15

I saw this beautiful, subtle & accessible problem a couple of days ago, and wanted to share it with you folks. (It's by the ground-breaking endgame theorist who e.g. invented the "Troitsky line" in KNNvKP analysis.)

[title "A.A.Troitsky - Bohemia 1910"]
[fen "1B6/bp3ppp/BR1pPp2/RPp1kP2/bpK5/3P2P1/1PP3PN/5nrr w - - 0 1"]  

White to play and mate in 2?

The forward play is pretty obvious. But, as is common in retrograde analysis, you need to justify the forward play by figuring out what was Black's last move.

Even if/when a speedybrain supplies a full answer, I hope that you can still enjoy solving the puzzle.

  • 2
    Why can't black's last move have been c6-c5 in which case bxc6 e.p. doesn't work? – Allure Feb 20 '18 at 5:13
  • 4
    @Allure: That is exactly the question to be answered! – Laska Feb 20 '18 at 5:34
6

Short answer: last move was -1. c7-c5, and before that -1...Rc6-b6, the white bishop returns to c1 and -d2-d3 unlocks the cage, enabling -Kd3-c4 and everything unlocks. -1. c6-c5 doesn't work because the white bishop will definitely end up trapped behind the black bishop in the northern dark squares and can't get out - -c6-c7 will leave the two bishops on d8/e7/f8 and not a7/b8 because of tempo concerns (retro-opposition), and they can't leave.

(Notice the use of minus signs, white moves noted with ..., and I said the white bishop "returns" to c1 - I'm retracting moves from the diagram position, which makes this kind of problem much easier.)


Long answer: Now how to explain this without fluff or skipping too many "obvious" details? From a pedagogical point of view, I will insist on talking about this problem using retractions rather than the usual forward chess moves; it's easier and far clearer to play moves backwards from the diagram than it is to create some arbitrary position, assert that it's legal then play forward moves to reach the diagram. It also ensures we don't miss anything - easier to check every possible retraction from the diagram than to check that every possible chess position leading to the diagram has ends with a certain move.

Those who already understand retraction can probably see the cage themselves, skip the first two bolded points and jump right to the section -1. c7-c5.

Pawn captures, basics of retraction explained

Standard approach is to look at pawn structure. Each side has 14 units and 8 pawns. White's captures were hxg3 and axb, black's were exf6 and axb. So from the diagram, we can't magically retract (take back) an uncapture immediately.

Also, the fact that all bishops are on the board and each side has their full 8 pawns means we can't retract exf6, d7-d6 or ...d2-d3 without the corresponding bishop being at home. (In the forward direction, it means that the pawn move was played while the bishop was at home and hadn't yet moved.)

A side can retract themselves into check but the other side must retract immediately to remove the check (once again this is seen in reverse - in the forward direction, one side checks and the king moves away.) For instance -1. c6-c5 Kb3-c4 would be possible, except that black can't remove the check from Ba4. -1. Kf4-e5 is impossible as white needs to retract -1...hxg3 to answer (pawn structure.) -1. Ne3 is impossible as one cannot deliver check by retracting (in the forward direction, that would be black moving away from check.)

What is a retrocage?

Usually, once all pieces have enough freedom of retraction, the position shouldn't pose any legality problems. We therefore want to retract moves with the goal of freeing as many pieces as possible. By considering all retractions (it's not hard once you're used to it, like seeing legal chess moves), we notice a few things:

  • The kings can't retract at the moment.
  • The southeast (bottom-right) corner unlocks with retracting a knight move, impossible for now because of the kings.
  • The west (left) side, specifically the 8 pieces in a rectangle except bBa7 and wRb6, is a cage that unlocks only with -Bb3-a4, which is again impossible because of the kings.

So to unlock the position, we need to free the kings by giving them a retraction. So a bishop has to go home to c8, f8 or c1 to enable a pawn retraction to (respectively) d7, e7 or d2. The black light bishop is trapped on a4, so that leaves just the two dark bishops.

Position is legal: -1. c7-c5

We retract for example -1. c7-c5 Rc6-b6 -2. Be3-a7 Bb8-a7 -3. Bh6-e3 Be3-a7 -4. Bg5-h6 Bc1-e3 -5. Bh6-g5 d2-d3 (the white dark bishop came home) -6. Bg5-h6 Kd3-c4 and the cage unlocks. The white king moving frees everything, as explained above.

En passant is legal: -1. c6-c5? is impossible

The crux of the matter. If we try to retract this, then the two dark bishops are locked in step on the 6 dark squares in the north. They can't pass each other, nor can they lose a tempo since there are no other legal retractions except -c7-c6, which incidentally commits the bishops to the dark squares to the left or right of c7. There are only 3 cases:

  1. Both bishops on the left (bBa7/wBb8). This is impossible - the bishops can't lose tempo, and with the bishops in such a position (i.e. right now, the only chance) it's white to retract, not black.
  2. Both bishops on the right (bBd8/wBf8). Once -c7-c6 is retracted, the bishops are trapped in these squares. The only way to let them out is to retract e7xf6 (impossible; the black bishop can't find itself on the right of the white one to be at home on f8), or to retract -d7-d6 (also impossible; the black light bishop on a4 can't come home before the cage is unlocked.)
  3. Black bishop on the left, white bishop on the right. The same problem as in (2), we can't get either bishop home as they're each trapped on the wrong side of the pawn wall.

With no other way to unlock the position, we conclude that -1. c6-c5 is impossible.

Conclusion: Therefore -1. c7-c5 was the last move (no other possibilities), so en passant is provably legal and forward mate is 1. bxc6 e.p.+ Bb5 2. Raxb5#.

  • 1
    A masterful solution, Remellion. This is very instructive in showing the right approach to tackling these kinds of problem. And you highlight the key feature of the problem which is parity. Normally knights and pawns are the pieces who can't make a triangle of moves to change an odd number of moves to even. But bishops when stuck in a long corridor such as a7 to f8 have the same limitation. In Troitsky's problem, not only do bishops become subject to parity, but it matters, and en passant and the checkmate hinge on it. – Laska Mar 11 '18 at 15:20
5

Pawn still being d2 allows us to complete the answer: c7-c5 must have been Black's move. Please see the added discussions below.

Some initial thoughts:

To get started, by a trivial process of elimination we can boil the choices down to only two black candidate moves: c6-c5 or c7-c5. Now we have to resolve the dilemma by stepping further back and figuring out what aspect of the position allows to distinguish between the two. It seems to me that the main twist lies in how the queen-side came to be, that is: which events and in which order took place for the two rooks to end up on b6 and a5, the bishop on a4 and the two pawns on b4 and b5. Some initial observations:

  1. It seems the only route for the king to get to c4 is via c3, which puts a constraint on when axb4 took place.
  2. Moreover, Kc4 must have been played after Ba4, as otherwise, if the king is on c4 the bishop can never reach a4.
  3. Easier points: Ra5 is played after axb4 (not to say immediately after), Ba6 is played before axb5 (because the capture closes all paths to a6) and Ba4 is played once axb5 has taken place.
  4. Point 2 suggests that the king must have the square c3 available in order for the bishop to reach a4. This would imply that axb5 capture happens first, the bishop gets to a4 then axb4 takes place.
  5. If 4 is true, then how the heck can the rook get to a5? :) If we can't resolve this, i.e., find a sequence of events that allow the above points together with the rook reaching a5, then point 4 contains a mistake/contradiction.
  6. Slightly different consideration would be: to assume the white bishop is for the moment on d7 for instance, so neither captures on b4 and b5 have taken place yet. The white rooks are on a6 and b6, (that they've reached e.g. via f4-c4-a6/b6) and the black knight (to be captured on b5) is on c7. We consider axb4 to happen first (assume it captures a white knight): then the rook gets to a5, the bishop goes to a6 via d7-b5, then the knight goes to b5 which leads to axb5 which in turn frees a4 for the black bishop, so Ba4 takes place. Everything seems fine at first glance, but in all of that, how the heck will the white king reach c4? :)

Quick recap: according to point 5., we cannot resolve how the rook gets to a5, whereas if we follow point 6., the rook problem is resolved but we cannot(?) account for how the king reaches c4 eventually. I believe if we can solve this dilemma of the queen-side, the distinction between whether c6-c5 or c7-c5 is correct will follow immediately.


A follow-up to complete the answer

A possibility that resolves the aforementioned queen-side problem:

Point 6 would actually work if white king had the d3 square available in order to allow the black bishop to get to a4. More precisely, these are the sequence of events:

  1. We are considering the case axb4 happens before axb5, the rooks are on a6 and b6, the white bishop is on d7, the white pawn is still on d2 so the king sits on d3.
  2. 7 implies that c1 bishop is still locked in and gets to reach b8 very late. axb4 frees a5 for the rook, so Ra5 takes place immediately after. Then d7 bishops returns to a6 via b5 after which axb5 happens (for instance capturing a knight, which could without any loss of generality have come from c7 to b5). After axb5, the black light square bishop gets to a4, e.g., via b3 directly. Now the king can hop to c4, allowing d2-d3.
  3. Now comes another crucial step: in the previous points we've assumed the black bishop is just sitting on a7, that means that if the c1 bishop is to ever reach b8, the a7 bishop must be able to move freely along the a7-f2 diagonal, which in turn implies the black c pawn cannot be on c6 as that would not only impede the rook from moving and freeing the a7 bishop's diagonal but also trap the white bishop on b5 or d7. Thus, the pawn must be on c7, which allows Rc6 and the aforementioned bishop maneuver, followed by a simple sequence (after kc4 and d2-d3) to get the two bishop to a7 and b8, e.g., black bishop to d4 and then c3 to let the white bishop to pass first, i.e., white bishop to e3-a7-b8, then the black bishop returns to a7, the rook on c6 returns to b6 and black plays c7-c5. Summary of that last dance of bishops: 1. Rc6 Bd4 2. Be3 Bc3 3. Ba7 Bd4 4. Bb8 Ba7 5. Rb6 c7-c5. It must be noted that the pawn still being on d2 does not interfere with the occurrence of any of the other events other than that of the dark squared bishops, which has been already resolved. Thus, it can safely be considered without leading to contradictions.
  4. In short, previous attempt in point 4 and 5 was deemed to be impossible as the rook on a5 could never be accounted for in those variations. Instead following up on the approach in point 6 in conjunction with the idea of the pawn still being on d2 before the exchanges on b4 and b5 have taken place and naturally in turn the moves Ra5 and Ba4, we seem to have found the only way the current queen-side situation could be reached, which required the black c pawn to be still on c7 throughout.
  • Good stuff so far! In the middle of an apparently open board we have a retro cage. Where can the cage be opened? – Laska Feb 20 '18 at 18:43
  • 50%! Yes you have the solution for c7-c5, with d2-d3 locking the cage. No I'm not convinced you said why c6-c5 couldn't have been preceded by a few bishop moves, with then c7-c6 before. Too much of your writing is about the detailed contents of the cage, but often with a cage as soon as you begin to unlock it, it all falls apart, as here. So maybe a sequence of moves for c7-c5 and a simple line of reasoning for c6-c5 are all that's necessary. The theme of the problem is "retro-opposition", if that's any help! Thanks again! :D – Laska Mar 1 '18 at 19:32
  • Dear user929304: hope the above is helpful. This kind of problem requires a slightly different approach to those of Smullyan, and it's a knack to focus on the right elements. Tally ho! :D – Laska Mar 1 '18 at 19:40
1

Nice.

Black has 8 pawns, so his bishops are original, so Black hasn't just played d6 or exf6. If Black has just moved a piece, it must've moved from a square where it was already checking; this is impossible. So Black has just moved the pawn now on c5. Why not c6-c5? The problem is that White can't give Black a penultimate move; wNh2 can't retract to retro-release bRh1 because wN would move from a square where it was checking. Black is missing only 2 units, so White played axb & h2xg3, so may not retract f2xg3. Bc7-b8 doesn't relieve Black's retrostalemate. White's only way to give Black a penultimate move is retro -1 ... c7-c5 -2 Rc6-b6 retro-releasing bBa7.

Except that

with a black pawn on c7, the pawns on c7, d6, f6 and g7 lock the bBa7 out and prevent it retracting to f8. So before bBa7 can retract to f8, bPd6 must retract to d7. This entails retracting bBa4 to c8. This entails moving wK. Seeing as wPd3 can't go back until wBb8 is home on c1, moving wK entails moving bK, which entails retracting f4-f5. So retro -1 ... c7-c5 -2 f4-f5 Kf5-e5 -3 Kd4-c4 Bb3-a4 and now bBb3 and bPd6 can go home, wRb6 can retract to c6, bBa7 can go home to f8 and finally it unravels.

  • 1
    "Bc7-b8 doesn't relieve Black's retrostalemate." is a bit of a shortcut. Before wBc7-b8, Black might have played bBb8-a7. You can even play several moves along b8-c7-d8-e7-f8 with those bishops without giving any additionnal retro-move to either side. But the trick remains: there is no way to release wB from the two last ranks; e.g., you cannot retro-play bPd7-d6 with the light squared bishop out of c8. – Evargalo Feb 20 '18 at 13:20
  • I agree with most of what Rosie said - need some key details, particularly about wBb8, for the win. – Laska Feb 21 '18 at 4:19

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