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After my most recent retrograde chess puzzle had turned out pretty difficult, I had intended to follow up with just a little refresher - some position in which the mystery of the last move was much more straightforward to uncover. But alas, after setting up the diagram, something must have gone wrong:

Final position

Huh? What happened here? It looks like the chess pieces in it were replaced by letters! From what I can vaguely remember about the position, I can tell you for a fact that letters of the same case are of the same colour and that the same letters (regardless of case) denote the same type of piece. Using this information, can you help to reconstruct my puzzle?

Can you determine the position and the last move?

Please provide your reasoning in your answer. Have fun solving this! :)

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    $\begingroup$ Is a light square and dark squared bishop considered the same piece here? $\endgroup$ Commented Oct 25, 2024 at 13:18
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    $\begingroup$ @PDT Yes, they are the same piece type in this puzzle $\endgroup$ Commented Oct 25, 2024 at 13:22

5 Answers 5

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First of all, let’s establish what letters we already know can’t be certain pieces.

B, C, and F can’t be pawns because they’re on the back rank. E can’t be the king because white doesn’t have one. A and B can’t be the king because there are duplicates in the same color. This will be more important when we start to get into the nitty-gritty of it.

Let’s now assume that nothing bizarre is going on and that a represents white pawns, and A is black pawns. However, this can’t be the case. Both white’s c1- and f1- bishops could not “escape” due to the unmoved pawns on b2, d2, e2 and g2, so they both had to be captured on their starting squares (this is plausible, since white has no “e” piece.) However, black has made two pawn captures on b6 and f6 in this scenario, so between the captures on b6, f6, c1 and f1 they must have made four captures. However, white is only missing three pieces.

Switching a to black pawns and A to white pawns does not help. White would have had to make at least one capture on b6 and f6, and at least two more captures for their a-pawn alone to pass Black’s (one to pass, and one to get it back to the a-file). Therefore, the letter a cannot represent pawns.

A also cannot represent knights. As I explained above, the king must be C or F. If A represented knights and the king was F, the king on h8 would be in an illegal double check. If A represented knights and the king was C, both kings would be in check.

A also cannot represent rooks or queens for a similar reason; whether the kings were C or F, both would be in check.

Therefore, the only valid piece for A is the bishop. If a is a bishop, then C cannot be the king due to an illegal double check, so F must be the king. This puts the king on d6 in check from the bishop on a3; this check must be a discovered check, since there’s no way for the a3-bishop to move there without having the king already in check. Keep that in mind.

B can’t be the queen because it would put the other king in check. It can’t be a knight due to an illegal double check from c8 and a3. Bishop and king are accounted for, and B can’t be a pawn (back rank), so B must be the rook.

From here, the rest is fairly simple. C can’t be the queen due to an illegal double check from e8 and a3. D can’t be the queen due to an illegal double check (having it move from c5 would have had the king in check already). Therefore, E must be the queen. C also can’t be the pawn (back rank), so d must be the pawn, leaving C to be the knight.

There’s still another problem. The pawn couldn’t have moved from c5, since it would have already had the black king in check. It couldn’t have taken a piece on c6, because then the king would have been in check already. If only there was a way for a pawn on d5 to take a piece from c5…

In conclusion, the final position is

enter image description here

We're not quite done, as @mathmandan pointed out:

Black's most recent move was c5; prior to that it had a pawn on c7, and the Black king was in check from the a3 bishop. Again, this must have been a discovered check. But what piece could have delivered it? Only the pawn: we must have had 1. b5+ c5 2. bxc6ep+.

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    $\begingroup$ Can you add an example game to show how it is possible to reach this position? I can't work out how you get that many pawns passed each other for promotion without more captures. $\endgroup$ Commented Oct 25, 2024 at 19:42
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    $\begingroup$ @fljx That's actually a great question that I hadn't considered! I'm not sure if it's possible; I actually made a similar argument to explain why A couldn't represent black pawns. This seems to be the intended solution due to how clever it is with the final solution, but I'm wondering if there's a way to actually get there. $\endgroup$ Commented Oct 25, 2024 at 19:46
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    $\begingroup$ Very nice answer (and welcome to Puzzling.SE)! :) There certainly is a way to reach this position, but it does require a bit of thought $\endgroup$ Commented Oct 25, 2024 at 20:11
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    $\begingroup$ The answer is a fun conclusion to the puzzle, but I believe it is incomplete without a proof game. $\endgroup$ Commented Oct 26, 2024 at 0:12
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    $\begingroup$ +1 But I think it was: rot13(okp6, abg qkp6. Oynpx'f zbfg erprag zbir jnf p5; cevbe gb gung vg unq n cnja ba p7, naq gur Oynpx xvat jnf va purpx sebz gur n3 ovfubc. Ntnva, guvf zhfg unir orra n qvfpbirerq purpx. Ohg jung cvrpr pbhyq unir qryvirerq vg? Bayl gur cnja: jr zhfg unir unq 1. o5+ p5 2. okp6rc+.) $\endgroup$ Commented Oct 26, 2024 at 4:46
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Example proof game for the position:

1. d4 e5 2. dxe5 f5 3. g4 fxg4 4. h4 g5 5. h5 Ne7 6. Nc3 Ng6 7. hxg6 b5 8. Ne4 b4 9. c4 h5 10. Nc3 bxc3 11. b4 a5 12. c5 a4 13. g7 Ra6 14. g8=B Rb6 15. cxb6 h4 16. Qb3 axb3 17. a4 h3 18. a5 b2 19. a6 b1=Q 20. a7 Qh7 21. a8=B d5 22. Rh2 c2 23. f4 Bh6 24. e4 g3 25. Re2 h2 26. f5 h1=B 27. f6 g2 28. Nf3 g1=B 29. Nd2 Bc5 30. f7+ Ke7 31. f8=B+ Ke8 32. e6 g4 33. e7 g3 34. Be6 g2 35. Bh3 g1=B 36. Bg7 Kf7 37. e8=B+ Ke7 38. Bh5 d4 39. Ba3 c1=B 40. b7 Nd7 41. b8=B d3 42. Ba7 Bd6 43. Bf2 Bgh2 44. e5 B2f4 45. Baf3 Bg2 46. Bfg4 Bgb7 47. Bhg2 Be3 48. B2h3 Ba7 49. Bhg2 Bd5 50. B2h3 Bcb7 51. Bhg2 Bf7 52. e6 Qe4 53. Bgb2 Qc4 54. B4h3 Qa6 55. Ne4 d2+ 56. Kd1 Nb6 57. Kc2 d1=B+ 58. Kc3 Bce3 59. B5g4 Ba4 60. B2f3 Beg5 61. B1g2 Kf8 62. e7+ Kg7 63. e8=B Kf8 64. Bb5 Ke7 65. Kd4 Bc5+ 66. Ke5 Bd6+ 67. Kf5 Bc5 68. Be1 Bfd5 69. Kg6 Rf8 70. Kh7 Qd7 71. Bf1 Rf7+ 72. Kh8 Rg7 73. Nf6+ Kd6 74. Nh7 Bf6 75. Nf8 Qf5 76. Re8 Qf4 77. Bhg2 Qh2 78. B1e2 Bf7 79. Bd2 Nd5 80. Bbd3 Nc3 81. Bc2 Bd7 82. Ra8 Bf5 83. Rc8 B5g6 84. Rc1 Bcb6 85. b5+ c5 86. bxc6+

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    $\begingroup$ Woah! I was wondering when someone would drop a proof game! +1, very nice work, and welcome! $\endgroup$ Commented Nov 3, 2024 at 2:29
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My response to this very nice puzzle builds on the great accepted answer, by looking in detail at the promotions.

An easy way to simplify the promotion analysis is to see that in the position prior to en passant, the two surviving pawns (wPb4 and bPc7) would be unblocked to promote to dark-squared bishops. Imagine that they do so. Then we would have at least 7 promotions by each side: 13 bishops & 1 black queen. This imaginary position is legal if and only if the actual position following en passant is legal. The missing material is White QNP & Black RNP.

A pawn capturing an officer enables two promotions on the same colour square, as one file is unblocked. A pawn capturing a pawn enables three promotions on the same colour square, as two files are unblocked (but one of the four pawns is captured).

So the missing material (4 officers and 2 pawns) is exactly enough to allow us to unblock the 8 files, but we don't have any spare captures to shift pawns to adjacent files before promotion. Nor can we afford to capture any of the original units (especially bishops), and replace them with promoted ones, as we need both of the two missing pawns in order to unblock two files each.

So if all the 14 promotions were to bishops, then the ratio of light/dark bishops is nearly the same for White as for Black. However the two pawn captures give us an extra white & black bishop of either tint.

After the hypothetical promotions of the wPb & bPc, we have 5 light to 4 dark White bishops, and 3 light to 5 dark Black bishops. If the black queen promotion occurred on a light square, then the bonus White bishop can be light and the bonus black bishop be dark.

This leaves us with 8 light/dark White/Black bishops who promoted by officer sacrifice. Given the mix of officers sacrificed, there is a final parity constraint that the number of such light bishops (of each colour) be even: here it is 2.

So this is achievable.

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  • $\begingroup$ Oh wait, I haven't read your claim correctly! Hm .. the position is certainly legal, but now I'm unsure, where your mistake lies $\endgroup$ Commented Oct 27, 2024 at 9:20
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    $\begingroup$ Ah, I see now: "So if all the 14 promotions were to bishops, then White's ratio of light to dark bishops would have to match Black's." This is not so! The promotions achieved by single PxP will offset this balance instead, by giving the capturing side an additional bishop of that colour. Indeed, your analysis then shows that we must use these possible offsets for both colours, to give white an additional lsb and black an extrta dsb. $\endgroup$ Commented Oct 27, 2024 at 9:43
  • $\begingroup$ Ok thanks for spotting the error. I’ve now fixed. It’s all possible as long as the black promotion to queen happens on a light square $\endgroup$ Commented Oct 27, 2024 at 14:47
  • $\begingroup$ So if it is achievable, can you write a proof game? $\endgroup$ Commented Oct 27, 2024 at 23:09
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    $\begingroup$ It's interesting that there's no engine to give us loose proof games, in contrast to the several engines (Stelvio, Popeye, Jacobi, Euclide, Natch) that will find tight proof games. It's the difference between arranging cornflakes & solving a jigsaw. Yet Nature will happily arrange cornflakes, while unable to figure out even a simple jigsaw $\endgroup$ Commented Oct 28, 2024 at 1:59
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A very clever puzzle! Raymond Smullyan would be proud!

However, I don't think the final position is legal, due to the promotion issue. White has one pawn remaining, and had to promote to 6 bishops to account to the existing 8 bishops (6 promoted, 2 original). This accounts for 7 White pawns, leaving one White pawn to "play with" to get out of the way of the advancing Black pawns.

Black, on the other hand, has to promote to 5 bishops to account for the existing 7 (5 promoted, two original). Also, one of Black's pawns had to be promoted to a queen, so that's 6 promotions. But remember that Black's c7 pawn has to remain there until the very end, allowing bxc6(ep), thus also leaving only one Black pawn to "play with."

There are two more White pieces that could have been captured (Q and N), and two more Black pieces (R and N). This means that there are a total of 3 White pieces and 3 Black pieces missing, and available for possible captures.

Thus, the question here is whether these 6 pawn captures are sufficient to allow the 12 required promotions to occur. I don't see how this is possible! For example, only one white pawn can be sacrificed to make way for a Black pawn. How can the 6 Black pawns sneak past the 7 White pawns? For example, after 1 e4 e5 2 d4 exd4 there can be two White promotions on the e-file, and two Black promotions on the d-file (after suitable pieces get out of the way). But no more White pawns can be sacrificed to let a Black pawn through. How can the Black h-pawn, say, score a touchdown, as it will be blocked by White's h-pawn?

Any problems with this reasoning?

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    $\begingroup$ Welcome to PSE. Suppose wPg (White pawn that started on g-file) captured bPh. Then wPg & wPh are free to promote on h8. Also bPg is free to promote on g1. Thus the sacrifice of one pawn enables 3 promotions. The sacrifice of an officer enables 2 promotions. So 4 officers & 2 pawns allows for 4x2 + 2x3 = 14 = 12 promoted units + 2 promotable pawns in the case of bPc7 & wPd5. $\endgroup$ Commented Oct 30, 2024 at 7:24
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    $\begingroup$ Welcome to Puzzling.SE! As already noted by Laska, the number of captures is (just) sufficient to achieve all these promotions, so your count is a bit off There are still some subtleties involved (arising from the fact that bishops cannot change their square colours), which are all adressed in Laska's answer - I recommend checking that out if you are curious. Building a proof game following their recipe is then straightforward enough if a bit tedious :) $\endgroup$ Commented Oct 30, 2024 at 10:51
  • $\begingroup$ You are right! I overlooked the "power" of the capture of the two extra pieces for both sides. $\endgroup$ Commented Oct 31, 2024 at 4:53
  • $\begingroup$ You are right! I overlooked the "power" of the capture of the two extra pieces for both sides. I have created a "proof of a proof" game using a starting position in which all queens, bishops, and rooks were removed. This left 8 pawns, 2 knights, and a king for each side in their starting position. The 2 knights are the "officers" to be captured; it doesn't matter for this conceptual proof that they are not the pieces that would need to be captured in an actual proof game. This "game" with my comments can be found at Lichess, Study, search Uommibatto and look for Retrograde. $\endgroup$ Commented Oct 31, 2024 at 5:02
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I have an idea.

I think that the E's are bishops, because they cover the diagonal and there's one on black and one on white. The A's are obviously pawns, there are tons of them and the front side has them lined up like pawns would be. I think the F's are the kings because there's only one of each and one of them could be getting attacked. I think the C's are the queens because there's two of them, but they can't be kings because of the pawns. I think the D's are knights, because if they were Rooks, the last available piece, the black King would be in check, meaning the black side either moved a price while in check or moving the King into check, which is illegal in chess. So that leaves the B's as Rooks, which leaves the position legal.

this is the last move

This is what I think was the last turn, since the queen is moved from checking the King, like the picture showed.

this is the next turn

And I think this is how the next turn played out, checking the King, since this is what I think the picture is of.

I could be wrong, although this seems to be quite correct.

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    $\begingroup$ "I could be wrong, although this seems to be quite correct." - have you checked the other answer above which has already been accepted (officially marked correct by the puzzle-setter)? $\endgroup$ Commented Nov 18, 2024 at 6:33
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    $\begingroup$ Welcome to Puzzling.SE :) Some of your heuristics here ("The A's are obviously pawns [...]" etc.) seem to have led you astray here. You can have a peek at the other solutions to see the correct logic. Or have another go at it yourself :) $\endgroup$ Commented Nov 18, 2024 at 11:28
  • $\begingroup$ @TimSeifert Yeah, I wasn't completely sure of this answer. I didn't get how the pawns weren't pawns, since there was so many of them, but now this just is an example of a wrong answer, I guess. $\endgroup$ Commented Dec 22, 2024 at 15:48

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