2263 problem(s) found in 875 milliseconds (displaying 100 problem(s)).

1 - P0000007
Andrej N. Kornilow
Andrey Frolkin
Dmitri W. Pronkin

5408 Die Schwalbe 97 02/1986
P0000007
(15+15)
BP in 33,0
1. Sc3 Sc6 2. Se4 Se5 3. c3 c6 4. Da4 Da5 5. b3 b6 6. La3 La6 7. Lc5 Lc4 8. Le3 Le6 9. Lh6 Lh3 10. gxh3 gxh6 11. h4 h5 12. Lh3 Lh6 13. Lf5 Lf4 14. Sh3 Sh6 15. Tg1 Tg8 16. Tg3 Tg6 17. Te3 Te6 18. f3 f6 19. Kf2 Kf7 20. Tg1 Tg8 21. Tg4 Tg5 22. Sg3 Sg6 23. Lb1 Lb8 24. Td4 Td5 25. Tdd3 Tdd6 26. Df4 Df5 27. Te4 Te5 28. Ke3 Ke6 29. Sf2 Sf7 30. Sg4 Sg5 31. Sh6 Sh3 32. Sg8 Sg1 33. Dh6 Dh3
play all play one stop play next play all
Henrik Juel: After a couple of hours Natch 3.1 had not even found the symmetrical solution (2018-12-7)
comment
Keywords: Unique Proof Game, Symmetrical position
Genre: Retro
Reprints: 581 Ukrainisches Album 1986-1990
111 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Nikolai Beluhov, 2011-5-6 more...
2 - 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...
3 - 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...
4 - P0000111
Andrey Frolkin
Leonid Lyubashevsky

6330 Die Schwalbe 112 08/1988
4. ehrende Erwähnung
P0000111
(28+0)
Färbe die Steine!
BP in 11,0
1. h4 f5 2. Th3 Kf7 3. Tb3 De8 4. Tb6 axb6 5. g3 Ta3 6. Lg2 Tc3 7. Lc6 dxc6 8. dxc3 Le6 9. Dd8 Lb3 10. axb3 Dd7 11. Ta8 Dd1+
play all play one stop play next play all
Henrik Juel: C+ by Natch 3.1
The colored position is 'natural' except for wTa8, wDd8, and sDd1 (2018-12-7)
comment
Keywords: Unique Proof Game, Coloring problem, Interchange
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2005-12-27 more...
5 - P0000127
Frank Christiaans
6573 Die Schwalbe 116 04/1989
P0000127
(8+11)
#3
1. Kg7xf6 Th6+ 2. Sxh6 ... 3. c8=D,T#
play all play one stop play next play all
s. P0000133
NN: It seems we fall short of one retro-move and fail to keep either the bK or bR still.
Let's analyse, the distribution of bPs indicate 5 captures (bPh7×g6×f5×e4×d3×c2) plus 3 captures (bPc7×d6, bPb7×c6, bg7×f6) equalling a total of 8 captures, thus accounting for all of white's missing pieces. Next, wPs captured at least 3 (fxexdxc) plus wPc2×b3. Now, in order to keep the bK & bR still, we require sufficient supply of retro-moves for black. Thus, we must strive to release the bNa1 (as pawns are not enough!) at the earliest. But this can only be done through uncapturing the bishop on c1 followed by retracting bPd3×c2, wPd2-d3. An optimal way of doing this may be: -1.bPb7×wNc6 wNc6-d4, -2.bPa4-a3 wPc6-c7, -3.bPc7×wBd6 wBd6-b4, -4.bPa5-a4 wBb4-d2 -5.bPe6-e5 wBd2-c1 and here we can't retract anymore moves without violating bK's right to castle. Retracting bPa6-a5 for instance would mean that bRa8 got out of the pawn to be captured by a wP (as bBc8 never left its homesquare) forcing the bKe8 to move.
Thus, Black can't castle and the solution is 1.Kxf6 Rh6+ (1...0-0 is illegal) 2.Nxh6 ... 3.c8R/Q# (2018-11-30)
Mario Richter: If we assume that Black can still Castle, White needs a fifth capture to provide wPh2 as a capture object.
Furthermore, before d2-d3 can be retracted, both wRa1 and wBc1 have to return home (to b1 and c1) (2018-12-1)
comment
Keywords: Castling (sk), Cant Castler
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2014-1-15 more...
6 - P0000129
Manfred Seidel
6575 Die Schwalbe 116 04/1989
P0000129
(13+12) cooked
BP in 18,5
1. c4 Sa6 2. c5 Sxc5 3. Sf3 Se4 4. Se5 Sg3 5. Sxd7 e5 6. Sb6 axb6 7. hxg3 Ta3 8. Txh7 Tf3 9. gxf3 Lf5 10. Lh3 Lxh7 11. Lc8 f5 12. d3 Kf7 13. Le3 Kg6 14. Sd2 Kh5 15. Tc1 g6 16. Tc4 Lh6 17. Te4 Df8 18. Ld4 Le3 19. Le6
play all play one stop play next play all
Henrik Juel: Here is one of the three similar cooks
1.Sg1-f3 Sb8-c6 2.Sf3-e5 Sc6-d4 3.Se5xd7 Be7-e5
4.Sd7-b6 Ba7xb6 5.Sb1-c3 Ta8-a3 6.Sc3-b1 Ta3-g3
7.Bh2xg3 Sd4-f3 8.Bg2xf3 Lc8-f5 9.Th1xh7 Lf5xc2
10.Lf1-h3 Lc2xh7 11.Lh3-c8 Bf7-f5 12.Bd2-d3 Ke8-f7
13.Lc1-e3 Kf7-g6 14.Sb1-d2 Kg6-h5 15.Ta1-c1 Bg7-g6
16.Tc1-c4 Lf8-h6 17.Tc4-e4 Dd8-f8 18.Le3-d4 Lh6-e3
19.Lc8-e6 (2018-12-7)
comment
Keywords: Unique Proof Game
Genre: Retro
Reprints: 146 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2004-8-27 more...
7 - 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...
8 - P0000143
Nikita M. Plaksin
6689 Die Schwalbe 118 08/1989
4. Lob
P0000143
(13+9)
Löse die Stellung auf!
Madrasi RI
R: 1. Se6xDd8 Bd2xTe1=L 2. Tb1-e1 c3xTd2 3. Tb8-b1 b4xTc3 4. b7-b8=T 5. a6xT(S)b7
play all play one stop play next play all
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
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-15 more...
9 - P0000172
Dmitri W. Pronkin
6991v Die Schwalbe 123 06/1990
P0000172
(13+13)
BP in 24,5
1. Sc3 Sf6 2. Se4 Sd5 3. Sg5 Sc3 4. Sxh7 Sb1 5. Sxf8 Th5 6. h4 Tg5 7. hxg5 Kxf8 8. Th6 Kg8 9. Tb6 Kh7 10. g6+ Kh6 11. e4 Kg5 12. Lb5 Kf4 13. Ke2 Ke5 14. Kf3 Kd4 15. Kf4 Kc5 16. b4+ Kxb4+ 17. La3+ Kxa3 18. Df3+ Kb2 19. a4 Kc1 20. Se2+ Kd1 21. Sc1+ Ke1 22. Sd3+ Kf1 23. Sb2+ Kg1 24. Lf1 Kh1 25. Dd3
play all play one stop play next play all
Henrik Juel: The long promenade is made by k, not K (2018-12-7)
comment
Keywords: Unique Proof Game, Promenade (K)
Genre: Retro
Reprints: 579 Ukrainisches Album 1986-1990
113 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2005-12-27 more...
10 - P0000198
Wolfgang Dittmann
Thomas Kolkmeyer

7170 Die Schwalbe 126 12/1990
P0000198
(12+12) cooked
KBP in 16.5, Stellung nach dem 17. Zug von Weiß.
Auf welchem Feld wurde die weiße Dame geschlagen?
Duellantenschach
b) Bg4 nach g2
c) Be7 nach g5
a) 1. Sc3 Sf6 2. Se4 Sd5 3. Sd6+ cxd6 4. Sf3 Sb4 5. Sd4 Sd3+ 6. cxd3 Sc6 7. Dc2 Sa5 8. Dxc8 Sb3 9. Dc4 Sxc1 10. Db3! Sxb3 11. Se6 Sd4 12. Sf4 Sf3+ 13. exf3 Da5 14. g4 Dg5 15. Sd5 Df6 16. Sxf6+ gxf6 17. b4
b) 1. Sc3 Sf6 2. Sa4 Sd5 3. Sb6 Sf6 4. Sxc8 Sd5 5. Sd6+ cxd6 6. Sf3 Sb4 7. Sd4 Sd3+ 8. cxd3 Sc6 9. Se6 Sd4 10. Sf4 Sf3+ 11. exf3 Da5 12. Da4! Dxa4 13. b3 Db4 14. Lb2 Dxf4 15. Lc3 Df6 16. Lxf6 gxf6 17. b4
c) 1. g4 g5 2. Sc3 Sh6 3. Se4 Sf5 4. Sf6+ exf6 5. Sh3 Sd4 6. Sf4 Sf3+ 7. exf3 Sc6 8. Sd5 Sa5 9. Sb6 Sb3 10. Sxc8 Sxc1 11. Sd6+ cxd6 12. De2+! Sxe2 13. b3 Sc1 14. b4 Sd3+ 15. cxd3 De7+ 16. Kd1 De1+ 17. Kxe1
play all play one stop play next play all
Cook: a) Dualistisch ab 7. Zug:
1. Sc3 Sf6 2. Se4 Sd5 3. Sd6+ cxd6 4. Sf3 Sb4 5. Sd4 Sd3+ 6. cxd3 Sc6 7. Se6 Sd4 8. Sf4 Sf3+ 9. exf3 Db6 10. b4 Db5 11. De2 Dc6 12. Dd1 Dxc1 13. Dxc1 Tb8! 14. Dxc8+ Txc8 15. Sd5 Ta8 16. Sf6+ gxf6 17. g4 (G. Wilts)
paul: b)1.Sc3 Sa6 2.Sb5 Sb4 3.Sd4 Sd3+ 4.cxd3 Sf6 5.Sf5 Sd5 6.Sd6+ cxd6 7.Sh3 Sb4 8.Sf4 Sc2+ 9.Qxc2 Qc7 10.Qb3 Qxc1+ 11.Qd1 Qc6 12.Qc2 Qf3 13.Qxc8+! Rxc8 14.exf3 Rd8 15.Sh5 Ra8 16.Sf6+ gxf6 17.b4 (Jacobi). So, wQ could be taken at c8. (2018-9-18)
paul: c)The position could be obtained even faster: 1.Sc3 Sh6 2.Se4 Sf5 3.Sf6+ exf6 4.Sh3 Sd4 5.Sf4 Sf3+ 6.exf3 g6 7.g4 g5 8.Sg6 Qe7+ 9.Sxe7 Sc6 10.Sxc8 Sd4 11.Sd6+ cxd6 12.Qe2+ Sxe2 13.b3 Sxc1 14.b4 Sd3+ 15.cxd3.(Jacobi) (2018-9-18)
more ...
comment
Keywords: Castling (wgsg), Single Combat, Where was piece x captured?, Unique Proof Game
Genre: Retro, Fairies
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-2-7 more...
11 - P0000231
Jasper van Atten
7376 Die Schwalbe 129 06/1991
3. Preis
P0000231
(16+13)
BP in 20,0
1. h4 a5 2. h5 a4 3. h6 Ta5 4. hxg7 Sh6 5. g8=T Sa6 6. Tg3 Tg8 7. c4 Tg6 8. c5 Lg7 9. c6 Kf8 10. cxb7 De8 11. b8=T Lb7 12. Td8 d6 13. Td7 Dd8 14. b3 Ke8 15. Lb2 Lf8 16. Lg7 Tf5 17. Tc3 Tf3 18. exf3 Sb8 19. La6 Lc8 20. Lb7 Sg8
play all play one stop play next play all
Henrik Juel: C+ by Natch 3.1 (5 min.) (2018-12-7)
comment
Keywords: Unique Proof Game, Non-standard material, Promotion (TT)
Genre: Retro
Reprints: 131 Shortest Proof Games 11/1991
feenschach 137 08-09/2000
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2006-1-9 more...
12 - 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...
13 - P0000372
Andrey Frolkin
8422 Die Schwalbe 145 02/1994
1. ehrende Erwähnung
P0000372
(14+12)
BP in 23,5
1. Sh3 g5 2. Sf4 g4 3. h3 g3 4. Th2 gxh2 5. Sh5 h1=S 6. f4 Sg3 7. f5 Sxf1 8. f6 Sh2 9. fxe7 f5 10. Kf2 f4 11. Dg1 f3 12. Kg3 Sf6 13. Dxa7 c5 14. Sc3 Sc6 15. Db8 Ta3 16. Sa4 Tc3 17. bxc3 f2 18. La3 f1=T 19. Lb4 Th1 20. Tf1 Db6 21. Tf3 Sd8 22. Te3 Sf1+ 23. Kf3 De6 24. exd8=T+
play all play one stop play next play all
"Uff, das war bannig oder sakrisch schwierig, je nach landschaftlicher Mundart. Daß die schwarzen UW-Steine den Platz tauschen, war wohl das Uranliegen des Autors. Aber dazu hat er sich doch ganz Hübsch etwas einfallen lassen: Falsche Fährten legen und Unerwünschtes verunmöglichen - das ist ihm bestens gelungen" (HHS) "Dieser Platztausch der umgewandelten Figuren fasziniert mich sehr" (RS) 3,5/III/6L.
Henrik Juel: Unfortunately cooked
1.Sg1-h3 Sb8-c6 2.Bf2-f4 Sc6-e5 3.Bf4-f5 Bg7-g5
4.Sh3-f4 Bg5-g4 5.Bf5-f6 Bg4-g3 6.Bf6xe7 Bf7-f5
7.Sf4-h5 Bf5-f4 8.Bh2-h3 Bf4-f3 9.Th1-h2 Bg3xh2
10.Ke1-f2 Bh2-h1=T 11.Kf2-g3 Th1xf1 12.Sb1-c3 Tf1-h1
13.Dd1-g1 Se5-g4 14.Dg1xa7 Sg4-h2 15.Da7-b8 Ta8-a3
16.Sc3-a4 Ta3-c3 17.Bb2xc3 Bf3-f2 18.Lc1-a3 Bf2-f1=D
19.La3-b4 Df1-f5 20.Ta1-f1 Df5-e6 21.Tf1-f3 Sg8-f6
22.Tf3-e3 Sh2-f1 23.Kg3-f3 Bc7-c5 24.exd8=T+ (2018-12-7)
comment
Keywords: Unique Proof Game, Interchange, Promotion (ts)
Genre: Retro
Reprints: H22 FIDE Album Annexe 1992-1994 2003
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2005-12-27 more...
14 - P0000393
Pascal Wassong
8552 Die Schwalbe 147 06/1994
2. ehrende Erwähnung
P0000393
(10+16) C+
BP in 23,0
1. h4 Sc6 2. Th3 Se5 3. Tb3 Sg4 4. Tb6 axb6 5. c4 Ta3 6. Da4 Th3 7. g3 Th1 8. Lg2 Sh2 9. Lc6 dxc6 10. Kd1 Lh3 11. Kc2 e6 12. Kb3 Lb4 13. c5 Lxd2 14. Db4 Lh6 15. Le3 Dxh4 16. Sd2 f6 17. Tc1 Kf7 18. Tc4 Kg6 19. Tg4+ Kh5 20. Tg6 hxg6 21. Lg5 fxg5 22. Ka4 Sf6 23. b3 Sh7
play all play one stop play next play all
more ...
comment
Keywords: Unique Proof Game, Ornament
Genre: Retro
Computer test: Henrik Juel: C+ by Natch 3.1 (4 min.) (2018-12-8)
Reprints: (I) Phénix 91 12/2000
Input: Gerd Wilts, 1995-6-3
Last update: Erich Bartel, 2018-12-8 more...
15 - P0000400
Waleri A. Liskowez
4295 Die Schwalbe 80 04/1983
P0000400
(14+9)
#2 BG
3 partial solutions in a specific order!

I 1. fxe6ep a try by the BG-logic (beruehrt-gefuehrt) with no
justification 1. ... 0-0-0/- 2. Qc4#/Rxg8# If rejected, the key must be
returned back:
II Unless the touched wP has a move:
1. Nxe5! (1. ... 0-0-0?? illegal)
If rejected again: the touched wP does have a move (ep!) and, thus, W must make it:
III 1. fxg6ep fxg6+ 2. f7#! (1. ... 0-0-0??)
NOT:
II 1. Nc5?? was illegal by the rule of touching chessmen.
(II 1. Kxe5?).
The last move g6(7)-g5 locks wRh8. ... Thus, castling implies ep
play all play one stop play next play all
A.Buchanan: Valery is one of the most rigorous and imaginative retro composers but very often I can’t follow his solutions as they omit too many intermediate steps. Here is an example. Can someone explain the details please? (2018-10-5)
VL: Indeed, some extra details need for the understanding, especially at present. Let me try to clarify a bit.
First of all, BG stands strongly instead of PRA. Restipulated as a PRA problem, this problem would have been soluble (of "Typ Oeffner", aka "a priori", with two partial solutions) but unsound since both 1.Sc5 and 1.Sxe5 solve if castling is illegal. BG always works with PRA-like problems (sound or not).
BG is a hybrid controversial genre that allows to find and justify partial solutions by means of some active actions (in the spirit of RS and AP!) that aren't necessarily legal moves at the moment when they are fulfilled. Namely, BG permits to introduce move attempts into the actual solution as if the moves were fulfilled OTB (of course only meaningful attempts are taken into account; in fact only ep captures or around). Such an attempt forces the referee/opponent to specify the prehistory: the attempt may be adopted (one partial solution), or it may be rejected as contradicting the prehistory of the position (another partial solution). However (contrary to usual PRA variations), in the latter case, the rejected move should be replaced by a move in accordance with the rule of touching pieces.
Now, it is assumed that any capture is made physically in exactly one way: both participating pieces are touched simultaneously...
It is worth adding to the provided solution that wrong (premature) would be the second try II 1.fxg6ep? since after rejection, White were forced to replace it by the capture of the touched bPg5 by an officer, with no mate!
In general, in my opinion, quite a natural logic. However, a very sophisticated scenario with the unexpected rejection of the second (not ep!) try of capturing Pe5. Simpler are my earlier BG-problems including ones considered in my old article Anwendung der `ber"uhrt-gef"uhrt'-Regel in Retroproblemen, Die Schwalbe, 1981, H.69. (2018-10-7)
A.Buchanan: Valery, I love you, but I don't find this sufficient. It is probably a good high level reminder if addressed to someone who basically understands BG,PRA,RS,AP intimately already. It has helped me to expand the definition of the Touch Move keyword. But it does not address my request to make the solution sufficiently detailed. My perspective is almost always to render our art comprenensible by intelligent newcomers from facebook, whatsapp, chess.com etc. It is insufficient to explain in terms of PRA,RS,AP, because those are insufficiently defined here in PDB. You said this is a complicated one: I will pick a simpler one, and have a go to solve that. If the best tutorial is in your 1981 article, maybe you can make that available online, or send it to me, and I will publicize it. Thanks so much! (2018-10-7)
comment
Keywords: Touch Move, Castling (sg), Promoted material on the board (L)
Genre: Retro, 2#
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-10-7 more...
16 - 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...
17 - P0000849
Gerd Rinder
184 Die Schwalbe 4 04/1970
P0000849
(2+16)
#2 (RV)
1. Dxg7 oder 1. Dc7
play all play one stop play next play all
Eine der beiden schwarzen Rochaden ist unmöglich, da entweder wBa2 oder wBh2 auf a8/h8 umgewandelt haben müssen, um als Schlagobjekt zur Verfügung zu stehen.
Sally: °
Das Diagramm zeigt zwei Problemstellungen,eine mit zulässiger
s0-0. Die gesamte schwarze Mannschaft steht auf dem Brett. Die
sBB schlugen von rechts nach links 12 mal, dazu axb = 1,
zusammen 13 mal. Weiß vermisst zwar 14 Steine. doch konnten seine Randbauern nicht von den sBB geschlagen werden. Einer davon müsste sich umwandeln, um als Schlagobjekt dienen zu können, der andere fiel duech eine schwarze Figur. So wurde die s0-0 bei Umwandlung des wBh verwirkt (Stellung a), die
s0-0-0 bei Umwamdlung des wB5b (Stellung b). Jeweils eine Rochade bleibt zulässig.
a) 1. Dc7! (2. De7#) - Kf8, 2. Df7#.
.. 1. - Lf8,2. Dd7#; 0-0 unzulässig!
b) 1. Dxg7!(2. De7#) - Kd8, 2. Dd7#.
.. 1. - Th7,2. Dg8#; 0-0-0 unzulässig. (2018-8-29)
comment
Keywords: Castling (sksg), Partial Retro Analysis (PRA)
Genre: Retro
Reprints: Problem 148-151 11/1972
D19 feenschach 27, p. 27, 04/1975
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-9-7 more...
18 - P0001110
Michel Caillaud
3396 Die Schwalbe 67 02/1981
4. Preis
P0001110
(14+12)
Wieviele Züge sind seit dem letzten Bauernzug mindestens geschehen?
Renny Bosch: I believe the minimum number of moves since the last pawn move is 68, based on the following PG (or SPG?): 1. g4 Nc6 2. Bg2 Na5 3. Bc6 Nb3 4. axb3 Nf6 5. Nc3 Ne4 6. Nd5 Nc3 7. dxc3 Rg8 8. Qd4 Rh8 9. Qb6 axb6 10. Be3 Rg8 11. Bc5 Rh8 12. Bb4 Rg8 13. Ra5 Rh8 14. Rc5 bxc5 15. Ba5 Rg8 16. Bb6 Rh8 17. Ba7 Rg8 18. Bb8 Rh8 19. Ba4 Ra6 20. Kd2 Re6 21. Kd3 Re5 22. Kc4 Re4+ 23. Kb5 Rc4 24. bxc4 Rg8 25. Bb3 Rh8 26. Nh3 Rg8 27. Ra1 Rh8 28. Ra7 Rg8 29. Nb6 Rh8 30. Na8 Rg8 31. Nf4 Rh8 32. Nd5 Rg8 33. Ndb6 h6 34. Na4 b6 35. Ba2 Bb7 36. Bb1 Be4 37. g5 Bh7 38. g6 Rh8 39. gxh7 { the last pawn move } g6 40. Rb7 Bg7 { move bB, bK, bQ out of row 8 } 41. Ba7 Be5 42. Rb8 Kf8 43. Ka6 Kg7 44. Kb7 Qg8 45. Rf8 Kf6 46. Kc8 { allow bR to pass wK } Qg7 47. Rg8 Ke6 48. Kd8 Qf6 49. Ke8 Qf5 50. Kf8 Bf4 51. Kg7 Qg5 { reverse order of the two rooks }52. Rb8 Rc8 53. Rb7 Rd8 54. Bb8 Rc8 55. Ra7 Rd8 56. Ra5 Rc8 57. Ba7 Rb8 58. Rb5 Rb7 59. Bb8 Ra7 60. Rb3 Ra5 61. Ra3 Rb5 62. Ra1 Rb3 63. Ba2 Ra3 64. Bb3 Ra2 65. Rd1 Ra1 66. Rd2 Re1 67. Rd1 Rf1 68. Ra1 Rb1 69. Ra3 Rc1 70. Ba2 Ra1 71. Bb1 Ra2 72. Rb3 Ra3 73. Rb5 Rb3 74. Ra5 Ra3 75. Ra7 Rb3 76. Rb7 Ra3 77. Ba7 Rb3 78. Rb8 Ra3 79. Rh8 Qf6+ 80. Kf8 { move bK past wR } Qg7+ 81. Ke8 Kf6 82. Kd8 Be5 83. Kc8 Qg8+ 84. Kb7 Kg7 85. Ka6 Kf8 86. Kb5 Ke8 87. f3 Kd8 88. Ba2 Kc8 89. Bb1 Kb7 90. Ba2 Qg7 91. Rg8 Qf6 92. Bb1 Qe6 93. Ba2 Bh8 94. Rg7 { put wR into pocket, and final cleanup } Kc8 95. Ka6 Kd8 96. Kb7 Ke8 97. Kc8 Kf8 98. Kd8 Rb3 99. Bb1 Rb5 100. Ba2 Ra5 101. Bb8 Ra7 102. Bb1 Rb7 103. Ba7 Qc6 104. Ba2 Qb5 105. Bb1 Qb4 106. Ba2 Qa3 107. Rg8# (2007-4-24)
Michel Caillaud: the exact stipulation ends "mindestens geschehen".
(in english : "minimum number of moves since the last Pawn move?").
the last Pawn move is f2-f3,that must be interposed before g7-g6 as otherwise the 50-moves rule would have been violated.
the answer is a minimum of 15,0 moves with a retro-play running as :
-1.Tg7-g8# ...-16.f2-f3! ... -65...g7-g6
with 49,5 moves between g7-g6 and f2-f3 the 50-moves rule has been respected (2007-9-12)
Michel Caillaud: more detail on intended retro-play :
-1.Tg7-g8# … -4…De6-c6 … -8…Tb3-b5 -9.Kb5-a6 Ta3-b3+ -10.La7-b8 … -13…Kb7-c8 -14.Tg8-g7 Ld4-h8
-15.Tf8-g8 Df6-e6 -16.f2-f3! Dg7-f6 -17.Th8-f8 Dg8-g7 -18.Lb1-a2 Ta1-a3 -19.La2-b1 … -24… Kf6-g7 -25.Kb7-a6 Dg7-g8 -26.Tb8-h8 Ke5-f6 -27.Kc8-b7 … -28… Dh4-f6 -29.Ke8-d8 … -35… Td2-d1 -36.Ta3-b3 … -37.Lb3-a2 … -39.Th1-a1 Td1-d2 -40… Ta1-d1 -41.Kg7-f8 Ta3-a1 -42.La2-b3 …-48… Th8-b8 -52.Tb7-a7 … -53.La7-b8 … -55.Tg8-b8 … -59.Kç8-d8 Kf6-e5 -60.Kb7-ç8 Kg7-f8 -61.Tb8-g8+ Kf8-g7 -62.Kc8-b7+ Ke8-f8 -63.Tb7-b8 Lg7-d4 -64.Lb8-a7 Lf8-g7 -65.Ta7-b7 g7-g6 (2007-9-18)
Henrik Juel: After -1.Tg7-g8# the main parts of the resolution are
(s/w mean black/white)
sDa3 out, sTb7 to a3, wKe8 to b5, sKf8 to b7, wTg7 to h8 and sLh8 to d4, sD to g8, sKb7 via g7 to e5, sDg8 out, wKb5 to g7, sTa3 to d2, wTh8 to e1, sTd2 to h8, wTe1 to g8,
(the rooks have switched place)
wKg7 to b7, wTg8 to b8, wKb7 to c8, sKe5 to e8, sLd4 to f8,
-65... g7-g6 -66.g6xLh7
(so -16.f2 was needed to avoid draw by the 50 moves rule)
wTb8 to h1, wKc8 to b5, sD to d8, sLh7 to c8,
-88... b7-b6
wSa8 out, wLa7 to a5,
-99... b6xTc5
wTc5 to a1, wLa5 to c1,
-111.d2xTc3
sTc3 to a8 and wSa4 to b1, wLa2 to a4
-222.b3xSc4 a7xDb6 -223.a2xSb3 (2018-12-5)
Henrik Juel: It is difficult to avoid typos in such a mess
in line 3, wKd8 to b5, not e8
Michel also seems to have made some typos:
49.0 moves between -16.f2 and -65... g7, not 49.5
-25... Kf6, not 24 (may have implications for the solution) (2018-12-5)
more ...
comment
Keywords: 50 move rule
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: hpr, 2008-8-26 more...
19 - 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...
20 - P0001404
Jean Roche
278 Europe Echecs 182 01/1974
Lob
P0001404
(7+13) C+
h#2
if white can castle: 1. Td7 0-0-0 2. Kd8 Dxd7#
if black can castle: 1. Td7 Txa7 2. 0-0-0 Ta8#
play all play one stop play next play all
NN: The above is basically two parts of a single solution depending on the history of the game and not two different solutions.

Proof of why castling rights are mutually exclusive: Let's assume black CAN castle. Now, the bPs piled up on h-file implies six captures of white units and these can only be wRBBNN & wPe2. Because all other possibilities involve promotion of wPs on d8/e8/f8 followed by their capture by bPs, interfering with black's castling right. Next, we observe that wRh1 getting captured by bPs implies one of the following two things: (i) wKe1 moved to let wRh1 out of the pawn-chain or (ii) wPg2,h2 cross-captured on h3,g3 respectively to facilitate the escape of wRh1. Furthermore, option (ii) necessitates bPd7 to be promoted to d1 so that it can be captured by wPs, implying wKe1 had to move anyway to let bPd7 promote.
Thus, Black can castle implies wKe1 moved and white can't castle. Contrapositively, white can castle will imply that black can't. (2018-11-29)
Henrik Juel: Nice analysis, Satanick (2018-11-29)
NN: Thanks Henrik Juel. Hope there are no mistakes. (2018-11-29)
Yoav Ben-Zvi: The claim that retaining Black castling requires an uncapture of a wR by Black's h-file pawns is not accurate. Consider the case that all 6 of these bP uncaptures are of wBs and wNs. In this case 2 of the uncaptured White pieces unpromote. Both could be unpromoted (on f8 or d8) by an uncapture from e7. This scenario requires a third wP uncapture (on the e file) so one of White's uncaptures is accounted for by [bPd7]. The direct uncapture of [bPd7] (on the e file) or its unpromotion without stepping on d2 requires this pawn to uncapture the wR, confirming the conclusion of mutually exclusive castling. A similar argument works for the case that the Black h file pawns play bPe7xwPf6 along with 5 uncaptures of WNs and wBs. (2018-12-3)
Henrik Juel: Now we can test
C+ by Popeye 4.61 (and analysis)
The main content is the retro analysis, of course, but as a helpmate problem the repeated 1.Td7 detracts (2018-12-3)
more ...
comment
Keywords: Partial Retro Analysis (PRA), Castling (wgsg), Castling, mutual exclusive
Genre: h#, Retro
Computer test: Popeye 4.61
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-3 more...
21 - P0001430
Luis Alberto Garaza
304 Europe Echecs 205 01/1976
P0001430
(13+14)
#3
if Black can castle long: 1. Dxc7! droht 2. Dxe7#
1. ... Sf5,Se7~ 2. Dd7+ Kf8 3. Df7#
1. ... Kf8 2. Dxe7+ Kg8 3. Dxg7#

if Black can castle short: 1. Dxg7! droht 2. Dxe7#
1. ... Sf5,Se7~ 2. Df7+ Kd8 3. Dd7#
1. ... Kd8 2. Dxh8+ Sg8 3. Dxg8#
play all play one stop play next play all
Satanick: Castling for black is legal only on one side. The above solution cover the two mutually exclusive cases.
NN: Solution: If Black can castle long: 1.Qxc7! (Threat: Qxe7#) 1...N moves (1...Kf8 2.Qxe7+ Kg8 3.Qxg7#) 2.Qd7+ Kf8 3.Qf7# & if Black can castle short: 1.Qxg7! (Threat: Qxe7#) 1...N moves (1...Kd8 2.Qxh8+ Ng8 3.Qxg8#) 2.Qf7+ Kd8 3.Qd7#

The bBe1 is obviously a promotee. This implies bPd7,e7 made one capture each namely, dxe & fxe. Consequently, either the missing wPg2 or wPb2 (but not both!) had to be captured by bP on the e-file. But neither wPg2 nor wPb2 could have gone to the e-file capturing white units because that, together with wPe2×d3, make three captures in all and black is missing only two units. Hence, either one of them promoted to h8/a8 first prior to be captured on the e-file. This shows exactly one of bRh8 and bRa8 has moved before facilitating the promotion and therefore, black has right to castle only on one side. Accordingly,1.Qxc7 is the key if black has the right to castle long and 1.Qxg7 is the key if otherwise. (2018-11-29)
more ...
comment
Keywords: Cant Castler, Castling (sksg), Promotion (l), Partial Retro Analysis (PRA), Promoted material on the board (sLe1)
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-11-30 more...
22 - 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...
23 - P0001613
Andrey Frolkin
483 Europe Echecs 343/344 07/1987
P0001613
(13+14) C+
BP in 21,5
1. a4 Sc6 2. Ta3 Sd4 3. Tg3 Sxe2 4. Kxe2 g5 5. Ke3 Lh6 6. Lc4 Kf8 7. La2 Kg7 8. c4 Kf6 9. Db3 Ke6 10. Db6+ axb6 11. b4 Ta5 12. Lb2 Tb5 13. Lg7 Sf6 14. Sf3 Dg8 15. Lf8 Dg6 16. Te1 Tg8 17. Te2 Tg7 18. Se1 g4+ 19. f4 gxf3ep+ 20. Tg5 Sg8 21. c5+ d5 22. cxd6ep+
play all play one stop play next play all
Ein schwarzer und ein weißer ep-Schlag.
more ...
comment
Keywords: Unique Proof Game, En passant
Genre: Retro
Computer test: Natch 3.1 (5 min.)
Reprints: 575 Ukrainisches Album 1986-1990
94 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-6 more...
24 - P0002005
George Hume
6 British Chess Magazine 20/12/1893
P0002005
(11+9) cooked
s#1
R: 0. Bg2xSh1=S, dann 0. ... Tgxh1,Thxh1 1. Lxg2+ Txg2#
play all play one stop play next play all
Original solution text: Black has just played P from K Kt 7, taking Kt on R sq, claiming Kt double ch; but as the thereby discovers ch th his own K, White calls upon him to retract his move, and take the Kt legally, viz.: either RxKt, then White 1 BxP ch, RxB mate. (Frank Müller)
In modern notation: Black just played g2xSh1=S illegally. White asks him to retract this and to capture the knight legally, so Black then plays Tgxh1 or Thxh1. Either way White forces selfmate by 1. Lxg2+ Txg2#. (Henrik Juel)
Cook: Question: is this cooked? In "Retrograde Analysis" by Dawson and Hunsdorfer, this is given as one of several examples where both kings are apparently in check. In the solution, they give no mention of the choice of piece captured on h1, and on the next page compliment the problem as being "more plausible" and "sounder" than the other examples, because "in the latter other suppositions would explain the matter just as well". However I think that D&H missed something here. The pieces outside the SE corner only serve to try to force uniqueness of the illegal move, indicated in the original solution.
A.Buchanan: Why was it S captured on h1? Could be anything even wL. E.g. R: 1. gxh1=S Sf3 2. ke1-f1 Dd2+ unwinds ok. Need e.g. to replace wBa4 with wLa5 to force gxSh1=S. Agreed? (2018-10-5)
Mario Richter: Andrew's suggestion doesn't force gxSh1, it's still possible to retract gxLh1. The resulting position (sPawn g2, wLa5+wLh1) can be explained by e.g.
wPawnCaps: b2-b4xBa5 d2xLe3 c2xSd3xLe4xDf5xTg6xSh7
wProms: a8=D a8=L
sPawnCaps: c7xDd6-d5xSe4xTf3xTg2
sProms: b1=T
White's last move after 1. ... g2xLh1 might have been 2. Db4-d2! Ke1-f1 3. Sd2-f3+ (2018-10-5)
A.Buchanan: Hi Mario, well nearly. So wSc2 instead of wLa5 will do it. (2018-10-5)
A.Buchanan: But no: R: 1. g2xh1=S S~f3 2. f3xg2 etc allows the retraction whatever the piece captured on h1. (2018-10-6)
A.Buchanan: I have a fix for this that Mario has verified (thank you!) and will try to publish. (2018-10-7)
more ...
comment
Keywords: Illegal position, Touch Move, Retract illegal move, Joke
Genre: Retro, s#
Reprints: 7 Retrograde Analysis 1915
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-10-7 more...
25 - P0002051
Frideswide F. Rowland
Chess Fruits 1884
P0002051
(15+10)
Nehme 3 Einzelzüge zurück, dann #2 durch Weiß.
Weiß am Zug.
R: 1. ... 0-0 2. Ld5-a8 b7-b6, dann 1. ... Txh8+ Kd7 2. Dxc7#
play all play one stop play next play all
One of the best tests for thoroughly understanding problems of this class is to play a game up to the position. If the solver succeeds in doing so he will find that he has made the moves which are to be retracted, as such must be made to arrive at the position.
A better way is to give the problem an analytical examination, count the numer of captures made on either side, and note where each of the men played from, when and how they played, was it by capturing or otherwise. In the above problem we note that White lost a R only, which must have been captured by Black's KP [e7xf6], as its position shows that it made a capture. Black lost six men, the number off the board. His QP at Q7 [sBd2] did not play from B6 [c3xd2?] ass it made no capture, and consequently could not have moved last, neither could his QKtP [sBb7] or KP [sBe7], otherwise White's Bs could not be where they are. His KRP [sBh7] is off the board - our analysis will presently account for how it was captured. It is obvious that his R at R3 [sTa6] could not have played last, neither could R at RBsq [sTf8], as then the White K would have been in check, and it is not allowable to suppose that it remained in check.
Now the question is, Could Black's last move have been K from Rsq to Ktsq [sKh8-g8], moving from a check given by the R? This, we find on a further analysis, could not be, as the White R did not capture Black's KRP, consequently it could not have given check at the square it stands on. None of the White's pieces made a capture, the men were taken by his Pawns only, QP at QR5 [wBd2xc3xb4xa5] took three, and KP, KBP, and KRP took one each [wBe2xd3,wBf2xe3,wBh2xg3], none of them could have taken a P, consequently it follows that six Black pieces off the board were captured. There are only five Black pieces off the board, also a P, that P is Black's KRP [sBh7], which moved down the file to R8 [h1], was promoted to a Piece, and duly captured.
After all this we find that there is only one possible move Black could have made on his last, and that was Castles, this we accordingly retract. Having retracted it, we look to see what move White could have played for his last. All his men appear to be free, that is, not confined as Black's are, and it may appear difficult to determine what his last move was. In this case we look to see what Black's second last move could have been.
He did not play K or KR [sTh8] as he was able to castle after, neither could he have played QR or QP [sTa6,sBd2], it is also obvious that he did not play KPxR [sBe7xTf6], in consequence of the B at Q8 [wLd8 (außerdem auch, weil sonst der sLf8 als Schlagobjekt fehlt)], but if the B at QR8 [wLa8] were not there he could have played P to Kt3 [sBb7-b6]; consequently to allow of this second last move of Black's, White must retract a move made with his KB. There is only one square it could have played from and that is Q5. Therefore B at Q5 to R8 [wLd5-a8] was White's last move, and P at QKt2 to Kt3 [sBb7-b6] was Black's second last move.
White for his last move could not have made a capture with any of his Ps as Kts stand behind QP and KBP, the KP played early in the game to allow KB to play out, and KRP did likewise to allow Black's RP to play to R8.
The conditions of the problem do not ask for more than two of Black's moves to be retracted, yet the solver could demonsterate what his third last move was, viz., R to R3 [sT-a6].
On word as to the line of play leading up to the position. It may be considered very bad play, and it would be such were it played as a game, burt no matter how improbable the line of play may be, as long as the moves are possible they are sufficient to prove that the position can be arrived at and consequently sound.
Having retracted the moves in the problem [1. ... 0-0 2. Ld5-a8 b7-b6] we give, White mates by 1 RxRch, I K to Q2, 2 QxBP mate.
Frideswide F. Rowland in 'illustrated Science Monthly': "Retraction Problems take the form of end games in which one or more moves are to be retracted prior to playing to effect the required mate. There are two classes, which we will term legal and illegal. In the former the position is constructed in strict conformity with the laws of the game, and proof is shown that the moves to be retracted are those which have been actually made. Problems of this class are, as a rule, most ingenoiusly set, instructive to the solver, and worthy of more attention than is given to themby the composer. There is some difficulty in their construction which may perhaps be the cause of their not being more frequently composed.
In the latter class the move to be retracted may be any, according to the whim of the composern, which will allow of the mate which follows it. There is no proof to show what the last move was, it is left with the solver to assume that the move to be taken back was the one made. These problems are neither ingenious nor abstruse, and very much inferior to all other classes. They are constructed by simply setting up an ordinary direct mate position and misplacing one of the men, the replacing of which constitutes the retraction. If the conditions of such a problem is 'Retract White´s last move and then White to play and mate', it simply means White is to make two moves in succession. We dismiss this class as one unworthy of further attention and give our solvers an illustration of the legal Retraction Problem."
Henrik Juel: -1... 0-0 -2.Ld5 b6, 1.Txh8+ Kd7 2.Dxc7#, although Black has the move! (2004-3-8)
A.Buchanan: Maybe the idea was that by moving b6, Black was playing out of turn and the error when detected later had to be unwound. Black’s prior move had to have been R-a6 from somewhere. (2018-8-29)
A.Buchanan: Her remarks about retractor types are surprisingly direct. Mrs Baird wouldn’t have been too pleased to read that! (2018-8-30)
Henrik Juel: Andrew, on your first comment:
I believe that Frideswide just did not care about irregular move order, which is surprising in view of her 'modern' view on retro problems
On your second comment:
Yes, apparently there was no love lost between the two women; Frideswide was the better analyst, but Mrs. B managed to produce the beautiful Twentieth Century Retractor (I have seen the book) (2018-8-30)
A.Buchanan: @Henrik I disagree that "Frideswide just did not care about irregular move order". Actually I think she cared very much, with two problems here revolving around the idea of unwinding play out of turn. (2018-10-7)
comment
Keywords: Castling in the retro play (sk), Help retractor
Genre: Retro
Reprints: Illustrated Science Monthly , p. 223, 06/1885
29B Retrograde Analysis 1915
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-9-7 more...
26 - P0002070
Joseph C. J. Wainwright
American Chess Bulletin 04/1912
P0002070
(7+2) C+
#3
1. b4+? Kb1 2. Lb3 Ka1 3. 0-0?
1. Lb3! Kb1 2. Kd1 Ka1 3. Kc2#
play all play one stop play next play all
Martin Ramsauer: Reprint: Schach-Echo 08/1936, Umschlag-Innenseite (Artikel "Die Scheinlösung" von Emil Ramin) (2018-8-23)
Henrik Juel: Kc1 entered via d2 or e1, so White may not castle
1.Lb3 Kb1 2.0-0?
C+ by Popeye 4.61 (2018-8-23)
comment
Keywords: Cant Castler, Castling (wk)
Genre: Retro
Computer test: Popeye 4.61
Reprints: 40 Retrograde Analysis 1915
Schach-Echo 08/1936
2 L'Echiquier de Paris 01/1956
RA7 diagrammes 13 02/1975
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-9-11 more...
27 - P0002188
Gerd Wilts
169 Shortest Proof Games 11/1991
P0002188
(0+16) cooked
Ergänze die 16 weißen Steine so, daß eine dualfreie BP in 10,0 ensteht!
1. g3 Sf6 2. Lg2 Sd5 3. Kf1 Sb6 4. Ld5 Sc6 5. Kg2 Se5 6. Df1 Sg6 7. Kf3 Tg8 8. Dg2 Sh8 9. Kf4 Tb8 10. Df3 Sa8
play all play one stop play next play all
Cook: NL
1. d3 Sf6 2. Le3 Sd5 3. Kd2 Sb6 4. Kc3 Sc6 5. Sd2 Tb8 6. Tb1 Sa8 7. Lb6 Se5 8. e3 Sg6 9. Kc4 Tg8 10. c3 Sh8
1. e3 Sf6 2. Lc4 Sd5 3. Ke2 Sb6 4. Ld5 Sc6 5. c4 Se5 6. Sc3 Sg6 7. Kf3 Tg8 8. De2 Sh8 9. Kf4 Tb8 10. Sf3 Sa8
(François Labelle)
François Labelle: Is the solution supposed to be unique? Some variations of the intention also work, for example 10.De4 instead of 10.Df3. Some completely different solutions also work, for example
1.d3 Sf6 2.Le3 Sd5 3.Kd2 Sb6 4.Kc3 Sc6 5.Sd2 Tb8 6.Tb1 Sa8 7.Lb6 Se5 8.e3 Sg6 9.Kc4 Tg8 10.c3 Sh8
1.e3 Sf6 2.Lc4 Sd5 3.Ke2 Sb6 4.Ld5 Sc6 5.c4 Se5 6.Sc3 Sg6 7.Kf3 Tg8 8.De2 Sh8 9.Kf4 Tb8 10.Sf3 Sa8
There are 16 solutions in total, according to my computer. (2018-10-30)
Henrik Juel: Strange that Gerd did not notice the Df3/De4 'dual' (2018-10-30)
comment
Keywords: Unique Proof Game, Add pieces
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-11-2 more...
28 - P0002206
Samuel Loyd
Le Sphinx 10, p. 145, 01/10/1866
P0002206
(16+16)
#4
Schwarz kopiert die weißen Züge
3 Lösungen
1) 1. c4 c5 2. Da4 Da5 3. Dc6 Dc3 4. Dxc8#
2) 1. d4 d5 2. Dd3 Dd6 3. Df5 Df4 4. Dxc8#
3) 1. d4 d5 2. Dd3 Dd6 3. Dh3 Dh6 4. Dxc8#
play all play one stop play next play all
Als Schachgedicht "Loyd und der Pigeon" von C. Kupffer im Deutschen Wochenschach 24.11.1907 (Quelle: Sam Loyd und seine Schachaufgaben, 1926, S. 469)

Loyds englische Formulierung der Forderung: "If both parties make the same moves, how can the first player mate in four moves?"

in 'Ruy Lopez' abgedruckt in der Rubrik "Partidas de fantasia"

Originalabdruck in 'Le Sphinx' in der Rubrik: "Questions du Sphinx" mit folgendem Text:
"Deux joueurs sont en présence et se disposent à commencer une partie. Celui qui les Noirs ne connaît rien du jeu; mais, après s'être donné comme très-fort, il n'a pu se dispenser d'accepter le défi. Il s'est dit en lui-même qu'il lui suffirait, pour se tirer dèmbarras, de copier exactement tous les couops de son adversaire. Le joueur qui a les Blancs, ayant flairé le manége, jouen en conséquence et donne le mat au quattrième coup.
Comment, les Blancs ayant le trait et les Noirs jouant les mêmes trois premiers coups que leur adversaire, celui-ci peut-il obtenir le mat en quatre coups?"
Frank Müller: Im American Chess Journal 1876, Seite 48, wird nur die Lösung mit 1. c4 genannt. (2015-2-23)
Henrik Juel: The third solution is similar to the first, so they are probably a sort of cooks, later promoted to solutions.
1. d4 d5 2. Dd3 Dd6 3. Dh3 Dh6 4. Dxc8# (2015-2-23)
Henrik Juel: After the re-numbering it is solutions 2) and 3) that look like cooks (2015-2-23)
A.Buchanan: I don't think this is a fairy: I think it's just a conditional problem. If it was a fairy, then it would be cooked by e.g. 1. Sa3 Sa6 2. Sb5 Sb4 3. Sc7# since Black cannot continue to imitate White. (2018-9-2)
more ...
comment
Keywords: Conditional problem, Construction task
Genre: n#
Reprints: American Chess Journal , p. 18, 15/06/1876
333 Chess Strategy [Loyd] , p. 174, 1878
Ruy Lopez 3 03/1897
Deutsches Wochenschach , p. 418, 24/11/1907
61-I Sam Loyd and his Chess Problems , p. 58, 1913
61/I Sam Loyd und seine Schachaufgaben 1926
feenschach 46, p. 81, 04-06/1979
K166 Problemkiste (94) 08/1994
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-9-2 more...
29 - P0002212
Anders Olson
7817 Springaren 57 06/1994
P0002212
(0+1)
h-retro-#2
Circe
b) wKg4, sKg1
c) wKb1, sKg6
d) wKe1, sKg7, sBf7
e) wKf8, wBh4, sKg6
A.Buchanan: There are today 1050 Circe problems in PDB which are not marked as fairies! Most are retros, and some poor person has already spent time and energy entering them under the mistaken assumption that a problem can only have 1 genre. And this is just 1 fairy condition. There is no point in us fixing them one by one because if Gerd ever returned to active maintenance of the site, he could fix the whole issue in a few minutes. Very frustrating. PDB is a great system but has a huge data quality issue. (2018-9-20)
Henrik Juel: This problem has the added difficulty that it is unclear what is expected by the solver
Possibly, in part a), the first white 'retraction' is just the addition of the white king somewhere (2018-9-20)
comment
Keywords: Help retractor, Circe, only Kings
Genre: Retro, Fairies
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-12-6 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 - P0002272
Andrey Frolkin
Themes-64 1985
Lob e.a.
P0002272
(12+14) C+
BP in 20.5
1. Sf3 a5 2. Sd4 a4 3. Sb3 axb3 4. a3 Ta4 5. Ta2 bxa2 6. h4 axb1=S 7. h5 Sc3 8. h6 Sxd1 9. hxg7 h6 10. Th5 Th7 11. Ta5 b5 12. Ta8 Ta7 13. f3 La6 14. f4 Dc8 15. f5 Db7 16. f6 Df3 17. gxf3 Kd8 18. Lh3 Kc8 19. Lg2 Kb7 20. Lh1 Kc6 21. f4+
play all play one stop play next play all
Henrik Juel: C+ by Natch 3.1 (2 min.)
Tempo loss at 13.f2-f3-f4 and 18.Lf1-h3-g2-h1 (2018-12-6)
Joost de Heer: IMO f2-f3-f4 isn't a tempo loss, as the pawn is needed for shielding. (2018-12-7)
Henrik Juel: I see no shielding by [Pf2], Joost
White could play 13.f4?, but he would run out of moves
White plays 17.gxf3, and this wP will later shield, so wLg2 does not check sKb7, but that is a different story (2018-12-7)
comment
Keywords: Unique Proof Game, Non-standard material, Loss of tempo (BL3), Promotion
Genre: Retro
Computer test: Natch 3.1 (2 min.)
Reprints: 63 Shortest Proof Games 11/1991
3397 U.S. Problem Bulletin 102/103 07-10/1995
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-7 more...
32 - P0002281
Andrey Frolkin
2320 diagrammes 94 07-09/1990
1. Preis
P0002281
(14+16) C+
BP in 15,5
1. d3 d5 2. Le3 d4 3. Sd2 dxe3 4. Sb3 Dd4 5. Dd2 Ld7 6. 0-0-0 exd2+ 7. Kb1 De3 8. Tc1 d1=T 9. d4 Td3 10. Td1 Tc3 11. Td2 Tc6 12. Kc1 Td6 13. Kd1 Lc6 14. Ke1 Sd7 15. Td1 0-0-0 16. Ta1
play all play one stop play next play all
Henrik Juel: No solution in 15.0 or 14.5, so C++... (2018-12-8)
more ...
comment
Keywords: Unique Proof Game, Castling (wgsg), Castling Paradox (wg hidden), Non-standard material (t), Promoted material on the board (t)
Genre: Retro
Computer test: Natch 3.1 (28 min.)
Reprints: 88 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-12 more...
33 - P0002284
Andrey Frolkin
Kalininskaja Prawda 1988
Spezialpreis
P0002284
(14+13) C+
BP in19,5
1. Sf3 Sc6 2. Se5 Sd4 3. Sc6 dxc6 4. b4 Dd5 5. b5 Ld7 6. b6 0-0-0 7. bxa7 Le8 8. a8=S Kb8 9. Sb6 Ka7 10. Sd7 Ka6 11. Sxf8 Td7 12. Sg6 hxg6 13. c4 Th5 14. e4 Te5 15. Le2 f5 16. 0-0 Sf3+ 17. Kh1 Sg1 18. f3 Sf6 19. c5+ b5 20. cxb6ep+
play all play one stop play next play all
Henrik Juel: C+ by Natch 3.1
No solution in 18.5 or 17.5, so C++ (2018-12-8)
comment
Keywords: Unique Proof Game, En passant, Castling, Promotion (S), Valladao Task
Genre: Retro
Computer test: C+ by Natch 3.1
Reprints: 96 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-12 more...
34 - P0002288
Ernest Clement Mortimer
Andrey Frolkin

Die Schwalbe 71 10/1981
P0002288
(15+13) C+
BP in 4,0
1. Sf3 e5 2. Sxe5 Se7 3. Sxd7 Sec6 4. Sxb8 Sxb8
play all play one stop play next play all
Version A. Frolkin. Die Originalfassung war mit sBd5 und e7.
Another version is reflected left-to-right.
YM: Possible symmetrical game:

BP in 4,0 (15+13) C+

1. Sc3 d5 2. Sxd5 Sd7 3. Sxe7 Sdf6 4. Sxg8 Sxg8

and even improved versions:

BP in 4,5 (13+15) C+

1.e3! (tempo) Sf6 2.e4 Sxe4 3.Se2 Sxd2 4.Sc3 Sxb1 5.Sxb1.

BP in 4,5 (13+15) C+

1.d3! (tempo) Sc6 2.d4 Sxd4 3.Sd2 Sxe2 4.Sf3 Sxg1 5.Sxg1. (2016-9-27)
A.Buchanan: These days, the style seems to be more common to mark as "BP in exactly 4.5", if a shorter proof game exists. (2018-10-1)
Olaf Jenkner: Isn't it anticipated by Schweig (P0002287) ? (2018-10-1)
A.Buchanan: Olaf: there is much to unwind here.
Firstly, this is Mortimer v Frolkin. There is no way to indicate the dependency in the Author fields. There is a field called "After" but there is no known way to search that field (see PDB FAQ examples). But putting this information in the Solution Text field is no answer, as this field too is unsearchable! I suggest that this information can best be placed in the Comment field (not UserComment).
Secondly, we don't know when Mortimer published his original: but his 12 publication dates here are between 1917 & 1935, while Schweig's composition is 1938.
It would be good to confirm the original publication date, but Schweig may have been anticipated. (2018-10-1)
comment
Keywords: Unique Proof Game, Homebase (2), Impostor (s)
Genre: Retro
Reprints: 100 Shortest Proof Games 11/1991
(5) diagrammes 103 10-12/1992
6 Introduction to Proof Games 03/2010
06b The Puzzling Side of Chess 8 18/08/2012
72 The Puzzling Side of Chess 141 16/12/2018
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-10-1 more...
35 - 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...
36 - P0002302
Kostas Prentos
5426 feenschach 89 10-12/1988
Ehrende Erwähnung
P0002302
(14+13)
BP in 18,0
1. c4 Sc6 2. c5 Se5 3. c6 dxc6 4. Db3 Lh3 5. Kd1 Sg4 6. Kc2 Dd3+ 7. Kxd3 0-0-0+ 8. Ke4 f5+ 9. Kxf5 e6+ 10. Kxe6 Lc5 11. Kf7 Td6 12. Kf8 Se7+ 13. Dg8 Sh6 14. b3 Ld7 15. La3 Le8 16. Lb4 Td8 17. Sa3 Ld6 18. Tc1 Txg8#
play all play one stop play next play all
paul: 2 x Indian theme. (2018-11-9)
Henrik Juel: Euclide 1.01 ran for more than an hour, producing nothing
But the first 17.5 moves were tested correct in 8 minutes (2018-11-9)
Henrik Juel: Good proof game, considering its age
The black bishops go Lc8-h3-d7-e8 and Lf8-c5-d6 with interferences (including a swithcback by Td8) after their first moves to let Kf8 through (2018-11-9)
François Labelle: The last move must be Th8xg8, so Henrik's test in 17.5 moves plus three similar tests with white S, L, T at g8 (returning no solutions) should be considered an HC+ proof of correctness. (2018-11-10)
Henrik Juel: Good point, Francois
What does HC+ stand for? (2018-11-10)
Henrik Juel: ... never mind
It must mean correct by Computer testing with some (correct) Human reasoning added
If so, HC+ is the label we can use with many other retros, where human reasoning rules out a castling, etc. (2018-11-10)
comment
Keywords: Bahnung, Räumung, Unique Proof Game, Castling
Genre: Retro
Reprints: 128 Shortest Proof Games 11/1991
H23 FIDE Album 1986-1988 1995
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2014-11-6 more...
37 - P0002312
Kostas Prentos
5673 feenschach 93 11-12/1989
P0002312
(14+11)
BP in 17,5
1. e3 b5 2. Ld3 b4 3. Lg6 b3 4. axb3 hxg6 5. Ta5 Th3 6. Tb5 Tg3 7. hxg3 a5 8. Th8 a4 9. Dh5 a3 10. Se2 Ta4 11. bxa4 gxh5 12. b3 g6 13. Lb2 Lh6 14. Lg7 Lg5 15. Lh6 Lh4 16. gxh4 a2 17. Sg3 a1=T 18. Txg8#
play all play one stop play next play all
Henrik Juel: The content is concealed pawn captures
Despite no visible captures White captured a2xPb3xTa4 to open the a-line for [Ta1,Pa7], and h2xTg3xLh4 with black h7xLg6xDh5 to open the h-line for [Th1,Th8]
This makes life difficult for the human solver and for Natch 3.1, who tested the first 15.0 moves correct in 40 sec.
Testing the entire proof game would need more than a couple of hours by Natch (2018-12-8)
comment
Keywords: Unique Proof Game, Promotion (t)
Genre: Retro
Reprints: 143 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2014-11-7 more...
38 - P0002313
Yaakov Mintz
The Problemist 1987-1988
3.-5. ehrende Erwähnung e.a.
P0002313
(12+11) cooked
BP in 17,5
1. f3 e5 2. Kf2 e4 3. De1 e3+ 4. Kxe3 Ke7 5. Dg3 De8 6. Kf2 Kd8 7. Ke1 Lc5 8. Kd1 Lxg1 9. De1 Lxh2 10. Txh2 d6 11. Txh7 Le6 12. Txg7 Th1 13. Th7 Lxa2 14. Th8 Lxb1 15. Txa7 b5 16. Tb7 Ta1 17. Ta7 b4 18. Ta8
play all play one stop play next play all
Cook: NL
1. f3 e5 2. Kf2 e4 3. De1 e3+ 4. Kxe3 b5 5. Dg3 b4 6. Kf2 Lc5+ 7. Ke1 Lxg1 8. Kd1 Lxh2 9. Txh2 d6 10. Txh7 Le6 11. Txg7 Th1 12. Th7 Lxa2 13. Th8 Lxb1 14. Txa7 Ke7 15. Tb7 Ta1 16. Ta7 De8 17. Ta8 Kd8 18. De1 (Mario Richter)
paul: C+ (Euclide 1.11) for the first 13 moves (in about 51 hours). I do not dare to check for 13.5 moves. (2018-3-8)
Henrik Juel: Tiny typo in Mario's cook: 10.Txh7 (2018-3-8)
Mario Richter: The typo comes from a "by hand conversion" of the output of several tools I'm using (here: rawbats and chessbase) - unfortunately each tool has it's own output format, incl. the PDB solution field format ... (2018-3-9)
A.Buchanan: I have a janitor's tool which allows me to change the representation of positions & solutions.
Try to copy the file here: https://www.dropbox.com/sh/ld64s00zm0f9tx8/AAA7r8BsA0vuLcZOOKNvKkmua?dl=0
Click on it, and then hit the download button on the right.
The most useful is the one which converts from FEN to PDB "pieces" format.
Stick the input string in the yellow cell, and pick up the result from the green cell.
If you want to see how it's done, unhide the hidden rows and columns: no VBA just simple Excel.
Only 1% as clever as rawbats, but it still saves me a lot of time.
Feel free to use and to improve. (2018-3-9)
Mario Richter: @Andrew: Thanks for your offer - but I have such tools myself. My fault was not to use them (I thought that I'll be able to transform such a short PG by hand without errors ...)
Would be nice if the PDB-System would be more helpful with respect to language support.
I've decided to use 'english' in the language settings, but that doesn't help here, since solutions still have to be given in german piece notation.
Might be difficult to change this, especially when thinking of fairy piece names, so I can live with that, but there are other aspect which sometimes drive me crazy, e.g. the handling of source names in different languages ... (2018-3-9)
A.Buchanan: I thought you might have such a tool, but I thought it was a nice opportunity to announce it for others who might find it useful. (2018-3-9)
more ...
comment
Keywords: Unique Proof Game, Interchange, Belfort
Genre: Retro
Reprints: 144 Shortest Proof Games 11/1991
Best Problems 47 07/2008
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-12 more...
39 - P0002314
Valentin Balanovsky
Andrey Frolkin

Mat (Belgrad) 1985
P0002314
(11+14)
BP in 18,0
1. h4 d5 2. Th3 d4 3. Ta3 Lh3 4. Txa7 e6 5. a4 Ld6 6. Ta3 Lh2 7. Tf3 d3 8. Tf6 Dd4 9. f4 Dxb2 10. Kf2 Dxc1 11. Ke3 Dxd1 12. Kd4 Dxf1 13. Kc3 Dxg1 14. Kb2 De3 15. Kc1 Sd7 16. Kd1 0-0-0 17. Ke1 Kb8 18. Txb7+ Ka8
play all play one stop play next play all
Version A. Frolkin
Henrik Juel: Cooked by
1.Ba2-a4 Bd7-d5 2.Ta1-a3 Bd5-d4 3.Bh2-h4 Lc8-h3
4.Ta3-f3 Be7-e6 5.Tf3-f6 Lf8-d6 6.Sb1-a3 Ld6-h2
7.Bf2-f4 Bd4-d3 8.Ke1-f2 Dd8-d4 9.Kf2-f3 Dd4xg1
10.Dd1-e1 Dg1xf1 11.De1-f2 Df1xc1 12.Df2xa7 Dc1xb2
13.Kf3-f2 Db2-e5 14.Da7-e3 Sb8-d7 15.Th1-b1 O-O-O
16.Kf2-e1 Kc8-b8 17.Tb1xb7 Kb8-a8 18.Sa3-b1 De5xe3
with lots of variations (2018-12-6)
comment
Keywords: Unique Proof Game, Castling
Genre: Retro
Reprints: 145 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2006-1-10 more...
40 - P0002315
Jasper van Atten
Probleemblad 1985
4.-5. ehrende Erwähnung
P0002315
(16+15) C+
BP in 23,0
1. d4 g6 2. Kd2 Lg7 3. Ke3 Lf6 4. Kf4 Le5+ 5. Kg5 Sf6 6. Kh6 Lf4+ 7. Kg7 Lh6+ 8. Kxh8 Lf8 9. Lh6 Sg8 10. e3 f6 11. Dh5 Kf7 12. Sf3 Ke6 13. Lc4+ Kd6 14. Td1 De8 15. Td3 Df7 16. Ta3 Dd5 17. Sc3 Ke6 18. Ta6+ Kf7 19. a3 De6 20. Dc5 Ke8 21. Se5 Df7 22. Le6 Kd8 23. Sd5 De8
play all play one stop play next play all
Henrik Juel: C+ by Natch 3.1 (59 min.)
The content is given by the diagram position: Get wKe1 to h8 and interchange the black royals
Lf8 makes the roundtrip f8-g7-f6!-e5-f4-h6-f8
[Dd8] goes d8-e8-f7-d5-e6!-f7-e8
Resentful of the dissipation by some of his officers [Ke8] wastes no time going e8-f7-e6-d6-e6-f7-e8-d8 (2018-12-8)
comment
Keywords: Unique Proof Game
Genre: Retro
Computer test: Natch 3.1 (59 min.)
Reprints: 149 Shortest Proof Games 11/1991
Best Problems 47 07/2008
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-12 more...
41 - P0002318
Andrey Frolkin
The Problemist 1989
3. Preis
P0002318
(31+0) C+
Färbe die Steine!
BP in 17,5
1. e4 c5 2. Dg4 Da5 3. De6 fxe6 4. Ke2 Kf7 5. Kf3 Kf6 6. Kg4 Ke5 7. Kg5 h5 8. Kg6 Kf4 9. Kf7 Sh6+ 10. Ke8 Sf7 11. Lb5 Th6 12. d3+ Dd2 13. g4 a5 14. Sh3+ Kf3 15. g5 Ke2 16. Sg1+ Ke1 17. h3 Dd1 18. Sf3++
play all play one stop play next play all
Henrik Juel: The capture was fxDe
If White made it, Dd1 is white and so is Ke1, but he would stand in an illegal check from sPf2
So Dd1 and Ke1 are black (and Ke8 is white), and the rest of the men have a 'natural' color because of the shortish proof game, which is
C+ by Natch 3.1 (2 sec.) (2018-12-8)
comment
Keywords: Unique Proof Game, Coloring problem
Genre: Retro
Computer test: Natch 3.1 (2 sec.)
Reprints: 155 Shortest Proof Games 11/1991
feenschach 137 08-09/2000
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-13 more...
42 - P0002319
Andrey Frolkin
Leonid Lyubashevsky

The Problemist 1988
1. Preis
P0002319
(30+0)
Färbe die Steine!
BP in 19,5
1. h4 h5 2. Th3 Th6 3. Tb3 Ta6 4. g3 g6 5. Lh3 Lh6 6. Le6 Le3 7. dxe3 dxe6 8. Ld2 Kd7 9. Lb4+ Kc6 10. Sc3 De8 11. Dd8 Sd7 12. 0-0-0 Sdf6 13. Kb1 Dd7 14. Ka1 Se8 15. Tb1 Dd1 16. Sd5 Kb5 17. Ld6+ Ka4 18. Tb6 Ld7 19. Sb4 Lb5 20. b3+
play all play one stop play next play all
Henrik Juel: The fastest captures are d2xLe3 and d7xLe6, so Lb5/Ld6 are black/white respectively
If Ka4 were white, so that Ka1,Tb1 would be black, more than 19 black moves would be needed
So Ka1, Tb1 (and Tb6) are white, needing white castling, which entails that Dd8 is white, because this saves a white move; then Dd1 and Ta6 are black, and the rest are 'naturally' colored, with Sb4/Se8 white/black
The colored position is C+ by Natch 3.1 (0.3 sec.)
The position after switching the colors of Sb4/Se8 cannot be reached in 19.5 moves (2018-12-8)
comment
Keywords: Unique Proof Game, Coloring problem, Castling
Genre: Retro
Reprints: 588 Ukrainisches Album 1986-1990
156 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2005-12-27 more...
43 - P0002320
Andrey Frolkin
Henri Nouguier

(8) diagrammes 92 01-03/1990
4. ehrende Erwähnung
P0002320
(31+0) C+
Bestimme die Art und Farbe der Steine für eine KBP!
1. h4 Sh6 2. Th3 Sf5 3. Te3 Sd4 4. Te6 dxe6 5. e3 Dd7 6. Lb5 Kd8 7. De2 De8 8. Ld7 Sbc6 9. Kd1 Tb8 10. De1
play all play one stop play next play all
Henrik Juel: The decoding is easy, with letters ABCDEF representing the men DKPLST; the coloring is natural, with Ld7 being white
The shortest proof game lasts 9.5 moves and is
C+ by Natch 3.1 (2018-12-8)
comment
Keywords: Letter problem, Unique Proof Game
Genre: Retro
Computer test: Natch 3.1
Reprints: 591 Ukrainisches Album 1986-1990
159 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-13 more...
44 - P0002333
Jasper van Atten
The Problemist 1987-1988
3.-5. Ehrende Erwähnung
P0002333
(16+15) C+
BP in 19,0
1. e4 Sf6 2. Lc4 Sd5 3. d3 f6 4. Lf4 Kf7 5. Dh5+ Ke6 6. Se2 De8 7. 0-0 Dg6 8. Kh1 Dg3 9. De8 De3 10. fxe3 Sc6 11. Tf3 Se5 12. Th3 Kd6 13. Sg3 Sb4 14. Lf7 Kc6 15. c4 Sg4 16. Sc3 Sh6 17. Tg1 Sg8 18. Th6 Sa6 19. h3 Sb8
play all play one stop play next play all
more ...
comment
Keywords: Unique Proof Game, Interchange (ss), Castling
Genre: Retro
Computer test: Natch 3.1 (1 min.)
Reprints: 70 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-13 more...
45 - P0002373
Alexander Kislyak
Die Schwalbe 136 08/1992
P0002373
(12+7) C+
BP in 15,5
1. d3 Sh6 2. Lxh6 c5 3. Lxg7 c4 4. Lxf8 c3 5. Lxe7 cxb2 6. Lxd8 0-0 7. c4 Kh8 8. c5 Tg8 9. c6 Txg2 10. c7 Txh2 11. Lh3 d5 12. Lxc8 f5 13. Lxb7 f4 14. Lxa8 f3 15. Lxd5 fxe2 16. Lf6#
play all play one stop play next play all
more ...
comment
Keywords: Unique Proof Game, Castling
Genre: Retro
Computer test: Natch 3.1 (3 min.)
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-13 more...
46 - P0002473
Waleri A. Liskowez
(5) Die Schwalbe 69 06/1981
P0002473
(7+3)
#2 BG

Quick solution:
Partial solutions in a specific order.
I 1. dxe6ep a try by the BG-logic (beruehrt-gefuehrt) with no
justification 1. ... 0-0-0 2. Bb7#. If rejected, the key must be returned back:
II 1. Nxe5 capturing the touched wP. (1. ... 0-0-0??)
NOT:
1. Nc5/Nxe5? 0-0-0!
II 1. Nc5/Ke6?? illegal by the rule of touching chessmen.
(II 1. Kxe5?)
Castling implies ep.

Detailed solution:
We don't know if Black castling or White e.p. capture are legal. We do know that if castling is legal, then so is e.p. because Black's last move was certainly e7-e5.
If e.p. is legal, then we have a solution, whether castling is legal or not.
1. dxe6ep! [2. Rf8#]
1. ... O-O-O 2. Bb7#
And we have a number of tries stopped only if castling is legal:
Try: 1. Ke6?/Sc5?/Sxe5? [2. Rf8#]
But 1. ... O-O-O!
So how to tie these facts together? We are going to try out moves, and this will force the opponent to specify the pre-history of the game in his favour. Each phase we will label with Roman numbers.
Suppose we try I) 1. dxe6ep. UNDER SO-CALLED "TOUCH-MOVE" BLACK CAN DEFEND BY DEFINING THE HISTORY OF THE GAME SUCH THAT A THREAT IS ILLEGAL, AND MUST BE RETRACTED. This way of resolving uncertainty is the key to the modern Touch Move convention. The confusing part is that it actually has relatively little to do with Touch Move! This is the secret that you weren't being told!
Now we have to retract 1. dxe6ep. What have we gained? Well we know that ep is illegal, and therefore castling doesn't work either. There is a price to be paid: that we must follow the touch move rule. So although it appears that all of II) 1. Ke6/Sc5/Sxe5 will work, in fact only the last, II) 1. Sxe5, is valid, by the touch-move rule. wPd5 can't move except for the e.p. which we know now isn't going to happen, but we must still capture bPe5 if possible.
Suppose that instead we had begun by trying one of I) 1. Ke6/Sc5/Sxe5. Black can defend by saying that castling is legal, and indeed just plays it I) 1. ... 0-0-0! There is no retraction possible, because none of the White moves were illegal.
A.Buchanan: See my solution. The penny has dropped and I will adjust the definition of Touch Move so that others aren't struggling in ignorance.
I still have no idea how "PR" ("PRA?") & "PF" fit into this problem: they don't seem to be necessary if BG is defined properly. Please can Valery confirm my solution, and explain precisely how "PR" & "PF" contribute. (2018-10-7)
VL: Well, Andrew, your description of the solution is correct.
The mark "PR&PF" (or so) should be removed from this and other problems stipulated BG. It was (and remains) significant only in the general context for the explanation of the genuine nature and place of this hybrid genre in the framework of the family of controversial RA-genres (that is, ones connected with interacting move legalities). In view of this, I recommend to preserve PF as a keyword and even add PRA as such... (2018-10-7)
A.Buchanan: Valery: I have adjusted the stipulations as requested. I have not yet touched the keywords.
If I had a catalogue of pets, and classified a loyal dog called Fido, I would assigned him the keyword "dog". Although there are interesting genetic, evolutionary & cultural relationships between dogs and cats, I would not assign him the keyword "cat", because that would obscure more than it would help, particular to an audience of new and prospective pet owners. The relationship between dogs and cats is much better discussed, once, in the keyword definitions.
I would also make sure that before launching in to such advanced comparative discussions there was a clear, self-contained definition of the concept of "dog". This we still do not have in this database, either for dog or BG. In particular:
- What constrains or motivates the referee/opponent to accept or reject moves?
- Do the regular conventions (accept castling, reject en passant) play any role?
- In an adversarial game (direct or self play), can I reject or accept my opponent's move?
- As the player, do I keep playing moves until I have exhausted all possible partial solutions?
- Is there anything else important to say about BG itself?
Thanks so much! (2018-10-9)
comment
Keywords: Touch Move, En passant, Castling (sg), Post Factum (PF)
Genre: Retro
Reprints: Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-10-9 more...
47 - P0002534
Dirk Borst
Markus 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...
48 - 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...
49 - 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...
50 - 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...
51 - 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...
52 - 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...
53 - 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...
54 - 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...
55 - 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...
56 - 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...
57 - 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...
58 - 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...
59 - 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...
60 - 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 (wkwg), Odds game
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-3 more...
61 - 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, Pawn endgame
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-3 more...
62 - 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 (wkwg), 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: Alfred Pfeiffer, 2019-1-3 more...
63 - 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...
64 - 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...
65 - 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...
66 - 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...
67 - 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...
68 - 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...
69 - 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...
70 - 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...
71 - 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...
72 - 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...
73 - 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...
74 - 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...
75 - 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...
76 - 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...
77 - 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...
78 - 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...
79 - 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...
80 - 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...
81 - 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...
82 - 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: Promoted material on the board (sLa5)
Genre: Retro
Reprints: Schach mit Sherlock Holmes 1982
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-6 more...
83 - 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...
84 - 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...
85 - P0002691
Per Olin
Suomen Tehtäväniekat 15/01/1992
P0002691
(1+1)
Ergänze 4 Bauern zu einem IC!
Wieviele Lösungen?

Typ 1: Dreieck wBa2-a4-c2: 3*12+5=41,
Typ 2: Dreieck wBd2-e2-e3-f2: 10-3=7,
Typ 3: Dreieck wBc4b3c3d3: 2*8=16.
Summe=41+7+16=64!
Per Olin: As published, the Black king was on d5. The problem was
dedicated to the Chess Day of Finnish Chess Federation 15 January 1992.

The solution as given by the composer:

There are three types of solutions.

Type 1) on the six squares triangle a2, a4 and c2 four white pawns can be placed in 15 different ways. Three out of these do not comply with the stipulation, e.g. pawns a2, a3, b2 and c2 is still illegal after removal of c2. Fulfilling the stipulation are 15 - 3 = 12 positions. Mirroring the positions is possible, as well as a similar arrangement of black pawns in NW corner. In the NE corner the white king decreases the amount of solutions to 5. Total of Type 1 is 3 x 12 + 5 = 41.

Type 2) Positions type white pawns d2, e2, e3 and f2 are, including Black's corresponding ones and excluding those included in type 1, altogether 7.

Type 3) A checking white pawn on c4 or e4, whose all starting squares are occupied. The occupying pawns can be 8 combinations of different colors. This gives 2 x 8 positions.

The total of Types 1 - 3 is 41 + 7 + 16 = 64. (2018-8-29)
comment
Keywords: Illegal cluster, only Kings
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-8-29 more...
86 - 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...
87 - P0003038
Luigi Ceriani
4338v The Fairy Chess Review 3 28/11/1939
T. R. Dawson zum 50. Geburtstag
P0003038
(16+16)
BP in 26,0
1. h4 h5 2. Th3 Th6 3. Tc3 Te6 4. e3 Te4 5. De2 e6 6. Da6 La3 7. Lc4 De7 8. d3 Db4 9. Sd2 d6 10. Sdf3 Sd7 11. Ld2 Sdf6 12. 0-0-0 Ld7 13. Tf1 0-0-0 14. Se1 Tf8 15. Sgf3 Se8 16. Th1 Sgf6 17. Th3 Th8 18. Tg3 Th6 19. Tg5 Tg6 20. Ta5 Tg3 21. Lb5 Th3 22. Tc6 Th1 23. Sh2 Tf1 24. f3 Tf2 25. Lc3 Td2 26. Kb1 Td1#
play all play one stop play next play all
corrected version by V. Onitiu
Henrik Juel: Natch 3.1 did find a strategy, but otherwise nothing happened in more than an hour (2018-12-6)
comment
Keywords: Unique Proof Game, Castling, Capture-free
Genre: Retro
Reprints: The Fairy Chess Review 08/1940
147 La Genesi delle Posizioni 1961
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2014-5-22 more...
88 - P0003049
Raymond Smullyan
The Chess Mysteries of the Arabian Knights 1981
P0003049
(14+15)
Ein sL hat einen weißen Stein geschlagen. Welcher sL?
sLg4 hat wS geschlagen
play all play one stop play next play all
Alfred Pfeiffer: Falsche Lösung. Richtige Antwort ist: sLe5 hat wL geschlagen. (2010-7-27)
Henrik Juel: Alfred is right, so please correct the official solution
The capture sequence must have been
g7xSh6 before d2xTc3 before Lx[Lc1] (2019-1-23)
comment

Genre: Retro
Reprints: Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-3-5 more...
89 - P0003050
Raymond Smullyan
The Chess Mysteries of the Arabian Knights 1981
P0003050
(5+1)
Im letzten Zug wurde kein Stein geschlagen. In welche Richtung spielt Weiß?
R: 1. Td2-f2 Kf2-f1 2. Be2xXd1=S+
play all play one stop play next play all
Henrik Juel: The stated retroplay is the only possibility
so White pawns move downwards (2019-1-23)
comment
Keywords: Conditional problem
Genre: Retro
Reprints: Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-3-5 more...
90 - P0003736
Giuseppe Brogi
1249 Sinfonie Scacchistiche 07-09/1972
P0003736
(12+12) cooked
h#2
b) wSa5
a) 1. f5 Le5 2. 0-0 Th8#
b) 1. Kd8 0-0-0 2. Te8 Lf8#
play all play one stop play next play all
Try to deduce the diagram error, as we do have the intended solutions!
a) wRh7 is original, but cannot have escaped from cage by white cross-capture as there are too many captures needed. On the other hand, Black only has to make 4 captures so the cross-capture is ok.
b) changing the colour of Sa5 shifts the capture balance: now White can cross-capture but not Black, so White but not Black can castle.
Thus the retro-logic works with the diagram as it is, but the forward logic does not. How can we repair the diagram?
A.Buchanan: Something odd going on here. There are numerous h#2 in both a&b. PRA not normally found in a problem with single solution for each twin. White has 8 pawns so trivially wRh7 must be original. (2018-10-13)
A.Buchanan: There are three similar problems by Brogi which are all twinned RS rather than PRA, but don't have diagram error. P0003743, P0003746 & P0003747. (2018-10-13)
comment
Keywords: Castling, Retro Strategy (RS)
Genre: h#, Retro
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2018-10-13 more...
91 - P0003821
Pierre Gre
RA23 diagrammes 18 11-12/1975
P0003821
(6+5)
h#2
1) 1. Lf8 0-0-0 2. Lb4 Sb1#
2) 1. Kb2 0-0 2. Lc3 Tfb1#
play all play one stop play next play all
NN: White can't have rights to castle on both sides as his only legal last moves are with wK or wRs. (2018-11-30)
more ...
comment
Keywords: Castling (wkwg), Partial Retro Analysis (PRA)
Genre: h#, Retro
Reprints: (3) StrateGems SG19, p. 156, 07/2002
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2017-12-4 more...
92 - P0003927
Herbert Hultberg
Springaren 1979
P0003927
(10+2)
-1w, dann #1
VRZ, Typ Proca
R: 1. g7xSh8=S? (illegale w Bauernstruktur)
R: 1. c7xSb8=S!, dann 1. c8=D#
play all play one stop play next play all
klären: # vor 1 Zügen!? - 1 Zug oder n Zügen! (VF)
Mario Richter: #1 vor 1 ist möglich: R: c7xSb8=S, dann 1.c7-c8=D#
Der spiegelbildliche Versuch R: g7xSh8=S, dann 1.g7-g8=D# scheitert an der Illegalität der wB-Struktur. (Das erklärt aber natürlich noch nicht das 'VRZ,Typ Proca'.) (2009-12-8)
Henrik Juel: Hultberg also made Høeg retractors, so maybe the type designation just underscores that this problem is not a Høeg retractor. If it were, it would be unsolvable: Black chooses to supplement some other officer than a knight on b8, and then there is no mate. (2009-12-11)
Mario Richter: Thank you, Henrik! Your explanation sounds very plausible to me. (2009-12-11)
paul: Actually, this is a help retractor. (2018-12-14)
comment
Keywords: Symmetrical position, Defensive Retractor, Type Proca, Asymmetrical solution
Genre: Retro
Reprints: RA135 diagrammes 49 01-02/1981
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-14 more...
93 - P0004043
Karl Fabel
Parallele 50 1950
P0004043
(2+1)
Weiß nimmt 1 Zug zurück, dann #2
R: 1. "0-0", dann 1. "0-0-0" Ka1 2. Ta3#
play all play one stop play next play all
Veröffentlicht in der Rubrik "Professeur Cosinus"
Mario Richter: Vorwärtsspiel m.E. zumindest teilweise vorweggenommen durch P1359470. (2019-1-12)
A.Buchanan: This one, in constrast to P1359470, remains a joke and a good one. (2019-1-12)
comment
Keywords: Odds game, Help retractor, Joke, Castling (wkwg)
Genre: Retro
Reprints: 490 La Composition Contemporaine [Martin] 1951
(9) diagrammes 86 07-09/1988
Input: Gerd Wilts, 1995-6-3
Last update: A.Buchanan, 2015-8-12 more...
94 - P0004066
S. N. Ravishankar
2069 diagrammes 89 04-06/1989
P0004066
(2+1)
#1 vor 6
VRZ, Typ Proca
S N Ravi Shankar: This problem is not mine. (2019-1-2)
Mario Richter: My first guess for the solution was:
R: 1. Ke2xLf1 2. Kd3xLe2 3. Kc4xLd3 4. Kb5xLc4 5. Ka6xLb5 Lc6-b5+ 6. Sd7xLb8, dann 1. Sb6#, but that doesn't work because of 5. ... Ld7-b5+!
Does someone see (or know) the correct solution? (2019-1-3)
Henrik Juel: No, because with 1... Lh3-f1+, 5.Ka6-b5 thr. 6.Sd7xLb8 fails similarly on 5... Ld7-h3 (2019-1-3)
comment
Keywords: Defensive Retractor, Type Proca, Rex solus (s), Aristocrat
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-4 more...
95 - P0004176
Leon Loewenton
(IV) Problem 33-36 02/1956
P0004176
(9+8) cooked
BP in 16,0
1. a4 Sf6 2. a5 Sd5 3. Ta4 Sb6 4. axb6 h5 5. bxc7 h4 6. cxd8=L h3 7. Lxe7 hxg2 8. Lxf8 gxh1=L 9. Lxg7 Lg2 10. Lxh8 Lxf1 11. Le5 Lxe2 12. Lxb8 Lxd1 13. Lg3 Lxc2 14. f4 Lxb1 15. Kf2 Lg6 16. Te4+ Kf8
play all play one stop play next play all
Cook: NL
1. a4 h5 2. a5 Th6 3. Ta4 Tb6 4. axb6 Sf6 5. bxc7 h4 6. cxd8=L h3 7. Lxe7 hxg2 8. Lxf6 gxh1=L 9. Lxg7 Lg2 10. Lxf8 Lxf1 11. Ld6 Lxe2 12. Lxb8 Lxd1 13. Lg3 Lxc2 14. f4 Lxb1 15. Kf2 Lg6 16. Te4+ Kf8
Henrik Juel: Jacobi got nowhere in more than an hour (2018-12-6)
Mario Richter: Cook found by rawbats:
1. a4 h5 2. a5 Th6 3. Ta4 Tb6 4. axb6 Sf6 5. bxc7 h4 6. cxd8=L h3 7. Lxe7 hxg2 8. Lxf6 gxh1=L 9. Lxg7 Lg2 10. Lxf8 Lxf1 11. Ld6 Lxe2 12. Lxb8 Lxd1 13. Lg3 Lxc2 14. f4 Lxb1 15. Kf2 Lg6 16. Te4+ Kf8 (2018-12-7)
Henrik Juel: Thanks for bringing your versatile program out, Mario
I was pretty sure that this proof game must be cooked, but could not show it by my own tools (2018-12-7)
comment
Keywords: Unique Proof Game, Promotion, under-promotion (L l), Promoted material on the board (wLg3 sLg6)
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-7 more...
96 - P0004263
George A. Teodoru
(11) Problem 52-54 08/1958
Lob
P0004263
(11+10)
BP in 19,0
1. a4 Sf6 2. a5 Sd5 3. Ta4 Sb6 4. axb6 Sc6 5. bxa7 Tb8 6. axb8=S Se5 7. Sxd7 c6 8. Sxf8 Lg4 9. Sxh7 Lxe2 10. Sg5 Lxf1 11. Se4 Le2 12. Sec3 Lxd1 13. Sa2 Lxc2 14. Ke2 Lf5 15. Ke3 Th3+ 16. Kf4 Ta3 17. Kg5 Lh3 18. Tg4 Ta8 19. Sa3 Kf8
play all play one stop play next play all
Henrik Juel: The given solution implies that wSg1 is missing in diagram
With this added, there are lots of cooks, e.g.
1.Sb1-c3 Sb8-c6 2.Ba2-a4 Ta8-b8 3.Ba4-a5 Sg8-f6
4.Ta1-a4 Sf6-d5 5.Sc3-a2 Sd5-b6 6.Ba5xb6 Bd7-d5
7.Bb6xa7 Bd5-d4 8.Ba7-a8=S Bd4-d3 9.Sa8-b6 Bd3xe2
10.Sb6-c4 Be2xf1=S 11.Ke1-e2 Tb8-a8 12.Ke2-f3 Sc6-e5
13.Kf3-f4 Sf1-e3 14.Kf4-g5 Bh7-h5 15.Dd1xh5 Se3xc2
16.Dh5xh8 Sc2-a3 17.Dh8xf8 Ke8xf8 18.Sc4xa3 Lc8-h3
19.Ta4-g4 Bc7-c6
With the given diagram position, the problem is also badly cooked (2018-12-6)
comment
Keywords: Unique Proof Game, Interchange
Genre: Retro
Input: Gerd Wilts, 1995-6-3
Last update: Gerd Wilts, 2006-1-10 more...
97 - P0004299
Luis Alberto Garaza
(30) Problem 73-78 06/1961
1.-2. Preis e.a.
P0004299
(12+11)
#3
Längstzüger
31. TT
Henrik Juel: 1.0-0 Ta4 2.Ta1 Ta8 3.Txa8#
C+ by Popeye 4.61
But the real content is probably a proofgame showing how the position could be reached with the maximummer condition in effect throughout (2018-9-23)
Henrik Juel: The magazine issue can be downloaded as
http://problem64.beda.cz/silo/problem_73-78_1961.pdf
On p.39 in solution 30 a proof game in 171.0 moves is given
but I have not checked it (yet) (2018-9-24)
Henrik Juel: The given proof game seems OK, albeit with several typos
The main critical positions are:
after 74.Da1
k6s/1s1ppppp/1p1p2l1/pS1LPt1t/1P3L1P/1PPT4/3P1PP1/D3K2T (15+14)
Here the black knights are controlled as are the other black officers, allowing black pawn captures and the surprising 79... a1=T
after 127.Ld5
1Dsk3s/3ppppp/1p4l1/3LTt1t/1P3ptP/1PP2PS1/3P2P1/4K2T (13+14)
Now the black king can move to g8, after he has been helping out on the board, and we see the need for three black rooks as [Th1] is unavailable
after 151.Dc8
2D1s1k1/3ppppp/1p5s/3LTtt1/1P3PtP/1PP3p1/3P2P1/4K2T (12+13)
Now we can get the black king all the way to h8, and the rest is rather easy... (2018-9-24)
hans: 1. a3 Nc6 2. e3 Ne5 3. Nh3 Ng6 4. Nc3 Nh4 5. Nb5 Ng6 6. Nd6+ cxd6 7. Nf4 Qa5 8. Ne6 Qh5 9. Nc7+ Kd8 10. Na6 Qa5 11. Nb4 Qh5 12. Nc6+ Kc7 13. Qg4 Qa5 14. Qc4 Qh5 15. Ne5+ Kb6 16. Nf3 Qa5 17. Qb3+ Qb4 18. axb4 Ne5 19. Qa3 Nf6
20. Qa6+ Kc7 21. Nh4 Nc6 22. Ng6 Nb8 23. Be2 Ng4 24. Bf3 Ne5 25. Qa5+ b6 26. Qb5 Ba6 27. Bb7 Nc4 28. Nxf8 Ne5 29. h4 Nc4 30. Ng6 Rc8 31. Nf4 Rh8 32. Qf5 Rc8 33. Nd5+ Kd8 34. Nf6 Rc5 35. Qf4 Rh5 36. Qf5 Nc6 37. b3 Nd4 38. Bb2
Ne2 39. Be5 Na5 40. Bf4 Bd3 41. Qe5 Bg6 42. Qf5 Nc4 43. c3 Nd4 44. Ng4 Ne6 45. Ne5 Ng5 46. Nc6+ Kc7 47. Nd4 Rh8 48. Nb5+ Kb8 49. Nc7 Rc8 50. Ne8 Rc5 51. Qe4 Rf5 52. Qb1 Ra5 53. Qc1 Bb1 54. Qc2 Rf5 55. Bd5 Ne5 56. Qb2 Ng6 57. Qc2 Nh8 58. Bc4 Ra5 59. Qb2 Bg6 60. Bd3 Rf5 61. e4 Ra5 62. Nc7 Rf5 63. e5 Ne6 64. Na6+ Ka8 65. Bc4 Nd8 66. Ra5 Ne6 67. Rd5 Nd8 68. Rd3 Ne6 69. Nb8 Nd8 70. Nc6 Ne6 71. Nd4 Nd8 72. Nb5 Nb7 73. Bd5 a5 74. Qa1 dxe5 75. Re3 exf4 76. Re5 a4 77. Qa2 a3 78. Qb1 a2 79. Qd1 a1=R 80. Nc7+ Kb8 81. Nb5 Ra8 82. Nd4 Ra1 83. Nc6+ Kc7 84. Nd4 Ra8 85. Bc4 Ra1 86. Nc2 Ra8 87. Ba6 Rg8 88. Ne3 Ra8 89. Nd5+ Kd6 90. Qe2 Rg8 91. Qc4 Ra8 92. Re4 Rg8 93. Rd4 Ra8 94. Qc7+ Ke6 95. Qc4 Rg8 96. Qd3 Ra8 97. Nf6 Rg8 98. Ne8 Ra5 99. Nf6 Rag5
100. Qg3 Bb1 101. Rd3 Ra8 102. Qh3+ Rg4 103. Nd5 Rg8 104. Rd4 Bg6 105. Nc7+ Kf5 106. Ne8 Nc5 107. Qf3 Na4 108. Nc7 Ra8 109. Bc8 Ra5 110. Nb5 Ra8 111. Qd1 Ra5 112. Qa1 Ra8 113. Qa3 Ra5 114. Ba6 Nc5 115. f3 Ne6 116. Qa4 Ng5 117. Nd6+ Ke6 118. Ne4 Rf5 119. Rd5 Nh3 120. Ng3 Nf2 121. Re5+ Kd6 122. Qa5 Ne4 123. Qd5+ Kc7 124. Qb7+ Kd8 125. Bc4 Nd6 126. Qb8+ Nc8 127. Bd5 fxg3 128. f4 Ke8 129. Qa8 Kf8 130. Qb8 Kg8 131. Qd6 Na7 132. Qxg6 Nb5 133. Qc6 Nd6 134. Qa4 Ne8 135. Bb7 Nf6 136. Ba6 Ng6 137. Rd5 Nf8 138. Rd4 Ra5 139. Rc4 Rhb5 140. Rc5 Nh5 141. Qa3 Ne6 142. Qa4 Nd4 143. Qa3 Nf5 144. Qa4 Nh6 145. Rc4 Rbg5 146. Re4 Raf5 147. Re5 Nf6 148. Bc4 Ne8 149. Bd5 Nf6 150. Qa8+ Ne8 151. Qc8 Kh8 152. Qc6 Nc7 153. Qe6 Nb5 154. Qxf5 Na3 155. Qxg4 Nf5 156. Bb7 Nd6 157. Qe2 Ne4 158. Re6 Ra5 159. b5 Ra8 160. Bc8 Ra4 161. Ba6 Rd4 162. Qc4 Nd6 163. Re2 Ne4 164. Qb4 Nc4 165. Bb7 Ng5 166. Be4 Ne6 167. Bd3 Ng5 168. Qd6 Ne4 169. Qe6 Ng5 170. Qh3 Ne4 171. Qh2 Ne5 position (2018-9-24)
Henrik Juel: Thanks for supplying the intended proof game, Hans
It is remarkable what is possible with the max condition, but with the advent of unique proof games it has no longer much attraction, at least to me (2018-9-24)
comment
Keywords: Maximummer, Castling (wk)
Genre: Retro
Input: Gerd Wilts, 1995-6-3
98 - P0004300
Luis Alberto Garaza
(22) Problem 73-78 06/1961
3. Preis
P0004300
(8+10)
#3
Längstzüger
31. TT
Henrik Juel: 1.0-0 Sc6/Se6/Sf7 2.Tf8+/Txe6/Txf7
C+ by Popeye 4.61
But, like the previous problem, how could the position arise? (2018-9-23)
comment
Keywords: Maximummer, Castling (wk)
Genre: Retro
Input: Gerd Wilts, 1995-6-3
99 - P0004348
Edward J.M.D.P. Dunsany
1 The Week-End Problems Book , p. 179, 1932
P0004348
(8+16)
#4
(Brettdrehung 180°)
1,2) 1. Sc6,Sd7 Sf3 2. Sb4,Sc5 Se5 3. Dxe5 ... 4. Sd3#
play all play one stop play next play all
Das Brett muss um 180 Grad gedreht werden, da sK und sD vertauscht sind.
AB: Two variant solutions.
(In any case, board must be rotated by 180 degrees, then 1 Nc6/d7 Nf3 2 Nb4/c5 Ne5 3 Qxe5 ~ 4 Nd3# (2002-4-3)
more ...
comment
Keywords: Joke, Board Rotation (180), Illegal position
Genre: Retro, n#
Reprints: Problem 101-102 09/1966
73 100 Classics of the Chessboard 1983
17 Die gefesselte Zeit , p. 137, 1987
(L) Die Schwalbe 151 02/1995
Karl 4/2018, p. 57, 12/2018
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2019-1-6 more...
100 - P0004367
Andrey Frolkin
560 Europe Echecs 403 07-08/1992
P0004367
(10+12) cooked
KBP
a) Weiß am Zug
b) Schwarz am Zug
a) 1. e4 h5 2. Dxh5 Txh5 3. h3 Txh3 4. a3 Txa3 5. Txa3 a6 6. Txa6 Sf6 7. Txf6 gxf6 8. Th6 Lxh6 9. d3 Lxc1
b) 1. d3 h6 2. Lxh6 Txh6 3. h4 Txh4 4. a4 Txa4 5. Txa4 a6 6. Txa6 Sf6 7. Txf6 gxf6 8. Th6 Lxh6 9. Dc1 Lxc1 10. e4
play all play one stop play next play all
Cook: Duale
a) 1. e4 h5 2. Dxh5 Txh5 3. h3 Txh3 4. a3 Txa3 5. Txa3 a6 6. Txa6 Sh6! 7. Tf6 gxf6 8. Txh6 Lxh6 9. d3 Lxc1
b) 1. d3 h6 2. Lxh6 Txh6 3. h4 Txh4 4. a4 Txa4 5. Txa4 a6 6. Txa6 Sh6! 7. Tf6 gxf6 8. Txh6 Lxh6 9. Dc1 Lxc1 10. e4
paul: Dual: a) 6.- Sh6! 7.Tf6 gxf6 8.Txh6 Lxh6 9.d3 Lxc1. (2010-7-30)
Henrik Juel: Part b) is also ruined, by the same dual (2018-12-9)
Henrik Juel: ... almost the same, namely
6... Sh6 7.Tf6 gxf6 8.Txh6 Lxh6 9.Dc1 Lxc1 10.e4 (2018-12-9)
comment
Keywords: Unique Proof Game
Genre: Retro
Reprints: (12) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-6-3
Last update: Alfred Pfeiffer, 2018-12-10 more...
next 100 problems
Show statistic for complete result.

The problems of this query have been registered by the following contributors:

Gerd Wilts (100)