Die Schwalbe

27 problem(s) found in 3271 milliseconds (displaying 27 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT S='Augsburger Allgemeine' AND K='Ergänze Steine'] [download as LaTeX]

1 - P0000324
Josef Haas
8259 Die Schwalbe 143 10/1993
P0000324
(7+5)
a) Wer setzt in 1 Zug matt?
b) Auf welchem Feld muß ein schwarzer Bauer eingefügt werden, damit die andere Partei als in a) mattsetzt?
b) (+sBc7) 1. ... Lg8xe6#
a) 1. Tg6#
1) R: 1. ... Kg6xBf6! 2. g5xf6ep++ f7-f5 3. La2-b1+
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"Vermutlich aus der Kleinkunstkiste des Autors hervorgekramt.
a) sollte einfach formuliert sein: 'Matt in 1 Zug' - denn wie es hier heißt, klingt es als ob nur einer mattsetzen kann. Das aber ist nicht der Fall, denn beide können's: 1. ... Lxe6# und 1. Tg6#. Üblicherweise hat Weiß das Prae und kann darauf pochen, den Schwarz hat einen altklassischen letzten Zug: 1. ... Kg6xBf6! (nebst 2. Bg5xBf6ep++ Bf7-f5 3. La2-(x)b1)" (HHS);
also ist Weiß am Zug und setzt matt mit 1. Tg6#.
b) Nach Einfügen eines sBc7 geht die o.g. Rückzugfolge nicht, weil der wK nicht auf die 8. Reihe gelangen kann. Also Schwarz am Zuge und 1. ... Lxe6#
"Allzubekanntes - kein Problem für Schwalbelöser" (HHS)
Wenn das alles so bekannt ist, erstaunt doch sehr, daß nur drei Löser die Autorintention nachvollziehen konnten. Alle anderen Löser (5) kamen zu genau entgegengesetzten Erkenntnissen (in a) setzt Schwarz matt, in b) Weiß), was wohl durch die nicht ganz konventionelle Formulierung suggeriert wurde. Ich find's ein interessantes Beispiel für Massenhypnose! (GL) 2/I/3L.
vergl. P0004915 (Hans Gruber, Schach 1979)
Brassaud: La solution proposée 1/Tg6# est possible
Mais il y a aussi le rétro jeu -1) Fa2-b1, Rg5g6 -2) Ta4-a5+, Rf4-f5 etc … et avec le trait aux noirs : 1) Fxe6 # est possible (2017-08-30)
A.Buchanan: @Brassaud: yes I agree. There is no reason why White should not have moved last. So both players can mate, but part (b) implies that the intended solution in (a) is 1 player. If the published stipulation for (a) was maybe just "#1", which by default is white to move, then there is a unique solution.
For (b) I am wondering about +sBg6, which would also stop the en passant trick, both by blocking sK from retreating there and also by locking sL in an impossible cage with sBf7. (2017-08-31)
Henrik Juel: Adding a black pawn on g6 of course prevents a black last move by Kf6, but it allows f7xg6 as last move; Lg8 is not locked, because Ph7 is white (2017-08-31)
A.Buchanan: Yes (2017-08-31)
Anton Baumann: vergl. P0004915 (Hans Gruber, Schach 1979) (2023-01-03)
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comment
Keywords: Add pieces, No legal last move for Black, En passant in the retro play
Genre: Retro
FEN: 4K1br/1p4pP/4Pk2/R7/3P4/8/8/1B4R1
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-01-03 more...
2 - P0000822
Josef Haas
1938 Die Schwalbe 41 10/1976
P0000822
(12+11)
Ergänze den wK, dann #2
Kees: +wKb5 1. Txc8 (2. Lxe7#)
0-0 is illegal for K or T must have made a move.
-1. a7-a5? Not possible with position of wL and bS (2022-11-23)
comment
Keywords: Castling (sk), Add pieces
Genre: Retro
FEN: 2nBk2r/3pp3/1p1p2P1/p4NN1/PP4p1/7b/PP2P1Pp/2R2B2
Input: Gerd Wilts, 1995-06-03
3 - P0001601
Filip S. Bondarenko
471 Europe Echecs 334 10/1986
P0001601
(16+0)
Ergänze die 16 schwarzen Steine so, daß kein Stein angegriffen ist!
KBP
1. Sf3 b5 2. Sd4 e5 3. Sc6 g6 4. e3 Sf6 5. c4 Sd5 6. d4 Sb6 7. c5 a5 8. b4 a4 9. Ke2 Sa6 10. Kf3 e4+ 11. Kg4 Ld6 12. Sb8 Ke7 13. Sa3 Tf8 14. Kg5 f5 15. g4 h5 16. h4 Lh2 17. c6 d6 18. Lb2 f4 19. Kh6 Ke6 20. g5 Kf5 21. Db3 Kg4 22. Tc1 Kf3 23. Tc5 Lh3 24. Te5 Sc5 25. Lc4 Sd3 26. Kh7 Ke2 27. Te8 f3 28. Le6 Ta7 29. d5 Sa8 30. Sc4 Tf4 31. Kh8 Tg4 32. Lg7 Df6 33. Tc1 Da1 34. Tc3 Dh1 35. a3 Tg1 36. Sb6 Lg2 37. S6d7 Se1 38. Db2+ Kd1
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Originalforderung: Compléter le diagramme par les 16 pièces noires de façon à ce qu'aucune pièce des deux camps ne soit en prise.
Plus courte partie justificative aboutissant à cette position.
hans: Kd1, Dh1, Tg1, Ta7, Lg2, Lh2, Se1, Sa8, a4, b5, c7, d6, e4, f3, g6, h5. (2013-07-27)
Mario Richter: @Alain: with black Bishop on f1 instead of g2 white pawn g5 is attacked by black Rook g1 (2021-08-23)
Alain Brobecker: @Mario, you are right of course, thank you for correcting this! (2021-08-23)
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comment
Keywords: Add pieces, Non-Unique Proof Game, Construction task
Genre: Retro
FEN: 1N2R2K/3N2B1/2P1B3/3P2P1/1P5P/P1R1P3/1Q3P2/8
Input: Gerd Wilts, 1995-06-03
Last update: Alain Brobecker, 2021-08-23 more...
4 - P0002214
anonymous
Rochade Europa 9 09/1994
P0002214
(14+10)
Ergänze wL, wB, 2sDD, sT, sS auf den markierten Feldern!
Ein "Kristufek"-Problem: der Autor wird nicht genannt!
Alfred Pfeiffer: Es besteht kein Grund daran zu zweifeln, daß die in Peter Krystufeks Rubrik "Kniffel-Schach" publizierten Retro-Aufgaben nicht von ihm (d.h. von Peter Krystufek) sind. (2011-08-30)
Henrik Juel: Add wLc8, wBf7, sDa1, sDh1, sTb4, sSa6.
White captured [Lc8], hxLg, and promoted on c8.
Black captured [Lf1], Tb3xb4++, a7xb6, c7xd6, fxexd, and promoted on d1 and h1.
The blackened squares have been included in the piece count, which should be (8+10). (2011-08-30)
Henrik Juel: It goes without saying that wK stands (or is added) on a4 (2021-10-31)
Mario Richter: I do not get the point here: f7=white pawn, c8=white Bishop are forced, but the rest of the pieces can be placed in many ways, e.g. sDa6,sDa4,sTh1,sSb4, wKa1.
So what am I missing? (2021-11-01)
comment
Keywords: Add pieces
Genre: Retro
FEN: 1nyk4/1p1p1yP1/yp1p3B/3R4/yy5r/P3p1p1/1P2P1P1/y2b3y
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-31 more...
5 - P0006661
Charles Masson Fox
The Chess Amateur 1930
P0006661
(3+4)
Ergänze den wK, dann h=1.5
a) +wKc8?: Schwarz hat keinen letzten Zug
b) +wKg7: 1. ... Gc8 2. Gf4 Gh8
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milan: +wKd6 wGc4-b4 bGe2-e1 1. bGe1-a5 K×c7 2.Ga-f5 Gb8 = all pieces cooperating.
h#2 +wGb8 wGe4 1.bGa5 Ke7 2.Gf5 Gf8# M.Frelih (2021-09-05)
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comment
Keywords: Add pieces
Pieces: du = Grasshopper (G)
Genre: Retro, Fairies
FEN: k7/p1*2q5/P7/4*2Q3/2*2Q5/8/4*2q3/8
Reprints: 73 C. M. Fox, His Problems 1936
Input: Gerd Wilts, 1996-07-17
Last update: A.Buchanan, 2022-10-22 more...
6 - P0007106
Josef Haas
1607 feenschach 29 08-09/1975
P0007106
(9+13)
Füge einen wT so ein, daß Weiß #1
Weiße Schlagfälle: a2xDb3, sLc8f8; schwarze Schlagfälle: c7xLb6, h7xg6xf5xe4xd3xBc2. +wTc1? (1. Txa1#) +wTa8! (1.Txa7#). Zur Auflösung der Stellung muß ein wL auf b6 entschlagen werden und nach c1 zurückgebracht werden. Dann kann nach Rücknahme von wBd2-d3, sBd3xBc2 und sBe4xDd3 die wD nach d1 zurückgebracht werden. Die schwarzen Türme müssen nach h2 und h3 gebracht werden. Da aber der wK nicht über g2 nach e1 kann, muß er vor Zurücknahme von d2-d3 auf die Grundreihe gebracht werden. Auf g1 läßt er dann die wD nach d1. Für einen weiteren wT ist im Korridor c1-g1g2h2h3 einfach kein Platz mehr!
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Hans-Jürgen Manthey: + wTh8, mögliche Zugfolge: 1. Sg1-f3 Sb8-c6 2. Sf3-g5 Sc6-b8 3. Sg5-e6 Sb8-c6 4. Se6xLf8 Sg8-f6 5. Sf8-g6 h7xSg6 6. Sb1-c3 Th8-h3 7. Sf3-d5 Th3-e3 8. Sd5-b6 Ke8-f8 9. Sb6xLc8 Kf8-g8 10. Sc8-d6 Kg8-h7 11. Sd6-f5 Dd8-h8 12. Sf5-d6 Kh7-g8 13. Sd6-f5 Dh8-h4 14. Sf5-d6 Kg8-h7 15. Sd6-f5 Ta8-h8 16. Sf5-d6 Kh7-g8 17. Sd6-f5 Dh4-b4 18. Sf5-d6 Th8-h3 19. Sd6-f5 Th3-f3 20. h2-h4 Sc6-d4 21. Ta1-b1 Kg8-h7 22. Sf5-g3 Kh7-h6 23. Sg3-e4 Kh6-h5 24. Se4-g5 Kh5-g4 25. Th1-h3 Kg4-f5 26. Sg5-e4 Kf5-e5 27. Th3-g3 Ke5-d5 28. Se4-d6 Kd5-c6 29. Sd6-f5 Kc6-b5 30. Tg3-g5 Kb5-a5 31. Tg5-h5 Tf3-h3 32. Th5-h8 Th3-h2 33. Sf5-d6 Te3-h3 34. Sd6-f5 Sf6-e4 35. Sf5-d6 Se4-g3 36. Sd6-f5 Sg3-h1 37. Th8-a8 Sd4-b3 38. g2-g3 g6xSf5 39. Lf1-g2 Sb3-a1 40. Lg2-e4 f5xLe4 41. Ke1-f1 Db4-b3 42. Kf1-g1 Db3-b6 43. Dd1-f1 Db6-b5 44. Df1-g2 Db5-b4 45. Dg2-f3 Db4-b3 46. Df3-d3 e4xDd3 47. Kg1-f1 d3xc2 48. d2-d3 Th2-g2 49. Kf1-e1 Th3-h2 50. Lc1-e3 Tg2-g1+ 51. Ke1-d2 Th2-g2 52. Le3-b6+ c7xb6 53. a2xDb3 Tg1-d1+ 54. Kd2-c3 Tg2-g1 55. Kc3-c4 b6-b5+ 56. Kd4-c5 Tg1-f1 (2021-06-30)
comment
Keywords: Add pieces, Corridor
Genre: Retro
FEN: 8/pp1pppp1/8/kpK5/7P/1P1P2P1/1Pp1PP2/nR1r1r1n
Input: Gerd Wilts, 1996-08-13
Last update: Gerd Wilts, 1996-08-14 more...
7 - P0007560
Bror Larsson
Tidskrift för Schack 1953
P0007560
(0+8)
Auf welchen Feldern darf der wK nicht stehen, weil die Stellung dann illegal wäre?
12 Felder
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Adrian Storisteanu: And for good measure, P1004380 (bK vs white officers)... (2021-05-07)
Henrik Juel: The reason for illegality is impossible (double, triple) check throughout
a7d7e7f7h7 = 5
a6d6h6 = 3
a5 = 1
h4 = 1
a3h3 = 2 (2021-05-07)
comment
Keywords: Add pieces
Genre: Retro
FEN: rnbqkbnr/8/8/8/8/8/8/8
Reprints: feenschach 46 04-06/1979
Input: Gerd Wilts, 1996-08-31
Last update: Gerd Wilts, 2005-12-03 more...
8 - P1000069
Henrik Juel
7019 Thema Danicum 86 04/1997
2. Preis
P1000069
(8+9)
Ergänze einen Stein!
Letzter Zug?
+sLf8
R: 1. ... g7xSf6 2. Sg4-f6+ Kd8-e8 3. Se5-g4 Kd7-d8 4. Sc6-e5+ Kc8-d7 5. Sd8-c6 Kb8-c8 6. d7-d8=S Kc8-b8 7. c6xTd7+ Td8-d7 8. b5xLc6 0-0-0 9. a4xTb5 Ld7-c6 10. a3-a4 Lc8-d7 11. c3xDd4 Dd8-d4 12. a2-a3 d7xLe6
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There is some choice in the retro path, but in the shortest resolution, the sK round trip is uniquely defined: e8-c8(castling)-b8-c8(switchback)-d7-d8-e8. Note that wTa, still stuck in its cage, was captured on c2.
Mu-Tsun Tsai: Can't believe that this one didn't get the 1st prize. The design is really brilliant! (2011-04-17)
Henrik Juel: Thanks for your nice words, Mu-Tsun. In 1997 there were few retro problems in Thema Danicum, and they were judged together with fairy problems, so the competition was stiff. (2011-04-17)
Henrik Juel: Thanks for posting the solution, Andrew
Two minor points:
The uncheck plusses should be R: 1. ... g7xSf6 2. Sg4-f6+ Kd8-e8 3. Se5-g4 Kd7-d8 4. Sc6-e5+ Kc8-d7 5. Sd8-c6 Kb8-c8 6. d7-d8=S Kc8-b8 7. c6xTd7+
sK could wander around on the queen side while wS pendulates on, say, g4 and h2; but in the shortest resolution the sK round trip is unique (2021-01-17)
A.Buchanan: welcome Henrik - also added the famous sLf8 to the animation (2021-01-18)
Henrik Juel: Six of the pawn captures happen to be uniquely determined, namely
b5xLc6xTd7, c3xDd4
d3xTc2, d7xLe6, g7xSf6 (2021-01-18)
A.Buchanan: An earlier sequence where timing was important: dxTc2+ Kc1 Td3/Sd3+ exd3 (2021-01-18)
A.Buchanan: Solved in YouTube: https://www.youtube.com/watch?v=sKi8ahju9Mc (2022-02-17)
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comment
Keywords: Add pieces (l), Last Move?, Type A, Kindergarten Problem, Volet Pawn, Ceriani-Frolkin Theme (S), Castling, Round Trip (k6)
Genre: Retro
FEN: 4k3/ppp1pp2/4pp2/8/3P4/3P2P1/1PpP1PP1/2K5
Reprints: I) Economy Records in Add Unit(s) Problems , p. 8, 03/2011
YouTube 17/02/2022
Input: Gerd Wilts, 2000-07-31
Last update: A.Buchanan, 2022-02-18 more...
9 - P1004380
Bror Larsson
Tidskrift för Schack 1953
P1004380
(8+0)
Auf welchen Feldern darf der sK nicht stehen, da die Stellung dann illegal wäre?
Es sind die 12 Felder a2,a3,a4,a6,d2,d3,e2,f2,h2,h3,h5,h6.
Auf Feld f3 steht der schwarze König legal weil das Doppelschach
Se2-g1 möglich wäre.
play all play one stop play next play all

Duplicate Diagram: P1386681

t/One-Liner 8-feldrig a1h1
Adrian Storisteanu: And not to be missed, P0007560 (wK vs the black home-based officers)... (2021-05-07)
comment
Keywords: Add pieces, Oneliner (8-feldrig a1h1)
Genre: Retro
FEN: 8/8/8/8/8/8/8/RNBQKBNR
Reprints: feenschach (46), p. 81, 04-06/1979
O4 Problemkiste 06/1999
Input: Gerd Wilts, 2003-01-10
Last update: A.Buchanan, 2023-04-13 more...
10 - P1011875
Andrej N. Kornilow
Andrey Frolkin

12111 Die Schwalbe 204 12/2003
3. Preis
P1011875
(13+12)
Ergänze einen weißen Stein auf der h-Linie!
a) Letzter Zug?
b) Letzter Zug des wK?
a) +wSh8 R: 1. Sg6xTh8 Th7-h8 2. Sc3-d1 Th8-h7 3. Sb5-c3 Th7-h8 4. Sd4-b5 Th8-h7 5. Sc6-d4 Th7-h8 6. Sb8-c6 Th8-h7 7. b7-b8=S Th7-h8 8. b6-b7 Th8-h7 9. b5-b6 Th7-h8 10. b4-b5 Th8-h7 11. Sh2-f1 Th7-h8 12. Sf3-h2 Th8-h7 13. Sd4-f3 Th7-h8 14. Sc6-d4 Th8-h7 15. Sb8-c6 Th7-h8 16. b7-b8=S Th8-h7 17. b6-b7 Th7-h8 18. b5-b6 Th8-h7 19. a4xBb5 Th7-h8 20. Dh8-g8 b6-b5 21. Kg8xSf8 Te8-e7 22. a3-a4 Le7-d8 23. a2-a3 Lc5-e7 24. b3-b4 e7-e6 25.
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Das Schach, in dem der sK steht, kann nur durch einen wSh8 oder einen wLh7 aufgehoben werden. Damit Schwarz vor dem Schach gezogen haben kann, muß die weiße Figur eine schwarze Figur entschlagen. Die beiden anderen fehlenden weißen Steine wurden von den sBBg und h geschlagen, d.h. die beiden wBa2b2 müssen sich beide auf b8 umgewandelt haben. Im letzten weißen Zug wurde auf der h-Linie ein sT entschlagen, da ein sL oder sS vorher nicht gezogen haben konnten und eine sD die Umwandlung des sBb7 auf b1 in eine D erfordert hätte, was unmöglich ist, da dann der wB mit zwei Schlägen um den sBb herumgeschlagen haben müßte, was aber insgesamt 5 weiße Schläge erfordert hätte. Als letzter Zug von Weiß ist also Sg6xTh8 oder Lg6xTh7 denkbar.
Um die Stellung aufzulösen, muß die wD nach h8 und der wK nach g8 ausweichen und ein Stein auf f8 gestellt werden, damit der sTe8 nach e7 kann. Da dazu einige weiße und schwarze Tempozüge benötigt werden kann dies nur wie folgt erreicht werden:
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comment
Keywords: Uncapture of pieces by pieces, Add pieces, Last move of a specific piece?, Last Move?
Genre: Retro
FEN: 3b1KQ1/p1pprp2/4pR1k/5pRq/6PB/4P1pb/2PP1PP1/3N1N2
Reprints: feenschach 155 2004
A3855 Phénix 147 01/2006
(8) Die Schwalbe 240 12/2009
Input: Gerd Wilts, 2003-12-30
Last update: A.Buchanan, 2024-01-15 more...
11 - P1012627
Filip S. Bondarenko
Themes-64 1986
P1012627
(16+0) cooked
Ergänze die 16 schwarzen Steine so, dass keiner angegriffen ist!
Rb1 Dd7 Ta6 et h6 Fh2 et g6; Ca1 et Ca7; Pa4, b5, c7, d5, e6, f5, g3, h5.
play all play one stop play next play all
Cook: Two solutions:
B4RK1/nNbq4/r1p1pBbr/1pPp1p1p/pP1P4/P3PPpP/3Q2P1/nk1N1R2
B4RK1/nNpq4/r3pBbr/1pPp1p1p/pP1P4/P3PPpP/3Q2Pb/nk1N1R2
"Add the 16 black units so that no piece is attacked."
Alain Brobecker: One way to correct this would be to move WRf1 to h1 and WNd1 to f1 ( B4RK1/1N6/5B2/2P5/1P1P4/P3PP1P/3Q2P1/5N1R ) , solution would then be B4RK1/nNbq4/r1p1pBpr/1pPp1p1p/pP1P4/Pb2PP1P/3Q2P1/nk3N1R (2021-09-21)
comment
Keywords: Add pieces, Construction task
Genre: Retro
FEN: B4RK1/1N6/5B2/2P5/1P1P4/P3PP1P/3Q2P1/3N1R2
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2021-09-21 more...
12 - P1080647
Dmitri Baibikow
503 Shakhmatnaya Kompozitsiya 74 2006
2. Preis
P1080647
(11+9)
Ergänze Steine!
R: 1. [+wTb2;+wSa4,+wSa2,+sLb5,+sSc5,+sSc3] d2-d3+ 2. Sd3-c5 Sc5-a4+ 3. La4-b5+ Tb5-b2+ 4. Sb2-d3+ Sb4-a2 5. Db1-a1+ Sa2-b4 6. Se4-c3+ Sc3-a2 7. Sg3-e4 g5-g6 8. Sf1-g3 g4-g5 9. f2-f1=S g3-g4 10. f3-f2
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Henrik Juel: the animation does not work (2023-04-13)
Mario Richter: Entering the diagram position as zero position and the solution Position [i.e. with the added pieces] as Position a) should make the Animation work correctly ... (2023-04-14)
comment
Keywords: Add pieces, Type C, Economy record
Genre: Retro
FEN: 8/1PPPp3/BpQp2P1/r2p4/2k5/K2PP3/2R1P3/q1bb4
Reprints: 6C) Economy Records in Add Unit(s) Problems , p. 7, 03/2011
Input: Gerd Wilts, 2009-01-19
Last update: Mario Richter, 2023-04-14 more...
13 - P1108458
Werner Keym
Main-Post 1968
P1108458
(14+0)
Ergänze den sK, dann #1
a) +sKd3 1. 0-0-0? Castling rights lost
b) +sKf7 1. g8=D#
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+sKd3? und 1.0-0-0# macht nicht, veil die Stellung ist illegal (saB & sbB).
Das Brett wird um 180° gedreht, dann +sKf7 und 1. g7-g8=D#
Henrik Juel: I believe that the try is
add sK on d3 and 1.0-0-0#
This addition is illegal, however:
the white h-pawns captured all 15 missing black men, but [Pa7,b7] need three captures to promote, and only two white men are missing (2023-04-13)
comment
Keywords: Joke, Board Rotation (180), Add pieces (sK), Cant Castler
Genre: Retro
FEN: 6B1/4R2P/P6P/B6P/7P/7P/1P5P/R3K1N1
Input: Gerd Wilts, 2010-06-23
Last update: A.Buchanan, 2023-04-15 more...
14 - P1215907
H. Hall
1097 Wiener Hausfrauen-Zeitung 3, p. 22, 21/01/1900
P1215907
(8+6)
#2
Wo muß der sK stehen, daß Weiß am Zug ein Matt in 2 Zügen erwingen kann?
+sKh1, dann
1. Kc1! droht 2. Dd1#
1. ... Lxa1 2. Dxc6#
1. ... Lb2+ 2. Kxb2#
1. ... Lxd2+ 2. Kxd2#
1. ... Lxd4 2. Dxc6#
1. ... f4 2. Dh7#
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WHZ schreibt 'Aus einem Problemturnier'

Forderung in der 'Bastion': "Wer komponiert mit? Wo muß der schwarze König stehen, damit Weiß ihn in zwei Zügen mattsetzen kann?"

Quelle und Forderung laut Albrecht-Sammlung (id=204694): 1. Preis "One King"-Problem Tourney 'Brighton Society' vor November 1899; Erschien ohne sK mit der Forderung "Find the place of the Black King"
Henrik Juel: After addition of sKh1, C+ Popeye 4.61 (2023-05-26)
comment
Keywords: Add pieces
Genre: Fairies
FEN: 2b5/8/2r1p3/4Bp2/1p1P4/2b1N3/2QPN3/R2K4
Reprints: 46 150 Schachkuriositäten 1910
47 Die Bastion 22/07/1934
Input: Felber, Volker, 2011-10-27
Last update: Mario Richter, 2023-05-26 more...
15 - P1227253
Andrew Buchanan
N1QCi) Economy Records in Add Unit(s) Problems , p. 4, 06/06/2011
P1227253
(5+5)
Add one unit.
+wDa2
R: Tb2-b1++
play all play one stop play next play all
Impossible treble check, so blocking unit needed on b2 or a2. If the former, then Bl just played R: 1. axb1=T++ & Wh only has a prior move if b2 unit is wS, but then illegal check to sK. So the unit must be added on a2. Last move was not b2-b1=R++ because of prior illegal check from sBb2. So it must be Tb2-b1++ (a capture implies retro pat). So what's on a2? Prior White move was not Kb1-a1, or the check from Tb2 is impossible. So it was L/D allowing L/Db1*a2. If the former, then there is a core cage comprising wKa1 wLb1 wBb3c2d2 sTb2 which can never unlock. So can only be +wDa2, with no problem for wD eventually sliding out.
Economy record for an "Add one unit" problem, type C, in which the unit to add is a Queen.
See http://abrobecker.free.fr/chess/addunits.pdf (2011-12-16)
more ...
comment
Keywords: Type C, Add pieces, Economy record
Genre: Retro
FEN: 8/8/8/8/1P6/rPb5/2PP4/Krnk4
Input: Alain Brobecker, 2011-12-16
Last update: A.Buchanan, 2023-04-13 more...
16 - P1227257
Andrew Buchanan
N18Ci) Economy Records in Add Unit(s) Problems , p. 6, 23/07/2011
P1227257
(9+5)
Add 18 units.
+wBa4b2d2f2h4 +wLc1 +wDc3 +sBa5b6c7d7e7f3g7h5 +sLf8 +sTc4 +sDg3
R: S-d4
play all play one stop play next play all
All pieces will be on board so pawns remain on their original files.
+BPc7d7e7g7 & +WPd2, implying +BBf8. Since BBd5 is out we've +BPb6 +WPb2, entailing +WBc1.
WKb3 is attacked by WNd4, so we need 4 units on the lines to the Ks and one of those lines is blocked by P, so +BPf3 +WPf2. Since all Rs must be out, we have +WPa4h4 +BPa5h5. Lastly, the remaining line interceptions are +BRc4 +WQc3 +BQg3.
Record of the highest number of units to add so that a position becomes legal (Type C, number of units specified).
See http://abrobecker.free.fr/chess/addunits.pdf (2011-12-16)
Henrik Juel: Good problem with well-explained solution (2021-01-17)
A.Buchanan: Thanks Henrik :) (2021-01-18)
comment
Keywords: Type C, Add pieces, Economy record
Genre: Retro
FEN: 8/8/2PNP1P1/1N1b4/1R1n2k1/1K1r4/6Rn/3B4
Input: Alain Brobecker, 2011-12-16
Last update: A.Buchanan, 2023-04-13 more...
17 - P1227636
Thierry Le Gleuher
5A) Economy Records in Add Unit(s) Problems , p. 7, 12/03/2011
P1227636
(15+8)
Add the missing unit(s)
b) +sBa3,sBb3,sBc3,sBd3,sBe3 R: 1. Sh3-g1 a4-a3 2. Dh1-d1 a5-a4 3. Tg1-b1 a6-a5 4. Lb7-a8 a7-a6 5. Lc8-b7 b4-b3 6. Kb1-a1 b5-b4 7. Kc1-b1 b6-b5 8. Kd1-c1 b7-b6 9. Ke1-d1 c4-c3 10. Kf1-e1 c5-c4 11. Kg2-f1 c6-c5 12. Ta1-g1 c7-c6 13. Dd1-h1 d4-d3 14. Kf1-g2 d5-d4 15. Ke1-f1 d6-d5 16. Sg1-h3 e4-e3 17. Lh3-c8 e5-e4 18. Lf1-h3 e6-e5 19. g2xDf3 Kg4-h5 20. Sf5-h6+
play all play one stop play next play all
WBc1 taken home, so no BP captured or promoted and it's impossible to retract WPe6xf7.
The only way to open the cage is to reorder white homebase and retract WPg2xf3, this takes 19 retro moves. Add all remaining BPs on third rank, and we can't retract both BPs on b7 and d7 to let the BBg8 get home, and we can't either retract BP to e7 since the BPg must retract to g7 in order to allow uncapture of dark squared BB by WPf7.
Record of the highest number of units to add so that a position becomes legal (Type A, number of units not specified, no promoted pieces).
See http://abrobecker.free.fr/chess/addunits.pdf (2011-12-22)
Alain Brobecker: See P1227635 for a version with promoted pieces. (2011-12-22)
Henrik Juel: Add five black pawns on a3,b3,c3,d3,e3 to provide the 18 tempo retractions needed.
-1.Sh3 a4 -2.Dh1 a5 -3.Tg1 a6 -4.Lb7 a7 -5.Lc8 b4 -6.Kb1 b5 -7.Kc1 b6 -8.Kd1 b7 -9.Ke1 c4 -10.Kf1 c5 -11.Kg2 c6 -12.Ta1 c7 -13.Dd1 d4 -14.Kf1 d5 -15.Ke1 d6 -16.Sg1 e4 -17.Lh3 e5 -18.Lf1 e6 -19.g2xDf3 Kg4 -20.Sf5 etc. (2011-12-22)
James Malcom: What formatting is required to add the five black pawns to the board with the solution? I know I have seen it done before somewhere. (2022-01-30)
A.Buchanan: You define another pieces list labelled e.g. b, then include that in the solution before you start moving pieces. Check out some other add pieces problems to see how it’s done. (2022-01-31)
James Malcom: Thanks! I have it figured out now. (2022-01-31)
comment
Keywords: Type A, Add pieces, Economy record
Genre: Retro
FEN: B5bn/5PrR/5prN/6pk/7p/5P2/PPPPPP2/KR1Q2N1
Input: Alain Brobecker, 2011-12-22
Last update: A.Buchanan, 2023-04-13 more...
18 - P1265741
Jesper Jespersen
Brighton Society
3. Preis
P1265741
(6+1)
#2
Wo muss der sK stehen?
+sKc5
1. De1! (Zugzwang)
1. ... Lxb2 2. Db4#
1. ... Lb4 2. Dxb4#
1. ... Kd4 2. Dc3#
1. ... Kd6 2. De7#
1. ... Kb6 2. Da5#
1. ... Kc4 2. Dc3#
1. ... Kd6 2. De7#
play all play one stop play next play all
Auszeichnung und Quelle laut 'Radlerin und Radler': 3. Preis 'Brighton Society'-Turnier. Nicht ganz klar, ob sich das auf die Stellung ohne oder mit sK bezieht.
Henrik Juel: C+ Popeye 4.61 also (2022-07-26)
comment
Keywords: Add pieces, Miniature
Genre: 2#
Computer test: Popeye WINDOWS-32Bit V4.37 (294312 KB)
FEN: 8/8/5P1P/8/K6Q/b4B2/1P6/8
Reprints: Radlerin und Radler 28/02/1902
117 150 exzentrische Schachaufgaben 1910
Input: Frank Müller, 2013-03-29
Last update: Mario Richter, 2022-07-26 more...
19 - P1305230
Thierry Le Gleuher
1 RIFACE 2015
P1305230
(11+1)
Ajouter des pièces.
+wSb2d2 +wLc4
R: 1. Dd1-c2+ Kc2-c3 etc
play all play one stop play next play all
A.Buchanan: How about wLa1 instead? (2020-10-06)
Henrik Juel: Yes, that would improve the economy slightly (2020-10-06)
Ladislav Packa: wBa1 is illegal... (2020-10-07)
A.Buchanan: Hi Ladislav why is wLa1 illegal, please? That was essentially my question before (2020-10-07)
Mario Richter: wLa1 instead of wDa1 is of course legal too: one of the dark-squared wBishops unpromotes on h8, one wQueen on g8. Black pawns g7+h7 let the wPawns g2+h2 pass by cross-capturing the 2 original wRooks. (2020-10-07)
more ...
comment
Keywords: Add pieces
Genre: Retro
FEN: 8/P7/P7/P1Q5/P3K3/P1k5/P1Q5/Q3B3
Input: A.Buchanan, 2015-07-08
Last update: A.Buchanan, 2023-04-13 more...
20 - P1305233
Alexandre Leroux
4 Retro Championnat de France RIFACE 2015
P1305233
(2+1) C+
KBP in 7.0
Ergänze die fehlenden Steine!
a) 1. a4 Sf6 2. a5 Se4 3. a6 Sxd2 4. Kxd2 h5 5. Kc3 Th6 6. Kb4 Te6 7. Ka5 Te4
play all play one stop play next play all
Originalforderung: Ajouter des pieces pour avoir une PCPJ en 7,0 Coups
Henrik Juel: Given the final position, the proof game is C+ by Euclide 1.01 (2015-07-08)
paul: Easily verified by Jacobi with the code: stipulation PG 7 pieces white Ka5 Pa6 black Re4 AddPieces (2021-07-06)
A.Buchanan: An essential linking word in the author's stipulation has been lost in translation: "Ajouter des pieces *pour* avoir une PCPJ en 7,0 Coups." I'm glad that on this occasion the Originalforderung was retained, but can we use a German template for these which is something like "Ergänze die fehlenden Steine um ein KBP in 7,0 zu haben". In English it would be "Add the missing pieces for an SPG in 7.0." (2021-07-07)
more ...
comment
Keywords: Unique Proof Game, Add pieces, Constrained problem
Genre: Retro
Computer test: rawbats, jacobi & Euclide
FEN: 8/8/P7/K7/4r3/8/8/8
Reprints: Phénix 2015
12 Phénix 296-297, p. 11536, 05-06/2019
1 Die Schwalbe 301, p. 440, 02/2020
Input: A.Buchanan, 2015-07-08
Last update: A.Buchanan, 2023-04-13 more...
21 - P1360105
Marken Foo
www.chess.com 2014
P1360105
(14+15)
PG in 4.0
Add the black queen
Author was inspired by a series of computer educational games entitled "Where on Earth is Carmen Sandiego", starring the eponymous villainess.
Moldenhauer: Computerprüfung: C+ Jacobi v0.7.5 in 5 Sekunden.
Notation: 1.h4 e6 2.Th3 De7 3.Ta3 Dxa3 4.b4 Dxc1 (2023-07-31)
comment
Keywords: Unique Proof Game, Add pieces (d)
Genre: Retro
FEN: rnb1kbnr/pppp1ppp/4p3/8/1P5P/8/P1PPPPP1/RN1QKBN1
Reprints: 1 Phénix 287, p. 11251, 07-08/2018
Input: A.Buchanan, 2019-01-25
Last update: A.Buchanan, 2019-01-25 more...
22 - P1360119
Marken Foo
www.chess.com 2016
P1360119
(14+12)
PG in 7.0 moves
Add the black queen
b) +wSb1
Marken Foo: There is a twin (b) +wSb1 pointed out by the chess.com user caveatcanis back in 2016, when accidentally misreading the diagram. It turns out to be a cute twin that fits the idea. (2020-10-21)
A.Buchanan: How many of these "Where in the world is Carmen Sandiego?" problems have you got, Marken? is it worth making a keyword? https://en.wikipedia.org/wiki/Carmen_Sandiego (2020-10-21)
comment
Keywords: Unique Proof Game, Add pieces (d)
Genre: Retro
FEN: rnbN1knr/pppppp2/8/8/8/8/PPPPPPP1/R1BQKB1R
Reprints: 9 Phénix 287, p. 11252, 07-08/2018
Input: A.Buchanan, 2019-01-25
Last update: A.Buchanan, 2020-10-21 more...
23 - P1375084
Edward Nathan Frankenstein
4 British Chess Magazine , p. 21, 1883
P1375084
(4+3)
Add one Black man, and White mates in three moves
(+sBd1, dann) 1. Kf5 Kd3 2. De4+ Kd2 3. Lh6#
play all play one stop play next play all
Add a black pawn on d1, then ...
Henrik Juel: After the addition of sPd1, C+ Popeye 4.61 (2022-01-26)
Henrik Juel: Other variations
1... Kc4/Ke3 2.Dc2+/Lb4 (2022-01-26)
comment
Keywords: Dummy Pawn, Golden Age (Dummy Pawn), Add pieces (b)
Genre: 3#, Retro
FEN: 5B2/8/6K1/1p6/3k4/6p1/1P4Q1/8
Input: James Malcom, 2020-04-27
Last update: A.Buchanan, 2023-08-04 more...
24 - P1384831
Lion Xray
Facebook 31/12/2020
P1384831
(9+9)
In how many ways can you add a White Queen and a Black Queen to this diagram to obtain a legal position?

2021 solutions
A.Buchanan: beautiful! (2021-01-01)
Henrik Juel: Yes, beautiful simplicity; finally one I can handle...
There are 64-9-9 = 46 empty squares, so the queens can be added in 46x45 = 2070 ways
The only illegality is that both kings are in check, which happens in 7x7 = 49 ways
Answer: 2070-49 = 2021 (2021-01-01)
A.Buchanan: Oh the other two today aren't hard, Henrik. But this one is a real gift from the Gods (2021-01-01)
comment
Keywords: Add pieces (Dd), Symmetrical position, Kindergarten Problem
Genre: Retro, Mathematics
FEN: 4k3/3ppp2/ppp3pp/8/8/PPP3PP/3PPP2/4K3
Reprints: Retros mailing list 01/01/2021
Input: A.Buchanan, 2021-01-01
Last update: A.Buchanan, 2021-01-01 more...
25 - P1384837
Joaquim Crusats
corr PDB Website 01/01/2021
P1384837
(10+3)
Add a black pawn inside the clock, then 9.5 moves to the same position but pawnless
a) +sBe5 1. d6 Lxd6 2. e7 Lxe7 3. f6 Lxf6 4. g5 Lxg5 5. f4 Lxf4 6. e3 Lxe3 7. d4 Lxd4 8. c5 Lxc5+ 9. Kxe5 Lb4 10. Ke4+
play all play one stop play next play all
Try is bP at "9pm"; actual solution is "12am" i.e. midnight of new year. Note that bL goes clockwise
Greeting card, inspired by the traditional Spanish 12 chimes to welcome the New Year.

Original version was cooked, due to FEN file confusion as composer had to compose it "against the clock" as he had been prioritizing deadline for releasing Problemas. Composing a clock problem "against the clock" is thematic though :D:D:D So there were cooks with +bPb2 or +bPc3
SCHRECKE: +sBe5
1.d6 L:d6 2.e7 L:e7 3.f6 L:f6 4.g5 L:g5 5.f4 L:f4 6.e3 L:e3 7.d4 L:d4 8.c5 L:c5+ 9.K:e5 Lb4 10.Ke4+ (2021-01-02)
A.Buchanan: What engine for C+, pls? (2021-01-02)
SCHRECKE: My brain solved it. (2021-01-03)
A.Buchanan: Ha me too! I will remove the C+ tick I put in anticipation (2021-01-03)
James Malcom: The question is, who's computer is smarter? :) (2021-01-03)
Mario Richter: Without long thinking I put the pawn on e5 for two reasons:
1. it shields the black King against white Bishop a1
2. e5 is the only square with property 1 where it doesn't prevent a White pawn from reaching a dark square.

I did some computer tests with 'rawbats' without noticing that the stipulation has changed [by adding the restriction "inside the clock"], the results being:
- no solution if the black pawn is added outside the diagonal b2-g7
- many solutions if black pawn is added on b2 (since it can promote on a1 or b1 and do all the "pawn elimination work")
- one principal solution if the black pawn is added on c3, starting with 1. Lb2 cxb2 2. d6,f6,Ke5 b1=L a.s.o.
- no solution if black pawn is added on d4,f6 or g7
- exactly one solution if black pawn is added on e5

So the "C+" can be added again, I think ... (2021-01-03)
comment
Keywords: Add pieces (b)
Genre: Retro
FEN: 7k/8/4P3/3P1P2/rbP1K1P1/3P1P2/4P3/B7
Input: A.Buchanan, 2021-01-01
Last update: A.Buchanan, 2021-01-03 more...
26 - P1400086
Alain Brobecker
R587 The Problemist 2022/03
Dedicated to Mario Richter
P1400086
(2+4) C+
Add all remaining units for a legal position with no unit attacking an enemy unit.
We must have one WP and one BP per column, the BPs being above the WPs since no capture occurred.
On the c column the WP cannot be on c2, where it would attack the square d3 where a black shield would be needed
to protect from BQe4, and BP cannot be on c7 to c5, thus we have WPc3 and BPc4.
This implies that on column b we have BPb3 and WPb2, then due to BPb3 the only possibility on
column a is WPa3 and BPa4.
Now let us look at the shield needed to protect WRb7 from the attack of BQ e4: a black piece on c6 is not
possible, so we must have a black piece on d5. It cannot be a BN which would attack WQb6 or a BB which
still would attack WRb7, and the BK and BRs already are elsewhere, so we have BPd5. We then deduce that
we have WPd2, and this forces WBc1 which has not moved from the game.
We need another shield to protect BPb3 from the attack of WQb6. Now that WBb4 is no more possible since this
bishop is still home, the only possibility is WNb5.
We cannot have a black piece on e3, hence no white piece on e2 and from this we deduce that we have
WPd6 and BPd7. The only possible shield to protect WPe6 from the attack of BQe4 is BNe5 (BBe5 would attack WPc3).
Now we also need a shield on c7 to protect BPd7 from the attack of WRb7, and the only possibility is WKc7.
On the f column we have WPf2 which is the only possibility.
More difficult to see, the only possibility for light squared WB is WBh5. To see this we must acknowledge
that g2 will contain a black piece as a shield and that other squares are attacked and no shield will help:
WBb1 is not possible because no shield can be put in-between it and BQe4, and WBe8 is not possible
since no B piece can be placed in b8-d8 as a shield.
Let's suppose that we have WPh4, then two black shields are needed on on g2 and h3 to
protects WPf2 and WPh4 from BRh2, but one of this shield would be the light squared BB
and the other one would be the remaining BN which would attack on of the WPs.
Thus our hypothesis is wrong, and we have WPh6 then BPh7.
We deduce then that we have BPg5 then WPg3.
What we said about the black shields on g2 and h3 (to protect WBh5 this time)
still holds, but this time we have a possibility which is BNg2 and BBh3.
BBf6 is the only possible square for dark squared BB. Thus we have BPf5.
WNd1 is the only possible square for the remaining WN.
Last we have two possibilities for the remaining WR, namely a1 and f1.
But since WPa3, WPb2, WPd2, WBc1 forms a well know retro-analytical cage
from which the WR cannot have escaped, we know that we have WRa1.
r6k/1RK1p2p/1Q2Pb1P/1N1pnppB/p1p1q3/PpP3Pb/1P1P1Pnr/R1BN4
play all play one stop play next play all
Alain Brobecker: The dedication was forgotten in the magazine! (2022-03-28)
comment
Keywords: Construction task, Add pieces, Aristocrat, Miniature
Genre: Retro
Computer test: Verified with homebrew program: http://abrobecker.free.fr/tools/bndrnk.zip
FEN: r6k/1R6/1Q6/8/4q3/8/7r/8
Input: Alain Brobecker, 2022-03-28
Last update: Alain Brobecker, 2023-06-16 more...
27 - P1402333
Harry Goldsteen
ChessCafe.com 12/2000
The Horse Concoction
P1402333
(5+0)
Add ten Black knights in a legal position such that White must either stalemate Black or be mated in twelve moves

Solution Diagram: White stalemates or Black wins
a) 1. Dxe1=

1. Kh2 Kf3! 2. Txd3+ Kg4!! 3. Thxg3+ Shxg3 4. Txg3+ S2xg3 5. Dxg3+ Sxg3 6. Dxe1 S2f3+ 7. Kg2 Sxe1+ 8. Kxf2 Sdc2 9. Lhxg5 Sd3+ 10. Kg2 Sce1+ 11. Kh2 Se5 12. ... S5f3#

b) 1. Sh3 h5 2. Sf4 h4 3. g3 h3 4. Sg2 hxg2 5. f3 Th4 6. gxh4 g5 7. h5 g4 8. h4 g3 9. h6 d5 10. h7 Lg4 11. fxg4 e5 12. Sc3 d4 13. e4 f5 14. a3 Lb4 15. axb4 Dd5 16. exd5 a5 17. d6 a4 18. Sd5 Ta5 19. Se3 Td5 20. c4 dxe3 21. cxd5 b5 22. Lc4 bxc4 23. g5 f4 24. g6 e4 25. g7 f3 26. d4 e2 27. Lh6 c5 28. Tb1 a3 29. b3 c3 30. d7+ Kf7 31. d6 c4 32. d5 a2 33. d8=L c2 34. d7 c3 35. d6 g1=S 36. Th2 g2 37. Kf2 e3+ 38. Kg3 f2 39. Lf6 e1=S 40. d8=L e2 41. d7 Sef3 42. Lb6 e1=S 43. d8=L Sg5 44. b5 Sef3 45. Lbd4 Sd2 46. L8b6 S1h3 47. Lbc5 g1=S 48. Kg2 f1=S 49. b6 Se2 50. Lg1 Sf2 51. Th3 Sfg3 52. Kh2 Sgh1 53. Kg2 Seg3 54. Kh2 a1=S 55. Kg2 c1=S 56. Kh2 c2 57. Kg2 Sce2 58. Kh2 c1=S 59. Kg2 Scd3 60. Kh2 Ke6 61. Kg2 Kf5 62. Lb2 Ke4 63. Tc1 Sc6 64. Tc3 Scd4 65. Lc1 Ke3 66. Df1 Sf6 67. g8=D Sf6-e4 68. Dg7 Sa1-c2 69. De5 Sc2-e1+

Shortest possible proof game of legality, found by Bader Al-Hajiri in 2/2001
play all play one stop play next play all
An earlier version was published at The Chess Cafe, December 2000

Article Reprint: https://timkr.home.xs4all.nl/tour/horse.htm

We return to the initial position of Goldsteen's problem. (See Diagram)

First the stalemate. Clearly, as many knights as possible must be pinned: eight, along all the diagonal and orthogonal lines the black King stands on. This means there must be eight pinning pieces; at least three promoted ones. That leaves two other knights to be bridled. One of them will be captured with the stalemating move, but the tenth knight can only be stalemated by being jammed. That can only happen on a corner square (the geometry of the chessboard forbids other squares; they would necessitate jamming knights on squares where they could not all be pinned) and this in turn means that both squares a knight's jump away from the corner square, must also be occupied by knights. Therefore, a8 and a1 are impossible. Square h8 also being impossible is the result of incredibly complex considerations (Goldsteen spent a month and a half on this problem; 'Not much, for me') that would be beyond the scope of even this article. But in the bottom right corner square, it can be done. And proudly, and with many thanks to Harry Goldsteen, I hereby announce the international premiere of his Horse Concoction. (See Diagram)

White is in check; With 1.Qxe1 he can stalemate Black. A sight to be savoured.
The first question that arises is whether the diagrammed position is legally possible. That is where the retrograde analysis comes in. Blacks has ten knights, so all of his pawns must have promoted. Therefore, his b- and h-pawns must each have captured at least once. As there are thirteen White pieces remaining, this leaves one other capture for Black. Black's remaining eleven pieces leave five captures in White's past. With two of them, his b- and h-pawns became doubled - therefore he has three past captures which, together with the remaining capture by Black, should explain that the white c2-f2 pawns and the black c7-f7 pawn could have passed each other on the way to their promotion squares. That is easily possible, but as we have four white Bishops of the dark squares, we still have to explain that at least three white pawns have promoted on black squares. Is that possible? Just. Black has captured dxe at one point, and White cxd, exd and fxg; precisely enough for three white Bishop promotions on d8, and one Queen promotion on g8.
But why so much trouble, one might ask, to promote to Bishops; wouldn't Queens have helped just as well in the stalemating? They could have promoted on white squares just as well. Sure, but we mustn't forget the second part of the stipulation: if White does not stalemate, Black must win. And with a Queen instead of even one of White's Bishops, Black could not do that.
So that is that. The position is possible, White can stalemate, and we now turn to Black's task of winning after any other move than 1.Qxe1. In fact, White only has one: 1.Kh2 The way in which the nine remaining horses, all of whom were pinned only a moment ago, now break loose massively, is almost as impressive as the whole Horse Concoction itself. Black plays 1...Kf3!, a move we would be tempted to call silent if we didn't hear the deafening whinnying of the four Knights that are now unpinned. Nxf1 mate is threatened, and after extensive analysis, Goldsteen has concluded that White will be mated no later than at move 12. The main variation is: 2.Rxd3+ Kg4!! (That beautiful move would also have followed 2.Qxf2+) 3.Rhxg3+ Nhxg3 4.Rxg3+ N2xg3 5.Qxg3+ Nxg3 6.Qxe1 N2f3+ 7.Kg2 Nxe1+ 8.Kxf2 Ndc2 9.Bxg5 Seven Knights have now perished, but the remaining three have concocted a mate: 9...Nd3+ 10.Kg2 Nce1+ 11.Kh2 Ne5 and mate next move by 12...Nf3.
There are numerous side variations.
James Malcom: It's rather hard to believe this gold of Goldsteen's wasn't in the database before...

Any and all help in cleaning up the stipulation is needed, please. (2022-06-29)
comment
Keywords: Add pieces, Miniature
Genre: Retro
FEN: 8/7P/1P6/8/7P/1P6/8/5Q2
Input: James Malcom, 2022-06-29
Last update: A.Buchanan, 2023-12-09 more...
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