twenty-eight
$\begingroup$

Here is my third retrograde puzzle—I hope you will enjoy it :) For this construction, I took some inspiration from shoopi's excellent puzzle series from a while back. It should also be a bit more approachable than my previous ones:

Is it possible to reach this position in such a way that White can still castle?

 Final position, FEN is: 1B1n1b1n/p1pppppp/8/P7/8/6P1/1P1PPP1P/R3K1kr

Please provide your reasoning in your answer.

$\endgroup$
one
  • five
    $\begingroup$ Another nice one. Keep them coming :-) $\endgroup$ May 18 at 16:59

3 Answers three

Reset to default
twenty-one
$\begingroup$

My answer is

Yes, it's possible!

To back it up,

Here's a proof game:
1. Nh3 Nc6 2. Nf4 Na5 3. Ng6 Nc4 4. Nxh8 Na5 5. Ng6 Nb3 6. Ne5 Nxc1 7. Nc6 bxc6 8. Nc3 Nb3 9. Nd5 cxd5 10. a3 Nc5 11. g3 Nf6 12. Bg2 Ba6 13. Be4 dxe4 14. Qb1 Bc4 15. Qa2 Be6 16. Qb3 Bh3 17. Qf3 exf3 18. Rg1 Bf1 19. Rg2 Nd5 20. Rg1 Nf4 21. Rg2 Ng6 22. Rg1 Nh8 23. Rg2 Qc8 24. Rg1 Qa6 25. Rg2 Kd8 26. Rg1 Kc8 27. Rg2 Kb7 28. Rg1 Kc6 29. Rg2 Kd6 30. Rg1 Ke6 31. Rg2 Kf5 32. Rg1 Kg4 33. Rg2 Kh3 34. Rg1 Rb8 35. Rg2 Rb5 36. Rg1 Ne6 37. Rg2 Nd8 38. Rg1 Qc6 39. Rg2 Qb6 40. Rg1 Qa6 41. Rg2 fxg2 42. c3 g1=R 43. c4 Rh1 44. cxb5 Kg2 45. bxa6 Kg1 46. a4 Bg2 47. a5 Bb7 48. axb7 Ne6 49. b8=B Nd8 50. O-O-O+
https://lichess.org/study/QqcwimPw/XzEKBTe2

To explain how I found this answer,

If white can castle, then the black rook on h1 is the promoted b7 pawn, having captured five white pieces to reach g2 and promoting on g1. This is how all of the white pieces were captured, except for the c1 bishop which never moved.

Also note that

When the black pawn promoted, a black knight or bishop was on f1. All the white pieces that could've stood there were already captured.

Now the difficulty is that

Working backwards from the final position, only white pawns have moved for the last several turns, and we only have a limited number of pawn moves available in the past. It looks like we have nine pawn moves available, three by the a5 pawn and six by the pawn that promoted to become the b8 bishop. But actually, if white had played those nine pawn moves in the turns immediately preceding the current position, there would be no way for the white queen to escape to get captured by the black pawn. So white can only have played eight consecutive pawn moves leading up to the current position.

How recently

Can black have captured a white piece, so that white had something else to do besides pawn moves? Well, that can only have happened before the black pawn promoted. So it depends on which black piece stood on f1 when the promotion happened. If a black knight stood on f1, then five knight moves, two king moves, two rook moves and a pawn capture must've happened, which is ten moves, one move too many. If a bishop stood on f1, then that bishop must've been captured by the pawn that promoted to the b8 bishop. If that capture was cxb7, or prior to the seventh rank, then the bishop doesn't get there in time. But if the capture was axb7, then things work out precisely: two waiting moves by black, two bishop moves, two king moves, two rook moves, and a pawn capture. Nine moves, exactly right. And fortunately, there are just enough missing black pieces, the queen and a8 rook, for the white pawn to get to a6. (The h8 rook never escaped). And with all that, a proof game can be made.

$\endgroup$
eight
  • two
    $\begingroup$ Very nice writeup! That's exactly the solution I was looking for :) $\endgroup$ May 18 at 19:28
  • two
    $\begingroup$ rot13("vs juvgr whfg cynlrq gubfr avar cnja zbirf, gurer jbhyq or ab jnl sbe gur juvgr dhrra gb rfpncr") I'm not sure I follow; isn't is more that rot13(vs juvgr unq cynlrq nyy avar bs gubfr cnja zbirf? v.r. vg'f abg gung n ynetre frg bs zbirf vf arrqrq, ohg gung bayl n fznyyre frg vf cbffvoyr.) $\endgroup$
    –  Rosie F
    May 19 at 13:42
  • $\begingroup$ @RosieF Sorry for the confusion, I've edited the passage. I meant "just played" as in "played in the immediate past", rather than "played no more than", I see why it was confusing. $\endgroup$
    –  isaacg
    May 20 at 3:03
  • $\begingroup$ Do you mean passed pawn rather than "past pawn"? $\endgroup$
    –  psmears
    May 20 at 20:00
  • $\begingroup$ @psmears No, by "past pawn moves" I meant "pawn moves that happened in the past". $\endgroup$
    –  isaacg
    May 20 at 20:10
ten
$\begingroup$

It

is possible.

Proof

game: 1. Nf3 Nf6 2. Nh4 Nd5 3. Ng6 Nc6 4. Nxh8 Nd4 5. Ng6 Nb3 6. Ne5 Nf4 7. g3 Nxc1 8. Bg2 Ng6 9. Be4 Nh8 10. Rg1 Nb3 11. a3 {let the queen out while preserving as many stalling moves as possible} Nd4 12. Nc3 Ne6 13. Qb1 Ng5 14. Qa2 Ne6 15. Qd5 Rb8 16. Qc6 bxc6 17. Nd5 Ba6 18. Bf3 Bd3 19. Nc4 Bf5 20. Ne5 Bh3 21. Be4 Bf1 {it must be the bishop, a knight would be too slow to reach their final position from here} 22. Rg2 Qc8 23. Nf3 Rb6 24. Rg1 cxd5 25. Rg2 dxe4 26. Rg1 Qa6 27. Rg2 Kd8 28. Rg1 exf3 29. Rg2 Kc8 30. Rg1 Kb7 31. Rg2 Kc6 32. Rg1 Kd5 33. Rg2 Ke4 34. Rg1 Kf5 35. Rg2 Kg4 36. Rg1 Kh3 37. Rg2 Rb3 38. Rg1 Nd8 39. Rg2 fxg2 {from here black needs to finish before white's remaining mov es are used up} 40. a4 g1=R 41. a5 Kg2 42. cxb3 Rh1 43. b4 Kg1 44. b5 Bg2 45. bxa6 Bb7 {just in time} 46. axb7 Nc6 47. b8=B Nd8 and now white could castle if they saw it fit

$\endgroup$
six
  • two
    $\begingroup$ Apologies for the simultaneous post - I was writing my answer when you posted yours. $\endgroup$
    –  isaacg
    May 18 at 16:50
  • three
    $\begingroup$ @isaacg No need to apologize. Great minds and all that ;-) $\endgroup$ May 18 at 16:52
  • two
    $\begingroup$ I suggest rot13(uvqvat gur jbeqf "cebbs tnzr" oruvaq fcbvyre znexhc orpnhfr gurl erirny jurgure gur nafjre vf lrf be ab). $\endgroup$ May 18 at 19:05
  • two
    $\begingroup$ Very well done! I see that you changed your mind for the better, thanks for keeping at it :) $\endgroup$ May 18 at 19:27
  • two
    $\begingroup$ @TimSeifert Shh, not so loud, please. Unlike our friend Gareth I do believe in hiding my mistakes ;-p $\endgroup$ May 18 at 19:55
one
$\begingroup$

I'm new to these and I may be wrong, but I'm gonna say:

Yes it is possible. The points below are a kind of attempt at disproving it that ultimately finds it doesn't seem it can be disproven.

1. White's dark-square bishop must be a promoted pawn to be where it is with black's a and c pawns unmoved.

2. For white's king to be unmoved with white's pawns where they are, black's rook must be a promoted pawn, and specifically originally the b pawn. Furthermore, under the same assumption, it can only have been promoted on g1 or h1.

3. Black's pawn cannot have been promoted on h1 because it would have captured 6 of white's pieces along its diagonal path, when there are only 5 pieces it can have captured: 2 knights, the queen, the missing rook and the light-square bishop. Additionally, these 5 pieces must have been captured by the black pawn before its promotion.

4. For black's pawn to have promoted on g1 before moving to h1 without white's king ever moving, there must have been a piece on f1 shielding white's king from the rook until black's king moved to g1. With the aforementioned 5 white pieces having already been captured, this must have been a black piece. Since it can't have checked the king, it must have been a bishop or a knight.

5. White's dark-square bishop can't have ever moved, as white's b and d pawns are unmoved.

6. For white's castlability to be intact, the only moves white can have made since the black pawn capturing onto g2 are the advancement of the a and c pawns including at least 1 capture by the c pawn, and the possibility of more captures by these 2 pawns is not to be neglected. For white's queen to make its way into the capture path of the black pawn, one of these white pawns must have already been moved at least once. Therefore maximally 8 moves can have been made by white since the g2 capture.

7. Since capturing on g2, black will need minimally 4 moves to promote and move the new rook to h1, then move the king to g1.

8. If the black piece on f1 were a knight, it would need 5 moves to get to its final position, totaling 9 moves for black, so it if this position is reachable where white can still castle, then it must have been a black bishop which was shortly thereafter captured by white's a or c pawn.

9. Black's light-square bishop needs 2 moves to get b7, which is a good place for it to get captured. This means that 6 moves must have already happened since black's pawn capturing on g2. After white's 6 pawn moves (and the earlier pawn move to free the queen), white's c pawn would already be in c6, blocking the bishop from b7--unless white's c pawn had already captured its way to the a file, that is.

10. If two of black's pieces, a queen and a rook for example, were lined up for the taking on b5 and a6, then the original c pawn would not be blocking black's bishop and the position is reachable.

Thinking about further,

While 1-8 are pretty invariable as far as I can tell, there seem to be a lot of ways it could shake out with the pawns. For instance, the a pawn might actually be the one that captures the bishop on b7, while the c pawn captures to the a file, or maybe the bishop doesn't even get captured on b7.

This was a fantastic puzzle!

New contributor
stumpgrump is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct .
$\endgroup$
one
  • $\begingroup$ Very good work, thanks for your solution! :) For your last point, you're right that rot13(gur rknpg ynfg zbirf bs gur juvgr cnjaf ner abg havdhr, abg rira gurve bevtva vf. Abarguryrff, gur ovfubc zhfg unir orra gnxra ba o7), which I'm sure you can verify yourself after coming this far :) $\endgroup$ 2 days ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy .

Not the answer you're looking for? Browse other questions tagged or ask your own question .