Henrik Juel: After a couple of hours Natch 3.1 had not even found the symmetrical solution (2018-12-7)
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Mario Richter: So wie abgebildet, sollte die Problemno. nicht '5467v' sein.
Die Verbesserung (also das 'v') scheint erst später erschienen zu sein.
Das Diagramm ist kurzlösig durch: R: 1. Tg1-f1 (dr. 2. Tf1-f3, dann 1. Th1#) Tf2-g2 2. Tf1-g1, dann 1. Tf3xf2#. (2011-8-18)
Alfred Pfeiffer: Dies ist schon eine Veränderung des Originals. Wo die (in der LB bereits erwähnte) "Korrekturfassung" erschien, ist unklar. (2018-1-16)
S N Ravi Shankar: Given Version of the problem is incorrect. bRg2 should be on f2 and a bP should be placed on g2. Stipulation is -12 and #1.
Solution is
-1. Rg1! Rb2 (say) -2. Rf2 Re2! -3.Re1! Rb2 (say) -4. Re2! Ra2!(-4 ... Rd2? -5. Rg1! as in text)
-5. Ra1! Rb2 -6. Rb1 Rc2 -7. Rc1 Rd2 -8. Rg1! Rc2 -9. Rd2! Rb2 -10. Rc2 Ra1 -11. Rb2 g3 -12. Bg2
and now 1. Rh1# (2019-1-2)
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Henrik Juel: C+ by Natch 3.1 (16 min.)
The colored position has black Ta1,h1 and Pb5,g6, and white Ta7,h8, otherwise 'natural' (2018-12-7)
François Labelle: Henrik, did you really test all 2^30 possible colorings with Natch, or did you use human deductions to reduce the possibilities (making this HC+), or did you only test the intended coloring (making this a partial C+)? (2018-12-9)
Mario Richter: I do not think that it is necessary to check all 2^30 colorings, since the color of the pawns on files c,d,e and f is completely determined. This and more restrictions on the set of potentially possible colorings may be derived from insights presented in an article "Aggregierte Schlagbilanz" by Frolkin & Kornilov, feenschach 130, p.411ff. Since the mechanisms presented there can at least partially be implemented in a little computer program, even a "full C+" label for this problem is not out of reach ... (2018-12-10)
Henrik Juel: I only tested the intended coloring, Francois,
so the C+ label is not justifiable
In the other coloring proof games I write something like
The colored position was C+ by Natch

Mario is right, of course, in that not all colorings need testing; but still the number is very large
This genre is somewhat messy; at first I thought that the solver could determine the coloring, but this is clearly not the case; also, the intention is not
'color the men such that a correct proof game in 19 results' (2018-12-11)
A.Buchanan: Henrik: so what exactly should the stipulation be for clarity? (2018-12-12)
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Henrik Juel: C+ by Natch 3.1
The colored position is 'natural' except for wTa8, wDd8, and sDd1 (2018-12-7)
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s. P0000133
NN: It seems we fall short of one retro-move and fail to keep either the bK or bR still.
Let's analyse, the distribution of bPs indicate 5 captures (bPh7×g6×f5×e4×d3×c2) plus 3 captures (bPc7×d6, bPb7×c6, bg7×f6) equalling a total of 8 captures, thus accounting for all of white's missing pieces. Next, wPs captured at least 3 (fxexdxc) plus wPc2×b3. Now, in order to keep the bK & bR still, we require sufficient supply of retro-moves for black. Thus, we must strive to release the bNa1 (as pawns are not enough!) at the earliest. But this can only be done through uncapturing the bishop on c1 followed by retracting bPd3×c2, wPd2-d3. An optimal way of doing this may be: -1.bPb7×wNc6 wNc6-d4, -2.bPa4-a3 wPc6-c7, -3.bPc7×wBd6 wBd6-b4, -4.bPa5-a4 wBb4-d2 -5.bPe6-e5 wBd2-c1 and here we can't retract anymore moves without violating bK's right to castle. Retracting bPa6-a5 for instance would mean that bRa8 got out of the pawn to be captured by a wP (as bBc8 never left its homesquare) forcing the bKe8 to move.
Thus, Black can't castle and the solution is 1.Kxf6 Rh6+ (1...0-0 is illegal) 2.Nxh6 ... 3.c8R/Q# (2018-11-30)
Mario Richter: If we assume that Black can still Castle, White needs a fifth capture to provide wPh2 as a capture object.
Furthermore, before d2-d3 can be retracted, both wRa1 and wBc1 have to return home (to b1 and c1) (2018-12-1)
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Henrik Juel: Here is one of the three similar cooks
1.Sg1-f3 Sb8-c6 2.Sf3-e5 Sc6-d4 3.Se5xd7 Be7-e5
4.Sd7-b6 Ba7xb6 5.Sb1-c3 Ta8-a3 6.Sc3-b1 Ta3-g3
7.Bh2xg3 Sd4-f3 8.Bg2xf3 Lc8-f5 9.Th1xh7 Lf5xc2
10.Lf1-h3 Lc2xh7 11.Lh3-c8 Bf7-f5 12.Bd2-d3 Ke8-f7
13.Lc1-e3 Kf7-g6 14.Sb1-d2 Kg6-h5 15.Ta1-c1 Bg7-g6
16.Tc1-c4 Lf8-h6 17.Tc4-e4 Dd8-f8 18.Le3-d4 Lh6-e3
19.Lc8-e6 (2018-12-7)
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paul: The stipulation in English: Show that, in the course of the retro game, the possibility for the en passant move existed! Madrasi (2019-1-13)
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paul: If 4. -b5-b4 5.a6xBishop b7? Bc8-b7, then Black will be retro-stalemated. To unlock the position, the retro-play must be: Rf2 - Rg8 - g7-g8=R - Ph6xRg7 - Ph5 -Pg4xSh5(bSg3-h5), Rh1 - Rg8 - g7-g8=R - Ph6xRg7 - Ph5, Sh3-g5, Pg7-g5, Kg6-f7, Pf4-f5+, bBf8 was captured at home. (2019-1-13)
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Henrik Juel: The long promenade is made by k, not K (2018-12-7)
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Henrik Juel: C+ by Natch 3.1 (5 min.) (2018-12-7)
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paul: a) is Jacobi+ (2019-2-23)
Mario Richter: b) is cooked:
1.Nb1-c3 Pb7-b5 2.Nc3xb5 Pc7-c6 3.Nb5xa7 Qd8-c7 4.Na7xc8 Ra8xa2 5.Nc8xe7 Ra2xa1 6.Ne7xc6 Ra1xc1 7.Nc6xb8 Rc1xc2 8.Nb8xd7 Qc7xh2 9.Nd7xf8 Qh2xg1 10.Rh1xh7 Rc2xb2 11.Rh7xh8 Rb2xd2 12.Rh8xg8 Rd2xd1+ 13.Ke1xd1 Qg1xf2 14.Rg8xg7 Qf2xg2 15.Rg7xf7 Ke8xf7 16.Bf1xg2 Kf7xf8 (2009-2-16, hidden)
paul: bK must be at g8. (2015-3-29, hidden)
Henrik Juel: No, Paul, in part b) the black king is on f8 (and the white bishop on g2) (2015-3-29, hidden) more ...
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Brassaud: Je suis d’accord avec la solution proposée jusqu’à 53) Wartezug, d7-d6 et la partie est considérée comme nulle.
La position intermédiaire obtenue après 53) … , d7-d6 est la suivante :
T1ft1R1c/pppp/PPFF/4PtPf/3pppp1/8/1P6/P2pr3/7T
Cette position est-elle légale ? Je ne le crois pas. Les pions noirs ont pris 4 fois. Le pion h2 a donc été promu. Cette promotion peut être encore sur l’échiquier sous forme de T ou de F, ou bien a été prise mais c’est sans réelle importance. Pour dénouer cette position il faut d’abord reprendre d6xCe7 puis c5xDd6 afin de ramener ce pion en c2. A ce moment la tour blanche et les 2 fous de cases noires peuvent sortir mais le FB de g8 et le roi blanc sont encore dans la cage. Le pion de c2 a effectué les 2 seules prises de pièces noires. Les autres PB venus de e2,f2, g2 n’ont pas fait de prise et le pion de h2 a été promu en h8 sans prise. Il faut ensuite reprendre d6-d5 et e7xd6 (pièce quelconque)puis reprendre la promotion du pion en h8. Mais là je ne vois pas comment en raison du cavalier bloqué par les 2 pions blancs de f7 et g6. Le retour d’une pièce promue (dame ou tour ou même le fou de g7) en h8 ne peut se faire, donc on ne peut reprendre le retour de h7-h8 puis h6-h7 et h5-6 ni bien entendu reprendre h6xg5 pour tout dénouer.
Ainsi je pense que le problème en lui-même est illégal avant même de faire intervenir la règle des 50 coups. Pouvez-vous me dire si quelque chose m’aurait échappé ? (2015-9-11)
Thierry LE GLEUHER: Sorry but the position is legal. The retro play can continue with d6-d5 and d5xXe6, and unlock the rest is easy. (2015-9-13)
Brassaud: Merci pour votre aide. (2015-9-17)
Thomas Volet: TV: The composition's only point is the extension of the retroplay by the interaction of the 3 Bs.
I do not accept anyone's "codex" that purports to alter the rules of chess by making the draw automatic or that imports castling. Under the chess rules, a draw is at the option of the moving player whenever the sequence of non-P, non-capturing moves is sufficiently long (and has nothing to do with castling).
If interested in positions that involve the 50-move rule, please see P0008399 (75-move sequence with several rook circuits), P1202286 (where the "codex" automatic draw rule presents the situation in which either player on the move can make a move that is simultaneously a draw and a checkmate), or P1331867 (move preceding the sequence must have been a capture of a non-P by a non-P). (2019-1-2)
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Henrik Juel: Unfortunately cooked
1.Sg1-h3 Sb8-c6 2.Bf2-f4 Sc6-e5 3.Bf4-f5 Bg7-g5
4.Sh3-f4 Bg5-g4 5.Bf5-f6 Bg4-g3 6.Bf6xe7 Bf7-f5
7.Sf4-h5 Bf5-f4 8.Bh2-h3 Bf4-f3 9.Th1-h2 Bg3xh2
10.Ke1-f2 Bh2-h1=T 11.Kf2-g3 Th1xf1 12.Sb1-c3 Tf1-h1
13.Dd1-g1 Se5-g4 14.Dg1xa7 Sg4-h2 15.Da7-b8 Ta8-a3
16.Sc3-a4 Ta3-c3 17.Bb2xc3 Bf3-f2 18.Lc1-a3 Bf2-f1=D
19.La3-b4 Df1-f5 20.Ta1-f1 Df5-e6 21.Tf1-f3 Sg8-f6
22.Tf3-e3 Sh2-f1 23.Kg3-f3 Bc7-c5 24.exd8=T+ (2018-12-7)
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Henrik Juel: C+ by Natch 3.1 (4 min.) (2018-12-8, hidden) more ...
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HBae: R: 1. ... Kh6xBg5[+sBg7], dann 1. Lb5# bzw. 1. Sf7#
R: 1. ... Kd3xBc4[+wBc2], dann 1. Dgd4# bzw. 1. Dbg2# (2019-2-4)
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Henrik Juel: The solution position is illegal, because Sh8-g6 is longer than g7-g6 and h7xg6, so these pawn moves could not happen
The try position is legal with a promoted Sh1 (2019-2-14)
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Henrik Juel: With the additional condition ruling out solutions like +wPa1, the problem now seems correct
Adding wPe2,g2 gives an illegal position, as [Lf1] could not have been captured on f1 in Mars Circe (2019-2-14)
Erich Bartel: Nachdruck:
1) G) Problemkiste (50) VII 1987 (2007-4-10, hidden) more ...
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Basically there are 1,6,6,6 moves, and choices of promotion to D/L in 2 cases. The difficulty is in managing the collisions on f2,f3&g2. The best way is probably consider in turn which pawn has priority to pass through the square. This partitions the solution set into 2^3=8 cases.
A.Buchanan: I see that the term "Pawn Endgame" is now the English translation of "Kindergarten Problem". To me, an endgame is a study. Does "Kindergarten Problem" carry that connotation in German, or does it refer more to general compositions such as this one? (2019-2-2)
Mario Richter: The latter is the case, we even have some famous Kindergarten Problems in the Retro genre (think e.g. of some of Thierry LeGleuher's works). "Kindergarten" refers to the fact that besides the two kings only "children" (aka pawns) are on the board.
So I do not understand the choice of "Pawn Endgame" as the english translation, even less because the english language has a word like "kindergarden" ... (2019-2-2)
A.Buchanan: Rainer Staudte has recently changed it. Out of respect, I don't want to unilaterally change it back but would like to hear his perspective. Apparently even in English, Kindergarten is the more common spelling. (2019-2-2)
A.Buchanan: Rainer Staudte has recently changed it. Out of respect, I don't want to unilaterally change it back but would like to hear his perspective. Apparently even in English, Kindergarten is the more common spelling. While we're here :) what does rawbats make of this problem by Hans Gruber, Mario? (2019-2-2, hidden)
Mario Richter: rawbats' Output:
911797992 Loesungen gefunden
time= 1.535000(s) (2019-2-2, hidden) more ...
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Paulo Roque: Solution:
1. c7 ~ 2. c8=D (or T) # (1. ...0-0 is illegal !)

Demonstration the illegality of the castle:

a) Logical elements in the diagram (LED):
(a.1) The bishops black and white were captured in original squares.
(a.2) sLa7 is pawn-promoted, note square f8. The promotion was in square g1.
(a.3) captured two black pieces (vide a.1, then two bishops ).
(a.4) captured seven white pieces (two rooks,one knight, two bishops, queen and one unknown).
The piece unknown: (uk1) is one pawn, or (uk2) is one pawn-promoted, or (uk3) is one piece original equal the pawn-promoted which is in the diagram.

b) Development:
(b.1) The white pieces were captured in: The pawn-h black was promotion in square g1, is necessary hxgxh2xg1 (three captures) + two bishops white captured in original squares xLc1 xLf1 (two captures) + c7xd6(one capture) + a7xb6(one capture) = seven captures pieces white.
(b.2) The structure in the diagram with sLa7,sBb7,sBb6,sBd7,sBd6 is only possible if the promotion of pawn-h black and -c7xd6- preceding -a7xb6-.
(b.3) Then pawn-a white was promotion in square a8 and the piece unknow (a.4) is uk3, only possibility a8=S.
(b.4) On the way Sa8 for e5, necessarily have to go in square c7,then impossible s0-0.

c) Conclusion: 1. ...0-0 is illegal.

Keywords: Cant Castler (2009-7-17)
Henrik Juel: The illegality of castling may also be shown by looking at the retroplay. La7 is [Ph7] promoted on g1, so all missing white men are accounted for, and [Pa2] must have promoted on a8. We cannot uncapture a white officer fast, so the retroplay must start by unpromoting Se5. But it needs to pass c7, so black king has moved. (2009-8-21)
Paulo Roque: Henrik, your solution is very elegant! (2009-8-21)
Yoav Ben-Zvi: As noted by Henrik above, proof that Black "Cant Castle" is based on showing that the White piece unpromoted on a8 is wNe5 rather than an officer uncaptured by one of the bPs. This is not affected by time pressure. Instead, note that retraction of bPc7xf6 must be preceded by exit of bBa7 from the North-West (via c7) which requires previous retraction of bPa7xb6 which implies previous unpromotion on a8 with the unpromoted wP retreating past a7 to allow its occupation by bP. This shows the unpromotion on a8 occurs before Black can uncapture on b6 or on d6 or by bP unpromoted on g1. It follows that the piece unpromoted on a8 must be wNe5. (2018-12-26)
Yoav Ben-Zvi: Typo bPc7xd6 (not xf6) (2018-12-26)
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Mark Thornton: In January 2011, a fictional Hungarian detective named "Tibor Orban" appeared in the BBC crime drama "Silent Witness". Tibor Orban was played by the actor Ivan Kamaras (http://en.wikipedia.org/wiki/Iv%C3%A1n_Kamar%C3%A1s). (2012-7-21)
Dirk Borst: The word "genau" can and should be removed. "BP in 4,0" historically and logically means: "Position after the 4th black move; how did the gane go?" So a shorter game cannot be a cook, by definition. (2019-4-18)
Henrik Juel: I disagree, Dirk
Usually the composer asks for a shortest proof game; the 'genau' addition is a handy way of emphasizing the few exceptions, like this one
1.e4 e6/c6 2.Lc4/Lb5 c6/e6 3.Lxe6/Lxc6 dxe6/dxc6 are two of the four shortest proof games without loss of tempo (2019-4-18)
Per Olin: The solution is unique also in 4.5 moves: 1.e3 e6 2.Lb5 Ke7 3.Lxd7 c6 4.Le8 Kxe8 5.e4 (2019-4-19)
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Renny Bosch: I believe the minimum number of moves since the last pawn move is 68, based on the following PG (or SPG?): 1. g4 Nc6 2. Bg2 Na5 3. Bc6 Nb3 4. axb3 Nf6 5. Nc3 Ne4 6. Nd5 Nc3 7. dxc3 Rg8 8. Qd4 Rh8 9. Qb6 axb6 10. Be3 Rg8 11. Bc5 Rh8 12. Bb4 Rg8 13. Ra5 Rh8 14. Rc5 bxc5 15. Ba5 Rg8 16. Bb6 Rh8 17. Ba7 Rg8 18. Bb8 Rh8 19. Ba4 Ra6 20. Kd2 Re6 21. Kd3 Re5 22. Kc4 Re4+ 23. Kb5 Rc4 24. bxc4 Rg8 25. Bb3 Rh8 26. Nh3 Rg8 27. Ra1 Rh8 28. Ra7 Rg8 29. Nb6 Rh8 30. Na8 Rg8 31. Nf4 Rh8 32. Nd5 Rg8 33. Ndb6 h6 34. Na4 b6 35. Ba2 Bb7 36. Bb1 Be4 37. g5 Bh7 38. g6 Rh8 39. gxh7 { the last pawn move } g6 40. Rb7 Bg7 { move bB, bK, bQ out of row 8 } 41. Ba7 Be5 42. Rb8 Kf8 43. Ka6 Kg7 44. Kb7 Qg8 45. Rf8 Kf6 46. Kc8 { allow bR to pass wK } Qg7 47. Rg8 Ke6 48. Kd8 Qf6 49. Ke8 Qf5 50. Kf8 Bf4 51. Kg7 Qg5 { reverse order of the two rooks }52. Rb8 Rc8 53. Rb7 Rd8 54. Bb8 Rc8 55. Ra7 Rd8 56. Ra5 Rc8 57. Ba7 Rb8 58. Rb5 Rb7 59. Bb8 Ra7 60. Rb3 Ra5 61. Ra3 Rb5 62. Ra1 Rb3 63. Ba2 Ra3 64. Bb3 Ra2 65. Rd1 Ra1 66. Rd2 Re1 67. Rd1 Rf1 68. Ra1 Rb1 69. Ra3 Rc1 70. Ba2 Ra1 71. Bb1 Ra2 72. Rb3 Ra3 73. Rb5 Rb3 74. Ra5 Ra3 75. Ra7 Rb3 76. Rb7 Ra3 77. Ba7 Rb3 78. Rb8 Ra3 79. Rh8 Qf6+ 80. Kf8 { move bK past wR } Qg7+ 81. Ke8 Kf6 82. Kd8 Be5 83. Kc8 Qg8+ 84. Kb7 Kg7 85. Ka6 Kf8 86. Kb5 Ke8 87. f3 Kd8 88. Ba2 Kc8 89. Bb1 Kb7 90. Ba2 Qg7 91. Rg8 Qf6 92. Bb1 Qe6 93. Ba2 Bh8 94. Rg7 { put wR into pocket, and final cleanup } Kc8 95. Ka6 Kd8 96. Kb7 Ke8 97. Kc8 Kf8 98. Kd8 Rb3 99. Bb1 Rb5 100. Ba2 Ra5 101. Bb8 Ra7 102. Bb1 Rb7 103. Ba7 Qc6 104. Ba2 Qb5 105. Bb1 Qb4 106. Ba2 Qa3 107. Rg8# (2007-4-24)
Michel Caillaud: the exact stipulation ends "mindestens geschehen".
(in english : "minimum number of moves since the last Pawn move?").
the last Pawn move is f2-f3,that must be interposed before g7-g6 as otherwise the 50-moves rule would have been violated.
the answer is a minimum of 15,0 moves with a retro-play running as :
-1.Tg7-g8# ...-16.f2-f3! ... -65...g7-g6
with 49,5 moves between g7-g6 and f2-f3 the 50-moves rule has been respected (2007-9-12)
Michel Caillaud: more detail on intended retro-play :
-1.Tg7-g8# … -4…De6-c6 … -8…Tb3-b5 -9.Kb5-a6 Ta3-b3+ -10.La7-b8 … -13…Kb7-c8 -14.Tg8-g7 Ld4-h8
-15.Tf8-g8 Df6-e6 -16.f2-f3! Dg7-f6 -17.Th8-f8 Dg8-g7 -18.Lb1-a2 Ta1-a3 -19.La2-b1 … -24… Kf6-g7 -25.Kb7-a6 Dg7-g8 -26.Tb8-h8 Ke5-f6 -27.Kc8-b7 … -28… Dh4-f6 -29.Ke8-d8 … -35… Td2-d1 -36.Ta3-b3 … -37.Lb3-a2 … -39.Th1-a1 Td1-d2 -40… Ta1-d1 -41.Kg7-f8 Ta3-a1 -42.La2-b3 …-48… Th8-b8 -52.Tb7-a7 … -53.La7-b8 … -55.Tg8-b8 … -59.Kç8-d8 Kf6-e5 -60.Kb7-ç8 Kg7-f8 -61.Tb8-g8+ Kf8-g7 -62.Kc8-b7+ Ke8-f8 -63.Tb7-b8 Lg7-d4 -64.Lb8-a7 Lf8-g7 -65.Ta7-b7 g7-g6 (2007-9-18)
Henrik Juel: After -1.Tg7-g8# the main parts of the resolution are
(s/w mean black/white)
sDa3 out, sTb7 to a3, wKe8 to b5, sKf8 to b7, wTg7 to h8 and sLh8 to d4, sD to g8, sKb7 via g7 to e5, sDg8 out, wKb5 to g7, sTa3 to d2, wTh8 to e1, sTd2 to h8, wTe1 to g8,
(the rooks have switched place)
wKg7 to b7, wTg8 to b8, wKb7 to c8, sKe5 to e8, sLd4 to f8,
-65... g7-g6 -66.g6xLh7
(so -16.f2 was needed to avoid draw by the 50 moves rule)
wTb8 to h1, wKc8 to b5, sD to d8, sLh7 to c8,
-88... b7-b6
wSa8 out, wLa7 to a5,
-99... b6xTc5
wTc5 to a1, wLa5 to c1,
-111.d2xTc3
sTc3 to a8 and wSa4 to b1, wLa2 to a4
-222.b3xSc4 a7xDb6 -223.a2xSb3 (2018-12-5)
Henrik Juel: It is difficult to avoid typos in such a mess
in line 3, wKd8 to b5, not e8
Michel also seems to have made some typos:
49.0 moves between -16.f2 and -65... g7, not 49.5
-25... Kf6, not 24 (may have implications for the solution) (2018-12-5)
Renny Bosch: Correction: The last pawn move is of course not 39. gxh7, but 39 ... g6. Therefore the answer is 67.5 moves since the last pawn move. (2007-9-4, hidden) more ...
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siehe P1291160
Sally: Der en passant-Schlag als einziger Schlüselzug lässt sich im Zweizüger erst mit acht Steinen darstellen. Dies ist die früheste Aufgabe und beste Darstellung, die mit Zweispänner
möglich ist. (2019-1-18)
Henrik Juel: C+ by Popeye 4.61 (2016-6-13, hidden) more ...
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NN: The above is basically two parts of a single solution depending on the history of the game and not two different solutions.

Proof of why castling rights are mutually exclusive: Let's assume black CAN castle. Now, the bPs piled up on h-file implies six captures of white units and these can only be wRBBNN & wPe2. Because all other possibilities involve promotion of wPs on d8/e8/f8 followed by their capture by bPs, interfering with black's castling right. Next, we observe that wRh1 getting captured by bPs implies one of the following two things: (i) wKe1 moved to let wRh1 out of the pawn-chain or (ii) wPg2,h2 cross-captured on h3,g3 respectively to facilitate the escape of wRh1. Furthermore, option (ii) necessitates bPd7 to be promoted to d1 so that it can be captured by wPs, implying wKe1 had to move anyway to let bPd7 promote.
Thus, Black can castle implies wKe1 moved and white can't castle. Contrapositively, white can castle will imply that black can't. (2018-11-29)
Henrik Juel: Nice analysis, Satanick (2018-11-29)
NN: Thanks Henrik Juel. Hope there are no mistakes. (2018-11-29)
Yoav Ben-Zvi: The claim that retaining Black castling requires an uncapture of a wR by Black's h-file pawns is not accurate. Consider the case that all 6 of these bP uncaptures are of wBs and wNs. In this case 2 of the uncaptured White pieces unpromote. Both could be unpromoted (on f8 or d8) by an uncapture from e7. This scenario requires a third wP uncapture (on the e file) so one of White's uncaptures is accounted for by [bPd7]. The direct uncapture of [bPd7] (on the e file) or its unpromotion without stepping on d2 requires this pawn to uncapture the wR, confirming the conclusion of mutually exclusive castling. A similar argument works for the case that the Black h file pawns play bPe7xwPf6 along with 5 uncaptures of WNs and wBs. (2018-12-3)
Henrik Juel: Now we can test
C+ by Popeye 4.61 (and analysis)
The main content is the retro analysis, of course, but as a helpmate problem the repeated 1.Td7 detracts (2018-12-3)
NN: Solutions: If white can castle: 1.Rd7 0-0-0 2.Kd8 Qxd7# & if black can castle: 1.Rd7 Rxa7 2.0-0-0 Ra8# (2018-11-29, hidden) more ...
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NN: Solution: If Black can castle long: 1.Qxc7! (Threat: Qxe7#) 1...N moves (1...Kf8 2.Qxe7+ Kg8 3.Qxg7#) 2.Qd7+ Kf8 3.Qf7# & if Black can castle short: 1.Qxg7! (Threat: Qxe7#) 1...N moves (1...Kd8 2.Qxh8+ Ng8 3.Qxg8#) 2.Qf7+ Kd8 3.Qd7#

The bBe1 is obviously a promotee. This implies bPd7,e7 made one capture each namely, dxe & fxe. Consequently, either the missing wPg2 or wPb2 (but not both!) had to be captured by bP on the e-file. But neither wPg2 nor wPb2 could have gone to the e-file capturing white units because that, together with wPe2×d3, make three captures in all and black is missing only two units. Hence, either one of them promoted to h8/a8 first prior to be captured on the e-file. This shows exactly one of bRh8 and bRa8 has moved before facilitating the promotion and therefore, black has right to castle only on one side. Accordingly,1.Qxc7 is the key if black has the right to castle long and 1.Qxg7 is the key if otherwise. (2018-11-29)
NN: Castling for black is legal only on one side. The above solution cover the two mutually exclusive cases. (2018-11-29, hidden) more ...
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Henrik Juel: Solution: Yes, the bK may never have been checked. -1.Bc4 b6 -2.Bb5 Bb7 -3.Ba6 Bc8 -4.Bb7 Rh2 -5.Ba8 Rh1 -6.B=a7 Rh2 -7.a6 a7xR -8.Rb3 Rh1 -9.Rg3 Bb7 -10.Bh2 Rb1 -11.Bg1 Rb3 -12.Re3 Bc8 -13.Re1 Rg3 -14.Bh2 Bb7 -15.Rh1 Bc8 -16.Bg1 Rb3 -17.Rh2 Rb8 -18.Rh1 b7xB -19.Bb5 Ra8 -20.Bf1 Rb8 -21.e2 Ke4 etc. Two screens on g3 help replacing bR by wR in the SE corner. (2003-5-15)
hans: NO,you can't retrack bK without a check. f7-f6 to let bK out makes Bf8 useless! (2010-5-18)
Henrik Juel: I only see your comment now, Hans, and I agree
Early in the game Sd6+ e7xd6 happened (2019-1-7)
Henrik Juel: The real content of this excellent retro is how to resolve the position, but the author chose to frame it with a question that fooled me some 16 years ago
Here is another tricky question he might have asked:
For which black pawn can the exact play not be deduced? (2019-1-8)
Henrik Juel: The black pawns played
a7xTb6-b5, b7xLc6, e7xSd6, and the rest never moved, except for [Pf7], which played f7-f6 or, if [Pb2] had already captured TLS to reach e5, ready for an ep capture, f7-f5 (2019-1-9)
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Henrik Juel: C+ by Natch 3.1 (5 min.) (2018-12-6, hidden) more ...
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klären: Quelle: Schach ohne Grenzen, S.35, Nr.3 = Magyar Sakkvilag, 1926 (VF)
Henrik Juel: -1.Q=c7 Bb4:S -2.Sc5, 1.Sb3#; -1.. Bc5:S -2.Sb4, 1.4Sc2#. It is illegal to supplement a checking bS on a6. (2003-10-9)
Sally: Weiß nimmt c7-c8D zurück. Nimmt Schwarz Lb4-a3 zurück, so ergänzt Weiß auf a3 einen Springer und nimmt Sc5-a6 zurück, um durch Sb3#. Nimmt Schwarz Lc5-a3 zurück, so ergänzt Weiß auf a3 wieder einen Springer und nimmt Sb4-a6 zurück, um durch Sc2 mattzusetzen. (2012-2-21)
Henrik Juel: Also 'On Retraction Chess Problems' gives Magyar Sakkvilag as original source (2019-2-10)
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paul: 1. – Ke7xBf8 2.Bg7-f8 Sd6xPf7 3.Bh8-g7 Kd8xRe7 4.Bg7-h8 Kd7xQd8 5.Bf8-g7 Kc6xPd7 & 1.Kxb5(Ke8)#

But cooked:

1. - Ke8-f8 2.Ka4-b5 Sd8-f7 3.Ka3-a4 Sb7xQd8 4.Qd1-d8 Kf8-e8 5.Qh1xQd1 Ke8-f8 & 1.Qb3#

(information obtained from Kjell Widlert) (2018-12-15)
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Henrik Juel: C+ by Natch 3.1 (2 min.)
Tempo loss at 13.f2-f3-f4 and 18.Lf1-h3-g2-h1 (2018-12-6)
Joost de Heer: IMO f2-f3-f4 isn't a tempo loss, as the pawn is needed for shielding. (2018-12-7)
Henrik Juel: I see no shielding by [Pf2], Joost
White could play 13.f4?, but he would run out of moves
White plays 17.gxf3, and this wP will later shield, so wLg2 does not check sKb7, but that is a different story (2018-12-7)
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Henrik Juel: No solution in 15.0 or 14.5, so C++... (2018-12-8)
Henrik Juel: C+ by Natch 3.1 (28 min.) (2018-12-8, hidden) more ...
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Henrik Juel: C+ by Natch 3.1
No solution in 18.5 or 17.5, so C++ (2018-12-8)
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Henrik Juel: Natch 3.1 tests the first 20.0 moves OK in 13 seconds, but the first 20.5 moves would take more than an hour, and the entire problem more than a couple of hours, because of the possibility of [Pd7] promoting
So we may never know for sure whether this proof game with 16 oscillations by Ta8 is correct (2018-12-8)
Mario Richter: Henrik, I think certainty about the correctness of this problem can be achieved, if we take into account that Black's last move must have been a capture. (2018-12-10)
A.Buchanan: Henrik: the times you propose for testing don't seem outrageous, even without Mario's sensible heuristic. Also, the last move must have been DxLd7 from one of 5 squares only. (2018-12-12)
paul: 18.Tca3 & 19.T3a4 (2010-4-30, hidden) more ...
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Henrik Juel: The content is concealed pawn captures
Despite no visible captures White captured a2xPb3xTa4 to open the a-line for [Ta1,Pa7], and h2xTg3xLh4 with black h7xLg6xDh5 to open the h-line for [Th1,Th8]
This makes life difficult for the human solver and for Natch 3.1, who tested the first 15.0 moves correct in 40 sec.
Testing the entire proof game would need more than a couple of hours by Natch (2018-12-8)
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paul: C+ (Euclide 1.11) for the first 13 moves (in about 51 hours). I do not dare to check for 13.5 moves. (2018-3-8)
Henrik Juel: Tiny typo in Mario's cook: 10.Txh7 (2018-3-8)
Mario Richter: The typo comes from a "by hand conversion" of the output of several tools I'm using (here: rawbats and chessbase) - unfortunately each tool has it's own output format, incl. the PDB solution field format ... (2018-3-9)
A.Buchanan: I have a janitor's tool which allows me to change the representation of positions & solutions.
Try to copy the file here: https://www.dropbox.com/sh/ld64s00zm0f9tx8/AAA7r8BsA0vuLcZOOKNvKkmua?dl=0
Click on it, and then hit the download button on the right.
The most useful is the one which converts from FEN to PDB "pieces" format.
Stick the input string in the yellow cell, and pick up the result from the green cell.
If you want to see how it's done, unhide the hidden rows and columns: no VBA just simple Excel.
Only 1% as clever as rawbats, but it still saves me a lot of time.
Feel free to use and to improve. (2018-3-9)
Mario Richter: @Andrew: Thanks for your offer - but I have such tools myself. My fault was not to use them (I thought that I'll be able to transform such a short PG by hand without errors ...)
Would be nice if the PDB-System would be more helpful with respect to language support.
I've decided to use 'english' in the language settings, but that doesn't help here, since solutions still have to be given in german piece notation.
Might be difficult to change this, especially when thinking of fairy piece names, so I can live with that, but there are other aspect which sometimes drive me crazy, e.g. the handling of source names in different languages ... (2018-3-9)
A.Buchanan: I thought you might have such a tool, but I thought it was a nice opportunity to announce it for others who might find it useful. (2018-3-9)
Mario Richter: The sequence of moves is not unique, e.g.:
1. f3 e5 2. Kf2 e4 3. De1 e3+ 4. Kxe3 b5 5. Dg3 b4 6. Kf2 Lc5+ 7. Ke1 Lxg1 8. Kd1 Lxh2 9. Txh2 d6 10. Lxh7 Le6 11. Txg7 Th1 12. Th7 Lxa2 13. Th8 Lxb1 14. Txa7 Ke7 15. Tb7 Ta1 16. Ta7 De8 17. Ta8 Kd8 18. De1 (2018-3-8, hidden)
Henrik Juel: this problem should be labelled cooked (2018-12-8, hidden) more ...
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Version A. Frolkin
Henrik Juel: Cooked by
1.Ba2-a4 Bd7-d5 2.Ta1-a3 Bd5-d4 3.Bh2-h4 Lc8-h3
4.Ta3-f3 Be7-e6 5.Tf3-f6 Lf8-d6 6.Sb1-a3 Ld6-h2
7.Bf2-f4 Bd4-d3 8.Ke1-f2 Dd8-d4 9.Kf2-f3 Dd4xg1
10.Dd1-e1 Dg1xf1 11.De1-f2 Df1xc1 12.Df2xa7 Dc1xb2
13.Kf3-f2 Db2-e5 14.Da7-e3 Sb8-d7 15.Th1-b1 O-O-O
16.Kf2-e1 Kc8-b8 17.Tb1xb7 Kb8-a8 18.Sa3-b1 De5xe3
with lots of variations (2018-12-6)
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Henrik Juel: C+ by Natch 3.1 (59 min.)
The content is given by the diagram position: Get wKe1 to h8 and interchange the black royals
Lf8 makes the roundtrip f8-g7-f6!-e5-f4-h6-f8
[Dd8] goes d8-e8-f7-d5-e6!-f7-e8
Resentful of the dissipation by some of his officers [Ke8] wastes no time going e8-f7-e6-d6-e6-f7-e8-d8 (2018-12-8)
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Henrik Juel: The capture was fxDe
If White made it, Dd1 is white and so is Ke1, but he would stand in an illegal check from sPf2
So Dd1 and Ke1 are black (and Ke8 is white), and the rest of the men have a 'natural' color because of the shortish proof game, which is
C+ by Natch 3.1 (2 sec.) (2018-12-8)
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Henrik Juel: The fastest captures are d2xLe3 and d7xLe6, so Lb5/Ld6 are black/white respectively
If Ka4 were white, so that Ka1,Tb1 would be black, more than 19 black moves would be needed
So Ka1, Tb1 (and Tb6) are white, needing white castling, which entails that Dd8 is white, because this saves a white move; then Dd1 and Ta6 are black, and the rest are 'naturally' colored, with Sb4/Se8 white/black
The colored position is C+ by Natch 3.1 (0.3 sec.)
The position after switching the colors of Sb4/Se8 cannot be reached in 19.5 moves (2018-12-8)
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Henrik Juel: The decoding is easy, with letters ABCDEF representing the men DKPLST; the coloring is natural, with Ld7 being white
The shortest proof game lasts 9.5 moves and is
C+ by Natch 3.1 (2018-12-8)
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Henrik Juel: C+ by Natch 3.1 (1 min.) (2018-12-8, hidden) more ...
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Henrik Juel: C+ by Natch 3.1 (3 min.) (2018-12-8, hidden) more ...
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Henrik Juel: C+ by Natch 3.1 (9 hours)
The two Belfort officers are wSg8 and sLc1
Here is the added feature that [Lf8] goes to c1 (to capture wLc1), then back to f8 (to make room for [Dd1] and [Ke1] to interchange, while clearing the way for [Ke8]'s march to a1), and then to c1 a second time (2018-12-9)
Mario Richter: Warum Genre h#? (2011-5-1, hidden) more ...
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Henrik Juel: Last move was g2xDTSh1=L+ (2019-1-1)
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Henrik Juel: The next retraction is also determined
2... Ka7xSa8 (3.Sb6-a8+ or Sb6xDLSa8+) (2019-1-1)
Henrik Juel: oops, that is not true
after 2... Ka7-a8, Lc5 can uncheck directly (2019-1-2)
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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 (2019-1-1)
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Henrik Juel: The pawn captures are determined, and also their sequence
d7xSe6 first, then a2xLb3, and finally g7xDh6 (2019-1-1)
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Henrik Juel: ... and wKb4 got out via h2 and g3 (2019-1-1)
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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) (2019-1-1)
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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 (2019-1-2)
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. (2019-1-2)
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Mario Richter: vgl. mit P1012380 (2010-4-17)
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 (2019-1-2)
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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 (2019-1-2)
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Alfred Pfeiffer: Auf a5 steht der wK. (2012-1-11)
Henrik Juel: No Mystery here (2019-1-2)
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.'" (2019-1-3)
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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#' (2019-1-2)
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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 (2019-1-2)
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?" (2019-1-3)
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 (2019-1-3)
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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 (2019-1-2)
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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 (2019-1-2)
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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 (2019-1-2)
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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++) (2019-1-2)
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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 (2019-1-2)
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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 (2019-1-2)
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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 (2019-1-2)
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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 (2019-1-3)
Alfred Pfeiffer: Im Original steht sLa6 (statt sLf5). Mit wBc4 kann der sLc8 nicht nach f5 gelangen. (2019-1-4, hidden)
Henrik Juel: Yes, the diagram is wrong
Please move Lf5 to a6
Thanks, Alfred (2019-1-4, 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 (2019-1-3)
Henrik Juel: diagram error: wPe4 should stand on f4 (2019-1-3, hidden) more ...
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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# (2019-1-3)
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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 (2019-1-3)
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) (2019-1-3)
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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. (2009-5-12)
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 (2019-1-3)
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). (2019-1-5)
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. (2019-1-5)
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.) (2019-1-5)
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Cooker: Illegal position. I don't see what black unit takes the white bishop f1. (2002-9-21)
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'. (2002-9-27)
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 (2019-1-3)
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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 (2019-1-3)
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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 (2019-1-3)
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. (2019-1-4)
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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 (2019-1-4)
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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 (2019-1-4)
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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 (2019-1-4)
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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 (2019-1-4)
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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 (2019-1-4)
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 (2019-1-4)
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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 (2019-1-4)
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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 (2019-1-5)
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 (2019-1-5)
Yoav Ben-Zvi: The convoluted stipulation is not needed as this is a standard #2 with solution supported by "Cant Castle" retroanalysis. (2019-1-7)
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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 (2019-1-5)
Henrik Juel: should be -6.Dd1, of course (2019-1-5)
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Henrik Juel: The solutions feature impossible check by T, L, and S (2019-2-14)
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Henrik Juel: In Grid Chess a white pawn cannot reach a6 (nor h6), as a5-a6 and b5xa6 are impossible
With No-capture Chess added as extra condition a white pawn can reach no higher than row 5 (2019-2-14)
Alfred Pfeiffer: Lösungspräzisierung:
a) 20: a1. ... h1, a8. ... h8, c2, e2, a6, h6 (2010-1-20, hidden)
Frank Müller: Dann wären es aber keine 20 Lösungen, Herr Pfeiffer (8+8+4) !?
Ich meine a6 bis h6 ist schon richtig. (2010-1-20, hidden) more ...
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Henrik Juel: The stipulation and solution are inconsistent
Assuming that 2 wBB is changed to 2 sBB, I cannot see how the solution position without Td6 could be legal
-1.Dc7xSb7+ Sd6xSb7 is not possible, as sSd6 checks wK (2019-2-14)
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Henrik Juel: C+ by Natch 3.1 (57 min.)
The author made 64 proof games, all ending in Black being mated, on the 64 squares of the board
This is a neat model mate (2018-12-9)
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corrected version by V. Onitiu
Henrik Juel: Natch 3.1 did find a strategy, but otherwise nothing happened in more than an hour (2018-12-6)
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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. (2010-7-27, hidden)
Henrik Juel: Alfred is right, so please correct the official solution
The capture sequence must have been
g7xSh6 before d2xTc3 before Lx[Lc1] (2019-1-23, hidden) more ...
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Henrik Juel: The stated retroplay is the only possibility
so White pawns move downwards (2019-1-23)
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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) (2019-1-23)
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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 (2019-1-23)
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." (2019-1-24)
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 (2019-1-24)
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. (2010-7-27, 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 (2019-1-23)
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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 (2019-1-24)
Henrik Juel: Alfred's addition was
Schwarz kann noch rochieren. (2019-1-24)
Alfred Pfeiffer: Forderung unvollständig, es fehlt: S kann noch rochieren! (2010-7-27, 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) (2019-1-25)
Henrik Juel: For a slightly clearer argument, please replace
'Lb7 also cannot be promoted'
with
La2 also cannot be the original [Lc8] (2019-1-25)
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 ... (2019-1-26)
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. (2019-1-26)
Alfred Pfeiffer: Forderungstext unvollständig. Zu ergänzen ist: Schwarz hat soeben rochiert, und es sind keine w UW-Figuren auf dem Brett. (2010-7-27, 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 (2019-1-24, 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. (2019-1-24, 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... (2019-1-25, hidden) more ...
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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 (2019-1-24)
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! (2010-7-27, 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 (2019-1-24)
Henrik Juel: for clarity, please replace the last parenthesis by
(because White must promote to S) (2019-1-24)
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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) (2019-1-26)
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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 (2019-1-26)
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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
. (2019-1-26)
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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
. (2019-1-26)
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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
. (2019-1-26)
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 (2019-1-26)
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Henrik Juel: solution: [Lf1] was captured on f1
-1... f4xg4ep -2.g2 f5
. (2019-1-26)
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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
. (2019-1-27)
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