by Karl Fabel - Solutions
1 - A. Troitzky
Black captures have been e7xd6xc5xb4xa3 and d7xc6 accounting for all 5 captured White pieces. White's 4 captures are hxgxh and fxgxh. White cannot retract b3-b4 before reconducting the wQR home. Hence he has only a few pawn moves before he is retropat. Bl. must free Wh. quickly with the retraction of d7xc6, but he must reconduct the bRa8 and the bBc8 before. These must be uncaptured by the wh. Pawns.
There is only one way to do this fast enough, e.g.
-1. g7-g5 g5xRh6
-2. Rd6-h6 g4-g5
-3. Rd8-d6 f3xBg4
-4. O-O-O f2-f3
-5. Bc8-g4 Kb5-a5
-6. d7xRc6+ etc.
Last move was g7-g5 and White wins with 1. h5xg6 !
2 - Dr. J. Sunyer
(Solution by mr.mip)
Assuming that black cannot castle, the solution is 1.Qxa4 ~ 2. Qa8#.
From the material balance and pawn structure it is evident that the Bd1 promoted at g8 from originally c2 pawn. Thus it has made 4 captures: c2/3xd3/4xe4/5xf5/6xg6/7. The remaining capture then has happened with b2xa3.
As no more wh. captures are available, black captures must account for pawns at h6 and h5. So black has captured the rooks with h7xg6xh5 and the bishop with e7xf6. Original f1,c8 bishops have been captured at home, of course.
The task then is to construct a position, where after the capture of the last black piece, there is as many black pawn moves available as possible in order to allow white to promote the bishop, retract it to d1 as well as put Kf5,Ne4 and Qc2 in place. Should this be possible without need to move K or KR the castling right will be preserved and there is no solution.
The optimum play starts from the position (or essentially similar pos.)
Note that either a5 or c5 has been necessary to get Ra8 and Qd8 out. Also b2xa3 has been necessary to get Bc1 and Ra1 out. Furthermore it is always possible to arrange tempos in the previous play so that it is white to move.
From the diagram position the game proceeds:
1. fxg7 hxg6 2. g8=B gxh5 3. Bh7 a4 4. Bc2 c6 5. Kf5 c5 6. Bf6 exf6 7. Bd1 c4 8. Qc2 c3 9. Ne4
and the problem position is reached. Unfortunately it is black to move and so it takes one move too long to restore white position. Hence black must have moved K or KR in the original position and cannot castle any more.
3 - Dr. J. Sunyer
(Solved by Henrik Juel)
Start from the following position, which can be reached easily (black captures are e7xNd6 and f7xBg6, white captures are g2xh3, b2xc3, c2xb3, and hxgxf. The pawn originally from h2 promoted to a bishop on f8, and went into the southwest cage.
1. b4 Ka4 2. Nb2 Ka3 3. Kf3 a4 4. Rg5 Ba5 5. Rhg1 b6 6. R1g4 Ba6 7. Qd5 Bc4 8. Qb5 Bb3 9. ab3 Ka2 10. Nc4 Kb1 11. Ba3 Ka2 12. Nb2 Kb1 13. Kg2 Kc1 14. Nc4 Kd1 15. Ne5 Ke1 16. Bc1 Kd1 17. Bab2 Ke1 18. Nf3 Kd1 19. Kf1 a3 20. Rg1 a2 21. Rh1 a1=B
In this position, the last pawn move was just played. If it'll take more than 49.5 moves to reach the original diagram from this position, black may claim a draw with his next move, due to the 50-move rule. The quickest play is:
22. Ba3 Bb2 23. Ng1 Ba1 24. Kg2 Ke1 25. Kg3 Kf1 26. Kg4 Kg2 27. Nf3 Bb2 28. Rd1 Ba1 29. Ng1 Kh1 30. Bcb2 Kg2 31. Rb1 Kh1 32. Bc1 Kg2 33. Rb2 Kf1 34. Ra2 Ke1 35. Bab2 Kd1 36. Ra4 Ke1 37. Kg3 Kd1 38. Kg2 Ke1 39. Ba3 Kd1 40. Kf1 Bb2 41. Nf3 Ba1 42. Rg1 Bb2 43. Rh1 Ba1 44. Ng1 Bb2 45. Kg2 Ba1 46. Kf3 Ke1 47. Ke4 Kf1 48. Nf3 Kg2 49. Rd1 Bb2 50. Ng1 Ba1 51. Bcb2 Kh1 52. Rb1 Kg2 53. Bc1 Kf1 54. Rb2 Ke1 55. Nf3 Kd1 56. Nd4 Ke1 57. Ra2 Kf1 58. Bab2 Ke1 59. R4a3 Kd1 60. Qa4 Ke1 61. Kd5 Kd1 62. Kc4 Ke1 63. Kb5 Kf1 64. Ka6 Kg2 65. Kb7 Kg1 66. Kc8 Kf1 67. Kd8 Ke1 68. Ke8 Kd1 69. Kf8 Ke1 70. Kg8 Kd1 71. Kh7
And we have arrived at the problem position with black to move as required. However, it has taken 70.5 - 21.0 = 49.5 moves without pawn moves or captures to do so. The game is drawn, if black now makes a non-pawn, non-capturing move, and the only move available for that is Ke1.
4 - H. Stempel
White must retract -1. Bh1-a8! and then play 1. Bh1-e4#
Since wBc1 was captured at home, the wBg7 must be a promoted bishop. The Wh. a-pawn must be the one that has promoted, since none of the other pawns currently on the board could be the pawn from a2. Since the promotion took place on a black square, the pawn couldn't have promoted on b8 or d8, since the bishop could not have escaped. Therefore it was f8 (can't be h8 since the wPa2 can't get there.) Therefore the wPa2 moved a2xb3xc4xd5xe6xf7-f8=B.
All these captures took place on light squares; therefore, the black KB was not captured by the wPa2, so that the wPa2 captured 2 bRs, 2 bNs, and the bQ. When e6xf7 was played, Black had already played f7-f6 but not yet d7xe6. Therefore, in order for the bRa8 to escape to be captured by the wPa2, b7-b6 must have been played (so that the bBc8 could get out, in turn freeing the bRa8). So both of the pawn advances, b7-b6 and f7-f6, occurred before the promotion.
Therefore, whatever White's last move was, Black's move before was not b7-b6 or f7-f6. If we try -1. N-d7 d7xQe6 -2. Q-e6+, Black is retropat. Therefore, Black's last move was not with any of his pawns. Can White uncapture the Black KB to allow Black to move it? No, because the only White pieces on dark squares that could have made this capture are wNf8 and wRh4, and both squares leave the resulting KB with no move and thus Bl. would be retropat. Therefore Black's last move was with his king from e4.
No matter what White's retraction is, the Black king will be in double check after Black retracts Ke4-f5. The only White retraction leading to a legal double check is -1. Bh1-a8! Ke4-f5 -2. Rg2-e2++ Kf3-e4+, so White's last move was -1. Bh1-a8.
5 - B. Sommer
White retracts -1. O-O! forcing -1...O-O-O -2. b4-b5 and then 1. Rh1-h8#.
Why are -1...O-O-O and -2. b4-b5 forced?
White captures are axb, g2xf3xe4xd5xc6. Since the wPg2 captured on light squares only, the black KB must have been captured by the a-pawn, so axb must be a3xb4. Black captures are d7xe6, g7xf6, h7xg6xf5xe4xd3xc2. These captures account for all missing pieces.
To extricate bPc2, we must undo d2-d3. Before that, we must bring the Wh. QB back to c1. Wh. QB must have been captured on f6. To undo g7xBf6 we must first undo a3xBb4. But in order to avoid locking out the white QR (the Wh. king can't let it back in "after" -1 O-O), we must undo d7xRe6, and before that we must uncapture the black QB along the sequence of captures by the wPg2 and bring it back to c8, and we must bring the Bl. QR back to a8 or b8. To reiterate, the sequence of events was: (1) d3xc2, (2) d2-d3, (3) wBc1-...-f6 and g7xBf6, (4) bBf8-...-b4 and a3xBb4, (5) wRa1-...-e6 and d7xRe6. Only then did the Bl. QR and QB escape.
Now "after" -1 O-O!, Wh. has only 4 moves, b4-b5 and 3 uncaptures with the wPc6 (it can't get all the way back to g2 without locking out the wBf1) before Bl. must uncapture d7xRe6 to avoid Wh. retropat, so we must conduct the bQR and bQB back to a8/b8 and c8, respectively, within that time. The only way this can be done is with -1...O-O-O -2. b4-b5 Qd8-d4 -3. d5xBc6 Bd7-c6 -4. e4xd5 Bc8-d7 -5. f3xe4 d7xRe6. (-2. d5xBc6? is premature; it blocks the Bl. Q's route back to d8. After -1...O-O-O, Bl. Q must also go back to d8 before d7xRe6 can be undone.)
6 - N. Plaksin
(Solution by mr.mip)
The inventory shows six (6) white pieces and one black knight missing. The bN has been obviously captured with f2xNe3. The unlucky position of the bBg8 and black pawn c2 indicate that this pawn has come from h7 and is accountable for the remaining 5 captures on diagonal h7-c2. Furthermore, before the last of these captures (d3xc2), the white pawn has still been at d2 and wB at c1 blocking the rook a1 from freedom. Since wK currently - after dxc2 - lives in a prison west from c1, the sixth man captured must be Ra1 heroically fallen at a1 or b1 before the d3xc2 capture.
As the white position is really bad, the only chance for a draw is to demand it based on the 50 move rule. This is the project we are set to prove.
Let us start from the fact that in order to return the current position to initial position, the white pieces must be uncaptured. This leads to the sequence of logic below, where the mark => should be read as : "But before that we must undo". Here goes:
Recover white pieces
=> Bc1 .. a7
=> Bf8 .. g1
before this reinstall : Rc8, Qd8,Ke8 in this order
before this bail out bQ and bR from their current prison
before this block the first rank by bringing bK to d1.
This will now form a "To Do"- list , although in reverse order, for counting moves. However we retract first the move Rc8-b8 in order to give white a chance to shuttle between a7 and b8 with his bishop. Now, playing in normal "forward" fashion and counting just black moves (white is just shuttling) until the pawn move g7-g6, we have :
Ke8 .. g7 (2)
Kg7 .. Kd1 (15)
Qd2 .. h4 (2)
Rh1 .. g3 (2)
Kd1 .. g7 (15)
Kg7 .. e8 (2)
Bf2 .. f8 (4)
And now undoing g7-g6 comes possible. However, summing up the moves in parenthesis gives a total of 52 moves - more than enough to claim a draw.
But why should white meekly just move the bishop? Why not undo, say somewhere around 49th retraction, a nice pawn move like g4-g5 or f2xNe3? (There are not others available).
The reason becomes apparent if we continue the To Do - list, until
wB has been released from its prison:
Return Bg8 to b7.
This happens in the phase when wB is at b8, losing a tempo if necessary. And now the critical phase:( As retro moves, black first)
b6-b7 Bd4-a7 etc.
The point is that it takes two pawn moves from white to get bishop out. If white had retracted one of these pawn moves earlier, it now had to retract g4-g3 as well. But this will close the way for the Rh1 to return home after resurrection, making the position illegal. (Any hope to sneak it via f-line by retracting g5-g4-g3-g2 is lost, as f2xNe3 must be undone to return Bc1, before the rook is reborn).
This means that no pawn moves can be undone before retracting g7-g6 and as shown earlier the only capture outside c2-pawn captures and f2xe3 is the capture of Ra1 that happened before d3xc2. So, the 52-move run has indeed happened without pawn moves or captures and white can prove the 50-move draw.
7 - N. Petrovic
Last 6 moves are: -1. d5xe6ep e7-e5 -2. d4-d5 Ke6xPb6 -3. e5xf6ep f7-f5
8 - N. Petrovic
Suppose white can castle. White captures were hxg, gxf, fxe, exd and c6xb7. Black captures were dxcxd (to let the white d-pawn pass without capture). So black's last move wasn't a capture. The only move that will give white a move that'll leave black with a new retro-move is c7-c5 (and white's last move was c6xb7). So if white may castle, black played c7-c5 as last move.
1. dc6! d3 2. OOO! (forced to prove the legality of the en passant capture) d2 3. Rd2 Be6 4.d7#.
9 - N. Petrovic
The position unlocks with
-2. a5-a4 Nc6-b8
-3. Kb8-a8 Nb4-c6
-4. Ka8-b8 Nd3-b4
-5. Kb8-a8 Nf2-d3
-6. Ka8-b8 h2-h3
-7. Kb8-a8 Nh3-f2
-8. Ka8-b8 Nf4xPh3
-9. Kb8-a8 Nd5-f4
and everything unlocks.
So White played last and Black wins with Rb7-? mate.
10 - Valerian Onitiu
Retract -1. O-O-O !!. Now it can be proven that Bl. last move is -1 ... g7-g5:
"After" the retraction of -1. O-O-O, single bl. cap is h6xBg5. Wh. caps include b2xc3, c2xd3, exf and Bc8 at home. bPe7 could not promote on e1 because Wh. may still castle. So that bPe7 was captured on its column. Same for bPa7, so that all caps are localized. bPc7 did promote on c1=B. We may also precise that, because wKe1 did never move, the bRf2 reached f2 through f3 so that exf really was e2xf3.
The position unlocks by reconducting wBh7 on f1 and retracting e2xf3: -1 ... g7-g5 -2. Be4-h7 g5-g4 -3. Bc6xPe4 e5-e4 -4. Ba4-c6 e6-e5 -5. Bd1xPa4 a5-a4 -6. Be2-d1 a6-a5 -7. Bf1-e2 a7-a6 -8. e2xQ/Nf3 and the position unlocks just before Bl. is retropat. Retracting -1 ... g6-g5?? -2. Bg8-h7 g7-g6 -3. Bh7-g8 is too slow.
-1 ... g7-g5 is the only possible last move (indeed, last 13 single moves are completely determined). So that forward 1. h5xg6 ep # works.
11 - M. Myllyniemi
Last moves are -1. b4xc3 ep+ c2-c4 -2. Ke6xNd6+ e5xf6 ep+ -3. f7-f5 Nf5-/xd6+.
12 - J. G. Mauldon
Suppose black can castle both king- and queenside. Then his last move was either b7-b5, d7-d5 or f7-f5.
Suppose now black can't castle queenside. Then #3 with 1. Bd6!, and (again) black has no proper defense against 2. Rh8 or 2. Ra8.
And the last case: Black can't castle kingside. Then #3 with 1. Bf6!
13 - S. Loyd
[This detailed analysis was written by T. Volet]
Under chess problem conventions, in a problem stipulating 'Mate in 4', White has the move. Only 1.PxPe.p. will suffice for a mate in four; the en passant capture can be justified only if Black's last move was Pf7-f5. This is what we now prove.
Captures generally. In the diagram as presented, White has 14 units (i.e. 2 were captured by Black) and Black has 10 (6 were captured by White).
White's captures. The White P at a7 could have originated only on f2. In order to get from f2 to a7 it accounted for 5 of the 6 White captures, each occurring on a dark square. The White Queen's Rook Pawn (QRP) must have captured from the a file to the b file to account for the configuration of the Ps at b3 and b4, which also accounts for the 6th White capture. As the Black Queen's Bishop is missing from the diagram it must have been captured; that B occupied only light squares, so it must have been captured by the White QRP on b3. All White captures were by Ps.
Black's captures. Black's 2 captures were made by its KRP, which explains the two Black Ps on the f file in the diagram. The Black KRP must have captured both of the missing White Bishops, one on a light square and one on a dark square. The only possible route requires that the Black KRP captured the White KB at g6 and the White QB at f4 (where that P appears in the diagram).
Black's QRP. As noted, all captures by both sides were made by Ps. The absent Black QRP must have been captured by the White KBP (the only other capturing White unit, the QRP, having captured the Black QB). Either the Black QRP (i) promoted at a1 (after the White QRP captured the Black QB at b3 and cleared the file), and is still on the board, or (ii) did not promote and was captured on the a file. (It could not have swerved to another file as that would entail a capture by Black in addition to the 2 required for the Black KRP to reach the f file). The only square on the a file reached by a White P is a7. But if the Black QRP was captured at a7 by the White KBP, then the square a7 was continuously occupied by a P, and the Bishop at b8 in the diagram could never have reached b8. Therefore the Black QRP promoted at a1, and it is still on the board, in its promoted form (so it must be the rook on g8)
The last move. The Ps at c7, d7, and g7 never moved. Neither the B at b8 nor the R at g8 could have made the last move, all squares from which they could have moved being occupied. As discussed above, the P at f4 came from g5 when it captured the White QB; thus it couldn't have made the last move. The P at b6 came from b7, but had it done so as recently as the last move the Black QB could not have left its original square (c8) to be captured, as it was, on b3. Similar reasoning shows that the P at e6 did not come from e7 on the last move, as then the Black KB could not have departed f8 (ultimately to reach b8, where it stands in the diagram). The P at e6 could not have come from f7 on the last move as that would entail too many captures by Black.
The only remaining possibilities are a move by the Black K, or a move by the P at f5. The Black K could have come from a square on which it would have been in check, but only if that check could have been delivered by White on the immediately preceding move. Therefore the Black K could not have come from h5, h4 or h3 on its last move, as in any of those cases the White Q could not have delivered the check on White's immediately preceding move. The Black K never could have occupied f3 or g3, as neither the White KP nor the White KRP, which deliver checks on those squares (respectively), has ever moved. Nor could the last Black move have been the capture by the Black K, while moving from h3 or h4, of a White B on g4 (which White B could have delivered a discovered check by having moved from h5 to g4 on White's immediately preceding move), as the 2 Black captures were made by the P at f4.
Therefore the P at f5 made Black's last move. That P could have moved only from f6 or f7. Assume that it came from f6 (and occupied that square when White made its last move). This confronts the crux of Loyd's innovation. In all previous retro problems with an e.p. key move the diagram position presented a K on the same rank and adjacent to the P that advanced two squares, in order to negate the possibility that the P had moved only one square (from a location where, illegally, it would have been moving while the other side was in check). Loyd's 1894 position pushed the analysis back another move, by showing that White would have had no possible immediately preceding move if the Black P had occupied the 6th rank.
Assume that the P at f5 were on f6, and examine the possible last moves for White. Ps at c2, d2, e2, and h2 have never moved. The P at a7 came from b6, so it has no last move. The P at g5 may not have swerved off the g file (which would require too many captures by White), and its path on the file is blocked, so it didn't move last. The P at b3 could not have come from b2 recently or the White QB would not have been able to depart from c1 (which it earlier did, having been captured at g6). Nor could the P at b3 have moved from a2 to capture the Black QB on the last move, because if it recently occupied the a file the Black QRP would not have been able, earlier, to advance along that file and promote on a1(as established above). No square is available from which any of the White K, N at a1, N at f8, or R at h8 could have moved. Any square from which the White Q could have come (h3, h4, or h5) would have involved its moving while delivering a check to Black; clearly White cannot be on the move if Black is in check. Finally, if Black had its KBP on f6 when White last moved, the R at g6, which otherwise could have come from f6, would have no available square from which it might have moved.
Therefore the Black KBP did not come from f6 on the last move, it came from f7, and White, immediately before, had moved its R from f6 to g6. Black's last move having been a 2 square P advance, W is free to capture en passant and mate in 4.
14 - J. Knöppel
(Solved by mr.mip)
White's only hope for win is 1.b5xc6 ep. This will need a proof that black's last move was c7-c5. The proof goes as follows:
White captures have been c2xb3, e2xd3xc4xb5 and f2xe3xd4xc5xb6. These 8 captures account for all missing black pieces, including the two promoted at f1 from black g- and h-pawns.
Black captures include axbxa to explain the a2-pawn and gxf, hxgxf by pawns on their way to promotion. So all black and white captures were by pawns.
A further consideration in the diagram position is providing a return path for the black king. For example white can retract c2xb3 or d3xc4 only after black king has managed to escape to the fourth row. The other problem is that white has to be careful not to block the f-file (by retracting f2xe3) before black pawns are unpromoted at f1 and returned beyond f2-square.
Because of the above considerations black's objective in the retroplay is either to quickly return the king to the fourth row or provide white with more moving men by an uncapture. It proves out that the former task is impossible without blocking the f-file, while the second succeeds with following retractions :
1. .. c7-c5 2. c5xRb6 Rf6-b6 3. d4xc5 Rf1-f6 4. e3xd4 f2-f1=R 5. c4xb5 g3xQf2 6. Qf6 etc, etc. Now that white has an extra piece, there is all the time in the world to unpromote the second pawn and wriggle out with the king via d4. If black starts retractions with 1 .. c6-c5 (or any other move) instead, it takes too long to reach f1 for the uncaptured piece.
So 1. .. c7-c5 is the only possible last move for black and starting the solution with en passant is justified.
15 - W. Keym
(solution by H. Juel)
(a): Black's last move can't have been c6-c5, because this would imply a retroplay like -1. ... c6-c5 -2. Qc7-b6 b6xNa5 -3. Be4-f3 Kg2-g1 -4. Bd3-e4 Kg1-g2 and now the only way to open the northwest cage is to uncapture a darksquared black bishop on b4, bring it to f8, and retract e7-e6; but then both black rooks would be caught in the cage. Thus the last move was c7-c5 with a retroplay like -1. ... c7-c5 -2. Qc6-b6 b6xNa5 -3. Kd1-e1 Kf1-g1, also showing that white can't castle. The only solution is 1. b5xc6 ep#
(b): Now the last move could be c6-c5, with retroplay like -1. .. c6-c5 -2. Qc7-b6 b6xNa5 -3. Be4-f3 Kg2-g1 -4. Bd3-e4 Kg1-g2 and the cage is opened by retracting the white rook to h8, the white queen to d8, and the black pawn on c6 to c7, and then uncapturing c6xb7. So white may not capture en-passant. But this retroplay also shows it's possible to preserve white's queenside castling rights. So #1 with 1. OOO#
16 - W. Keym
There are 8 possible en-passant captures half-finished.
So the only legal en-passant capture was f4xg3. So Black mates in 1/2 by removing the white pawn on g4.
16a - W. Keym
The white pawns on the queenside captured nine times, these are all the missing black pieces. This also means that the g5 pawn is the original g-pawn. Black captures are b5xa4, [h/f]5xg4 and gx[h/f]. The captures on a4 and g4 are on white squares, so the missing white bishop was captured by the g-pawn. This also means that the white h-pawn promoted before it got captured. This means that the capture on g4 was done by the black h-pawn. So the black f-pawn promoted on f1, breaking both white's castlings. The promotion on h8 breaks black's kingside castling. So the only castling allowed in this position is the black queenside castling.
Add the white king on d6, and the black king on c8, and mate with 'finish 000'
17 - Dr. Niels Høeg
White captures are c6xd7 (the c-pawn must have promoted to a rook, since Rh1 was captured on h1 or g1) and bxa. Black captures are Rh1, axb, and h7xg6xf5.
During the first phase, white makes waiting moves with his king between b8 and a8, while black retracts:
-1. Kd8-c8 -2. Ke8-d8 -3. Kf8-e8 -4. Bh7-g8 -5. Kg8-f8 -6. Bg8-h7 -7. Kh8-g8 -8. Kh7-h8 -8. Kh6-h7 -9. Kg5-h6 -10. Kf4-g5 -11. Ke4-f4 -12. Kd4-e4 -13. Kc4-d4 -14. Kb3-c4 -15. Kc2-b3 -16. Kd1-c2 -17. Ke1-d1 -18. Kf2-e1 -19. Kg1-f2 -20. Kf2xRg1
During the next phase, white makes waiting moves with his rook between g1 and h1, while black retracts:
-21. Ke1-f2 -22. Kd1-e1 -23. Kc2-d1 -24. Kb3-c2 -25. Kc4-b3 -26. Kd4-c4 -27. Ke4-d4 -28. Kf4-e4 -29. Kg5-f4 -30. Kh6-g5 -31. Kh7-h6 -32. Kh8-h7 -33. Bh7-g8 -34. Kg8-h8 -35. Kf8-g8 -36. Ke8-f8 -37. Kd8-e8 -38. Kc8-d8 -39. Rd8-d7
and now white plays forward 1. Rc7#.
18 - Dr. Niels Høeg
Bl. did not capture. All wh. caps are by Ps: exd, fxe, gxf, hxg & two cross-captures a-b or b-c to get a Pb7 beyond the barrier of bPs. With these constraints, last 13 singles moves are forced:
-1. Rd8-d7+ d7-d6 -2. f5xe6 ep+ (e5-e6+?? bl. retropat) e7-e5 -3. f4-f5+ Kd6-c7 -4. b5xc6 ep+ c7-c5 -5. b4-b5+ Ke6-d6 -6. g5xf6 ep+ f7-f5 -7. g4-g5+ and Bl. avoids retropat.
19 - Dr. Niels Høeg
If black still can do both castlings, his last move must've been d7-d5 or f7-f5, since a pawn capture on d5 or f5 would take too many captures.
Now suppose black can't castle queenside. Then #3 with 1. Ng6!
The last case is: Black can't castle kingside. Then #3 with 1. Nc6!.
20 - Dr. Niels Høeg
Here, castlings are mutually exclusive. To see this assume, by way of contradiction, that both sides may castle. Then wQ and wQR have been captured by bNs (together with the wQB) and wQf4 is the promoted wPf2 on c8 or b8. So that bl. caps f7xg6, b7xa6, e7xd6 consumed the 2 wNs and the wBf1. There are now three cases:
Mate in 3 is with 1. Qxd6!! (threat. 2. Rf1 & 3. Rf8#) Bb7 2. O-O !! (not 2. Rf1??) forbidding the now-illegal defense 2 ... O-O-O??, and allowing 3. Rf8#.
21 - A. Hazebrouck
There are four possibilities:
So all four castlings are legal, but they're not legal at the same time!
22 - J. Haas
(solution by Henrik Juel)
(a) Last move could be Nc2xBa1+. Earlier in the game White captured bBc8 on c8, b6xNa7, c6xNb7, and gxPf, and promoted Pe2 on e8, while Black captured the h-pawn on its file and exPdxcxbxPa, and promoted Pg7 and Ph7 to bishops on g1 and h1. Solution: Retract Nc2xBa1, and play forward 1. Nca3+ Kb2,Kc1 2. Qc2#
(b) Last move could be gxf6 e.p.+, and it is possible to preserve White's castling right: White captured bBc8 on c8, b6xNa7, and gxPhxNg, and promoted on e8, while Black captured exPdxPcxbxa and g2xf1=B. (Or White captured the g-pawn on its file, axNb, b6xRa7-a8=R, and b6xNa7, and promoted on e8, while Black captured exPdxPcxbxa and hxg-g1=R). Solution: Retract g5xf6 e.p., and play forward 1. 0-0+ Bd1 2. Rxd1#.
The (b) solution doesn't work in (a), because there it is impossible
to construct a game where wRh1 hasn't moved.
The (a) solution doesn't work in (b), because there promoting to B on e1 requires too many captures.
23 - J. Haas
Start from the following position:
1. ed3 Kd8 2. Be2 Kc8 3. Bh5 Kd8 4. Re1 Kc8 5. Re4 Kd8 6. Rd4 Kc8 7. Rd7 h6 8. Rb7 Kd8 9. Kh7 Kc8 10. Qg8 Kd8 11. Nf8 Kc8 12. Nd6 Kd8 13. Nc8 Ke8 14. Rb8 Kd8 15. Ra8 Ke8 16. Bb8 Kd8 17. f3 Ke8 18. Nd6 Kd8 19. Ne8 Kc8 20. Ng6 Kd8 21. Nh4 Kc8 22. Nf5 Kd8 23. Bg4 Kc8 24. Bh3 Kd8 25. Qf8 Kc8 26. Kg8 Kd8 27. Rh7 Kc8 28. Nd4 Kd8 29. Ne6 Kc8 30. Nd8 Kxd8 31. Bg3#
In this part of the solution, black played 29 king-moves. To release the white pieces in the northeast corner, black needs another thee king-moves (OOO, Kc8-b8, white pieces enter the northeast cage, Kb8-c8) The whole position now unlocks.
So black played at least 29+3=32 king moves.
24 - B. Giöbel
We will show that the BK has moved and thus that Black cannot castle. White can then mate via 1. Rd1, anything; 2. Rd8++.
The Wh. KB was captured on its own square, so a WR was captured on e6. The Bl. KB was captured on its own square, so the Bl. QB, the Bl. Q, and a Bl. R were captured on b3, c3, and e3.
The capture on e6 must have happened before the Bl. c-pawn moved to c6, otherwise the Bl. QB could not have gotten out to be captured on b3. Thus, to allow his R out, Wh. made a capture before Bl. did. What piece could Wh. have captured? It couldn't have been a Bl. R, since neither R could get out until after the capture on e6 created an opening in the Bl. pawn structure. It couldn't have been the Bl. QB, so it must have been the BQ. But, before the capture on e6, the Bl. Q-side pawns had not moved, so the only way the BQ could get out was via f7. This means that the BK must have moved, so Bl. cannot castle.
Now, the BQ was captured through d2xc3, because a capture through f2xe3 would not have let the Wh. QB out. Hence White's O-O-O is broken too because the captured Wh. Rook really is from a1, or had to disturb its King. This explains why the dualistic solution 1. O-O-O?? fails.
[Thanks to Roger Lipsett for spelling this out.]
25 - L. Garaza
Start from the following position: (unlocking the position from this one is fairly easy)
and play forward: 1. Rc4 Qd4 2. Ng6 Ng2 3. Nh4 Ne1 4. Ng2 Nc2 5. Ne1 Na1 6. Nc2 Nb3 7. Na1 Nc1 8. Nb3 Ne2 9. Nc1 Ng1 10. Ne2 Nh3 11. Ng1 Nf2 12. Nh3 Nh1 13. Nf2 c6 14. Nd1 Nf2 15. Nb2 Nh3 16. Nd1 Ng1 17. Nf2 Ne2 18. Nh3 Nc1 19. Ng1 Nb3 20. Ne2 Na1 21. Nc1 Nc2 22. Nb3 Ne1 23. Na1 Ng2 24. Nc2 Ne1
So black moved last. #1 with 1. Rh4#.
26 - W. Frangen
Suppose black moved last. Then his last move must've beem e7-e5. White's last move was either Bh2xg3 or Kf4-f5. If it was the latter, black's move before that must've been g6-g5. But that implies that the h-pawn promoted on g8 to a knight, not on h8. So in both cases, white captured a piece on the g-line. If black's last move was e7-e5, this also means that Bd4 is promoted. The only pawns that could promote on a black square without capturing are the a- and the c-pawn. If the a-pawn promoted, the white a-pawn must have captured 3 times on its way to b8 or d8 (where it promoted to a bishop). Together with the 2 captures on the c- and d-file, the capture on the g-file, and the bishop on f8, this is 7 captures. So black's a-pawn didn't promote. If the c-pawn promoted, white must've captured b2xc3 and cxd. Together with a7xb8, Bf8 and the promotion on g8 this equals 5 captures. But none of these captures happened on the a-file. So the c-pawn couldn't have promoted either. So black didn't move last. So #1 with 1. ... Nh6!#
27 - E. Fielder
Retract -1. .. Rg6-f6 -2. Na3-b1 Rf6-g6 -3. Nc4-a3 Rg6-f6 -4. Ne3-c4 Rf6-g6 -5. Nd5-e3 Rg6-f6 -6. Nf6-d5 b7-b6 -7. Nh5-f6 Rh6-g6 -8. f6-f7 Rh8-h6 -9. e5xRf6 h6xNg5 and the position unlocks. So black moved last. #1 with 1. Be1#.
28 - Dr. Karl Fabel
Bl. caps are a7xb6, bxc and wPf on its f-file. For balance, the wPg2 did promote on h8 after g6xh7. Other wh. caps are d2xc3, a2xb3xc4 and bBf8 at home. This accounts for all captures.
The position unlocks by (1) unpromoting a wR on h8, (2) resurrecting bl. QB through g6xBh7, (3) reconducting this bB to c8, (4) retracting b7-b6 and later b6xc5, (5) thus freeing wNa4 and then wBb1.
Unpromoting a wR requires that it uses the d1-a1-a3-b3-b5-a5-a7-b7-b8-h8 corridor. But there is a bR trapped in the corridor right now and the only way to eventually unpromote a wRh8 is to assume that wKe1 did move, so that the bRa1 can be freed. In conclusion, Wh. cannot castle.
Solution: 1. Tf1# and not 1. 0-0??
29 - K. Fabel
Suppose white can still castle. First white played f6-f7, after that black played g7xf6, and played his bishop to g1/h2, over e3, then white played e2-e3. So the white bishop had to go to e8 via c8 and d7. But then there is no room left to squeeze in all the black pieces and the white rook into the northwest corner. So white may not castle (and the bishop entered the southeast corner over e1,f2). So #1 with 1. Rd1!# (OOO?# illegal).
30 - K. Fabel
(Solution by H. Juel)
Earlier, the position must have been something like this: (or bRg3, or wBg4 anywhere on the diagonal c8-g4)
From this, forward play is: 1. hg3 h2 2. Bh3 h1=N 3. g4 Ng3 4. b4 Nf1 5. Rg3 a6 6. Ref3 Ne3 7. b5 Nd1 8. Re3 ab5 9. Ref3 Qe3 10. Nd3 Nb2 11. Ne5 b4 12. Ng6 Qd3.
Now only the black queen, and the white knight can move to reach the problem position with a white knight added. If the knight is added on a white square, white has the move, but can't give mate. Therefore the knight must be added on a black square, and black has the move. The only black square on which the knight can be added to give black the opportunity for a #1 is e3, and black has a #1 with Qd4# then. This position is reached by continuing the forward play 13. Nh4 Qe3 14. Nf5 Qd3 15. Ne3 Qd4#.
31 - K. Fabel
(Solution by mr.mip)
As both parties have mate in one, the question becomes: Whose turn it is to move?. The unfortunate location of the wBa3 and pawn formation around it tells us that the white captures have been: a2xQb3 and d2xBc3 accounting for all missing black pieces. The black captures must have been e7xRf6 and two pawn captures, one after a promotion of e-pawn.
Concerning the promotions, following must have happened:
The task then is to unpromote the two black knights, retract the resulting pawns beyond a2 and then retract the white move a2xQb3, after which the position releases without much trouble. However, there is a further problem: It is not possible to directly take both "free" black knights (Nb1 and Nb6) to the unpromotion square a1, as they need to go through b3 square, needing a white tempo manouvre each time. But there readily exists only one such manouvre : Ng3 Re4 Ne2.. Fortunately there is an indirect way to accomplish this:
Now suppose black has the move. Then the retro-play is:
-1. Rb3-c3 Na8-b6 -2. Rc3-b3 Nc7-a8 -3. Rb3-c3 Ne6-c7 -4. Rc3-b3 Nf8-e6 -5. Rb3-c3 Ng6-f8 -6. Rc3-b3 Ne5-g6 -7. Ne3-g4 Ng4-e5 -8. Nf1-e3 Nd2-b1 -9. Ne3-f1 Nb3-d2 -10. Nf1-e3 Na1-b3 -11. Ne3-f1 a2-a1=N -12. Rb3-c3 Nf4-g2 -13. Ng2-e3 Ne6-f4 -14. Rc3-b3 Nc5-e6 -15. Rb3-c3 c7-c6 -16. Rc3-b3 Nb3-c5 -17. Ng3-e4 Re4-e2 -18. Ne2-g3 Na1-b3 -19. b3-b4 and black is retro-pat!
So white has the move. The retro-play is:
-1. ... Na8-b6 -2. Rb3-c3 Nc7-a8 -3. Rc3-b3 Ne6-c7 -4. Rb3-c3 Nf8-e6 -5. Rc3-b3 Ng6-f8 -6. Rb3-c3 Ne5-g6 -7. Ne3-g4 Ng4-e5 -8. Rc3-b3 Nd2-b1 -9. Nf1-e3 Nb3-d2 -10. Ne3-f1 Na1-b3 -11. Nf1-e3 a2-a1=N -12. Ne3-f1 Nf4-g2 -13. Ng2-e3 Ne6-f4 -14. Rb3-c3 Nc5-e6 -15. Rc3-b3 Nb3-c5 -16. Ng3-e4 Re4-e2 -17. Ne2-g3 Na1-b3 -18. b3-b4 c7-c6 -19. Bd6-a3 a3-a2 -20. Bf4-d6 a2-a1=N -21. Bc1-f4 a4-a3 -22. Bd2-c1 a3-a2 -23. Bc1-d2 b4xRa3 -24. Ra1-a3 b5-b4 -25. a2xQb3 Qa3-b3 -26. Rb3-c3 b6-b5 -27. Rb5-b3 Qb4-a3 -28. Re5-a5 Qa5-b4 -29. c3-c4 Qa8-a5 -30. d2xBc3 Bb4-c3 -31. Re8-e5 Bf8-b4 -32. e7-e8=R Qd8-a8 -33. e6-e7 e7xRf6 and the whole position is finally unlocked.
So, it must be white to move and win with 1. Rxd3#
32 - K. Fabel
(Solution by mr.mip)
The pawn formation around bBh6 tells us that the black captures must have been h7xQg6 and e7xBf6, accounting for all missing white men. White, on the other hand, has made pawn captures bxa and gxh, the first taking a pawn and the latter a rook. Both these pawns, as well as the original h-pawn, have then promoted. Promotion to a bishop has taken place on a8, while both h-pawns have been promoted to a knight, presently existing on g1 and h3.
The task is then to unpromote wNg1 and wNh3 on h8, retract the resulting pawns beyond h7 and then retract black's h7xQg6, after which the position releases comparatively easily.
Suppose now that white has the move. Retroplay would be:
-1. ... Rf6-g6 -2. Nf4-h3 Rg6-f6 -3. Nh3-g1 Rf6-g6 -4. Ng6-f4 Nd6-b5 -5. Rb5-b7 Nb7-d6 -6. Nh8-g6 Rg6-h6 -7. h7-h8=N Rf6-g6 -8. Nh3-f4 Rg6-f6 -9. f2-f3 Rf6-g6 -10. Ng6-f4 Nb6-d5 -11. Rd5-d7 Nd7-b6 -12. Nh8-g6 g6-g5 and white is retro-stalemate.
So black has the move. Retroplay is then -1. Nf4-h3 Rf6-g6 -2. Ng6-f4 Nd6-b5 -3. Rb5-b7 Nb7-d6 -4. Nh8-g6 Rg6-f6 -5. h7-h8=N Rf6-g6 -6. Nh3-g1 Rg6-f6 -7. Nf4-h3 Rf6-g6 -8. Ng6-f4 Nb6-d5 -9. Rd5-d7 Nd7-b6 -10. Nh8-g6 g6-g5 -11. f2-f3 Bf4-h6 -12. h6-h7 Be3-f4 -13. h5-h6 Bf4-e3 -14. h7-h8=N Bd6-f4 -15. h6-h7 h7xQg6 etc, etc.
The solution must then be: Black to move and win with 1. ... Re6 2. Ne6 Ne7#
33 - H. Eriksson
Last moves are
-1. h4xg3 ep+ g2-g4
-2. Rf5-c5+ b5xc6 ep+
-3. c7-c5 b4-b5+
-4. Kc6-d6 c5xb6 ep+
-5. b7-b5 Nb5-c3+
and everything now unlocks.
34 - Lord Dunsany
What was black's last move? It can't have been with Bb1 or Pa2. It
can't have been d6xe5 either, because this would require 8 captures
by black. If the last move was d7xc6, then the bishop on b1 is promoted.
For this, the g-pawn needs to capture once, and promote on f1. Bust
this requires 7 captures. It can't have been e7-e5: pawn on c6 and the
pawn on a2 account for all black captures. But all those captures were
on white squares, and white misses his black-squared bishop too. So
black's last move was with his king or his rook.
1. Ke6 ~ (OOO?) 2. Rh8#
35 - T. R. Dawson
The only way to give mate in 1 is to add the wh. Rook on b1 or c3 and play 1. b3#. Now, what is the right square for the Rook?
Counting missing White pieces: including the Rook which is to be placed, and the Wh. KB which fell on its home square (f1), there are two other missing pieces: a wR and a wN. One of these fell on f6.
Now, White's h-pawn made one capture, on a black square, to reach g3. White's c-pawn captured the remaining Bl. pieces via c2xd3xe4xf5xg6xh7, all on white squares. So White's h-pawn must have captured the Black KB, and White's c-pawn captured a bP (or its promoted replacement), two N's, a Q and a R. For the Bl. b-pawn to be captured by W's c-pawn, it would have to have made at least two captures (or else it made one capture to get to the c-file and the c pawn made one capture to get to the d-file). In either case, there is only one remaining White piece to be captured (whichever one was not captured on f6). So the Black b-pawn must have promoted and made at most one capture in doing so. It must therefore have captured a piece on a2 and promoted on a1.
So, the wN and wR were captured on a2 and f6 (though not necessarily in that order). However, Bl.'s KB could not have gotten out to be captured on g3 until after the capture on f6. As the Wh. QR was permanently confined to a1-a2-b1, neither wR could have gotten out to be captured until after the capture on f6. Thus, the wN was captured on f6, so a wR (necessarily the QR) was captured on a2. So it must be the Wh. KR that is supporting the pawn delivering mate. Since the Wh. KR cannot get to b1, it must be on c3.
36 - T. R. Dawson
Black captures are axb and bxc. Either one of these captures is the
white e-pawn, or this pawn promoted and is still on the board. In both
cases, this pawn had to promote, since there's only one capture left
(axb and d6xc7 are the other two captures). This pawn had to capture
a piece on d7. To unlock the position, black needs to retract d7-d6,
after bringing his bishop (which has to get uncaptured) back to c8.
The whole cluster on the left of the board was already completely on
its place when black played d7-d6, so the promoted piece is still on
the board. The only piece that could've escaped from d8 is the queen
on f2. To prevent this queen from giving a check, when it was on d8,
a screen is needed on c8 or b8. The only piece which can give this screen
is the white king. It entered the 8th rank via h6 and g7. So white may
1. Kd1 Ra3 2. Kc1 Ra1#.
37 - T. R. Dawson
(Solution provided by Mr. Mip)
White's captures are g6xNf7 and h6xBg7. Black captures are c5xd4, b5xc4 and a7xb6, in this order. So the white a-pawn promoted and is still on the board. The only white piece that can go to a8 and unpromote is the knight on g8.
Suppose black has the move. Then the retroplay is:
-1. Ba7-b8 Bb8-c7 -2. Rc7-c6 Rc6-d6 -3. f3-f4 Rd6-e6 -4. Re6-e7 Ne4-g5 -5. Re7-e6 Ng3-f4 -6. Re6-e7 Nf5-g3 -7. Re7-e6 h4-h3 -8. Re6-e7 Ne7-f5 -9. Nh6-g8 Ng8-e7 -10. Re7-e6 Re6-d6 -11. Nf5-h6 Rd6-c6 -12. Rc6-c7 Bc7-b8 -13. Ne3-f5 Bb8-c7 -14. Nc2-e3 Bc7-b8 -15. Nb4-c2 Bb8-c7 -16. Na6-b4 Bc7-b8 -17. f2-f3 Bb8-c7 -18. Nc7-a6 h5-h4 -19. Na8-c7 Bc7-b8 -20. Bb8-a7 h6-h5 -21. a7-a8=N and black is retro-pat.
With white to play, the retro-play is almost similar, except that black doesn't have to lose a valuable pawn tempo early in the retro-play, because, due to the changed tempo, Ng8-e7 can get retracted without the pawn retraction earlier.
So white has the move: #1 with 1. Bc7#
38 - T. R. Dawson
Suppose white may still castle. The whole position unlocks when black
can uncapture axb. The only way this can be done is if the black rook
goes to d2, and the white knight from d2 unpromotes on a8, and goes
to a3 or a2. Retro-play would be:
-1. ... Rb7-c7 -2. Ba2-b1 Ra7-b7 -3. Bb1-a2 Ra1-a7 -4. Ba2-b1 Rc1-a1 -5. Bb1-a2 Rc2-c1 -6. Ba2-b1 b6-b5 -7. Nb1-d2 Rd2-c2 -8. Na3-b1 b7-b6 -9. Nb5-a3 d6-d5 -10. Nc7-b5 e6-e5 -11. Na8-c7 e7-e6 -12. a7-a8=N f6-f5 -13. a6-a7 f7-f6 -14. a5-a6 g6-g5 -15. a4-a5 and black is out of retro-moves, since bothe d7-d6 and g7-g6 must have been played earlier to let out the bishops. So unlocking the position, assuming white may still castle, fails on 1 tempo.
So white may not castle. Unlocking the position is then easy. The white king came from f1, and the knight can unpromote on a8 without problems then.
h#1 with 1. Ba2 Nd3#
39 - Harold Cross
Wh. caps are axb & bxc. Bl. caps are axb & dxe. Wh. cannot retract c2-c3 because of his wBd1. The cage unlocks by pushing the three Rooks one square to the right and inserting something on e1 so that the wBd1 can be extracted. Only a wN, uncaptured by bPdxe can go to e1.
-1. d2-d4!! d5xNe4
-2. Ng5-e4 Rg2-g1
-3. Nf3-g5 Rg1-f1
-4. Rf1-e1 d7-d6
-5. Ne1-f3 h7-h6
-6. Bc2-d1 Kd1-c1
and everything unlocks just in time to avoid bl. retropat.
We see that last move must have been -1. d2-d4!! (and not -1. d3-d4??) so that -7. Bc2+ can be retracted later. Therefore, the en-passant capture is legal.
40 - Harold Cross
Bl. caps are a3xb2, g3xf2 & h4xg3. Wh. only cap is axb. For balance, the wPh2 had to promote on h8. Position unlocks by reconducting bNh4 on c1, thus freeing the wNb1 that can be unpromoted on h8. When the promoted wPh2 is reconducted past h4, we can unlock by retracting bPh4xg3.
-1. Ng6-h4 Rc1-d1
-2. Nf4-g6 R..
-3. Nd3-f4 R..
-4. Bh3-f1 Rd1-c1
-5. Nc1-d3 Nc3-b1
-6. Bf1-h3 Nd5-c3
-7. B.. Nf4-d5
-8. B.. Ng6-f4
-9. B.. Nh8-g6
-10. B.. h7-h8=N
-11. B.. h6-h7
-12. B.. h5-h6
-13. B.. h4-h5
-14. Bf1-h3 h2-h4
-15. h4xg3 etc.
Starting with a Wh. retraction would lead to an unsurmountable retro-opposition on h4 between wP and bB. It is not possible to swap side to move. Therefore Bl. played last and Wh. wins with 1. Nc3!! b1 2. Rxb1# (not Bl. with 1 ... Bg2#??)
41 - L. Ceriani
W is missing a R, Q, and the KB. The QR was obviously captured on b1, since the WN on a1 was there before bxa3, and prior to bxa3, there was no way for the WQR to get out. Thus, the WKB was captured on g6 and the WQ on f6. B is missing a P and a R. The BR was clearly captured on b3, since there was no way for W's a-pawn to get to the b-file, and it could not have promoted prior to axb3.
The BN on h8 was there prior to hxg6 (for similar reasons to the WN on a1). Thus, the WR on h7 and the BR on h6 got there via h5 and h6, not via h8. This means that the WN on h5 got there after the two rooks were on h6 and h7. In order for the WR to get to h7, two things must have happened first: 1) the WP on f4 must have been there in order for the WKR to get out, and 2) the WP on g3 must have been there in order for the WKB to get out to be captured on g6, freeing h7. Thus, before the WN got to h5, there were pawns on f4 and g3. So the WN must have gotten to h5 via f6.
Now, when the WN moved to f6, B's c, d, e, and f-pawns were all on the 7th rank (the e-pawn was there because the WN is on f6; the d-pawn was there because if it had already moved, the BKB could never have gotten to b8). Both B bishops were on their home squares. So when the WN moved to f6, the BK must have moved out of check (or not have been on e8). The only place for it to go is to d8. This means that the BQ was not on d8, but it could not have moved from d8, so it was captured there.
Thus, the original BQ was captured on d8, then the B a-pawn promoted on b1, capturing the WQR in the process. The original BQ never moved, the actual BQ is a promotee, its first move was Qb1-a2 (since both a1 and c1 were occupied).
42 - L. Ceriani
If black may castle, then his last move must've been b7-b5, because c6xb5 requires too many captures (Bc1 was captured on his home-square). So #2 with 1. cb6 Kd8/c6 2. Qf8#. If black may not castle, then #2 with 1. OO ~ 2. Rf8#.
43 - L. Ceriani
Analysis: Three wh. caps are hxg, the bQR in the NW corner, & the bKB at home. One bl. cap. (at least) is required by bPb2.
The only way to unlock the South cage is to move the bK to c1. But this needs interposing something on b1, between wQa1 and bKc1. Which unit can we bring back to b1? Nothing can go there from outside the cage ... After some time, we see that the only possibility is to uncapture on c1 something that can go to b1. But only a bl. Queen will do, and only a wh. Knight can reach c1 and uncapture a bQ there. Now is all this possible?
We need a wh. Knight. It can be uncaptured by Black. But retracting e.g. -1. h4xPg5 h6xNg5 will fail because Bl. has not enough waiting moves. We can get bl. retromoves by uncapturing the bQR but the uncaptured wNg5 needs too much time to reach a8 or b8. The trick is that we can first quickly uncapture the bQR and only then let it uncapture our wN !
Here is the way: retract
-1. g4-g5 f6-f5
-2. g3-g4 f7-f6
-3. h2xPg3 h4xBg3
-4. Bb8-g3 h5-h4
-5. Be5xRb8 Ra8xNb8
-6. Na6-b8 R~
-7. Nc5-a6 R~
-8. Nd3-c5 R~
-9. Nc1-d3 R~
-10. Nd3xQc1 Qb1-c1
-11. Nc5-d3 Kc1-c2
and now the position unlocks e.g. with -12. c2-c3.
(Note that the 5 uncaptures form some sort of AUW: PNBRQ are uncaptured!)
44 - L. Ceriani
(Solution by Henrik Juel)
The position unlocks when black can retract a6xb5. For this, the white knight from c7 needs to unpromote on a8, and the pawn should go back to a5. This means that a black knight on c7 is needed to prevent a check by the white bishop on b8. This means that the white king should go to f8.
The retro-play is: -1. ... Nc8-a7 -2. Ba7-b8 Nb8-d7 -3. Na8-c7 Na6-b8 -4. Nc7-a8 Nb4-a6 -5. Na8-c7 Nc2-b4 -6. Nc7-a8 Ne2-c2 -7. Na8-c7 Ng4-e3 -8. Nc7-a8 Nh6-g4 -9. Na8-c7 Rg8-g7 -10. Nc7-a8 Rh8-g8 -11. Bb8-a7 Ng8-h6 -12. Ba7-b8 Bh6-f8 -13. Bb8-a7 Bg7-h6 -14. Ba7-b8 Bf8-g7 -15. Na6-c7 Nh6-g8 -16. Nb4-a6 Ng4-h6 -17. Na2-b4 Ne3-g4 -18. Nc3-a2 Nc2-e3 -19. Nd1-c3 Nb4-c2 -20. Ne3-d1 Na6-b4 -21. Ng4-e3 Nb8-a6 -22. Nh6-g4 Nd7-b8 -23. Ng8-h6 Bh6-f8 -24. h3-h4 Nf8-d7 -25. Nh6-g8 Rg8-h8 -26. Ng4-h6 Rh8-g8 -27. Ne3-g4 Bh6-g7 -28. Nc2-e3 Rg8-h8 -29. Nb4-c2 Rg7-g8 -30. Na6-b4 Nd7-f8 -31. Nc7-a6 Nb8-d7 -32. Kf8-e8 Na6-b8 -33. Na8-c7 Nc7-a6 -34. Bb8-a7 Na7-c8 -35. h2-h3 Nc8-a7 -36. a7-a8=N d4-d3 -37. a6-a7 Na7-c8 -38. a5-a6 a6xBb5
and finally the position unlocks. Very surprising and hard to find tempo-maneuver with the black bishop!
45 - L. Ceriani
(Solution by Henrik Juel)
Start from the following position:
and play forward:
1... Rg7 2. Rg8 d5 3. Ref8 d4 4. Ke8 Ne6 5. d3 Nc5 6. c3 Ne4 7. de4 d3 8. c4 d2 9. c5 d1=R 10. Ke7 Rd8 11. Kf6 Re8 12. Kg5 Rd8 13. Kf4 Re8 14. Kf3 Rd8 15. Kg2 Re8 16. Kf1 Rd8 17. Ke1 Re8 18. Kd1 Rd8 19. Kc2 Re8 20. Kb3 Rd8 21. Kc4 Re8 22. Kd4 Rd8 23. Ke5 Re8 24. Kd6 Rd8 25. Ke7 Re8 26. Re8
An impressive king-cycle!
46 - B. P. Barnes and M. Lipton
(a) Suppose white can castle. Ba7 is a promoted bishop. The
only two squares it could've promoted on are c1 and g1. Promotion on
g1 requires 6 captures, so that's 1 too many. Promotion on c1 requires
only 4 captures. But two of these captures took place on the b-line,
and on c1. This implies that at least two white pawns promoted, without
capturing. This isn't possible, unless the d-pawn captured twice. But
this means too many captures again. So white can't castle. So #2 with
1. Kd2! (Nd1~#) Qg2/Qg3/Qg5/Rb5 2. Ndf2/Qg3/Ne3/Nb2#, and not with 1.
(b) Now the black pawn could've promoted on c1 without capturing, so white can castle. #2 with 1. Nb2! (Kd2#) Qg2/Qg3/Qg6/Qg5 2. OOO/Qg3/OOO/Ke2#.
47 - H. A. Adamson
White captures are a6xb7, b6xc7 and h2xg3. Black capture is f3xe2. Suppose black just castled. When h2xg3 was played, the whole northeast corner must have been there already: f2-f4 was already played to let in the rook on f3. Bg5, Qh5 and Bh4 must've been there too, because otherwise the only route the black bishop could've gotten in was over f6 or h6, delivering an impossible check to the king, and the white pieces were there too. This means that the piece that was created on h1 by promotion, is either still on the board, or was captured on c7. Since all black pieces were already on their positions they were now, it was captured on c7. So the piece that was created on c8 by promotion is still on the board. It can't have been the knight on d5, since it couldn't get away without giving check, so it was the rook on b7. The knight on d5 arose from promotion on b8 then, and this promotion happened before the promotion on c8.
So the last moves were, if black just castled:
-1. ... OOO -2. Nb4-d5 Rf2-f3 -3. Nc2-b4 Rf3-f2 -4. Nd4-c2 Rf2-f3 -5. Ne6-d4 Rf3-f2 -6. Nd8-e6 Rf2-f3 -7. Rb8-b7 Rf3-f2 -8. Rc8-b8 Rf2-f3 -9. c7-c8=R Rf3-f2 -10. Ne6-d8 Rf2-f3 -11. Nc5-e6 Rf3-f2 -12. Na6-c5 Rf2-f3 -13. Nb8-a6 Rf3-f2 -14. b7-b8=N Rf2-f3 -15. a6xBb7 Bc8-b7 -16. a5-a6 b7-b6 -17. b6xNc7 Nd5-c7 -18. a4-a5 Nb4-d5 -19. a3-a4 Nd3-b4 -20. a2-a3 Rf3-f2 -21. c2-c3 Nf2-d3 -22. b5-b6 Nh1-f2 -23. b4-b5 h2-h1=N -24. b3-b4 h3-h2 -25. h2xNg3
But now the rooks can't go back to behind the pawn barrier! So retracting the position, assuming black just castled, isn't possible. So black didn't castle as last move.
Solution: Retract Re8-d8, and h#1 with 1. Kd8 Rb8#
48 - H. A. Adamson
[Solution by Thomas Volet]
There is alot of time pressure. The position will unlock once white can retract e2xd3, and for this, the white bishop needs to go to f1 (and thus a white rook has to go to h1 too)
Retro play is: 1. Be5-h8 Pe4-e3 2. Bg3-e5 Pe5-e4 3. Qb1-c1 Pe6-e5 4. Rc1-f1 Pe7-e6 5. Bb8xPg3 Ph4xNg3 6. Nf1-g3 Ke1-e2 7. Bd1-c2 Ph5-h4 8. Rc2-c1 Ph6-h5 9. Nc1-a2 Ph7-h6 10. Bg4-d1 Kd1-e1 11. Bc8-g4 Ke1-d1 12. Qa2-b1 Kd1-e1 13. Ne2-c1 Ke1-d1 14. Ng3-e2 Ke2-e1 15. Ne4xPg3 Kd1-e2 16. Ng5-e4 Ke2-d1 17. Nf3-g5 Kd1-e2 18. Ne1-f3 Ke2-d1 19. Ne3-f1 Kd1-e2 20. Nd5-e3 Ke2-d1 21. Nf3-e1 Pg4-g3 22. Rc1-c2 Pg5-g4 23. Rf1-c1 Kd1-e2 24. Ne1-f3 Kc1-d1 25. Ne3-d5 Pg6-g5 26. Nd1-e3 Kc2-c1 27. Nf3-e1 Kc1-c2, the WRs leave the first rank, the WK exits, the WKR returns, the WKB returns, then the WKP uncaptures to e2 and the positions unwinds.
49 - H. A. Adamson
(solution by Henrik Juel)
The position is released by quickly uncapturing a wN on the b-file, which is then brought to f1 to screen. Then wBc2 may be extracted and later a wR, followed by retracting wB to f1 and opening the cage by uncapturing e2xBd3. The whole thing is rather complicated (both wR's must go via the queen side), and all available tempo moves by bP's are used. Start the forward play from the following position.
Play forward: 1. . Nb2 2. Kc2 Be4 3. Kc1 Bd3 4. exd3 h6 5. Be2 h5 6. Bf3 Kf1 7. Be4 Ke1 8. Kc2 Kf1 9. Re5 Ke1 10. Nf3 Ke2 11. Rc1 g6 12. Ne1 Kf1 13. Bf3 Kg1 14. Be2 Kh1 15. Bf1 Kg1 16. Nf3 Kh1 17. Nd4 Kg1 18. Ree1 Kh1 19. Red1 Kg1 20. Nf3 Kh1 21. Ne1 Kg1 22. Be2 Kh1 23. Bf3 Kg1 24. Be4 Kf1 25. Nf3 Ke2 26. Rh1 g5 27. Rcg1 f6 28. Ne5 f5 29. Ng4 f4 30. Nd3 d6 31. Nf1 Ke1 32. Kc1 Ke2 33. Bf3 Ke1 34. Bd1
(a) e5 35. Bc2 Ke2 36. Bg3 Ke1 37. Ne3 Ke2. Now wN can only
reach the b-file in exactly two moves: 38. Nd5 h4 39. Nb6 axb6 40. Rf1
b5 41.Rfg1, so Black has the move and wins by e.g. 41. .. cxd2#.
(b) d5 35. Bc2 Ke2 36. Bg3 Ke1 37. Ne3 Ke2 38. Nf5 h4 39. Nd6 a6 40. Nb5 ab5 , so White has the move and wins by 41. Re1#.
50 - H. A. Adamson
Start from the following position:
and play forward: 1. Ng7 Kh6 2. d5 Nf4 3. d6 Qf3 4. b3 Rh5 5. b4 Bh4 6. b5 Qg3 7. hg3 h2 8. a3 h1=Q 9. a4 Qd1 10. a5 Qd5 11. a6 Qb7 12. ab7 Nd3 13. b8=N Ne1 14. Na6 Nd3 15. Nc5 Ne1 16. Na4 Nd3 17. Nc3 Ne1 18. Nd1
So white moved last. #1 with 1. ... Ne4#