybe this is wrong, but I'd love to hear your thoughts.

  • $\begingroup$ @RewanDemontay You are correct; I will keep mulling over this... $\endgroup$
    – Cloudy7
    Commented Apr 3, 2020 at 18:10

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