Die Schwalbe

5197 problem(s) found in 1391 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20181209]

1 - P0000033
S. N. Ravishankar
5467v Die Schwalbe 98 04/1986
P0000033
(5+3) cooked
#1 vor 14
VRZ, Typ friedlich
R: 1. Tg1-f1 2. Tf2-f3 Te2- 3. Te1-g1 4. Te2-f2 Ta2- 5. Ta1-e1 Tb2-a2 6. Tb1-a1 Tc2-b2 7. Tc1-b1 Td2-c2 8. Tg1-c1 Tc2-d2 9. Td2-e2 Tb2-c2 10. Tc2-d2 Ta2-b2 11. Tb2-c2 g3-g2 12. Lg2-h3+, dann 1. Th1#
play all play one stop play next play all
Cook: Kurzlösung:
R: 1. Tg1-f1 droht 2. Tf1-f3, dann 1. Th1#
R: 1. Tg1-f1 Tf2-g2 2. Tf1-g1, dann 1. T3xf2#
(Mario Richter)
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)
comment
Keywords: Defensive Retractor, Type Pacific
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-1-17 more...
2 - P0000108
Andrey Frolkin
6272 Die Schwalbe 111 06/1988
1. Lob
P0000108
(30+0) C+
Färbe die Steine!
BP in 19,0
1. h4 d6 2. Th3 Sd7 3. Tb3 Sdf6 4. Tb6 axb6 5. f4 Ta3 6. Kf2 Th3 7. a4 Th1 8. Ta3 Lh3 9. Tg3 Dd7 10. Tg6 hxg6 11. a5 Th5 12. a6 Ta5 13. h5 Ta1 14. h6 Da4 15. h7 b5 16. h8=T Sh7 17. a7 f6 18. a8=T+ Kf7 19. Ta7 Dd4+
play all play one stop play next play all
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)
comment
Keywords: Unique Proof Game, Coloring problem, under-promotion (TT), Phoenix
Genre: Retro
Computer test: Natch 3.1 (16 min.)
Reprints: 158 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-7 more...
3 - P0000132
Nikita M. Plaksin
6578 Die Schwalbe 116 04/1989
2. ehrende Erwähnung
P0000132
(8+12)
Zeige, daß im Verlauf des Retrospiels zwingend die Möglichkeit zum ep-Schlag bestand!
Madrasi
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)
comment
Keywords: Madrasi, En passant
Genre: Retro, Fairies
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2012-7-8 more...
4 - P0000139
Leonid M. Borodatow
6685 Die Schwalbe 118 08/1989
P0000139
(15+10)
Löse die Stellung auf!
R: 1. Dh3-h6# Kf6-g6 2. Tg7-a7 Kg6-f6 3. Tc7-g7+ Kf6-g6 4. Tg7xBc7+ Kg6-f6 5. Td7-g7+ Kf6-g6 6. Tg7xBd7+ Kg6-f6 7. Te7-g7+ Kf6-g6 8. Tg7xBe7 Ke6-f6 9. Tf5-a5+ Kd6-e6 10. Lg5-f4+,Lh6-f4+ ...
play all play one stop play next play all
Nikolai Beluhov: Unplaying 1. Qh3# Kf6 2. Rg7+ Kg6 3. Rc7+ Kf6 4. Rg7:Pc7+ Kg6 5. Rd7+ Kf6 6. Rg7:Pd7+ Kg6 7. Re7+ Kf6 8. Rg7:Pe7+ Ke6 9. Rf5+ Kd6 10. B~f4+... seems to release the position. Three black pawns are uncaptured on their home squares. (2010-7-30)
Henrik Juel: Following Nikolai's retroplay the further resolution is easy:
White captured PaxPb, gxf, and hxg, and promoted on b8,b8,f8,f8,g8
Black captured fxSg and promoted on a1 (2019-9-9)
comment
Keywords: Non-standard material, Promotion in the retro play (LLLLT)
Genre: Retro
Reprints: RC/49 feenschach 192 03-04/2012
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-9-10 more...
5 - P0000143
Nikita M. Plaksin
6689 Die Schwalbe 118 08/1989
4. Lob
P0000143
(13+9)
Löse die Stellung auf!
Madrasi RI
R: 1. Se6xDd8 d2xTe1=L 2. Tb1-e1 c3xTd2 3. Tb8-b1 b4xTc3 4. b7-b8=T ... 5. a6xTb7,a6xSb7
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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)
comment
Keywords: Madrasi (Rex inklusive)
Genre: Retro, Fairies
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-2-6 more...
6 - P0000169
Leonid M. Borodatow
6988 Die Schwalbe 123 06/1990
4. ehrende Erwähnung
P0000169
(11+8)
#3
b) ohne Bauern
a) 1. Dg2! droht 2. h7 ... 3. h8=D#
1. ... Sd2 2. Kxd2
1. ... Sa3 2. Txa3
1. ... Kxh6? 2. Dg6#
b) 1. Dg8! droht 2. Ta7 ... 3. Th7#
1. ... Sd2 2. Kxd2
1. ... Sa3 2. 0-0-0!
1. Dg2? Kh6!
play all play one stop play next play all
In a) darf Weiß nicht mehr rochieren, da sich der sBa7 auf a1 umgewandelt haben muß.
Henrik Juel: A couple of errors in a):
1... Kxh6? 2.Dg6#, not 2.Ta8? Kh7!
[Pa7] promoted on a1, not [Ph7] on h1 (actually Black captured hxg with promotion on g1) (2019-9-9)
comment
Keywords: Cant Castler, Promotion, Castling
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-9-10 more...
7 - P0000236
Karlheinz Bachmann
7437 Die Schwalbe 130 08/1991
P0000236
(3+1) cooked
BP in 16,0
Zeroposition
a) Kf8 nach g8
b) Lg5 nach g2
a) 1. Sf3 e5 2. Sxe5 Df6 3. Sxd7 Dxb2 4. Sxf8 Dxb1 5. Sxh7 Txh7 6. Txb1 Txh2 7. Txb7 Txg2 8. Txa7 Txf2 9. Txc7 Txf1+ 10. Txf1 Txa2 11. Tfxf7 Txc2 12. Txc8+ Kxf7 13. Txb8 Txd2 14. Txg8 Txd1+ 15. Kxd1 g5 16. Lxg5 Kxg8
b) 1. Sc3 d5 2. Sxd5 Dd6 3. Sxe7 Dxh2 4. Sxc8 Dxg1 5. Sxa7 Txa7 6. Txh7 Txa2 7. Txh8 Txa1 8. Txg8 Txc1 9. Txg7 Txc2 10. Txf7 Txb2 11. Txc7 Txd2 12. Txb7 Txd1+ 13. Kxd1 Dxf2 14. Txb8+ Ke7 15. Txf8 Dxg2 16. Lxg2 Kxf8
play all play one stop play next play all
Cook: NL
b) 1. Sb1-c3 b7-b5 2. Sc3xb5 c7-c6 3. Sb5xa7 Dd8-c7 4. Sa7xc8 Ta8xa2 5. Sc8xe7 Ta2xa1 6. Se7xc6 Ta1xc1 7. Sc6xb8 Tc1xc2 8. Sb8xd7 Dc7xh2 9. Sd7xf8 Dh2xg1 10. Th1xh7 Tc2xb2 11. Th7xh8 Tb2xd2 12. Th8xg8 Td2xd1+ 13. Ke1xd1 Dg1xf2 14. Tg8xg7 Df2xg2 15. Tg7xf7 Ke8xf7 16. Lf1xg2 Kf7xf8 (Mario Richter)
paul: a) is Jacobi+ (2019-2-23)
more ...
comment
Keywords: Unique Proof Game, Massacre PG, Rex solus (s)
Genre: Retro
Reprints: 119 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2017-4-25 more...
8 - P0000307
Thomas Volet
8130 Die Schwalbe 141 06/1993
P0000307
(12+14)
Ist die Stellung legal?
Die kürzeste Zugfolge bis zum letzten B-Zug sieht prinzipiell so aus ('...' = Wartezug):
R: 1. Ta8-b8 Kd1-c1 2. Kb8-c8 Ke2-d1 3. Lf8-g7 Lg7-h6 4. Lg8-h7 ... 5. Tb1-b2 ... 6. Th1-b1 ... 7. Th7-h1 Lh6-g7 8. Tg7-h7 ... 9. Lh7-g8 ... 10. Tg8-g7 ... 11. Lg7-f8 ... 12. Tc8-g8 ... 13. Lg8-h7 Td8-d7 14. Lh7-g8 Tg8-d8 15. Lf8-g7 Tg7-g8 16. Lg8-h7 Th7-g7 17. Te8-c8 Lg7-h6 18. Kc8-b8 Th1-h7 19. Lh7-g8 Lh6-g7 20. Lg7-f8 Tg1-h1 21. Tg8-e8 Tg4-g1 22. Lf8-g7 ... 23. Tg7-g8 24. Lg8-h7 25. Th7-g7 Lg7-h6 26. Th1-h7 Lh6-g7 27. Kd7-c8 ... 28. Lh7-g8 ... 29. Lg7-f8 ... 30. Tg8-a8 ... 31. Lf8-g7 ... 32. Tg7-g8 ... 33. Lg8-h7 ... 34. Th7-g7 Lg7-h6 35. Th3-h7 Th4-g4 36. ... Th7-h4 37. ... Lh6-g7 38. ... Tg7-h7 39. Lh7-g8 Tg8-g7 40. Lg7-f8 Ta8-g8 41. Lf8-g7 Lg7-h6 42. Lg8-h7 ... 43. Th7-h3 Lh6-g7 44. Tg7-h7 Ld3-c2 45. Lh7-g8 ... 46. Tg8-g7 ... 47. Lg7-f8 ... 48. Td8-g8 ... 49. Ke8-d7 ... 50. Kf8-e8 Lb5-d3 51. ... Ld7-b5 52. ... Lc8-d7 53. ... d7-d6!
play all play one stop play next play all
"Nach 'unserer' Ausdrucksweise ist die Stellung also illegal, weil nur unter Außer-acht-Lassung der 50-Züge-Regel erspielbar." (HHS)
"Schöne Rangierarbeit mit guter Ausnutzung der LL-Schleuse" (DB).
Auf derartige Forderungsfragen erwartet man ja i.a. ein JA als Antwort und so wäre eine Verkürzung der Zügezahl auf genau 50 Züge bis zum letzten B-Zug angebracht (und auch leicht zu bewerkstelligen). So vermutet jeder einen besonderen - aber nicht vorhandenen - Trick des Autors, um die 50-Züge-Beschränkung nicht zu überschreiten. Mehrere Löser kamen beim Rangieren offensichtlich so ins Schwitzen, daß sie gar nicht an die 50-Züge-Regel gedacht haben und so fälschlicherweise 'Legalität' behauptet haben! Ebenso falsch lagen einige Löser, die 'Illegalität' attestierten mit der Begründung sSh8 oder sLh6 seien nicht zu realisieren. 3/III und 5L., welche die Stellung aufgedröselt haben (von denen kam nur HHS zu der Erkenntnis: illegal!).
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)
comment
Keywords: 50 move rule
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2015-9-17 more...
9 - P0000407
Günter Glaß
4360 Die Schwalbe 81 06/1983
5. Lob
P0000407
(7+9)
Weiß nimmt 1 Zug zurück, so daß Weiß und Schwarz #1
Circe
Duplex
w) R: 1. Kh6xBg5[+sBg7], dann Sf7#
bzw. R: 1. Kh6xBg5[+sBg7], dann 1. Lb5#

s) R: 1. ... Kd3xBc4[+wBc2], dann 1. ... Dgd4#
bzw. R: 1. ... Kd3xBc4[+wBc2], dann 1. Dbg2#
play all play one stop play next play all
HBae: R: 1. ... Kh6xBg5[+sBg7], dann 1. Lb5# bzw. 1. Sf7#
R: 1. ... Kd3xBc4[+wBc2], dann 1. Dgd4# bzw. 1. Dbg2# (2019-2-4)
comment
Keywords: Circe, Help retractor
Genre: Retro, Fairies
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-2-4 more...
10 - P0000434
Hans Gruber
4700 Die Schwalbe 86 04/1984
2.-3. Lob e.a.
P0000434
(1+1)
Ergänze einen sS und 2 sBB zu einem IC!
Längstzüger
(+sBf7,sBg6,sSh8)
play all play one stop play next play all
Nicht +sSh1, sBf2, sBg3

Duplicate Diagram: P1187718, P1347424

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)
comment
Keywords: Illegal cluster, Maximummer, only Kings
Genre: Retro, Fairies
Reprints: 110 Die Schwalbe 244, p. 582, 08/2010
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-2-15 more...
11 - P0000456
Hendrik Prins
5027 Die Schwalbe 91 02/1985
P0000456
(11+6)
Welches war der letzter Zug?
b) sLh1 statt Sg1
a) R: 1. f2xSg1=S+
b) R: 1. g2xDf1=D+
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Henrik Juel: The twinning is not very nice, but clear enough, I think:
remove Sg1 and add sLh1 (another double check) (2019-10-5)
more ...
comment
Keywords: Last Move? (BxS=S), Type C, Last Move? (BxD=D), Promotion in the retro play
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-10-15 more...
12 - P0000549
Hans Gruber
René Jean Millour

3464 Die Schwalbe 68 04/1981
P0000549
(1+1)
Ergänze weiße Steine minimaler Gesamtqualität zu einem IC
Keine Bauern auf der 1./8. Reihe
Mars-Circe
+wBe2,wBg2
play all play one stop play next play all
Die Lösung des ursprünglichen Alleinautors HG (+wBe2 f2 g2) ist nicht die Lösung, da die Stellung nach Entfernen des wBf2 immer noch illegal ist. Die Lösung von RJM, der damit auf Wunsch von HG zum Koautor wurde (+ wBe2 g2), ist aber auch nicht richtig: NL +wB auf 1. oder 8. Reihe. Dies muß also ausgeschlossen werden.
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)
more ...
comment
Keywords: Circe (Mars), Illegal cluster, only Kings
Genre: Retro, Fairies
Reprints: G) Problemkiste (50) 08/1987
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-2-15 more...
13 - P0000726
Hans Gruber
2873 Die Schwalbe 58 08/1979
P0000726
(3+5) C+
ser-h#19
Wieviele Lösungen?
Eine Lösung ist: 1. e5 2. f5 3. f4 4. f3 5. f2 6. f1=L 7. Ld3 8. g5 9. g4 10. g3 11. g2 12. g1=D 13. De3 14. h5 15. h4 16. h3 17. h2 18. h1=D 19. Dhe4 c3#
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911,797,992 Lösungen
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)
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comment
Keywords: Path enumeration, Series mover, Promotion in forward play, Kindergarten Problem, Excelsior
Genre: Mathematics, Fairies
Computer test: Mario Richter: rawbats' output: 911797992 Loesungen gefunden time= 1.535000(s) (2019-2-2)
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-2-3 more...
14 - P0000764
Leonid Makaronez
1160 Die Schwalbe 24 12/1973
P0000764
(9+14)
#2
1. c7! ... 2. c8=D,T#
1. ... 0-0? is illegal!
play all play one stop play next play all
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)
comment
Keywords: Castling (sk), Cant Castler
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-27 more...
15 - P0000811
Tibor Orban
1791 Die Schwalbe 38 04/1976
Lob
P0000811
(15+15) C+
BP in genau 4,0
1. e4 e6 2. Lb5 Ke7 3. Lxd7 c6 4. Le8 Kxe8
play all play one stop play next play all
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)
comment
Keywords: Unique Proof Game, Switchback, Loss of tempo
Genre: Retro
Computer test: Resolution time : 1.21 s., Natch 2.3 Copyright (C) 1997,98,99,2001,2002,2003,2004 Pascal Wassong
Reprints: 32 Probleme für Tiger 3 03/1984
(I) Schachkalender 1987
7 Shortest Proof Games 11/1991
(3) feenschach 123 01-06/1997
(1) Probleemblad 03/1999
4 Introduction to Proof Games 03/2010
4b 64 Proof Games 2012
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2013-4-20 more...
16 - P0000888
Bruno Ebner
638 Die Schwalbe 13 02/1972
P0000888
(5+10)
#1
a) -wDc6: 1. Sc6#
b) -wSd4: 1. Td4#
c) -wTd3: 1. Sd3#
d) -wSc5: 1. Dc5#
play all play one stop play next play all
Mario Richter: Illegale Stellung (mehr schwarze Bauernschläge notwendig als weiße Steine fehlen).
Offensichtlich ein Scherzproblem, Lösung vermutlich: entferne eine weiße Figur, dann:
- wTd3: Sd3#
- wSd4: Td4#
- wSc5: Dc5#
- wDc6: Sc6# (2010-2-24)
Henrik Juel: Writing the above solution in this order
-Td3, Sc5-d3#
-Sc5, Dc6-c5#
-Dc6, Sd4-c6#
-Sd4, Td3-d4#
discloses a cyclical pattern. (2010-2-26)
A.Buchanan: I guess it would spoil the joke to ask why we're not removing a black pawn? (2019-10-22)
Alfred Pfeiffer: Die Entfernung eines beliebigen schwarzen Bauern erlaubt kein #1. (2019-10-22)
A.Buchanan: Exactly, Alfred (2019-10-22)
comment
Keywords: Illegal position, Joke, Remove pieces, Cycle
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-10-22 more...
17 - P0000947
Theophilus Harding Willcocks
Werner Keym

Die Schwalbe 1978
P0000947
(3+9)
Letzter Zug?
R: 1. Da8xTb8 Tb7-b8 and e.g. 2. Lb8xSc7
play all play one stop play next play all
siehe auch P0000948
more ...
comment
Keywords: Last Move? (DxT), Type A, Economy record, Type B (a fortiori)
Genre: Retro
Reprints: 1.9A Eigenartige Schachprobleme , p. 174, 2010
1.9B Eigenartige Schachprobleme , p. 174, 2010
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2019-10-30 more...
18 - P0000957
Theophilus Harding Willcocks
Die Schwalbe 1959
P0000957
(4+8)
Letzter Zug?
R: 1. a7xTb8=L
play all play one stop play next play all
Yoav Ben-Zvi: WBd8 can be replaced with a BN after moving BPg7 to f7 and BPh5 to h6. (2018-8-8)
A.Buchanan: Yes after all these years, I found this a few months ago, but when I told Thomas Brand, he said that Werner Keym had found it - one of a series of modifications through seeking to avoid non-standard material. (2018-8-12)
Henrik Juel: Werner's improvement can be found in his 'Eigenartige Schachprobleme' from 2010, p.196, dia 1.68 (2018-8-12)
A.Buchanan: I thought it was more recent (2018-8-12)
A.Buchanan: In fact Werner was not trying to avoid non-standard material, but to prefer "cheaper" knights over bishops. But this is not the canonical ordering, which regards knights and bishops as equivalent. So the older record will win out in classical terms. (2019-10-4)
A.Buchanan: And non-standard material is no defect, according to the 1977 grading criteria, so old Theophilus reigns supreme (2019-10-4)
Henrik Juel: The further retroplay is
-1... h6 -2.a6 and e.g. -2... Ka7 -3.a5 Ta8 -4.Tb8 Ka6 -5.Kc8
and wK out via g6 (2019-10-4)
more ...
comment
Keywords: Type A, Last Move? (BxT=L), Promotion in the retro play (L), Economy record, Non-standard material
Genre: Retro
Reprints: 1.49A Eigenartige Schachprobleme , p. 190, 2010
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2019-10-4 more...
19 - P0001141
George Hume
Jamaica Gleaner 12/1891
Weihnachtsturnier 1891
1. Preis
P0001141
(9+9)
Auf welche Gedanken kommen Sie bei dieser Stellung?

Der Ld6 ist keine UWF und der sBg7 wurde auf seinem Ausgangsfeld geschlagen. Nach dem Autor muss der letzte Zug also Lf8-d6 gewesen sein, also illegale Stellung.
Datum der Originalpublikation nicht 100% sicher, laut ACM aus der "Jamaica Gleaner Christmas column".

Originalforderung: How has the position been arrived at and who is the winner, and in how many moves?

From the Jamaica Gleaner: "White mates in two moves. The last move made was by Black playing his Bishop and announcing mate. As it can be demonstrated that the Bishop is not a promoted Pawn and that Black's King's Knight's Pawn was captured on its original square by White's Queen's Knight's Pawn the Black Bishop must have been played from Bishop's square (f8) to Q3 (d6). This being an illegal move, White enforces the penalty of compelling Black to retract it and move his King whereupon White plays 1 PXB(Q) ch (1.gxf8=Q+) and mates next move by 2 Q-B4 (Qf4#). The following is a brief but pointed analysis, demonstrating the false move: White's Pawns have made six captures all on black squares. The Q Kt P (Pb2) made five of these and consequently captured the Kt P on the square upon which it now stands (g7). They could not have captured the Q B which is also lost. The White Bishop is the QRP (Pa2) promoted, the original KB having been captured on its own square as the unmoved Pawns show. To allow this promotion Black's QRP (Pa7) made two captures, the QKtP (Pb7) one, and the QBP (Pc7) two. The KRP (Ph7) has also made a capture, which accounts for the seven pieces White has lost. The Black Bishop is not a promoted Pawn, as if the Black KBP (Pf7) had played to the 7 th square (f2) and then captured a White piece on K or Kt square (e1 or g1) the captures by White Pawns cannot be accounted for without including the Black QB or KRP neither of which is available. As it can be demonstrated, then that the Black Bishop is not a promoted one, and that the KKtP was captured by the White Pawn which now stands on that square, in order to reach Q3 the Black Bishop must have an impossible move.

Der Kolumnist des ACM merkt aber zurecht an:
ACM: The above is a very fine piece of analytical work; but there is a slight flaw in connection with the minor condition, 'mate in two'. In a position of this kind we believe only that which can be proved; thus we do not think that White has any right to enact a penalty, as neither the analysis nor the conditions show that the Black Bishop came from Bishop's square on his last move; indeed, that Bishop may have played outside the Pawns on the very first move of the game which, being played, brought about the position.
A.Buchanan: I don't think this is cooked, I think it's just a very bad joke. Most time when moves are retracted it's because of self-check or illegal castling, occasional illegal en passant. But here we are meant to conclude that the Bl bishop has hopped over Bl pawn. There are problems by Hieronymus Fischer e.g. P1309484 where we are meant to conclude that a unit has escaped its game array cage somehow, but the piece is simply replaced: there is no move to be retracted and supplanted. But as long as such a problem is true to its own (joke) internal logic, it can't be described as cooked. If it fails (e.g. P1246715) it can indeed be cooked and repaired. (2019-10-22)
HBae: White plays 1 PXB(Q) ch (1.gxf8=Q+). Muß der sK nicht auf f5 stehen? (2019-10-22)
Henrik Juel: Last move (supposedly) was Lf8-d6#, which is obviously impossible and hence illegal
The penalty for this is that Black must replace Ld6 on f8 AND instead make an arbitrary move with his king
So the forward play is
0... Kf5,Kf6,Kh6 1.gxf8=D+ Kg5,Ke5 2.Df4# (2019-10-22)
comment
Keywords: Illegal position, Joke, Retract illegal move (stuck at home), Touch Move
Genre: Retro
Reprints: American Chess Monthly 1, p. 11, 03/1892
Jamaica Gleaner 30/04/1892
17 Europe Echecs 14 10/1959
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-10-23 more...
20 - P0001275
Henry Anthony Adamson
The Chess Amateur 03/1926
P0001275
(14+10)
#2
1. Kf2+ Dc1 2. Txc1#
play all play one stop play next play all
Henrik Juel: Not 1.0-0?, e.g. -1... Dc2 -2.Lc1 Db1 -3.Td2 Dc2 -4.Td1 Db1 -5.Ld2 Dc1 -6.Lc3 D=c2 -7.Td2 ?? (2004-2-23)
Yoav Ben-Zvi: [bBf8] was captured at home by an officer, the other 5 missing Black pieces were captured by the wPs (2 on the b file, by [wPc2] and [wPa2], and 1 each on the d, e and g files). It follows that [bBc8] was captured by a wP outside it's home, after being released by bPb7-b6, and that each of [bPc7], [bPf7] and [bPh7] accounts for a wP capture. Since none of the captures by wPs is on the c, f or h files, each of these bPs either promotes or else captures to arrive at a file where a wP can capture it directly.
Assuming White retains the right to castle implies the following:
[bPh7] could not move directly (without capture) to promotion on h1 as this would require a move by [wRh1]. [bPf7] could not move directly to promotion by bPf2-f1=X as this would require a prior move by wK. Each of these bPs must therefore have captured one of the missing wNs. No other White pieces are missing so [bPc7] did not capture. In order to account for a wP capture it must have moved directly to promotion on c1. Prior to the promotion on c1, the way to promotion on the c file must be cleared by [wPc2] capturing on the b file. Promoting [bPf7] entails a move by wK even if the bP captures (promoting from f2 or on e1) so, to preserve White castling, it must be captured directly by wPf2xPe3 (after capturing a wN on the e file).
Looking backwards from the diagram position:
The pieces on the a and b files form a locked West cage. All officers on the a file are immobile until the cage is released. The West cage remains locked until the retraction of wPc2xb3 or wPc3xb4 or bPb7-b6, not bPc7xb6? as [bPc7] does not capture (see above). Retraction of wPc2xb3 requires the previous release of the cage to allow exit of bKa1 via c2, d3, since alternative exit paths of the bK require a move by wK to clear the way. Retraction of bPb7-b6 requires previous uncapture and return home of [bBc8]. The uncapture of [bBc8], by a wP, occurs on a light-colored square, b3 or d3. Each of the alternatives for uncapturing [bBc8] requires previous release of the cage: wPc2xb3 since it is preceded by exit of bKa1(as noted above), wPa2xb3 or wPe2xd3 since these are preceded by exit of wBa2 to remove the obstruction on a2 or to retract home [wBf1]. It follows that the uncapture of [bBc8] and therefore the retraction of bPb7-b6 requires previous release of the cage. The remaining alternative for release of the West cage under the assumption that White may castle, retraction of wPc3xb4, is next to be considered.
Prior to the unpromotion by bPc2-c1=X, the following is known: the cage remains locked as it can be released only after the unpromoted bP retreats to c4 allowing wPc3xb4 to be retracted without blocking the retreat of bP to c7, the officers on the a file are immobile, retraction of bPb7-b6 locks out [bBc8], retraction of wPc2xb3 or wPc3xb4 blocks the retreat of Black's unpromoted pawn on the c file, retraction of wPd2-d3 requires prior retraction of wBd2 to c1 leading to permanent obstruction of the unpromotion on c1, retraction of wPe2-e3 or wPe2xd3 locks out [wBf1], retraction of wPf2xe3 prevents subsequent retraction of bBa5 to g1 and retraction of wPh2xg3 prevents subsequent unpromotion of bBa5 by bPh2xNg1=B. The rejection of all of these alternatives (prior to the unpromotion on c1) means that neither side can uncapture or play a pawn retraction. This means that the retroplay preceding the unpromotion on c1 consists only of moves by bQc7, wBd2 and wRe2, proving that the Black piece unpromoted on c1 must be bQc7. Since wK and wPs are immobile, wBd2 is restricted to the squares c1, d2 and c3. Since it has been shown that [bPc7] promotes to queen, not bishop, and [bPf7] does not promote at all, the promotion to bBa5 is by [bPh7] playing bPh2xNg1=B.
Since the wK stands immobile on e1, the unpromotion of the bQ by bPc2-c1=Q occurs with a shield standing on d1 to protect the wK from check by the promoted bQ. This shield can only be provided by wRe2 since d1 is a light-colored square that cannot be occupied by wBd2 and (as noted above) no uncaptures can be played prior to the unpromotion so the sheilding piece must be present in the diagram. If the unpromotion on c1 occurs with wB standing on c3 then White responds to the unpromotion by wRd2-d1+ and Black is in Retro-Stalemate caused by wBc3 obstructing the retreat of the unpromoted bP and the lack of Black tempo retractions that would allow White to remove the obstruction (this is shown in the Try noted by Henrik).
Another possibility that needs to be considered is that, when the unpromotion on c1 occurs, wB stands on d2 (instead of c3). In this case White responds to the unpromotion by retracting wBc1-d2+ (shielding bK from wRd1) followed by retreat of the unpromoted bP to c4 and the retraction of wPc3xb4 releasing the West cage. This shows that the position with bQc7 on c1, wRe2 on d1 and all other pieces as in the diagram, and White to play, can be reached with White retaining the right to castle. However, due to Retro-Opposition between bQ and wB colliding on c1 and the lack of tempo retractions by either side, other than moves by wK or wRh1, to overcome parity of bQ shuffling between b1 and c1 and wB shuffling between d2 and c3, there is no valid retraction from the diagram position with castling rights preserved. This is demonstrated in Henrik's Try if Black delays unpromotion of bQ, retracting -6...bQb1-c1 but it is simpler to prove by considering forward play. It can be seen that in the relevant position (with wKe1 and wRh1 immobile) White must play 1.wBd2-c3 (c1 is obstructed by bQ) followed by bQc1-b1 (the shield for the bK must be maintained) 2.wBc3-d2, bQb1-c1 repeating the position. There is no way to advance to the diagram position and preserve White's right to castle so in this case too (wB standing on d2 when bQ is unpromoted) White can't castle.
It has been shown that the assumption that White can castle leads to contradiction in every case thus proving (by "Reducto ad Absurdum") that White cannot castle. (2019-8-5)
A.Buchanan: Beautiful solution, with a dramatic tempo-based finale - thanks for posting such an authoritative solution. Which tempo keyword is meant to be used for this? :) (2019-8-6)
Yoav Ben-Zvi: The reprint of this problem in Ceriani's book appears in the chapter devoted to the Retro-Opposition phase of a resolution, under the heading "Defferred R.O.".
The analysis contains several instances of repeated thematic content (Echo) such as: "Retro-Shield" (by bQ, wR and wB, on b1, c1, d1), "Retro-Opposition" (temporary obstructions by wB and bQ on c3 and c1) and locking of various Home and Dynamic cages.
The #2 stipulation reflects the preference, at the time this was published, for RA problems to be presented as orthodox. This is no longer considered important so the analysis can be used to reject castling in the last move, as in the following version:
wBd2 on c3, wRh1 on f1, wRe2 on f2, wKe1 on g1 - "Last Move?". (2019-8-14)
A.Buchanan: (1) Have you got both of Ceriani's books? How did you get hold of them? I guess they are still under copyright. Retro Corner has been saying for 20 years that you need to write to Ceriani's son. Is that still true? Hasn't he got an email address now.
(3) I think your statement "This is no longer considered important" is too black and white. Generally I find it satisfying if the composer is able to tie future and past together in this way. And a check in the diagram position is to my mind an unpleasant defect, to be avoided if possible. (2019-8-15)
Henrik Juel: There are two ways of telling the solver immediately who has the move:
By having a king in check or by specification, directly by saying so or indirectly by a forward stipulation
It has got out of fashion to cast retro problems as forward ones, but there are still examples of imprecise stipulations, like #1: sometimes it should be interpreted as a condition, saying that White has the move, and sometimes it means 'Who mates in 1?' (2019-8-15)
A.Buchanan: (1) Yes #1 is a special stipulation. There are 4 possible interpretations: who moves first (Codex Article 15, "the official joke") but also who gets mated (maybe should be added to Codex or certainly to PDB glossary somehow).
(2) I get that RA has kind of come of age: it no longer needs to justify itself behind an orthodox stipulation. Hurray, I suppose.
(3) I am not objective in liking these old-style stipulations, because DP is all about interaction between forward and retro play. One of my heroes, Guus Rol, is also a big enthusiast of combining these two modalities, which he terms "retro-activity".
(4) And it is cool to read the same diagram in two ways: both forward and backwards. It's like a cryptic crossword clue where the same text has two meanings. Even if the forward play is not very thematic or interesting.
(5) So I hope that the RA community can feel sufficiently confident in its status now that it doesn't need to "fix" a classic problem with a perfectly natural #2, by making a second-rate Type C position out of it.
(6) Any info about the Ceriani books please? (2019-8-16)
Henrik Juel: You can buy a used copy of '32 characters and 1 author' on an italian homepage for €350
which seems very expensive to me
Most of the problems from the book are on the PDB (2019-8-16)
comment
Keywords: Cant Castler, Castling (wk), Promotion (l)
Genre: Retro
Reprints: 155a 32 personaggi e 1 autore 1955
147 Europe Echecs 86 03/1966
Input: Gerd Wilts, 1995-6-3
21 - P0001343
Friedrich Amelung
Düna-Zeitung 1897
P0001343
(5+3) C+
#2
1. hxg6ep Kh5 2. Txh7#
play all play one stop play next play all
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)
more ...
comment
Keywords: En passant, En passant as key
Genre: Retro, 2#
Computer test: Popeye 4.61
Reprints: 55B Retrograde Analysis 1915
169 Allgemeine Zeitung Chemnitz 04/12/1927
(VIII) Problem 37-40 09/1956
(1) Problem 103-105 01/1967
215 Europe Echecs 124 04/1969
D12 feenschach 27, p. 27, 04/1975
(2) Mat Plus 12 10-12/2009
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-17 more...
22 - P0001438
Niels Høeg
1862 The Problemist FCS 12, p. 129, 06/1935
P0001438
(14+8)
Wer gewinnt?
1. ... bxc2#?
1. Tc1+ Kxa2 2. Sc6 fxe1=D+ 3. Txe1 ... 4. Sb4#

R: 1. ... Ka1-b1 2. Sc6-b8 Kb1-a1 3. Sd8-c6 Ka1-b1 4. Se6-d8 Kb1-a1 5. Sf4-e6 Ka1-b1 6. Sh3-f4 Kb1-a1 7. Sf4xBh3 Ka1-b1 8. Sg2-f4 Kb1-a1 9. Se3-g2 Ka1-b1 10. Sb6-a4 Kb1-a1 11. Sc4-b6 Ka1-b1 12. Sa3-c4 b4-b3 13. Sb1xBa3 a4-a3 14. Tc1-c2 a5-a4 15. Kc2-d1 a6-a5 16. Kd3-c2 b5-b4 17. Dd1-e1 h4-h3 18. Db3-d1 h5-h4 19. Te1-c1 h6-h5 20. Sd1-e3 h7-h6 21. Sa3-b1 Kb1-a1 22. Sc4-a3 Ka1-b1 23. Kd4-d3 Kb1-a1 24. Da3-b3 Kc1-b1 25. Sa5-c4 Kc2-c1 26. Sb3-a5 Kb1-c2 27. Sc1-b3 Ka1-b1 28. Se3-d1 Kb1-a1 29. Sg4-e3 Kc2-b1 30. Sb3-c1 d6-d5 31. Ta1-e1 d7-d6 32. Tb1-f1 Kd1-c2 33. Sc1-b3
play all play one stop play next play all
8/2bbb1bb/b7/1b6/3K2S1/D1B2B2/BB1BBb1B/TTSk2Lt (14+10)
A possible proof game from this intermediate position is:
1. Sb3+ Kc2 2. Tf1 h6 3. Tae1 h5 4. Sc1 Kb1 5. Se3 Ka1 6. Sd1 Kb1 7. Sb3 Kc2 8. Sa5 Kc1 9. Db3 Kb1 10. Kd3 Kc1 11. Sc4 Kb1 12. Sa3+ Ka1 13. Sb1 h4 14. Se3 h3 15. Tc1 d6 16. Dd1 d5 17. De1 a5 18. Kc2 a4 19. Kd1 a3 20. Tc2 b4 21. Sxa3 b3 22. Sb5 Kb1 25. Sb8 Ka1 28. Sxh3 Kb1 32. Sa4 Kb1
Black captured the white bishops on c1 and f1. White captured Lf8 on f8 and g2xf3 and promoted his f-pawn on b8 to L, which was brought to g1, implying that a7, b6, and b7 are unoccupied. Between moves 13 and 21 white has no time for a tempo move, and after this neither side can change the tempo, so white has the move in the problem position and wins.
Henrik Juel: Moving Sb8 to b4 would allow a more contemporary stipulation: #1 (who?)
with the solution: 1.Tc1#, not 1... bxc2? (2016-10-12)
Henrik Juel: If you prefer retroplay, the plan is to uncapture black pawns on h3 and a3, screen with a knight on b1, retract Tc1-c2, extract Kd1 and De1, retract Te1-c1, screen with the other knight on d1, retract the first knight to c1, extract Sd1, hide the white rooks on a1 and b1 while black king waits on c2, and then he can get out via g2 and h3. The details might be
-0... Ka1 -1.Sc5 Kb1 -2.Se4 Ka1 -3.Sg5 Kb1 -4.Sh3 Ka1 -5.Sf4xPh3 Kb1 -6.Sg2 Ka1 -7.Se3 Kb1 -8.Sc6 Ka1 -9.Sa7 Kb1 -10.Sb5 Ka1 -11.Sa3 b4 -12.Sb1xPa3 b5 -13.Tc1 a4 -14.Kc2 a5 -15.Kd3 a6 -16.De1 d6 -17.Db3 d7 -18.Te1-c1 h4 -19.Sd1 h5 -20.Sa3 Kb1 -21.Sc4 Ka1 -22.Kd4 Kb1 -23.Da3 Kc1 -24.Sa5 Kc2 -25.Sb3 Kb1 -26.Sc1 Ka1 -27.Se3 Kb1 -28.Sg4 Kc2 -29.Sb3 h6 -30.Ta1 h7 -31.Tb1-f1 Kd1 -32.Sc1 (2016-10-12)
Henrik Juel: The source with more details is
1862 The Problemist Fairy Chess Supplement no.12, June 1935, p.129
This magazine was later renamed Fairy Chess Review (2019-5-28)
more ...
comment
Keywords: Whose move?, Obvious obtrusive promoted material (L)
Genre: Retro
Reprints: 220 FIDE Album 1914-1944/III 1975
312 Europe Echecs 213/214 10/1976
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-5-28 more...
23 - P0001455
Alexander Kislyak
Die Schwalbe 1976
1. Preis
P0001455
(12+14)
#1 (wer?)
1. SxcjSxb2#?)
R: 1. Se3-c4 Sd3-b2 2. Sg2-e3 Se5-d3 3. Se1-g2 Sg6-e5 4. Sg2-e1 Sf8-g6 5. Se1-g2 f7-f8=S 6. Sg2-e1 f6-f7 7. Se1-g2 f5-f6 8. e2-e1=S f4-f5 9. e3-e2 f3-f4 10. f4xLe3 Ld4-e3 11. f5-f4 Lb2-d4 12. f6-f5 Lc1-b2 13. f7-f6 b2-b3 14. Sb3-a1 h2-h1
Unglaublich, wie die 4 weißen Bauern e-h mit ihren Schritten haushalten müssen, damit am Ende der Weg für den weißen Königsturm und den umgewandelten schwarzen Läufer a7 nicht versperrt wird.
play all play one stop play next play all
Dr. K. Fabel-Gedenkturnier
Henrik Juel: Typos:
The solution should read something like
1.Sxc4# (not 1... Sxb2??)
(because the retroplay shows that White has the move)
The last retraction in the given retroplay should read h2-h3, not h2-h1
Finally, the current PDB standard is to write the white retraction first, so the retroplay should read something like
-1... Se3 -2.Sd3 Sg2 ... -14.b2 Sb3 -15.h2
(Yes, I know that the animation machine requires full retro notation, but abbreviated notation is so much easier to write and to read...) (2019-11-10)
Henrik Juel: This fine retro deserves a closer look, I think
White captured [Lf8] and axb (to promote on b8, as Ld8 cannot get home to c1)
Black captured bxc, c7xd6, a third pawn capture to promote [Pf7] or [Ph7] to L on a black square, and a fourth capture to promote the other missing black pawn (since White did not capture a black pawn on the f- or h-file)
So Black did not capture [Lc1] on c1, and the plan for the retro play is to unpromote wSb2 on f8 and retract the resulting wP to f3 (not to f2, as sLa7 must unpromote on g1), allowing sSc4 to unpromote on e1 and letting the resulting sP uncapture on e3 a dark-squared wL, which goes to c1, permitting b2-b3 and Sb3-a1
The two knights cannot change the tempo, h2-h3 is needed at the end, and d3-d4 would prevent a white rook to get home to h1, so we can determine which camp starts the retroplay (2019-11-11)
comment
Keywords: Whose move?
Genre: Retro
Reprints: 329 Europe Echecs 225/226 10/1977
feenschach 41 01-03/1978
632 FIDE Album 1974-1976 1980
(2) Die Schwalbe 103 02/1987
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2002-10-12 more...
24 - P0001576
Alexander Kislyak
446 Europe Echecs 316 04/1985
T. R. Dawson et W. Hundsdorfer zum Gedenken
P0001576
(13+11)
Konnte die Stellung ohne Schach dem sK erreicht werden?
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)
comment
Keywords: Promotion, Volet Pawn
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-8 more...
25 - P0001884
Thomas R. Dawson
Aachener Anzeiger 1926
P0001884
(8+2)
#1 vor 2
VRZ, Typ Hoeg
R: 1. c7-c8=D Lb4xSa3 2. Sc5-a6, dann 1. Sb3#
R: 1. c7-c8=D Lc5xSa3 2. Sb4-a6, dann 1. Sc2#
play all play one stop play next play all
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)
comment
Keywords: Promotion, Defensive Retractor, Defensive Retractor, Type Høeg
Genre: Retro
Reprints: 127 Magyar Sakkvilag 04/1926
(31) On Retraction Chess Problems 1927
3 Chess unlimited , p. 35, 1969
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-2-12 more...
26 - P0001951
Luigi Ceriani
Karl Fabel

Am Rande des Schachbretts 1947
P0001951
(9+16)
KBP?
(AL: 183,0)
1. a4 c5 2. a5 c4 3. a6 c3 4. b3 d5 5. Lb2 cxb2 6. Sa3 Kd7 7. Sc4 Kc6 8. Se5+ Kb5 9. Ta5+ Kb4 10. Tb5+ Ka3 11. Sef3 Ka2 12. c3 d4 13. Dc1 e5 14. Kd1 e4 15. Kc2 e3 16. Kd3 g5 17. Ke4 g4 18. Kf4 d3 19. Dc2 dxc2 20. h3 c1=T 21. Th2 Ta1 22. Th1 Kb1 23. Th2 Kc1 24. Th1 Lf5 25. Th2 Lb1 26. d3 f5 27. Sd2 exd2 28. Sf3 Sf6 29. Kg5 Ld6 30. Kh6 Sd5 31. Kg7 Se3 32. Kf7 h5 33. Kg7 Ta4 34. Kf7 La2 35. Tb6 d1=L 36. Kg7 Lc2 37. Kf7 Lcb1 38. Kg7 Sd1 39. Kf7 g3 40. Kg7 gxh2 41. e3 h1=T 42. Le2 Sd7 43. Sh2 Sc5 44. Lg4 Se4 45. Kf7 hxg4 46. Kg7 g3 47. Kf7 gxh2 48. Kg7 Te1 49. Kf7 Te2 50. Kg7 Tc2 51. Kf7 h1=T 52. Kg7 Te1 53. Kf7 Lh2 54. Kg7 Sg3 55. Kf7 Sh1 56. Kg7 Tah4 57. Kf7 Kd2 58. Kg7 Ke2 59. Kf7 Tc1 60. Kg7 Lc2 61. Kf7 Ta1 62. Kg7 Lab1 63. Kf7 Ta5 64. Kg7 La2 65. g3 Kf3 66. Tb4 Tb5 67. Kf7 Te2 68. Kg7 Lcb1 69. Kf7 Tc2 70. Kg7 Tc1 71. Kf7 Lc2 72. Kg7 Ta1 73. Kf7 Lab1 74. Kg7 Ta5 75. Ta4 bxa6 76. Ta1 La2 77. Tc1 Lcb1 78. Tc2 f4 79. Te2 Lc2 80. Te1 Lab1 81. Tg1 Ta1 82. Tg2 La2 83. Kf7 Tc1 84. Kg7 Lcb1 85. Kf7 Tc2 86. Kg7 Te2 87. Kf7 Te1 88. Kg7 Tg1 89. Kf7 Lc2 90. Kg7 Lab1 91. Kf7 Ta5 92. Kg7 Ta1 93. Kf7 La2 94. Kg7 Tc1 95. Kf7 Lcb1 96. Kg7 Tc2 97. Kf7 Te2 98. Kg7 Tee1 99. Kf7 Tef1 100. Kg7 Lc2 101. Kf7 Lab1 102. Kg7 a5 103. Kf7 a4 104. Kg7 a3 105. Kf7 a2 106. Kg7 a1=T 107. Kf7 La2 108. Kg7 Tc1 109. Kf7 Lcb1 110. Kg7 Tc2 111. Kf7 Te2 112. Kg7 Tee1 113. Kf7 Lc2 114. Kg7 Lab1 115. Kf7 a5 116. Kg7 a4 117. Kf7 a3 118. Kg7 a2 119. Kf7 a1=T 120. Kg7 La2 121. Kf7 Tc1 122. Kg7 Lcb1 123. Kf7 Tc2 124. Kg7 Ke2 125. Kf7 Kd2 126. Kg7 Kc1 127. Kf7 Tce2 128. Kg7 Kd2 129. Kf7 Lc2 130. Kg7 Lab1 131. Kf7 Ta1 132. Kg7 La2 133. Kf7 Tc1 134. Kg7 Lcb1 135. Kf7 Tc2 136. Kg7 Kc1 137. Kf7 Tcd2 138. Kg7 Lc2 139. Kf7 Kb1 140. Kg7 Ka1 141. Kf7 Lab1 142. Kg7 Ka2 143. Kf7 Ka3 144. Kg7 La2 145. Kf7 Lcb1 146. Kg7 Tc2 147. Kf7 Tc1 148. Kg7 Lc2 149. Kf7 Ta1 150. Kg7 Lcb1 151. Kf7 Tc2 152. Kg7 Dd5 153. Kf6 Df3 154. Kf7 De2 155. Kg7 Dd2 156. Kf7 Dc1 157. Kg7 Tce2 158. Kf7 Dd2 159. Kg7 Lc2 160. Kf7 Tc1 161. Kg7 Lcb1 162. Kf7 Tc2 163. Kg7 Dc1 164. Kf7 Tcd2 165. Kg7 f3 166. Kf7 Lc2 167. Kg7 Da1 168. Kf7 Lcb1 169. Kg7 Tc2 170. Kf7 Tc1 171. Kg7 Lc2 172. Kf7 Lab1 173. Kg7 Ka2 174. Kf7 Ted2 175. Ke7 Tee2 176. Kd7 Tfe1 177. Kc7 Tgf1 178. Kb7 Lg1 179. Ka7 T4h6 180. Th2 Tb6 181. Tg2 Th7+ 182. Ka8 Thb7 183. Th2 Tb8+
play all play one stop play next play all
Yoav Ben-Zvi: The Move Length Record has been surpassed, after more than 70 years, by Goldsteen, see P1345778. (2019-7-7)
more ...
comment
Keywords: Non-Unique Proof Game, konsekutive Umwandlungen 6 (tltttt), Non-standard material, under-promotion, Move Length Record
Genre: Retro
Reprints: 127 32 personaggi e 1 autore 1955
1438 FIDE Album 1945-1955 1964
Die Schwalbe 3, p. 81, 02/1970
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-6-13 more...
27 - P0001964
Poul Anker Rasch Nielsen
5705 Skakbladet 02/1946
Dr. N. Høeg zum 70. Geburtstag
P0001964
(2+1) cooked
Hilfs-Subretropatt vor 2 Zügen
R: 1. Kh2xTh1 Td1xTh1 2. Kg1xTh2 0-0-0+ (einziger legaler Rückzug nun) 3. g2xSh1=T
play all play one stop play next play all
Cook: (Mario Richter & rawbats):
R: 1. Kg2xTh1 h2-h3 2. Kf2xBg2 Kd2-c1 (einziger legaler Rückzug nun) 3. Kf1xLf2
wegen der Forderung s. P1366564

N. Høeg: "Noget af det smukkeste, jeg har set paa Skakbrættet; at en Stilling med 3 Brikker kan skjule saa nydeligt og genialt et Indhold, er forbavsende." [Eine der schönsten Aufgaben, die ich auf dem Schachbrett gesehen habe; es ist erstaunlich, dass eine Stellung mit drei Steinen einen so schönen und brillanten Inhalt verbergen kann.]
Henrik Juel: In the stipulation, the word 'retropatt' is incorrect. It should be 'sub-retropatt', meaning that Black has just one legal retraction, which is g2xSh1=T. (2003-2-21)
Mario Richter: Inspired by Henrik's question (s. P1366564) I let my program 'rawbats' make some tests, and it claims that there is at least one alternative solution: R: 1. Kg2xTh1 h2-h3 2. Kf2xB(auer)f2 Kd2-c1 (with the only legal retraction now being 3. Kf1xLf2 (which reminds me of P0000454).
If Henrik can confirm the correctness of the claim, the problem should be labeled "cooked". (2019-8-23)
Henrik Juel: It is certainly a nice solution (2. Kf2xBg2, not Bf2)
Well done, rawbats
I am uncertain whether the 'Hultberg conditions' were meant to exclude a solution like this, which involves retracting into a check, but I would consider it a cook (or a second solution...), because the only black retraction is not an uncheck (2019-8-23)
Henrik Juel: I found Høeg's presentation of subretrostalemate in Skakbladet 07/1945 p.107
http://www.skak.dk/images/skakbladet/1945/1945-07.pdf
(I was born before 07/1945, so I should have known...)
The 'Hultberg conditions' are a misnomer, because they are included in Høeg's definition
Much as I admire Niels Høeg's contributions, I find that a stipulation which includes two exceptions is rather dubious
Apparently subretrostalemate problems only appeared for less than a year, and only in Skakbladet; in his seminal work 'Der Blick zurück' Wolfgang Dittmann does not mention them
Høeg' presentation includes two examples, one of which is cooked, and the other is very simple:
Diagram: wKc3, sKf3 (1+1); Help-subretrostalemate in 2
-1.Ke2xTf3 Kc2-c3 -2.Ke1xDe2 Dg2xPe2 allowing -3.e3-e2 only (2019-8-23)
Mario Richter: Since the "official" solution starts with a self-check (Kh2xTh1), it would make the stipulation even more "dubious", as Henrik put it, if self-checks were allowed in the retractions leading to the subretrostalemate position, but not in that position itself.
Conclusion: there is a second solution (in the meantime, rawbats has finished its investigation [computation time 9 hours 46 min]), and it is the only one rawbats found, so a possible correction could be to simply add "2 solutions". (2019-8-24)
Mario Richter: Just to make the literature list a little bit more complete: Høeg's presentation of retrostalemate can be found in Skakbladet 01/1939 p.10ff (2019-8-24)
Henrik Juel: Thanks, Mario
The result of the mini-tourney is in Skakbladet 04/1939 p.67-68
Most of the retrostalemate problems were cooked, but not IV (which is already in the PDB) and I (which is a nice introduction to the genre)
I particularly like the tricky retractor A, which also ought to be in the PDB... (2019-8-24)
Mario Richter: Done, s. P1366626 (2019-8-24)
Henrik Juel: Thanks, Mario; is retrostalemate I also in the PDB? (2019-8-24)
Mario Richter: Now it is - s. P1366629 (2019-8-24)
Henrik Juel: Thank you again (2019-8-24)
comment
Keywords: Help retractor, Castling in the retro play (wg), Kindergarten Problem, Rex solus (s), Promotion in the retro play (t)
Genre: Retro
Reprints: 1423 FIDE Album 1945-1955 1964
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-8-30 more...
28 - P0001972
Thomas R. Dawson
Pittsburgh Post 1923
P0001972
(16+11)
h#1
1. cxd3ep Kd4#

R: 1. d2-d4 Tf5-f6 2. Sf6-g4+ Td5-f5 3. Sf5-h4+ Td3-d5
play all play one stop play next play all
Cook: Henrik Juel (2005-2-18): R: 1. Le5-g3 Tf5-f6 2. Lf6-e5+ Td5-f5 3. d2-d4 Td3-d5 etc.
auch in der Zeitschrift 'The Chess Problem' exakt mit dieser Stellung nachgedruckt, als Quelle ist dort (in der Rubrik "Selected From Early Days") genannt: "Italia Scac. 1917".

Falls tatsächlich ein Druckfehler vorliegt, und der wLg3 nach f2 gehört, scheint er sich weit verbreitet zu haben ...
Henrik Juel: 0... cxd3ep 1.Kd4#. -1.d2 Tf5 -2.Sf6 Td5 -3.Sf5 Td3 etc. (2004-3-18)
Henrik Juel: The problem is incorrect, because the retroplay could also be -1.Le5 Tf5 -2.Lf6 Td5 etc. (2005-2-18)
Mario Richter: Since there's also the even more trivial retroplay 0. ... f2-f3 Tf5-f6 1. Sf6-g4+ Tf3-f5 2. Sf5-h4+, my guess is, that the diagram is misprinted and that the wLg3 should be placed on f2. Doing so, the ep-key is justified, because in this case White's last move must have been d2-d4 (to provide d3 as a square for the black rook). (2011-8-19)
Mario Richter: A sin from my youth - today I do not see how the retroplay can be continued after 0. ... f2-f3 Tf5-f6 1. Sf6-g4+ Tf3-f5 2. Sf5-h4+ ???
So Henrik's cook is the only one, but I still think that my suggested correction is valid. (2019-10-18)
A.Buchanan: I agree with Mario. And it feels like a bug rather than a typo. This problem does not appear in WinChloe, either in source or number of pieces, so I can't check. (2019-10-19)
comment
Keywords: En passant, Last Moves?, Retro Opposition, En passant as key
Genre: h#, Retro
Reprints: The Chess Problem (Whitburn)
Input: Gerd Wilts, 1995-6-3
Last update: Mario Richter, 2019-10-18 more...
29 - P0002178
Niels Høeg
Deutsches Wochenschach 02/06/1907
P0002178
(14+11)
#3 (RV)
1) R: 1. f7-f5: 1. gxf6ep Txa1 2. Sg7+, 1. ... Txh5+ 2. Dxh5+
2) R: 1. d7-d5: 1. cxd6ep Ta5+ 2. Txa5
3) R: 1. T-a8: 1. Sg6
4) R: 1. T-h8: 1. Sc6
5) R: 1. K-e8: 1. Sg6/1. Sc6
play all play one stop play next play all
Henrik Juel: The solution text above should be construed to sit in two columns, reading: If last move was f7-f5, White mates in three by 1.gxf6ep Txa1 2.Sg7+, 1... Txh5+ 2.Dxh5+; if last move was d7-d5, White mates in three by 1.cxd6ep Ta5+ 2.Txa5; etc. (2004-4-26)
Henrik Juel: The pawn captures were
White: axb, d4xc5, fxe, hxg (plus one capture with an officer)
Black: b3xc2 (and possible one more, but not with Pc4)
so last move was not b5xc4 (2019-8-22)
comment
Keywords: En passant, Partial Retro Analysis (PRA), Castling (sksg)
Genre: Retro
Reprints: The Chess Amateur 11, p. 340, 08/1907
99 Retrograde Analysis 1915
34 32 personaggi e 1 autore 1955
(D) Die Schwalbe 13 02/1972
RA65 diagrammes 27 05-06/1977
(19) Die Schwalbe 68 04/1981
Input: Gerd Wilts, 1995-6-3
Last update: Mario Richter, 2019-8-29 more...
30 - P0002213
Anders Olson
Kjell Widlert

7818 Springaren 57 06/1994
P0002213
(2+1)
h-retro-#5
Circe RI
Weiß beginnt
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)
comment
Keywords: Help retractor, Circe, Aristocrat, Rex solus (s)
Genre: Retro, Fairies
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-6 more...
31 - P0002215
Ronald Turnbull
The Problemist 14 07/1994
P0002215
(4+4)
h#2
Fuddled Men
Ceriani Ethics
b) from mating position of a)
a) 1. Ka3 Lf8 2. Tb2 Lg8#
b) 1. Tb4 Lg7 2. Kb2+ Lc4#
play all play one stop play next play all
Author's solution:
"Ceriani Ethics": In the diagram, each piece must be considered as potentially retaining all its powers.
a) Black is in check. 1.Ka3 Bf8 2.Rb2 Bg8.
b) Again, Black is in check. 1.Rb4 Bg7 2.Kb2+ (check to wK from reviving bR!) Bc4. Such paradoxical stipulations have been the subject of an article Quantum Fairies by Peter Fayers and me.
In each twin, the actual past is forgotten, which explains any paradox.
a) we forget that bK can't move (to run away to b1 or c2) at the end of b).
b) we forget that bR can't move (to block on b4) at the end of a).
A.Buchanan: The chess is clear, but I don't really understand the meaning of "Ceriani Ethics" (a kind of PRA?). How would this problem break if the stipulation was under RS rather than PRA? This is so annoyingly vague. (2019-10-8)
comment
Keywords: Fuddled Men, Partial Retro Analysis (PRA)
Genre: h#, Retro, Fairies
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-10-15 more...
32 - P0002289
Andrey Frolkin
Andrej N. Kornilow

2235 diagrammes 92 01-03/1990
P0002289
(14+14)
BP in 21,0
1. f3 Sc6 2. Kf2 Sd4 3. Kg3 Sxe2+ 4. Kg4 Sg3 5. hxg3 Tb8 6. Th5 Ta8 7. Ta5 Tb8 8. Lb5 Ta8 9. c4 Tb8 10. Dc2 Ta8 11. Dg6 Tb8 12. d3 Ta8 13. Ld2 Tb8 14. Le1 Ta8 15. Sd2 Tb8 16. Tc1 Ta8 17. Tc3 Tb8 18. Tca3 Ta8 19. T3a4 Tb8 20. a3 Ta8 21. Lxd7+ Dxd7+
play all play one stop play next play all
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)
more ...
comment
Keywords: Unique Proof Game
Genre: Retro
Reprints: 101 Shortest Proof Games 11/1991
58b 64 Proof Games 2012
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2012-12-26 more...
33 - P0002337
Michel Caillaud
4019 Die Schwalbe 76 08/1982
Dr. Karl Fabel zum Gedenken
1. Preis
P0002337
(11+15)
BP in 47,0
1. Sf3 a5 2. Sd4 a4 3. Sb3 axb3 4. a3 Ta4 5. Ta2 bxa2 6. Tg1 axb1=S 7. Th1 Sc3 8. Tg1 Sd5 9. Th1 Sb6 10. Tg1 Sa8 11. Th1 b6 12. Tg1 La6 13. Th1 Dc8 14. Tg1 Db7 15. Th1 Df3 16. gxf3 h5 17. Lh3 h4 18. Le6 h3 19. La2 Lc4 20. Lb1 La2 21. b3 Thh4 22. Lb2 Thb4 23. Lf6 gxf6 24. Kf1 Lh6 25. Kg1 Kf8 26. Df1 Kg7 27. Dg2+ hxg2 28. h4 Kg6 29. Kh2 g1=S 30. Kg2 Lf4 31. h5+ Kg5 32. Th4 Lh2 33. Td4 Sh3 34. f4+ Kh4 35. Kf3 Sg5+ 36. Ke3 Kh3 37. h6 Kg2 38. h7 Kf1 39. h8=T Ke1 40. Th5 Sh7 41. Ta5 Kd1 42. Ta7 Kc1 43. Tb7 Ta7 44. Kd3 Sa6 45. Tb8 Kb2 46. Tf8 Ka1 47. Kc3 Lg1
play all play one stop play next play all
paul: Cooked: 1.a3 g5 2.Sh3 a5 3.Sf4 a4 4.Sg6 hxg6 5.Ra2 Rh4 6.h3 Rb4 7.h4 Ra5 8.h5 Bg7 9.Rh4 Be5 10.Rd4 Bh2 11.g3 Kf8 12.Bg2 Kg7 13.Bc6 Kf6 14.Kf1 Qf8 15.Kg2 Qh6 16.Kf3 g4+ 17.Ke4 Qf4+ 18.Kd3 Kg5 19.gxf4+ Kh4 20.Qe1 Kh3 21.Qd1 g3 22.Ra1 g2 23.Qh1 g1=S 24.Qd5 Sf3 25.Qc5 Kg2 26.Qc4 Kf1 27.Qc5 Ke1 28.Qc4 Kd1 29.Qd5 Kxc1 30.Qb3 axb3 31.Ra2 bxa2 32.h6 axb1=S 33.h7 Sc3 34.h8=R Sa4 35.Rh5 Sb6 36.Rb5 Sa8 37.Rb6 Sg5 38.Ra6 Sh7 39.Ra7 b6 40.Rb7 Ra7 41.Bd5 Sa6 42.Rb8 Bb7 43.Ba2 Bd5 44.Bb1 Ba2 45.b3 Kb2 46.Rf8 Ka1 47.Kc3 Bg1.[N. Plaksin, Die Schwalbe 130/1991] (2019-9-9)
paul: Comparing the obtained positions, it was observed that the pawn in the g6 must be at f6. A mistake by the publishers, but mine! (2019-9-10)
Henrik Juel: So Michel's proof game is not cooked; yet... (2019-9-10)
comment
Keywords: Unique Proof Game, Length Record, Non-standard material, Promotion (ssT)
Genre: Retro
Reprints: diagrammes 69 01-02/1985
989 FIDE Album 1980-1982 1988
81 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2006-1-10 more...
34 - P0002534
Dirk Borst
Johannes M. Ott

FIDE Kongress 37 Belfort 07/1994
2. Preis
P0002534
(13+15) C+
BP in 15,0
1. h4 e5 2. Th3 Df6 3. Ta3 Lxa3 4. b3 Lxc1 5. Sc3 La3 6. Db1 Lf8 7. Kd1 Ke7 8. Kc1 Kd6 9. Kb2 Kc5 10. De1 Kb4 11. Kc1 Ka3 12. Kd1 Kb2 13. Sd5 La3 14. Se7 Kxa1 15. Sxg8 Lc1
play all play one stop play next play all
(note: location of BK is not thematic for Belfort Theme).
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)
more ...
comment
Keywords: Unique Proof Game, Belfort
Genre: Retro
Computer test: Natch 3.1 (9 hours)
Reprints: (24) diagrammes 110 07-09/1994
8 feenschach 114, p. 414, 12/1994
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-13 more...
35 - P0002554
Unto Heinonen
Probleemblad 1991
1. Preis
P0002554
(16+16) cooked
BP in 43,0
1. h4 a5 2. Th3 Ta6 3. Tf3 Tg6 4. Sh3 Tg3 5. Sf4 Th3 6. g3 Th2 7. Lh3 Sh6 8. Le6 Sf5 9. Lc4 d5 10. Kf1 Th1+ 11. Kg2 Tg1+ 12. Kh3 Kd7 13. Kg4 h5+ 14. Kg5 Th6 15. a4 Tb6 16. T1a3 Kc6 17. Tac3 Tb3 18. Sg6 Ta3 19. b3 Ta2 20. La3 Dd6 21. Lc5 Sd7 22. La7 Sb6 23. Sh8 Kc5 24. La6+ Kb4 25. Tc5 g6 26. Sc3 Lg7 27. Se4 Lc3 28. Da1 Sd4 29. Kh6 Ld7 30. Kg7 Lb5 31. Kf8 Ld3 32. Ke8 Dc6+ 33. Kd8 Sc4 34. Kc8 Ka3 35. Kb8 Lb4 36. Ka8 De8+ 37. Lb8 Sb6+ 38. Ka7 Sa8 39. c4 Sc2 40. Df6 Sa1 41. Db6 Kb2 42. Sf6 Ta3 43. Sd7 Ka2
play all play one stop play next play all
Cook: NL
1. h4 a5 2. Th3 Ta6 3. Tf3 Te6 4. g3 Te3 5. Lh3 Sh6 6. Le6 Sf5 7. Lc4 d5 8. Kf1 Dd6 9. Kg2 Da6 10. Kh3 Kd7 11. Kg4 h5+ 12. Kg5 Th6 13. a4 Tb6 14. Ta3 Kc6 15. Tc3 Tb3 16. Sh3 Ta3 17. b3 Ta2 18. La3 Sd4 19. Lc5 Lf5 20. La7 Db6 21. Sf4 Kc5 22. Sg6 Kb4 23. Sh8 g6 24. La6 Lg7 25. Tc5 Db5 26. c4 Sc2 27. Sc3 Ka3 28. Se4 Lc3 29. Da1 Lb4 30. Df6 Tc3 31. Kh6 Kb2 32. Kg7 Ta3 33. Kf8 Sd7+ 34. Ke8 Ka2 35. Kd8 Sa1 36. Kc8 Sf8+ 37. Kb8 Sd7+ 38. Ka8 Tc1 39. Lb8 Sb6+ 40. Ka7 Sa8 41. Db6 Tg1 42. Sf6 De8 43. Sd7 Ld3 (Sebastian Natschke)
Henrik Juel: Very good, Sebastian (although not for the author...) (2019-6-17)
paul: Probably the cook of the year, because the problem is well known. Which is now the length record for a capture free proof game? (2019-7-16)
paul: The length record for a capture-free PG is now 40.5 moves by Th. le Gleuher (Phenix50/1997) (2019-7-16)
Mario Richter: The problem Paul was referring to is P0006623. So far it has resisted all my and rawbats' attempts to cook it, and I think that chances are high that it is indeed correct ... (2019-7-17)
more ...
comment
Keywords: Unique Proof Game, Length Record, Capture-free
Genre: Retro
Reprints: H17 FIDE Album 1989-1991 1997
feenschach 137 08-09/2000
Input: Gerd Wilts, 1995-6-3
Last update: Olaf Jenkner, 2019-6-18 more...
36 - P0002605
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002605
(7+1)
In welche Richtung ziehen die Bauern?
von oben nach unten
play all play one stop play next play all
Henrik Juel: Last move was g2xDTSh1=L+ (2019-1-1)
comment
Keywords: Promotion
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-26 more...
37 - P0002606
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 13, 1979
P0002606
(7+1)
In welche Richtung ziehen die Bauern?
R: 1. exd6ep+ d7-d5 2. e4-e5+
play all play one stop play next play all
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)
comment
Keywords: En passant, Rex solus (s)
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-3 more...
38 - P0002607
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002607
(3+1) C+
Welches war der letzte Zug?
Weiß am Zug
R: 1. Ka7xSa8
play all play one stop play next play all
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)
comment
Keywords: Last Move? (KxS), Type B, anticipated (P0001033)
Genre: Retro
Computer test: RSP 1.2
Reprints: Schach mit Sherlock Holmes 1982
(1) diagrammes 86 07-09/1988
Outrageous Chess Problems 2005
R1 Phénix 275-276, p. 10810, 07-08/2017
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-3-9 more...
39 - P0002608
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002608
(12+14)
Auf welchem Feld wurde die wD geschlagen?
h6
play all play one stop play next play all
Henrik Juel: The pawn captures are determined, and also their sequence
d7xSe6 first, then a2xLb3, and finally g7xDh6 (2019-1-1)
comment
Keywords: Where was piece x captured?
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-26 more...
40 - P0002609
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002609
(3+1+1)
Welche Farbe hat der Bg3?
Monochromes Schach
sBg3 (Weiß hat rochiert)
play all play one stop play next play all
Henrik Juel: ... and wKb4 got out via h2 and g3 (2019-1-1)
comment
Keywords: Monochrome, Castling (wk)
Genre: Retro, Fairies
Reprints: feenschach 50 04-06/1980
Schach mit Sherlock Holmes 1982
R4 Phénix 275-276, p. 10812, 07-07/2017
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-26 more...
41 - P0002610
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002610
(3+1)
Der wK hat weniger als 14 Züge gemacht. Weise nach, daß eine UW stattgefunden hat!
Monochromes Schach
Der sSb8 kann nur von einem Bauern geschlagen worden sein!
play all play one stop play next play all
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)
comment
Keywords: Monochrome, Promotion
Genre: Retro, Fairies
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-26 more...
42 - P0002612
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 30, 1979
P0002612
(11+10)
Welcher Stein steht auf h4?
(+wLh4!)
R: 1. c7xSd8=T,c7xLd8=T+
(sBhxXg2)
play all play one stop play next play all
Auf h4 steht ein wL
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)
comment
Keywords: Promotion
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
TLG/4 feenschach 173, p. 303, 07-09/2008
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-3 more...
43 - P0002614
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 45, 1979
P0002614
(12+11)
Schwarz ist am Zug. Darf er rochieren?
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)
comment
Keywords: Castling (wksk), Cant Castler
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-3 more...
44 - P0002615
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 46, 1979
P0002615
(6+4)
Weder Weiß noch Schwarz haben mit ihrem letzten Zug einen Stein geschlagen. Schwarz ist am Zug. Darf er rochieren?
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)
comment
Keywords: Castling (wksk)
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-3 more...
45 - P0002616
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 49, 1979
P0002616
(9+9)
Welcher Stein steht auf a5?
(+wKa5)
play all play one stop play next play all
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)
comment

Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-3 more...
46 - P0002618
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 53, 1979
P0002618
(8+10)
Weiß setzt in einem Zug matt.
a) Brettdrehung 90° in Uhrzeigerrichtung (a1->a8) 1. f7xe8=D,L#
b) Brettdrehung 90° gegen die Uhrzeigerrichtung (a1->h1) 1. g4#
play all play one stop play next play all
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)
comment
Keywords: Board Rotation
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
R3 Phénix 275-276, p. 10811, 07-08/2017
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-3 more...
47 - P0002620
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 56, 1979
P0002620
(13+13)
Keine der Damen hat jemals ihre Felderfarbe gewchselt.
Welche Rochaden sind zulässig?
b) ohne wTg1
c) wTg1->h1
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)
comment
Keywords: Castling (wgsg)
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-3 more...
48 - P0002621
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 58, 1979
P0002621
(12+14)
Weiß hat die wD vorgegeben und die beiden wSS sind Originalspringer. Welche Rochaden sind zulässig?
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)
comment
Keywords: Castling (wb), Odds game
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Rainer Staudte, 2019-8-11 more...
49 - P0002625
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 69, 1979
P0002625
(3+2)
Ergänze einen wB auf c4, c5, d4 oder d5, so daß die Partie (mit Schwarz am Zug) nicht remis enden muß!
+wBc4, dann 1. bxc3ep
play all play one stop play next play all
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)
comment
Keywords: Add pieces, En passant, Kindergarten Problem
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-3 more...
50 - P0002626
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 73, 1979
P0002626
(8+1)
Darf Weiß rochieren?
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)
comment
Keywords: Castling (wb), Rex solus (s)
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
R6 Phénix 275-276, p. 10813, 07-08/2017
Input: Gerd Wilts, 1995-6-3
Last update: Rainer Staudte, 2019-8-11 more...
51 - P0002627
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002627
(5+1)
Weiß am Zug. Ist die Stellung legal?
R: 1. Kc6-b6 oder 1. Ka6-b6
play all play one stop play next play all
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)
comment
Keywords: Last Move?, Promotion
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-28 more...
52 - P0002630
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002630
(15+15)
Welche Farbe ist Schwarz, welche Weiß?
Schwarz ist Weiß, Weiß ist Schwarz
play all play one stop play next play all
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)
comment
Keywords: Parity Argument
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-28 more...
53 - P0002631
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002631
(6+9)
Schwarz am Zug darf rochieren. Ergänze einen wB auf f2 oder g2!
+wBf2
play all play one stop play next play all
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)
comment
Keywords: Add pieces, Castling (sk)
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-28 more...
54 - P0002632
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002632
(14+15)
1. Weiß hat einen Turm vorgegeben,
2. die wSS haben noch nicht gezogen,
3. es fanden keine UW statt,
4. der letzte Zug von Weiß war e2-e4.Darf Schwarz rochieren?
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)
comment
Keywords: Conditional problem, Odds game
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-28 more...
55 - P0002633
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 93, 1979
P0002633
(13+15)
Die KK haben noch nicht gezogen. Ergänze einen wB!
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)
more ...
comment
Keywords: Add pieces, Conditional problem
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
56 - P0002634
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 95, 1979
P0002634
(15+14)
Schwarz ist am Zug. Weiß hat zuletzt mit dem Bf4 gezogen; woher kam dieser Bauer? Es befindet sich keine UWF auf dem Brett,
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)
more ...
comment
Keywords: Conditional problem
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
57 - P0002637
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002637
(15+14)
Befinden sich UWF auf dem Brett?
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)
comment
Keywords: Promotion
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-28 more...
58 - P0002638
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes 1979
P0002638
(14+14)
Weiß hat gerade rochiert. Befindet sich UWF auf dem Brett?
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)
comment
Keywords: Promotion, Castling (wksk), Conditional problem
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-28 more...
59 - P0002639
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 113, 1979
P0002639
(11+11)
Weiß setzt matt. In wieviel Zügen?
1. axb6ep#!
play all play one stop play next play all
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)
comment
Keywords: En passant
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
60 - P0002640
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 115, 1979
P0002640
(12+11)
Der weiße König ist unsichtbar.
Schwarz am Zug setzt matt!
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)
comment

Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
R8 Phénix 275-276, p. 10810, 07-08/2017
R8 Phénix 275-276, p. 10814, 07-08/2017
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
61 - P0002641
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 118, 1979
P0002641
(15+15)
Beide Seiten dürfen rochieren.
Ergänze einen wL auf a3 oder a4!
(+wLa3)
R: ?
play all play one stop play next play all
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)
comment
Keywords: Add pieces
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2019-1-5 more...
62 - P0002642
Raymond Smullyan
v The Chess Mysteries of Sherlock Holmes , p. 120, 1979
P0002642
(6+4)
Zeige, daß eine UW und ein ep-Schlag stattgefunden haben!
Weiß am Zug
Monochromes Schach
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)
comment
Keywords: Monochrome, En passant, Promotion
Genre: Retro, Fairies
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
63 - P0002643
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 123, 1979
P0002643
(6+4)
Monochromes Schach?
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)
comment
Keywords: Monochrome
Genre: Retro, Fairies
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
64 - P0002645
Raymond Smullyan
M2 The Chess Mysteries of Sherlock Holmes , p. 146, 1979
P0002645
(3+2)
In den letzten 5 Zügen haben weder der wK noch die Dame einen Zug ausgeführt. Auch sind keine Steine geschlagen worden. Was war der letzte Zug?
R: 1. Lg5-h6 a5-a4 2. Ld8-g5 a6-a5 3. d7-d8=L a7-a6 4. d6-d7 Kg8-h8 5. d5-d6+
play all play one stop play next play all
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)
comment
Keywords: Last Move?
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
65 - P0002646
Raymond Smullyan
M3 The Chess Mysteries of Sherlock Holmes , p. 146, 1979
P0002646
(3+8)
In den letzten 5 Zügen hat kein Bauer gezogen, und es wurden auch keine Figuren geschlagen. Versehentlich fiel der sK zu Boden. Auf welchem Feld muß er stehen?
(+sKc8)
R: 1. ... Td8-h8 2. Kg8-f7 0-0-0 3. Kh8-g8 Sc1-a2,S~ 4. Kg8-h8 Sa2-c1,5. Kh8-g8 Sc1-a2,S~ 6. Kg8-h8
play all play one stop play next play all
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)
comment
Keywords: Add pieces, Castling in the retro play
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
66 - P0002648
Raymond Smullyan
M5 The Chess Mysteries of Sherlock Holmes , p. 147, 1979
P0002648
(14+14)
Keiner der beiden KK hat bisher gezogen.
a) Weisen Sie nach, daß einer der 4 Springer schon einmal gezogen hat, wenn sich auf dem Brett keine weißen UWF befinden.
b) Weisen Sie nach, daß zwei der SS schon einmal gezogen haben, wenn sich auf dem Brett keine schwarzen UWF befinden.
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)
comment

Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
67 - P0002649
Raymond Smullyan
M6 The Chess Mysteries of Sherlock Holmes , p. 148, 1979
P0002649
(14+15)
Weiß darf rochieren. Ist die wDd2 die Originaldame?
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)
comment

Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
68 - P0002650
Raymond Smullyan
M7 The Chess Mysteries of Sherlock Holmes , p. 148, 1979
P0002650
(16+12)
Beide Seiten dürfen noch rochieren. Auf c6 steht eine unbekannte schwarze Figur, aber kein Turm.
a) Angenommen, es ist ein Springer, auf welchem Feld ist dann die fehlende schwarze Dame geschlagen worden?
b) Angenommen, es ist die Originaldame, auf welchem Feld ist dann der fehlende sS geschlagen worden?
c) Angenommen, es ist eine umgewandelte Dame, wo ist dann der schwarze Springer geschlagen worden?
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)
comment

Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
69 - P0002651
Raymond Smullyan
M8 The Chess Mysteries of Sherlock Holmes , p. 149, 1979
P0002651
(15+14)
Schwarz hat im ersten Zug Bd7-d5 gezogen. Der auf f5 stehende Springer hat genau dreimal gezogen. sD, sK und sTh8 haben noch nicht gezogen.
Teil I
A) Weisen Sie nach, daß drei Figuren, die jetzt noch auf ihrem Ausgangsfeld stehen, schon einmal gezogen haben.
B) Hat der Bauer h5 ein- oder zweimal gezogen?
C) Stünde der h-Bauer statt auf h5 auf h6, wäre die Stellung dann noch legal?
Teil II
Angenommen, man entfernt den sLc8 und stellt folgende Bedingungen: Der Läufer soll auf irgendeinem anderen Feld stehen. Er darf aber noch nie auf f7 gestanden haben, er darf auch nicht b7 oder c6 überquert haben, und er darf nicht vor dem weißen Läufer g4 gezogen haben. Auf welchem Feld muß der Läufer dann stehen?
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)
comment

Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-6 more...
70 - P0002652
Raymond Smullyan
M9 The Chess Mysteries of Sherlock Holmes , p. 149, 1979
P0002652
(13+14+1)
Keiner der beiden Könige hat bisher gezogen. Auf f2 steht ein schwarzer oder ein weißer Bauer. Ein weißer Springer steht entweder auf f3 oder f4.
A. Welche Farbe hat der Bauer auf f2?
B. Steht der weiße Springer auf f3 oder f4?
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)
comment
Keywords: Obvious obtrusive promoted material (sLa5)
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-6 more...
71 - P0002653
Raymond Smullyan
M10 The Chess Mysteries of Sherlock Holmes , p. 150, 1979
P0002653
(11+10) C+
Weiß ist am Zug. Kann er in zwei Zügen mattsetzen?
1. Dd6 droht 2. Dxe7#
1. ... exd6 2. Sg7#
1. ... exf6 2. Dxf8,Txf8# (Mattdual)
1. ... Kd8 2. Txf8#
1. ... 0-0-0? (illegal!)
play all play one stop play next play all
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)
comment
Keywords: Castling (sg)
Genre: Retro
Computer test: Popeye 4.61
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-6 more...
72 - P0002654
Raymond Smullyan
The Chess Mysteries of Sherlock Holmes , p. 122, 1979
P0002654
(5+4)
Monochromes Schach
Der wK hat erst zweimal gezogen.
War das Feld h8 mehr als einmal besetzt?
R: 1. g3xTh4 Th8-h4 2. Kf2-e3 Kh7xDg8 3. Dg4-g8+ Tf8-h8 4. h2xLg3 Lh4xTg3 5. Ke1-f2 Kg8-h7 6. Dd1-g4 0-0
play all play one stop play next play all
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)
comment
Keywords: Monochrome, Castling
Genre: Retro, Fairies
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-6 more...
73 - P0002661
Henry Tanner
5094 Thema Danicum 63 07/1991
P0002661
(1+1)
Ergänze sT, sL und sS zu einem IC!
3 Lösungen
a) +sTb4,+sLc4,+sSb3
b) +sTc2,+sLb3,+sSc4
c) +sTd1,+sLb1,+sSc3
play all play one stop play next play all
Henrik Juel: The solutions feature impossible check by T, L, and S (2019-2-14)
comment
Keywords: Illegal cluster, no 8x8 board, only Kings
Genre: Retro, Fairies
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-11-12 more...
74 - P0002678
Hans Gruber
18) feenschach 45, p. 37, 01-03/1979
P0002678
(1+1)
Ergänze 1 wB zu einem IC!
Wieviele Lösungen?
Gitterschach
b) +Ohneschlag

a) 20 Lösungen: a1, ..., h1, a8, ..., h8, c2, e2, a6, h6
b) 34 Lösungen: a1, ..., h1, a8, ..., h8, c2, e2, a6, ..., h6, a7, ..., h7
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)
more ...
comment
Keywords: Illegal cluster, No-capture chess, Grid Chess, only Kings
Genre: Retro, Fairies
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2015-3-20 more...
75 - P0002730
Maryan Kerhuel
991 Phénix 14 09/1991
Michel Caillaud gewidmet
P0002730
(1+3)
Ergänze wD, wL, 3 wBB, sT und 2 sBB zu einem IC!
Patrouilleschach
(+wDb7,wLc6,wBb6d5d7,sTd6,sBb5e7) R: 1. Dc7xb7, aber keine Auflösung
play all play one stop play next play all
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)
comment
Keywords: Illegal cluster, Patrol Chess, Kindergarten Problem, Rex solus (w)
Genre: Retro, Fairies
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-2-15 more...
76 - P0002795
Alexander Kislyak
32 Powerschennii Monarch 1993
P0002795
(9+13) C+
BP in 15,5
1. a4 Sc6 2. Ta3 Sd4 3. Tf3 Sxe2 4. Txf7 Sxc1 5. Txf8+ Kxf8 6. Lc4 De8 7. Lxg8 Txg8 8. f4 Kf7 9. Sf3 Ke6 10. 0-0 Kd5 11. c4+ Kxc4 12. Te1 Kd3 13. Te6 dxe6 14. De2+ Kxe2 15. a5 Sd3 16. Sc3#
play all play one stop play next play all
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)
comment
Keywords: Unique Proof Game, Castling
Genre: Retro
Computer test: Natch 3.1 (57 min.)
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-13 more...
77 - P0003020
Luigi Ceriani
129 La Genesi delle Posizioni 1961
P0003020
(14+14)
Welches war der erste Zug des sBa?
Schwarz darf rochieren.
s.a. 32Pe1A
Yoav Ben-Zvi: Corrects P0005157. (2019-9-2)
Henrik Juel: White captured SxDd8 and an original S by a2xSb3; Black captured [Th1] in its corner and a2xTb1=S
The forward sequence in the NE corner was Sh6-g8, h7-h6, Th8-h7, Sg6-h8; in the SW corner it was Ta1-b1, Sb3-a1, a2xSb3, a3-a2xTb1=S, so the release must start with unpromoting an S on b1
But White has a single retraction available only right now, so the retroplay must begin with Black: -1... Sh1 -2.h2 Sg3xTh1 -3.Tg1
The parity of the number of white moves from game start to diagram position is: TT uneven, SS even, PP even, for a total of uneven
As Black made last move, the black parity must also be uneven: [Th8] uneven, SSS uneven (no matter which S is promoted and whence the other diagram S came), [Ph7] uneven, for a subtotal of uneven, so [Pa7] made an even number of moves to promotion, i.e. 6, with first move a7-a6 (2019-9-2)
comment
Keywords: Parity Argument, First Move?, Excelsior
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Yoav Ben-Zvi, 2019-9-2 more...
78 - P0003049
Raymond Smullyan
The Chess Mysteries of the Arabian Knights , p. xii, 1981
P0003049
(14+15)
Ein sL hat einen weißen Stein geschlagen. Welcher sL?

Der sLe5 hat den wL geschlagen.
H.Juel: The capture sequence must have been g7xSh6 before d2xTc3 before Lx[Lc1]
more ...
comment

Genre: Retro
Reprints: Die Schachgeheimnisse des Kalifen , p. 10, 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
79 - P0003050
Raymond Smullyan
The Chess Mysteries of the Arabian Knights , p. xiii, 1981
P0003050
(5+1)
In welche Richtung spielt Weiß?
Im letzten Zug wurde kein Stein geschlagen.
R: 1. Td2-f2 Kf2-f1 2. Be2xXd1=S+
play all play one stop play next play all
Weiß spielt von oben nach unten!
Henrik Juel: The stated retroplay is the only possibility
so White pawns move downwards (2019-1-23)
comment
Keywords: Conditional problem, Aristocrat, Rex solus (s)
Genre: Retro
Reprints: Die Schachgeheimnisse des Kalifen , p. 12, 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
80 - P0003053
Raymond Smullyan
3a The Chess Mysteries of the Arabian Knights 1981
P0003053
(12+11)
Ersetze einen der Steine durch den wK, so daß die Stellung legal ist!
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)
comment
Keywords: Add pieces
Genre: Retro
Reprints: 3a Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
81 - P0003054
Raymond Smullyan
3b The Chess Mysteries of the Arabian Knights 1981
P0003054
(12+10)
Ersetze einen Stein einer anderen Art als in der vorigen Aufgabe 3a durch den wK, so daß die Stellung legal ist!
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)
more ...
comment
Keywords: Add pieces
Genre: Retro
Reprints: 3b Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
82 - P0003055
Raymond Smullyan
4 The Chess Mysteries of the Arabian Knights 1981
P0003055
(11+11)
Schwarz ist am Zug. Welcher Stein steht auf g4?
Darf Schwarz rochieren?

Steinkontrolle wegen des 'x' nicht exakt.
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)
comment

Genre: Retro
Reprints: 4 Die Schachgeheimnisse des Kalifen 1984
R10 Phénix 275-276, p. 10810, 07-08/2017
R10 Phénix 275-276, p. 10815, 07-08/2017
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
83 - P0003058
Raymond Smullyan
7 The Chess Mysteries of the Arabian Knights 1981
P0003058
(12+13)
Der schwarze Damenturm ist von einem Bauern geschlagen worden, der weiße Königsläufer wurde vorgegeben, Die beiden wT sind auf der selben Reihe geschlagen worden. Schwarz kann noch rochieren.
Wo wurden die wTT geschlagen?
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)
more ...
comment
Keywords: Conditional problem, Odds game
Genre: Retro
Reprints: 7 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
84 - P0003059
Raymond Smullyan
8 The Chess Mysteries of the Arabian Knights 1981
P0003059
(13+13+1)
Schwarz hat soeben rochiert und es sind keine weißen Umwandlungsfiguren auf dem Brett.
Ist der Ta5 weiß oder schwarz?

Ta5 is black
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)
more ...
comment
Keywords: Non-standard material (l)
Genre: Retro
Reprints: 8 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
85 - P0003061
Raymond Smullyan
10 The Chess Mysteries of the Arabian Knights 1981
P0003061
(14+13+1)
Kein Bauer hat bisher mehr als einen Stein geschlagen, die Dh5 ist eine Originaldame. Zeige, daß sich die Farbe der Dh5 bestimmen läßt, wenn bekannt ist, ob die fehlende Dame auf ihrer eigenen Linie geschlagen worden ist!
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)
more ...
comment
Keywords: Conditional problem
Genre: Retro
Reprints: 10 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
86 - P0003062
Raymond Smullyan
11 The Chess Mysteries of the Arabian Knights 1981
P0003062
(11+10)
Woher kommt der wBg3?
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)
comment

Genre: Retro
Reprints: 11 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
87 - P0003063
Raymond Smullyan
12 The Chess Mysteries of the Arabian Knights 1981
P0003063
(13+15)
Weiß darf noch rochieren. Hat sich der sBb7 umgewandelt?
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)
comment
Keywords: Conditional problem
Genre: Retro
Reprints: 12 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
88 - P0003064
Raymond Smullyan
13 The Chess Mysteries of the Arabian Knights 1981
P0003064
(14+15)
Beide Könige haben nur je einmal gezogen.
Der sBa7 hat sich umgewandelt; in welche Figur?
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)
comment
Keywords: Conditional problem
Genre: Retro
Reprints: 13 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
89 - P0003065
Raymond Smullyan
14 The Chess Mysteries of the Arabian Knights 1981
P0003065
(13+14)
Die schwarze Dame wurde auf ihrer eigenen Reihe und der weiße Damenturm auf seiner Linie geschlagen, keiner der beiden Könige hat gezogen, Schwarz darf rochieren. Was geschah mit dem weißen Bauern h2, der sich umgewandelt hat?
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)
comment
Keywords: Conditional problem
Genre: Retro
Reprints: 14 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
90 - P0003066
Raymond Smullyan
15 The Chess Mysteries of the Arabian Knights 1981
P0003066
(15+13)
Welcher der beiden wL ist der Originalläufer?
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)
comment
Keywords: Non-standard material (L)
Genre: Retro
Reprints: 15 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-28 more...
91 - P0003067
Raymond Smullyan
16 The Chess Mysteries of the Arabian Knights 1981
P0003067
(14+13)
Der Th8 ist auf auf seinem Ursprungsfeld geschlagen worden, keiner der beiden KK hat gezogen, auf dem Brett befinden sich keine schwarzen Umwandlungsfiguren. Welches ist der weiße Originalläufer?
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)
comment
Keywords: Conditional problem, Non-standard material (L)
Genre: Retro
Reprints: 16 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-30 more...
92 - P0003068
Raymond Smullyan
17 The Chess Mysteries of the Arabian Knights 1981
P0003068
(4+4)
Was geschah mit dem weißen Königsläufer?
Henrik Juel: solution: [Lf1] was captured on f1
-1... f4xg4ep -2.g2 f5
. (2019-1-26)
comment
Keywords: Non-standard material (L)
Genre: Retro
Reprints: 17 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-28 more...
93 - P0003069
Raymond Smullyan
18 The Chess Mysteries of the Arabian Knights 1981
P0003069
(11+11)
Welcher weiße Stein wurde vorgegeben?
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)
comment
Keywords: Odds game
Genre: Retro
Reprints: 18 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-3-5 more...
94 - P0003070
Raymond Smullyan
18b The Chess Mysteries of the Arabian Knights 1981
P0003070
(11+11)
Es fanden keine Umwandlungen statt.
Welcher weiße Stein wurde vorgegeben?
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]) (2019-1-27)
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. (2019-1-28)
comment
Keywords: Conditional problem, Odds game
Genre: Retro
Reprints: 18b Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-28 more...
95 - P0003072
Raymond Smullyan
20 The Chess Mysteries of the Arabian Knights 1981
P0003072
(14+14)
Es befinden sich keine UWF auf dem Brett.
Zeige, daß der weiße Königsspringer bereits gezogen hat!
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
. (2019-1-27)
comment
Keywords: Conditional problem
Genre: Retro
Reprints: 20 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-28 more...
96 - P0003073
Raymond Smullyan
21 The Chess Mysteries of the Arabian Knights 1981
P0003073
(13+10)
Einer der beiden weißen Springer hat noch nicht gezogen. Welcher?
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?)
. (2019-1-27)
comment
Keywords: Conditional problem
Genre: Retro
Reprints: 21 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-3-5 more...
97 - P0003074
Raymond Smullyan
22 The Chess Mysteries of the Arabian Knights 1981
P0003074
(5+12+1)
Schwarz ist am Zug. Ist der Sa1 weiß oder schwarz?
Henrik Juel: solution: Sa1 is white
Last move was Kd1xSd2, so the black knights are accounted for
. (2019-1-27)
comment

Genre: Retro
Reprints: 22 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2019-1-28 more...
98 - P0003075
Raymond Smullyan
23 The Chess Mysteries of the Arabian Knights 1981
P0003075
(12+10)
Ein weißer und ein schwarzer Springer haben die Farbe getauscht. Welche Springer?
Henrik Juel: solution: Sa1 and Sa3
White pawns captured all missing black men, so last move was Sc2-a3+, preceded by b2xa1=S
. (2019-1-27)
more ...
comment

Genre: Retro
Reprints: 23 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-28 more...
99 - P0003083
Raymond Smullyan
30 The Chess Mysteries of the Arabian Knights 1981
P0003083
(12+14)
Eine Figur hat die falsche Farbe. Welche?
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
. (2019-1-27)
comment

Genre: Retro
Reprints: 30 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-3-7 more...
100 - P0003084
Raymond Smullyan
31 The Chess Mysteries of the Arabian Knights 1981
P0003084
(13+11)
Weiß ist am Zug, beide KK haben noch nicht gezogen. Zeige, daß der wLg7 bereits die 8. Reihe betreten hat!
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) (2019-1-29)
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 (2019-1-29)
comment
Keywords: Conditional problem
Genre: Retro
Reprints: 31 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-3-7 more...
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