hans: Yes, for Lg3 is a promoted piece. (2013-02-20)
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Henrik Juel: Last move was g2xDTSh1=L+ (2019-01-01)
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Henrik Juel: The next retraction is also determined
2... Ka7xSa8 (3.Sb6-a8+ or Sb6xDLSa8+) (2019-01-01)
Henrik Juel: oops, that is not true
after 2... Ka7-a8, Lc5 can uncheck directly (2019-01-02)
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AB: Anticipated by P0001033. I feel almost guilty pointing this out, because Smullyan did so much to popularize RA. (2001-10-30)
Henrik Juel: The author of P0001033, Jan Mortensen, was no slouch either
For some 40 years he was the driving force in DSK, the danish chess problem society (2019-01-01)
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Henrik Juel: The pawn captures are determined, and also their sequence
d7xSe6 first, then a2xLb3, and finally g7xDh6 (2019-01-01)
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Henrik Juel: ... and wKb4 got out via h2 and g3 (2019-01-01)
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Henrik Juel: Strictly speaking, [Sb8] could also have been captured by DTL, e.g. h4xPg5xPf6epxPe7xDd8=DxSb8
But White must have promoted (on a dark square) (2019-01-01)
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Forderungs-Alternative: Einfarben-Schach. Auf dem Brett steht ein unsichtbarer Läufer. Steht er auf einem schwarzen oder einem weißen Feld?
Yoav Ben-Zvi: In the book (page 26) the WB is placed between e3 and e4. It seems simpler to represent this with WB on e4, the stipulation "Monochrome chess. Is the position legal?" and a Twin b) WBe4 to e3. Proof of illegality with WB on e4: There is no piece that could perform the last capture on a dark square since no piece on the board is capable of a dark-squared capture and it could not be a piece that is missing since such a piece would itself have been captured later on a dark square. Smullyan's version uses poetic language including the metaphor of a Bear and a Lion fighting until each has entirely devoured the other. Essentially the argument uses the well-known mathemtical technique of "Infinite descent" leading to "Reductio ad Absurdum": It can be shown that assuming the existence of a "minimal" instance that has some property (such as most recent capture on a dark square) leads to the conclusion that it is not really minimal (there is a more recent capture on a dark square) so the assumption leads to self contradiction and is therefore not true. A formal definition of this method was developed by Fermat (17th Century) but it was used as far back as Euclid's "Elements" (300 B.C.). (2017-12-14)
Henrik Juel: It might be simpler to consider the retroplay:
With Le4 there is no way of producing a dark-squared man by uncapture (2017-12-14)
Brassaud: With Le4 it seems possible to producing a dark-squared man with the retroplay : 1) Lc2-e4, Kd7-e8 2) Ld3xPc2, b3xPc2 3) Ld3-e4, a4xb3 ep 4) b2-b4... But the position remains illegal as shown with the metaphor of the lion and the bear. (2017-12-15)
Henrik Juel: Good point, Francois
But the en passant trick still cannot create a full retroplay back to the initial position with 32 men
I should have written 'officer' instead of 'man' (2017-12-15)
Yoav Ben-Zvi: Henrik's modified claim that "with WBe4 there is no way of producing a dark-squared officer by uncapture" is correct but I do not see a simple way to prove it directly. A "lion and bear" type argument with retroplay could ask "Which of the pieces that move on the dark squares was the first to uncapture?". It should be noted that this piece could not be a pawn that was itself uncaptured e.p. (since the history of such a pawn does not include an uncapture) and that proving that such a piece does not exist proves illegality (since it implies that dark-squared officers such as BQd8 could not be uncaptured). A similar clarification is needed when applying the argument to forward moves.
The Homebase position of the diagram is an IC. It could be presented (optionally adding a WB on f1 and/or a BB on c8) with the stipulation "Is the position legal?" and the Twin b)-WPf2. (2017-12-17)
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Henrik Juel: White captured [Lf8] with an officer, and the remaining four missing black men by fxexdxc and c7xLSd8=T+
So Black captured b7xa6, f7xe6xd5xc4, and h3xg2 on light squares, leaving only the dark-squared [Lc1] to be disclosed on h4 (2019-01-02)
Yoav Ben-Zvi: The detailed solution in Smullyan's book (p29) is reprinted in "Four lives, 2014 Ed. Jason Rosenhouse" (a tribute to Smullyan published shortly before his death). Another version appears in the wonderful chapter by Bernd Gräfrath in "Philosophy looks at chess, 2008 Ed. Ben Hale" where it is stated that Smullyan composed this problem at age 16 and considered it his best Retro problem. The following analysis is adapted from these sources:
Last move was -1.wPc7x(B or N, not Q or R)d8=R. The unpromoted wP implies 3 captures by [wPf2] (not 4 captures by [wPg2] plus 1, on g3, by [wPf2] since [bBf8] was not captured by a wP) implying that wPg3 comes from g2. It also implies promotion of [bPh7] (to bB or bN) after capturing on g2 (behind wPg3), a fifth capture by bP. The piece on h4 could not be bQ or bR (both kings would be in check) or bB or bN (requires a second Black promotion) therefore the piece on h4 must be White. A White piece on h4 other than wB would imply that [wBc1] was captured by a bP but all 5 captures by bPs are on light colored squares so:
Solution - White bishop stands on h4. (2019-01-02)
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Mu-Tsun Tsai: In the book it claims that black can't castle, but I just constructed a counter example:

1.d4 e5 2.Nf3 Be7 3.Nc3 Bh4 4.d5 Bg3 5.hxg3 Qe7 6.Nd4 Qd6 7.Nc6 Qxd5 8.Qd3 Qf3 9.exf3 dxc6 10.Qa6 Bg4 11.fxg4 Nf6 12.Bb5 Nd5 13.Ke2 Nd7 14.Rh6 Nb4 15.Ne4 Rf8 16.Rg6 fxg6 17.c3 Nc5 18.cxb4 Rf5 19.Qa3 Nb3 20.Bd2 Na5 21.bxa5 Rh5 22.Ba6 c5 23.Bb4 cxb4 24.Ke3 Rg5 25.Nc5 Rh5 26.Nd3 Rg5 27.Ke4 Rh5 28.Bc4 Rf5 29.Kd5 Rh5 30.gxh5 bxa3 31.Nb4

This leads to the identical position, but black is able to castle. In fact, the fallacy in the solutino provided in the book claims that the black Pawn at a3 must have moved a4-a3, but as we can see in the above example, it can actually be bxa3. In this case, there's no way to deduce the claimed result. I think this problem is cooked. (2009-05-11)
Mario Richter: This problem is embedded in a little story. Sherlock Holmes enters the room, sees the position as given in the diagram and observes, that White is just making a move.
If one takes this into account, a stipulation that would preserve the problem's correctness could be:
"White to move. Does Black still have the right to castle?" (2009-05-11)
Mu-Tsun Tsai: Indeed, that should correct the problem, since in this case Black's last move must be a4-a3, as the provided solution claimed.
However, since the story stated that "White just finished a move with the Knight", the problem is then cooked. (2009-05-11)
Yoav Ben-Zvi: Smullyan's book "The Chess Mysteries of Sherlock Holmes" has recently been reprinted by Dover. The story of this problem begins on page 38. As noted above the observation that White made the last move with the Knight does not support the "Can't Castle" analysis. The Bishop (or Rook) should have been specified instead. The problem could be presented in conventional form as:
12+10 r3k3/ppp3pp/4Kp2/P3p2P/1P4B1/pRN3P1/P4PPR/8
Can Black Castle
OR
11+10 r3k3/pppN2pp/6p1/P2Kp2P/1N6/p4RP1/PP3PP1/8
Mate in 2 (2012-07-18)
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Mario Richter: vgl. mit P1012380 (2010-04-17)
Henrik Juel: Last move was a2-a3, so to prevent white retro-stalemate Black must uncapture a wS at once
Tb8xSa8 or LxSc8 do not give White a previous move, so the uncapture happened on e8 or h8, and Black may not castle (2019-01-02)
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Henrik Juel: Last move was not white castling, because the no-capture stipulation for Black would imply that the only previous white move available could be e2xf3??, rendering the position illegal (because d2-d1=L would be impossible)
So last move was Kh1-g1 or Kh2-g1, Th8 has moved to check, and Black may not castle (2019-01-02)
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Alfred Pfeiffer: Auf a5 steht der wK. (2012-01-11)
Henrik Juel: No Mystery here (2019-01-02)
Mario Richter: No mystery, but a little joke ("Sir Reginald's Jest"): the position was given as a trap to Sherlock Holmes, who immediately started analysing: Let's see now; Black is in check. What could have been White's last move? Obviously a rook from b7 capturing a Black piece on a7. I guess the next question now is, what was the Black piece? If it was a rook, then Black has promoted earlier..."
A this point the spectators could resist any longer laughing out loud pointing out that the piece obviously has to be the White King. Holmes concluded: "I have really been given a good dose of my own medicine. How many times have I not told Dr. Watson: 'In looking for the subtle, be careful not to overlook the obvious.'" (2019-01-03)
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vgl. P0004348
AB: 1.Nxc3#. White King & Queen are swapped, impossible unless board is rotated by 180 degrees. (2002-04-03)
Mario Richter: at least partially anticipated by the famous Dunsany problem P0004348, I think. (2016-09-13)
A.Buchanan: I much prefer Dunsany's original, which has some real chess, doesn't dress the board, and is closer to "found art". (2016-09-14)
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Henrik Juel: In the diagram, light and dark squares should be inverted (e.g., h1 should be dark), implying that the board must be rotated 90 degrees; this inversion is probably impossible to achieve in the PDB, otherwise Gerd would have done it
Clockwise rotation: 1.'Pb6xTa5=DL#'
Counter-clockwise: 1.'Pc2-d2#' (2019-01-02)
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Yoav Ben-Zvi: Appears on page 54 of the Book. WPb4 and BPg5 both captured Rooks. Neither BR could have escaped from the bottom rows, to be captured by [WPa2], before capture of the WR cleared the h column. It follows that the WR was captured first with WP still on column a and therefore the captured WR could not have come from a1 and was originally from h1 escaping through the h column. WRh1 is a "Sibling Impostor" and White "Can't Castle". (2018-03-15)
Henrik Juel: It seems that Lc1 is unnecessary
Even without it, White cannot castle, by a very similar argument; the only difference is that Th1 might be [Th1], but if so, it must have moved to make way for [Ta1] (2018-03-15)
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Henrik Juel: An important condition is missing:
The queens never changed square color
With this condition added, we can analyze
White captured all missing black men with exfxgxh
Black captured [Lc1] on c1 and [Dd1] on a light square, so he captured another original white officer with a7xb6, after which White promoted [Pa2] on a8, so Ta8 has moved and Black may not castle
The promoted officer must be Ta1 or Tg1, so White may not castle either (2019-01-02)
Mario Richter: Originalforderung:
It is given that neither queen has ever moved off her own color
a) Which side, if either, can castle?
b) If the rook on g1 is removed, would that affect the answer?
c) If the rook is replaced on h1, then what would the answer be?" (2019-01-03)
Henrik Juel: Thanks for the update, Mario
b)
No need for promotion, as Black could have captured the missing T by a7xTb6
Black may castle, White may not
c)
Like a), but if the promoted officer is Ta1
White may castle short, Black may not castle (2019-01-03)
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Henrik Juel: With g7xh6 Black did not capture [Lc1], nor [Dd1], nor [Lf1], so [Pe2] must have promoted
If g7xh6 captured the promoted man, the promotion needed two white pawn captures, exPf and either f7xe8 or f7xg8, so Black must have played b3xa2-a1=Y, and White may not castle long
If g7xh6 captured an original man, this was also not [Sb1,Sg1], so [Th1] must have moved, and White may not castle short
White may castle short or long, but we cannot say which (2019-01-02)
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Rosalie Fay: The only Black man with a previous move is bK. bK came not from d2 (illegal check from wBe1) but from b2 out of check from wRb5.

Suppose the check was discovered by a man X. X is not wK (impossible check by bPa5). So Black's last move was KxXc2. X was not N (entails a White underpromotion) or B (entails a promoted wB as [Bf1] never moved). No other discovery is possible.

So this check was direct. Then it was a capture (because of wRb6). The captured man was not [Ra8] (which never escaped rank 8), a B or N (entails Black underpromoting), or P (bPa5 is [Pc7] so a bPb5 has no home). So it was Q. For ...Q(x)b5+ to be possible, bQ came from c6 (wRc5 shielded wK), so +wPd6. (2015-09-23)
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Rosalie Fay: The only Black man with a previous move is bK. bK came not from d2 (illegal check from wBe1) but from b2 out of check from wRb5.

Suppose the check was direct. Then it was a capture (because of wRb6). The captured man was not [Ra8] (which never escaped rank 8), or a B or N (entails Black underpromoting), or P (bPa5 is [Pc7] so a bPb5 has no home), or Q (...Q(x)b5+ is impossible).

So the check was discovered by a man X. X is not wK (impossible check by bPa5). So Black's last move was KxXc2. X was not N (entails a White underpromotion), thus B. This was [Bf1], so +wPh2. (2015-09-23)
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Henrik Juel: Add the pawn on f5
The dark-squared bishops were captured by officers, so the pawn captures were c7xSb6, d7xLc6 by Black and exLf, f2xSg3
by White (when the pawn has been added)
The forward sequence was e2-e3 (or e2-e4) before d7xLc6 before exLf on a light square, so [Pe2] could not reach f4, and the undetermined capture was e4xLf5
This simple idea was later shown more elegantly in P1012384 (2017-12-15, hidden) more ...
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Henrik Juel: tiny typo in stipulation: mit Schwarz, not Scharz
A wP on c4 does not imply that last move was c2-c4, so the ep capture is not legitimized as such; but it is the only way to prevent the game from ending in a draw (2019-01-02)
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Henrik Juel: No, White may not castle right now, because it is Black to move, as he has no last move
Or (if you allow joking) the board is turned upwards down, in which case White could have the move, but castling would still not be possible (2019-01-02)
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Henrik Juel: Yes, the position is legal, as the retroplay could be -1... Kc6-b6 -2.b7-b8=S++ etc. (or Ka6-b6 -2.a7xb8=S++) (2019-01-02)
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vgl. P1349637
Mario Richter: Could be, inspired by P1349637, presented as twin: a) diagram, b) +sSb8 (2018-05-02)
Ladislav Packa: Mario: Position b) would be enough. H#1*: 1...bxa8=R#, 1.Sd7 b8=S# (2018-05-02)
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Henrik Juel: The number of white moves in the game is uneven, and the number of black moves is even, implying that White just moved (assuming that White started; this is the standard parity argument); but this contradicts that Black just checked on c3
If Black started the game, there is no contradiction (2019-01-02)
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Henrik Juel: The diagram white pawns captured 6 times, and black pawns 8 times, so after adding a white pawn there is room for just one more capture by each side
Only addition on f2 permits a legal position, where the remaining captures were black hxPg (with promotion on g1) and white h7xg8=DLS (2019-01-02)
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Henrik Juel: Yes, Black may castle, as he had no time to move his king or rooks during the 6.0 moves game (but it is White to move)
Proof game (the black moves could be shifted):
1.a3 Sc6 2.a4 Sh6 3.a5 e5 4.a6 Dg5 5.axb7 Lb4 6.e4 Lxb7
Too many conditions for my taste in this problem (2019-01-02)
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Henrik Juel: White captured g2xTh3, so to get [Ta8] out, Black cross-captured b7xTc6 and c7xLb6
The sequence was: [Ta1] to c6 before b7xc6 before g2xh3 before [Th1] to a1
Add wPc4 to permit the wT moves (2019-01-03)
Alfred Pfeiffer: Im Original steht sLa6 (statt sLf5). Mit wBc4 kann der sLc8 nicht nach f5 gelangen. (2019-01-04, hidden)
Henrik Juel: Yes, the diagram is wrong
Please move Lf5 to a6
Thanks, Alfred (2019-01-04, hidden) more ...
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Henrik Juel: with wPe4 moved to f4:
White captured [Lc8] on a light square with an officer and d2xc3, so Black captured the light-squared [Lf1] either by dxLc or by d3xLe2 (and promotion on e1)
In both cases the sequence must be
e2-e3 (to release [Lf1]) before dxL before d2xc3 before [Lc1] could reach g5 traversing f2
So last move was f3-f4 (2019-01-03)
Henrik Juel: diagram error: wPe4 should stand on f4 (2019-01-03, hidden) more ...
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Mu-Tsun Tsai: In the book Smullyan explain that this position is illegal because there's no way the a4 bishop can appear over there. But in fact that's not really the point; even if we change the diagram by putting the bishop somewhere else (d3, say), the position is still illegal, because there's no way the original c8 bishop can be captured at f3. (2011-05-03)
Yoav Ben-Zvi: The problem appears on page 97 and the solution on p 154
With WB on d3 the position is legal since BPb7 captured WRa1 on c6 releasing BBc8 to be captured on f3 releasing WBf1.
The released Bishop cannot reach a4 because WPc4 and WPb3 arrive there, in oreder to release WRa1, prior to the release of WBf1. (2013-04-21)
Henrik Juel: I agree with Mu-Tsun.
A sequence of moves could be: b2-b3 (to release Lc1), c2-c4 (to release Ta1), b7xTc6; now Lc8 is released, but it cannot reach f3 (2013-04-21)
Yoav Ben-Zvi: Henrik's comment clarifies for me Mu-Tsun's observation that the obstruction preventing WBf1 from reaching a4 also prevents BBc8 from reaching f3 so there is a dual refutation of legality. I think this can be corrected by moving BPd7 to b7. To show both obstructions: a)a4 to d5 (illegal, BBf8 can't reach f3) b)a + d7 to b7 (legal) c)b + d5 to a4 (illegal, WBc1 can't reach a4). (2013-04-22)
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Mu-Tsun Tsai: In the solution provided in the book, it claimed that this is an RV problem, but it is actually not, for 1. Ke6 always does the trick, since 1...0-0-0 will be followed by 2. a8=Q#, and all the other respond for Black will result in 2. g8=Q#. It appears the Smullyan didn't notice the possiblity of 2. a8=Q#, and instead he would play 2. b7# after 1...0-0-0 (and that's why he placed an extra bishop is the problem), which is unnecessary. In short, the problem is cooked.
One possible way to correct this problem is to replace the a7 Pawn by a bishop, I think. (2009-05-12)
Mu-Tsun Tsai: Later I saw this page: http://www.janko.at/Retros/Misc/SmullyanErrata.htm
And indeed Smullyan later corrected the problem as I predicted :) (2009-05-12)
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Yoav Ben-Zvi: The Black piece captured on c3 could not be [bBc8] (captured on a light square) or [bPh7]. Therefore [bPh7] promoted on g1 after capturing [wBc1] (on g3) which was released previously by wPb2xc3. This implies that the piece promoted on g1 could not have been captured on c3 as the capture on c3, releasing [wBc1], occurred before the Black could promote. Therefore the promoted Black piece must be present on the board, replacing the piece captured on c3. The idea is developed further in P0002638. (2018-12-29)
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Yoav Ben-Zvi: The Black piece that accounts for the capture on a3 could not be [bBc8] (captured on a light square) so it was a missing bP that promoted. The bP promotion could not have occurred on h1 ([wRh1] did not move before castling) so the promotion was on g1 or f1. wPg3 came from g2, it could not capture on g3 as this would be the second capture by White on a dark square and Black's 2 missing pieces include [bBc8] which was captured on a light square. This means that the promoting pawn could not have been [bPg7] as this would require 2 captures on the way to promotion (to bypass [wPg2]) plus a third capture by bPh7xg6. Therefore the promoted bP comes from h7 and, to reach the promotion square, it must have captured on column g behind wPg3. If the piece captured on a3 was an "original" (not promoted) Black piece (could be any officer other than [bBc8]) then the piece promoted by Black is present in the diagram, replacing the piece captured on a3. Henceforth we consider the alternative: that the piece captured on a3 was the promoted piece itself. This implies that the capture by the bP on its way to promotion came before the capture on a3. If wBg5 is not promoted then, prior to wPb2xa3, the wB was locked at c1 which implies that [wRa1] had not yet escaped its home corner and that [wBc1] subsequently moved to g5 via b2 crossing c3. This means the White pawn was still standing on c2 and therefore, since the wK had not moved, that the wQ could not have escaped from d1. It follows (for the case of Black promotee being captured on a3) that the White piece captured by the Black pawn on its way to promotion could not have been [wBc1] or [wRa1] or [wQd1] as it has been shown that, in this case, all 3 were locked when this capture occurred. The White piece captured by promoting bP also could not have been a wP since, as noted above, wPg3 came from g2 and the capture occurred behind it. Since the White piece captured on column g could not have been one of the missing original pieces, [wPh2] must have promoted. On its way to promotion [wPh2] could not capture on g7 (with bP waiting on h7) for the same reason as given above for the capture on g3 (not all White captures were on dark squares). Therefore wP must wait on h2 until its way to promotion is cleared by bPh3xg2 so it can subsequently promote on h8 to replace the captured piece (known to be a knight). In conclusion: for every alternative scenario there is a promoted piece on the board.
The stipulation, which might be defined more accurately as "Must there be a promoted piece on the board?", weaves together the alternative strands of the solution.
The presentation of the problem begins on page 106 of the book (chapter "Shades of the Past") relaying the condition that the last move was -1.0-0 and that Sherlock Holme's nemesis, Professor Moriarty, solved the problem in less than 4 minutes while Holmes admits it took him more than 20 minutes. The description of the solution begins on page 155. (2018-12-29)
Mario Richter: @Yoav: Quote: "... to replace the captured piece (known to be a knight)." This is not completely correct.
Just a small thought to think about: Let's assume that Black answers White's castling by playing 1. ... 0-0! In what way does this change the RA? (2018-12-29)
Yoav Ben-Zvi: The conclusion that the piece captured on g2 was a wN is reached under the assumptions that Black's promoted piece was captured on a3 (therefore it is not present in the diagram) and that wBg5 is not promoted (implying that [wBc1] exited after the capture on a3 and before [wQd1] could be released by wPc2-c3). It has been shown that, under these assumptions, [wBc1], [wRa1] and [wQd1] could not have escaped in time to be captured on g2. The piece captured on g2 also could not be a wP ([wPg2] moves to g3) or a wB (wBg4 could not be promoted on the dark square h8) or a promoted piece (the path to promotion of [wPh2] is open only after the capture on g2) so I think the conclusion that it had to be a knight is justified, under the assumptions (although it is not needed to meet the stipulation).
The suggested addition of the condition: "Black retains the right to castle 0-0" prevents any resolution that has [wPh2] promoting on h8. It has been shown above that capture of Black's promoted piece on a3 requires the White promotion on h8 (either to a wB that replaces the one captured on c1 or to a wN that replaces the one captured on g2). This contradiction (promotion on h8 is required but not allowed) means that the piece captured on a3 could not have been Black's promoted piece so it was an original (not promoted) piece. The piece captured on a3 could not be [bBc8] (captured on a light colored square) or a bP so it is a bQ or bR or bB or bN that is replaced in the diagram by the promoted piece, proving that there must be a Black promoted piece present in the diagram position. A fine idea that improves the accuracy of the resolution. (2018-12-30)
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Henrik Juel: White pawns captured bxa, cxb, f6xe7 (to promote on e8), and g2xh3 (the remaining missing black man was captured by an officer)
Black captured h7xg6xf5xe4xd3 and dark-squared [Lc1] with an officer
Last move was not a6xb5, c4xd3, d4-d3, d7xe6, nor d7-d5, e7-e6 (as White moved f6xe7-e8=L and promoted L to c6 or c8), so last move was b7-b5
White mates by axb6ep# (2019-01-03)
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Henrik Juel: The stipulation might be written more clearly:
Black to move mates in 1
In the Reprints it looks strange that the problem appeared as number R8 in Phenix twice on pages 10810 and 10814 (2019-01-03)
Henrik Juel: White captured [Lf8] on f8, cxd, f2xg3, g2xh3, and once more
Black captured b7xTa6, f7/h7xTg6, and [Lc1] on a dark square
With Black to move, wK could not stand in check on e4, f3, nor g2, so Kc6 stands in check from wLh1, and wK must be ready to uncheck (possibly following the retroplay -1.c5xPd5ep+ d7-d5)
The discovered check by wK could not be from e4 (Lxd3+ is impossible) or f3 (impossible double check), so it was from g2, and wK must stand on g1
Black mates with Sf3#
(On g2 wK would stand in double check explained by f2xLe1=S++, so Black earlier captured h7xTg6, not f7xTg6) (2019-01-03)
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Mu-Tsun Tsai: Actually to solve this problem there's no need to know if Black can castle. When it comes to the part of explaining why the Pawn originated in e2 can't promoted, it suffice to notice that it can only go straight forward and therefore cannot pass the Pawn originated in e7. (2009-05-12)
Henrik Juel: White's only capture was a2xb3, so Black must have captured [Pe2] with an officer and played e2xLd1=LS or e2xLf1=LS
Add the dark-squared wL on a3 (2019-01-03)
Yoav Ben-Zvi: Smullyan thought that if bK could vacate e8 then [wPe2] could promote there. However, as shown in the solution on page 158, with wK immobile the promotion of [bPe7] occurs from e2 not from d2 and, since Black does not have the captures needed to promote from e2 after clearing the way on the e file, [wPe2] is prevented from moving straight forward to promotion. Therefore (as noted by Mu-Tsun) White retaining castling rights is a sufficient condition . In addition to removing the condition on Black castling, I think wPd5 can be removed ([bPe7] could then capture the wP and arrive at e2 but it needs a further capture to promote). Another idea, closer to Smullyan's intention, is to remove bQd7 and retain the condition on Black castling (preventing wP promotion from d7). (2019-01-05)
A.Buchanan: White is marked as having 15 units - which means that both of the x marks are being counted as White! I am not sure I can edit these in any case. But conventionally the x+y indicates the number of pieces visible in the diagram, for checking purposes, and does not include units to be added. It hardly matters, but it interests me. It would be possible to replace the x marks with lines, which are definitely available for users to edit. (2019-01-05)
A.Buchanan: Curious that the x marks are coded as *Black*: sXa3a4, yet are counted as White. Also, even after units are added to positions, the counter (y+z) is never updated. (Here hit the "play" button above. The "?" after the R: is necessary to get the position to update.) (2019-01-05)
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Cooker: Illegal position. I don't see what black unit takes the white bishop f1. (2002-09-21)
Joost de Heer: In the second (or perhaps even later?) version of this book, a white bishop is added on f1.
See http://janko.at/Retros/Misc/SmullyanErrata.htm for a list of errata in 'Sherlock Holmes'. (2002-09-27)
Henrik Juel: Black has castled and his king exited via h7 and g6, so last move was done by Ka2
The only possibility is Kb3xTa2, preceded by Tc2-a2+
The uncaptured wT is a pawn promoted after four captures; three of these captured [Ta8,Lc8,Pb7], and as [Sg8] never moved, the remaining capture was en passant, e.g.
a2-a4xPb5xPa6epxLb7xTa8=T (2019-01-03)
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Henrik Juel: Suppose the monochrome condition is in effect
Then everything is honky-dory except for Da6, which is [Pd7 ] or [Ph7] promoted on f1 or h1, e.g., h7-h5xYg4xPh3epxDg2xLf1=D
Here Y is [Pa2] promoted to DTL on c8 or e8; this seems possible, as [Ta8,Lc8,Pd7] and an ep-capture are available, but it does not work, since [Ta8] cannot reach b5: a2-a4xZb5xPc6epxPd7xLc8??, and a2xVb3xWc4xPd5xPc6epxLd7xTc8?? is even worse
So the monochrome condition is not in effect (2019-01-03)
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+sBg7
Mario Richter: http://www.chessbase.com/newsroom2.asp?id=2906 gibt die Stellung wie folgt an: 2B5/8/6P1/6Pk/3P2qb/3p4/3PB3/2NrNKQR, wurde vielleicht bei Nachdrucken das fehlerhafte Original in dieser Weise korrigiert (d.h. wBg2 nach g6)? (2011-05-27)
Mu-Tsun Tsai: In the webpage http://www.janko.at/Retros/Misc/SmullyanErrata.htm
it is claimed that the problem is later revised as 2B5/8/6p1/6Pk/3P2qb/3p4/3PB3/2NrNKQR,
but it still doesn't prevent the cook -1.g7. So I think 2B5/8/6P1/6Pk/3P2qb/3p4/3PB3/2NrNKQR is probably the right one instead. (2012-07-03)
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Henrik Juel: With the stringent conditions the only possibility is to unpromote Lh6 and let the resulting pawn uncheck, when Black must retract Kg8-h8
Unpromotion on b8 is too slow
Last move was Lg5-h6 (2019-01-03)
Mario Richter: Originalforderung: "Neither the White king nor queen has moved during the last five moves, nor has any piece been captured during that time. What was the last move?"
Ich schlage vor, das "Figuren" in der deutschen Version der Forderung durch "Steine" zu ersetzen, weil "Figuren" im Deutschen die Bauern nicht mit einschließt, was aber hier offensichtlich bezweckt zwar. (2019-01-04)
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Henrik Juel: The only way to avoid retrostalemate under the conditions is to put sK on c8 and retract
-1... Td8 -2.Kg8 0-0-0 -3.Kh8 S~ 4.Kg8 S~ -5.Kh8 S~ -6.Kg8 (2019-01-04)
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Mu-Tsun Tsai: One way to fix this problem is to add a black queen somewhere. Then exf2 will then be impossible. (2012-07-04)
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Henrik Juel: I believe the stipulation is incorrect.
If there are no promoted men on the board:
Black played g3xLh2-h1=Y (DS), so Sg1 has moved
White played dxPc-c7xYb8=S-a6 (or =D), so Sb8 has also moved
If there are no white promoted men on the board:
White captured [Lc8] on c8 with Dd1 and promoted to DTS on c8, later captured on a6
Black promoted to L on h1, so Sg1 has moved; this L is now on a4 (2019-01-04)
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Henrik Juel: The missing men are a black T, the white pawn from a2 or c2, and the white L from f1
Black crosscaptured to let a T out to b3 ([Lf1] was captured on the light square) and allow a white promotion to replace the captured original officer (capturing a white pawn on the dark square does not work)
The details are:
1. a7xDb6
2. sT gets out to be captured by c2xTb3, and Pa2-a8=D
3. b7xLa6
Promotion to T would not work, because Ta1 is immobile and Tg1 is inaccessible by the castling condition; a light-squared L is not on the board now; an S could not leave a8
Crosscapture on d6,e6 (or a2xTb3) would not permit promotion
Dd2 is promoted (2019-01-04)
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Henrik Juel: The missing [Ta8] was captured in its corner
White captured the other two missing black men with b2xc3 and g2xh3
The only black capture was c7xDb6, so [Pg7] must have promoted to D, T, or L on g1 after g2xh3 (promotion to S is ruled out, because Sg1-f3 would ruin white castling)
b2xc3 (letting [Dd1] out) happened before c7xDb6, which lets [Dd8] out
Here follow the pawn captures and the promotion, and their sequence, for the three situations; this is easily found by considering the few possibilities
A. Black S on c6
1. b2xLc3
2. c7xDb6
3. g2xDh3
4. g1=L, now on g7
The D was captured on h3
B. [Dd8] on c6
1. g2xSh3
2. g1=Y
3. b2xYc3
4. c7xDb6
The S was captured on h3
C. Promoted [Pg7] on c6
1. b2xSc3
2. c7xDb6
3. g2xDh3
4. g1=D, now on c6
The S was captured on c3 (2019-01-04)
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Henrik Juel: The first condition could be formulated more clearly:
Black's first move in the game was 1... d7-d5 (2019-01-04)
Henrik Juel: I
Black has moved Sb8-c6-d4-f5
White captured [Lf8] on f8 with an S
The sequence was something like
d2-d3 and d7-d5, c7xLd6, b7-b6 and Sb8-c6-d4-f5, e2xTf3
Now the route by [Lf1] can be determined:
Lf1-e2-d1-b3-c4-a6-c8-e6-g8-h7-g6 (with [Sg8] screening on f7) -h5-g4
A. Dd1, Lc8, and Sg8 must have moved
B. [Ph7] went h7-h6-h5
C. With Ph5 on h6, the position would be illegal, because Sg8 could not get home
II
With the additional conditions Lc8 would have to run ahead of [Lf1] and now stand on h3 (2019-01-04)
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Henrik Juel: La5 must be a pawn promoted on g1
A.
If Pf2 were white, [Ph7] must have promoted on g1, as [Pf7] would need too many captures
Black could play h7xLg6xPh5-h2xLg1=L, but then a7xb6 could not happen (promoting [Ph2] on h8 makes no difference)
Or Black could capture [Ph2] and play h7-h2xLg1, but this would leave no dark-squared white men for a7xb6; if Black played h3xLg2-g1=L (after g2-g3), the L could not reach a5
So Pf2 is black, and it just checked by f3-f2+, because e3xf2+ would require too many black captures, as [Pf2] or [Ph2] was not captured by a black pawn
B.
wS is on f4, since Black just left f3 (2019-01-04)
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Henrik Juel: If Black cannot castle, White can mate in 2 moves
1.Dd6 thr. 2.Dxe7#
1... exd6/exf6/Kd8 2.Sg7/DTxf8/Txf8
C+ by Popeye 4.61, except for the dual (2019-01-05)
Henrik Juel: Let us show that Black may not castle (with White to move)
White captured [Lc8] and four of the five remaining missing black men with his pawns on the king-side
Black Pg2 captured four of the five missing white men, which include [Pa2,Pb2], so one of these must have promoted
h7xg6 is illegal as last move (requiring six black captures), and so is h3xg2 (illegal black pawn position)
If Black made last move with Ta8 or Ke8, he may obviously not castle
1. Suppose last move was Dg7-f8
Then White must retract Th8xYf8+, exhausting the five possible captures by White, so White could not promote [Pb2] as this would require another capture; instead he promoted [Pa2] on a8 to supply four capturable white men, meaning that Ta8 has moved, so Black may not castle
If last move was g7-g6 (so [Lf8] was not captured by a pawn), the same analysis applies
2. Suppose last move was Dg7xYf8
This leaves just the needed four white men captured by Pg2, so White must have captured the last available black man with [Pb2] to promote it, and hence promoted [Pa2] on a8, so Black may not castle
If last move was g3-g2 (so Pg2 did not capture the missing light-squared [Lf1]), the same analysis applies
Conclusion: Black may not castle, because he has moved his K or T (2019-01-05)
Yoav Ben-Zvi: The convoluted stipulation is not needed as this is a standard #2 with solution supported by "Cant Castle" retroanalysis. (2019-01-07)
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Henrik Juel: Black has castled, so [Th8] has visited f8.
Where was it captured? Not on f8 or d8, because by the condition wK could not go that far, [Ta1] could not reach row 8, [Lc1] never moved, and White did not promote a dark-squared pawn
So after the castling [Th8] must have moved back to h8
The retroplay could start like this:
-1.g3xTh4 Th8 -2.Kf2 Kh7xDg8 -3.Dg4 Tf8 -4.h2xLg3 Lh4xTg3 -5.Ke1 Kg8 6.Dd1 0-0 (2019-01-05)
Henrik Juel: should be -6.Dd1, of course (2019-01-05)
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Mu-Tsun Tsai: There's a second part to this problem: "Remove the pawn at e2. Given that it's white's move and the Black king didn't came from f1, then what was the last move?" The answer is Ke3xRf2. (2011-05-15)
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H.Juel: The capture sequence must have been g7xSh6 before d2xTc3 before Lx[Lc1]
Alfred Pfeiffer: Falsche Lösung. Richtige Antwort ist: sLe5 hat wL geschlagen. (2010-07-27, hidden)
Henrik Juel: Alfred is right, so please correct the official solution
The capture sequence must have been
g7xSh6 before d2xTc3 before Lx[Lc1] (2019-01-23, hidden) more ...
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Henrik Juel: The stated retroplay is the only possibility
so White pawns move downwards (2019-01-23)
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Henrik Juel: The precursor by Pavlovic is P0001095 . (2010-07-19)
Sally: W König auf : Kc3!
Zuletzt geschah .Kb3 x Bc3+, Bb4 x c3 en passant,Bc2 -c4, L-d5#.
Nach Branco Pavlovic: Sahovski Vjesnik 1950. (2010-07-10, hidden)
Sally: S. 233 Sammelsurium (H. Kastner ) Humbold Buch)
S. 152: 300 rompecabezas de Ajrdrez (Espana)L. Barden.
S. 106 Eigenartige S. Probleme (W. Keym).
S. 641 Schwalne Dez. 2008. (2010-07-19, hidden) more ...
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Mu-Tsun Tsai: Black king need not be at d7, it might as well be at c8. It's just that cxb8=N# takes care of both possibilities. (2011-05-15)
Mu-Tsun Tsai: The original stipulation to the problem is "Black king is invisible, white to move and mate in one no matter where he is". (2011-05-15)
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Henrik Juel: Solution
0. Replacing wK on g8, h7, or h8 obviously does not work; neither does a4, a6, a7, b5, b7, b8, c4, c6, d7, f7, or g6, having both kings in check
1. What about replacing the white man on d5, e4, g4, or h5? This gives four missing white men, three of which were captured on light squares by b7xa6 and d7xc6xb5, while the fourth, [Lc1], was captured on a dark one
Hence the retroplay -1.Df8xYg8 Yxg8 does not work, because g8 is light, nor does -1.Df8xSg8 Sf6 -2.Dh6, because Sf6 would check wK (on d5, e4, g4, or h5)
(-1... Kg7 is illegal, and no other black men can retract)
2. Replacing a black man on a5, a8, e7, or e8 does not work, because the three missing white men include the dark-squared [Lc1]
3. This leaves the solution: putting wK on c7; now the black pawn play could have been b7xa6 and c7xb6-b5, so many retroplays are possible (after the uncheck -1.Df8xg8) (2019-01-23)
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Henrik Juel: Replacing Pa7 or Pc7 with wK would work, but this is ruled out by Alfred's stipulation clarification: wK may not take the place of a black pawn
Compared to P0003053, Da8 has been removed, but an extra missing black man makes no difference, so the arguments from there still apply
b7 is no longer attacked by Black, however, so the solution is putting wK on b7 (2019-01-23)
Mario Richter: Imho, Alfred's stipulation doesn't rule out Pa7, it should be: "Ersetze einen Stein einer anderen Art als in der vorangegangenen Aufgabe 3a (P0003053) durch dem wK, so daß die Stellung legal ist". This fits the original intention, where the problem is embedded into a story about Caliph Haroun al Rashid (the wK), who wents out one night masquerading as some other piece (=3a), and then again the next night (=3b), quote: "Only on this second night, he costumed himself differently than he had on the first." (2019-01-24)
Henrik Juel: Thanks for your clarification clarification, Mario
At first I thought that a7 was a cook, but then I interpreted Alfred's stipulation more broadly (2019-01-24)
Alfred Pfeiffer: Forderung ungenau. Exakt wäre: Ersetze einen anderen Stein als in der vorangegangenen Aufgabe 3a (P0003053) durch dem wK, so daß die Stellung legal ist. (2010-07-27, hidden) more ...
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Henrik Juel: Solution
Black captured [Lc1] on c1 and played e7xd6xc5xb4xa3-a2xb1=L, capturing all six missing white men, so [Pf7,g7,h7] never captured, and Yg4 is black
Last move was not Te1-d1, so it was 0-0-0, and Y is not L
To let [Th1] out to be captured, [Pg2,h2] crosscaptured, so Lh2 is [Pg7] promoted on g1, and Y is not P
The forward sequence was: g2xh3 before [Th1] got out before g1=L before h2xg3, so there were not any more black promotions, hence the hidden black man is Sg4, and [Pf7,h7] were captured on their files
The white captures were a2xLb3, g2xPh3, and h2xLg3, leaving only [Pf7] who was captured on f2, f3, .. , or f7
[Pe2] must have promoted, either on e8, or on f8 via f7, so Ke8 has moved, and Black may not castle (2019-01-23)
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hans: + wTh7
bCap: [Sb1], [Lc1], [Sg1], dxTe6, exTf6, (dxSe6 and exSf6 is illegal due the stipulation)
wCap: [Ta8], hxLg3/5, gxLh7-h8=T (2013-02-19)
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Henrik Juel: Too many conditions for my taste
Alfred's addition shows that Ke8 (and Th8) never moved
The only white pawn capture was f2xe3, i.e., f2xTe3, so [Pb7,c7] cross-captured to let [Ta8] out
At first, only [Lc1] could get out, so the first black pawn capture was c7xLb6
But [Lc8] was in the way for [Ta8], so it was captured on c8, and La6 is a pawn promoted on f1
g2-g3 happened early to let promoted Lf1 reach a6, so Sf1 is also a promoted pawn
Hence Black captured gxf, and since the white rooks fell on the same row, they were captured on c6 an f6 (2019-01-24)
Henrik Juel: Alfred's addition was
Schwarz kann noch rochieren. (2019-01-24)
Alfred Pfeiffer: Forderung unvollständig, es fehlt: S kann noch rochieren! (2010-07-27, hidden) more ...
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Henrik Juel: First some observations
Black has captured b7xa6 and played g3xh2-h1=L (other ways to promote [Pg7] on a light square need to many captures)
Assume that Ta5 is white
Then the black captures were b7xYa6 (Y is [Pe2] promoted) and g3xLh2
La2 cannot be promoted, because White must play b2-b3 before [Lc1] can reach h2 to enable the black promotion
Lb7 also cannot be promoted, because Black must play c7-c6 before [Dd8] can get out to enable [Pe2] to promote on g8
So Ta5 is black
(Then one of the two needed black pawn captures could be b7xTa6 or g3xTh2) (2019-01-25)
Henrik Juel: For a slightly clearer argument, please replace
'Lb7 also cannot be promoted'
with
La2 also cannot be the original [Lc8] (2019-01-25)
Mario Richter: My first attempt was: +wTa5, then R: 1. ... Lc8-b7 2. ~ b7xSa6 3. Sb8-a6 ~ 4. c7xTb8=S ~ 5. d6xDc7 ~ 6. e5xLd6, before I realized that afterwards it is not possible to unpromote sLa2 ... (2019-01-26)
Alfred Pfeiffer: For the sLL both is possible, each of them can be the original or the promoted piece. Beside the above discussed solution White could play first wTa1 to a6. After bxa6 Black could move its Ta8 to a5 and play its Lc8 to a2. Then White plays b2-b3 and the wLc1 to h2. After the promotion of sBg7 at h1 the promoted bishop could move to b7. Then Black moves c7-c6. In this case there is no need to promote the wBe2. It could be captured somewhere, the sDd8 also. (2019-01-26)
Alfred Pfeiffer: Forderungstext unvollständig. Zu ergänzen ist: Schwarz hat soeben rochiert, und es sind keine w UW-Figuren auf dem Brett. (2010-07-27, hidden)
Henrik Juel: Both positions seem illegal
The number of missing men is 5 (2+3 or 3+2)
La2 and Lb7 imply that [Pg7] promoted on a light square
Black captured b7xa6, so Pg3,h3 must have cross-captured to allow g2xf1/h1=L (otherwise a total of 4 black captures would be needed)
Black also captured [Lc1] on a dark square, so all 5 captures are accounted for
But then the absence of [Pe2] cannot be explained (2019-01-24, hidden)
Alfred Pfeiffer: Hints:
1) The promotion of [bPg7] to a bB could be done via g3xh2 (no cross-capture at g3/h3)
2) The [Pe2] promoted at g8 and then it was captured on a6. (2019-01-24, hidden)
Henrik Juel: Thanks, Alfred
Your first hint clearly shows the error of my ways
Now I can only hope that some kind soul will hide these three comments; otherwise I shall probably be barred from commenting, because of gross incompetence... (2019-01-25, hidden) more ...
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hans: sTe5 is Damesturm, for Königsturm is captured on f8 for promotion of the e-pawn. This is needed, for white has only 1 Springer for the other capture. (2013-02-20)
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Henrik Juel: .
If [Dd8] was captured by d2xDc3, Dh5 is white
If not, the capture was d2xTSc3, and Black later played exDd-d1=TS, so Dh5 is black (the sequence was d2xc3 before the promotion before Se3-d1 before e2-e3 before [Lf1] could exit)
Thus, Dh5 is black if the missing D was captured on the d-line, and white otherwise (2019-01-24)
Alfred Pfeiffer: Letzte Satz der Forderung unpräzise. Richtig: Zeige, daß sich die Farbe der Dh5 bestimmen läßt, wenn bekannt ist, ob die fehlende Dame auf ihrer eigenen Linie geschlagen worden ist! (2010-07-27, hidden) more ...
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Henrik Juel: The retroplay was not
-1... Kf6 -2.d5xPe5ep e7 -3.d4
because this would require too many black pawn captures (incl. promotion of Ld8)
So it was -1... Kf6xSg6 -2.Se5
If Pg3 came from f2, too many white pawn captures would be required (incl. promotion to S), so Pg3 came from h2 (2019-01-24)
Henrik Juel: for clarity, please replace the last parenthesis by
(because White must promote to S) (2019-01-24)
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Henrik Juel: Black is missing one man ([Pb7]), so the only white capture was b2xc3, and the obvious explanation of the position is that White captured an original black officer, which was later replaced by promotion
Still, suppose Black did not promote
Then he must have captured a white officer with bxc, and White might promote [Ph2] on h8 after h7xg6 (later Black would have to complete the cross-capture with g7xh6)
In his first pawn capture Black must have captured an original white officer, but we can rule out all five possibilities:
Not S, as a promoted S could not leave h8
Not [Lf1], as it could not be replaced by promotion on h8
Not [Lc1], as it could not get out
Not T, as [Ta1] could not get out, and Th1 never moved
Not D, as Black would have to capture [Lc1] to let her out, and a8=L-h4 would require a total of four black captures
So [Pb7] did promote
(The possibility of White capturing the promoted officer after a2-a3 and b3xTa2-a1=Y will not work, as [Lc1] could not leave the SW corner, and b3xSLa2 are ruled out as above, so the promoted [Pb7] is still on the board) (2019-01-26)
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Henrik Juel: solution: the promoted pawn is Dd7
Both kings have castled
White castled early to let out [Th1], so Black could capture axTb
To avoid a check on b2, Black must capture again bxTa and promote on a1
Let us consider the possible promotions:
S or L could not move without checking
T could not reach d8,e8
So the promotion was to D (2019-01-26)
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Henrik Juel: Too many conditions for my taste
The move sequence must have been
h7xg6 before h6-h7 before g7xh6 before b2xLa3 before b7xTa6 before h7xDg8=D! (with a screen on f8)
(A promoted S would check on f6 and a promoted L or T could not exit)
So the promoted pawn is Dd1
. (2019-01-26)
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Henrik Juel: solution: La2 is original
[Ta8] was captured in its corner, so White could not promote before g7xf6
Hence the sequence must have been
Lf1-a2 before b2-b3 before g7xLf6 before g8=L-g2
. (2019-01-26)
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Henrik Juel: solution: Lb2 is original
Both h-pawns promoted
The sequences must have been
c2-c3 before c7xDd6 before h6xTg7xTh8=L-h2 before g2-g3
and hxTg before h1=Y before b2xYa3
. (2019-01-26)
Henrik Juel: If [Ta8] was captured on a3, [Ph7] need not promote, and White could just capture [Th8,Ph7] with officers
The result is unchanged (2019-01-26)
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Henrik Juel: solution: [Lf1] was captured on f1
-1... f4xg4ep -2.g2 f5
. (2019-01-26)
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Henrik Juel: solution: [Lc1]
The retroplay was -1.Dd8xYe8 Yxe8 (-1... b7xa6/c6xb5 are illegal as they would prevent wK from reaching c8), so Black has captured the other missing white men on a light square
. (2019-01-27)
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Henrik Juel: Seems cooked
As in P0003069 the retroplay shows that Black captured on a6,b5,c6,e8
The captured men must be wTTLS, as no wP could reach any of these squares (White also captured Dd8xe8)
But any one of [Pd2,e2,h2] could have been given as odds (one of Pd7,e6 could be [Ph2]) (2019-01-27)
Mario Richter: Smullyan wrongly argues: "... four White pieces must have been captured (one on e8, and three by the pawns on a6 and b5). The pawn from h2 could certainly not have been captured by either of the two Black pawns, nor could it get to e8 without promoting, which we are given it did not. Hence this pawn must have been given as odds."
As Henrik explained, Smullyan overlooked the simple possibility that wPh2 is still on the board (on d7 or e6).
So the problem is indeed cooked. (2019-01-28)
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Henrik Juel: .
[Pb7] did not promote, because otherwise it would be on the board
The capture sequence must have been bxTa before b2xPa3
So [Sg1] must have moved to let a wT out
. (2019-01-27)
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Henrik Juel: solution: Sg1
The retroplay was -1.Pc5xPd5ep d7 -2.c4 Ke8, so [Lc8] was captured on c8
The other white captures were b2xc3, e2xd3xc4, fxg, while Black played f7xTg6xPh5, axLb-b1=Y, so Sb1 must have moved
(The problem could have formulated without a condition by asking
Which wS must have moved?)
. (2019-01-27)
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Henrik Juel: solution: Sa1 is white
Last move was Kd1xSd2, so the black knights are accounted for
. (2019-01-27)
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Henrik Juel: solution: Sa1 and Sa3
White pawns captured all missing black men, so last move was Sc2-a3+, preceded by b2xa1=S
. (2019-01-27)
Alfred Pfeiffer: Forderung genauer: Ein weißer und ein schwarzer Springer haben die Farbe getauscht. Welche Springer? (2010-07-27, hidden) more ...
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Henrik Juel: (With sKe8) replacing sSa6 with wTa6 allows the retroplay -1.c5xPb5ep b7 -2.Tc6. If sK had not moved, wK could not have entered the top rows, so Black may not castle. (2010-07-28)
Alfred Pfeiffer: Diagrammfehler: Korrekt ist sKe8 (statt sKd8) (2010-07-27, hidden) more ...
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Henrik Juel: solution: Lf1
The pawns are true by stipulation (Figur means officer) and the kings are obviously true
[Lc1] was captured on c1, so Lh4 is true and checking Ke1
Hence Sf3 and Dd8,Th8,Sc7,Sf6 are true
Ta8 is true, because a wT could not have left row 1
Dc4 is true, because a black promotion would require four captures, but only wDS and one wT are available
So Lf1 is black, explainable by Black capturing [Lf1] and playing c5xTd4xTe3xSf2=L
. (2019-01-27)
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Henrik Juel: .
White captured [Ta8,Lc8,Dd8] with an S in the NW corner, g2xTh3, and [Lf8] somewhere
So Pb3,c3 did not cross-capture, [Ta1] was captured in the SW corner, and Pf5 never captured
Sg8 entered via h6, and last move was f6-f5 (not Sc2-a1 or Sd3-c1), allowing these (forward) sequences:
L-g7 before f7-f6
White captured [Lf8] on f8 before Th8-f8 before Sh6-g8 before g7xDh6 before g2xTh3
So [Th8] went Th8-f8-f7-g7-h3, showing that Lg7 visited h8 or f8 to make way (and that last move was not f7-f5)
(The penultimate move was g2xTh3 or L-e2) (2019-01-29)
Henrik Juel: It may be clearer to reverse the two sequences and rewrite:
g7xDh6 before L-g7 (or L-h8) before f7-f6 before Tf8-f7 (2019-01-29)
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Henrik Juel: Dd8 is [Pg7] promoted on g1
White captured L or S on b3 and some original black officer (not S) on c3, before [Ta1] could get out to help [Pg7] promote on g1
Promoting to LT does not work, so after d2xDc3 Black played gxT...xLg2-g1=D (with a screen on f1)
. (2019-01-29)
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Henrik Juel: Ta2
The only feasible sequence is
a2xLb3 before Ta1-f6 before g7xTf6 before Th8-a2 before b2xLa3 before c7xLb6 before Ta8-h3
. (2019-01-29)
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Henrik Juel: Silly problem
Retroanalysis requires legality throughout
Silly answer:
[Ta1] illegally reached e6 early, allowing d7xTe6 and then e.g. d2xDc3
. (2019-01-29)
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Henrik Juel: [Lc1] was captured on c1
The sequence must have been
d7xDc6 before Lc8-a2 before b2-b3
. (2019-01-29)
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Henrik Juel: [Lc1] was captured on b2
[Lc8] could not get out before d7xDc6, so La2 must be promoted: a3xLb2-b1=L
. (2019-01-29)
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Henrik Juel: Add wLh8, [Lf8] was captured on h6
[Lc1] was captured on c1, so the added wL must be [Pg2] playing g5xLh6-h8=L
. (2019-01-29)
Hans-Jürgen Manthey: R: 1. Td7 h8=L 2. The8 h7 3. O-O-O gxLh6 4. Df6 Lg2 5. Sg6 g5 6. Lh6 b3 7. La2 g4 8. Le6 g3 9. dxDc6 Dc6 10. Se7 De6+ 11. Sa6 Dc4 12. Lg5 Da6 13. h5 Dc6 14. Le7 Da4 15. e5 Da2 16. Sb4 Db1 17. Sa2 a5 18. SxLc1 Sf3 19. Sb3 Sc3 20. Sa5 a4 21. Sc6 a3 (2021-07-27)
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Henrik Juel: Yes
The sequence must have been
c2-c3 before e7xDf6 before Sb3xLc1 before Ta1-c1 before Sa3-b1 before b2xLa3 before b7xTa6
Only then could [Lc8] get out (it was not captured at home, because [Lc1] was), but White cannot have captured it, so it is still on the board somewhere
. (2019-01-29)
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Henrik Juel: .
The sequence must have been
b2xLc3 before c7xDb6 before g2xDh3 before g1=L
and f2-f3 before Lg1-g7
A wS could not reach g1 after Lg1 had left, so Sg1 must be removed
. (2019-01-29)
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Henrik Juel: On b6
The sequence must have been
b7xSc6 before g2xTh3 or g3xTh4 before [Th1] could get out
So it was captured on b6
. (2019-01-29)
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Henrik Juel: Lf1
No S was captured by a pawn
The pawn captures were f2xLe3, b7xTc6, g2xTf3, in this order, so Lf1 must be removed
. (2019-01-29)
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Henrik Juel: .
Seems to be unsolvable, as no white man could be captured on c6 to release [Ta8,Lc8]
. (2019-01-29)
Mario Richter: Wrong stipulation. The correct one is:
""This time, neither king has moved, and neither queen has been under attack. Where is the genie [the piece to be removed to make the position legal] now?" (2019-01-30)
A.Buchanan: I suppose sometime we should put English & French translations of these wretchedly long Smullyanesque stipulations. It's the kind of thing that novice visitors might come for. (2019-01-30)
Henrik Juel: With Mario's condition modification:
Remove Lc8, so that Dd8 could step aside to c8, allowing b7xSc6 (2019-01-30)
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Henrik Juel: Lh4
Suppose that Black may castle
Then [Ta8] was captured in its corner, leaving a light-squared L and a h-pawn as missing men for White and for Black
d2xc3 and e7xf6 were captures on dark squares, so both missing pawns must have promoted
White could not capture h6xg7 on a dark square, so first Black captured h3xLg2 allowing White to play h2-h7xLg8 later
But then we cannot explain Lh2
So remove Lh4 and play 1.Da6 thr. 2.Dc8#; Black cannot defend, because he may not castle
(The position without Lh4 is clearly legal, because Black need not promote)
. (2019-01-29)
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