Die Schwalbe

37 problem(s) found in 4486 milliseconds (displaying 37 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT A='Gross, Otto' AND K='Bedingungsaufgabe'] [download as LaTeX]

1 - P0001273
Luigi Ceriani
145 Europe Echecs 84 01/1965
P0001273
(6+7)
h#1 (wer? wie?)
Die F.P.I. wurde zweimal erreicht: das erstemal nach einer geraden, das zweitemal nach einer ungeraden Anzahl von Zügen.
Vertikaler Zylinder
Henrik Juel: The FPI (the initial array, but with Black to move) may be reached, e.g., by playing the white knights out, playing Ta1 to b1 and Th1 to g1, then correct the tempo by playing Tb1-a1-h1-b1, and finally moving rooks and knights back into the initial array.
The stipulation condition implies that all four castlings are illegal. Solution: 1.Txa7 Txh8#, not 1.0-0-0? Tc1 nor 0... 0-0-0? 1.Tc8. I do not see any tries involving king-side castling, although they would be nice to have also. (2010-12-08)
A.Buchanan: I am evidently being slow again: please explain. "The FPI was achieved twice: the first time after an even number of moves, the second time after an odd number of moves." To achieve FPI requires an odd number of single moves. Should the stip read "even number of *White* moves"? And then what does it mean to achieve FPI for the *second* time: does that mean game array with White to move again? I agree that the bottom line is to have eliminated all castling rights. (2020-12-25)
Henrik Juel: In P0001272 the FPI was reached once, implying that either White or Black has lost castling rights
I also do not understand this problem (2020-12-26)
Hans-Jürgen Manthey: da der lezte Zug b5x~a6 bzw b4x~a3 gewesen sein kann, sind alle Rochaden, sowie die Farbwahl zulässig.
Oder kann mir mal einer erklären was zum Teufel F.P.I. auf Deutsch heißt ?? (2020-12-26)
A.Buchanan: Siehe die Beschreibung des Schlüsselworts. Aber hier macht es nicht viel Sinn. Ich veröffentliche in ein paar Tagen einen Artikel in Problemas, der besser erklärt und einige neue Beispiele enthält. (2020-12-27)
Hans-Jürgen Manthey: Habe mich über F.P.I. schlau gemacht und FPI, Vertikale Zylinder sowie viele weitere Märchenschach-Bedingungen sind alsolut nicht mein Fall... (2020-12-28)
A.Buchanan: @H-JM: Ich bin kein großer Fan von Feenschach, aber es hat seinen Platz in der Welt der Komposition. Es vervielfacht den Gestaltungsraum für Schachkompositionen erheblich, und viele coole künstlerische Effekte können nur mit Feenschach erzielt werden. FPI ist bislang kein gutes Beispiel: Aber die zugrunde liegende Idee ist gut und reif für eine breitere Nutzung. (2020-12-28)
Henrik Juel: My goodness, Andrew, you are approaching Mario's command of all languages...
I believe, however, that Feenschach is an old-fashioned name for Märchenschach
It came about as a literal translation of Fairy Chess, but now lives on only in the magazine name feenschach
We did almost the same in Denmark: feskak in the 1930s, but now fantasiskak (2020-12-28)
A.Buchanan: Hi Henrik: thanks for your kind feedback. I try to draft in German, and then back translate in google translate to English to see what it thinks, and then go forwards again to correct dumb errors and give me better word choices. But there's still errors, which I am always keen to hear about, so thanks for reminding me about Märchenschach. I now understand that Märchen are "fairy tales" not "fairies" and that the term is used in English by folklorists. Encyclopedia Britannica characterizes them by: "Their usual theme is the triumph over difficulty, with or without supernatural aid, of the one least likely to succeed." I like the detail at the end, which suggests that the key move should be an unlikely one! (2020-12-29)
Henrik Juel: I had three years of german in middle school, but never used the language actively
I believe that most german problemists can read english as well as we can read german, so I do not make the extra effort to communicate in german; besides remembering to capitalize nouns, I would have to check masculinum, femininum, or neutrum (gender?), and check which prepositions entail accusative, dative, or both (kasus?) (2020-12-29)
Olaf Jenkner: (:-) (2020-12-29)
more ...
comment
Keywords: Cant Castler (wbsb), Fake game array, Castling (wbsb), Constrained problem, Vertical Cylinder (Vertikaler)
Genre: Retro, Fairies
FEN: r3k2r/P3p3/P3P3/8/8/p3p3/p7/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2020-12-26 more...
2 - P0001653
Philippe Leroy
522 Europe Echecs 373 01/1990
P0001653
(12+13)
BP in 60.0
Es fand ein ep-Schlag statt. Welcher?
1. a3 b5 2. a4 b4 3. a5 c5 4. a6 Lb7 5. axb7 a5 6. c4 bxc3ep 7. e4 f5 8. e5 Sf6 9. exf6 c2 10. De2 d5 11. De4 dxe4 12. g4 h5 13. g5 Th6 14. gxh6 g5 15. Se2 h4 16. Sg3 hxg3 17. f7+ Kd7 18. d4 c4 19. d5 Ta6 20. h7 g2 21. h8=L g1=L 22. Lc3 a4 23. La5 g4 24. h4 Lh2 25. h5 g3 26. h6 g2 27. h7 g1=L 28. h8=L a3 29. Lhc3 a2 30. f3 Ld4 31. Lcb4 Lh8 32. La3 c3 33. Le3 c1=L 34. Lg1 c2 35. Sc3 Tf6 36. Sa4 Sa6 37. b8=L Lch6 38. Lba7 Lb8 39. Th4 c1=L 40. d6 Kc6 41. d7 Db6 42. d8=L Lcg5 43. Lc7 e3 44. Lch2 e5 45. Sc5 Sc7 46. Tb4 e4 47. Kd1 e2+ 48. Kc2 e1=L 49. f4 Leh4 50. Kc3 e3 51. Kd4 e2 52. Ke5 e1=L 53. Td1 a1=L 54. Td7 Leg3 55. Sa4 Df2 56. Sb6 Lfg7 57. Lb5+ Kc5 58. Lc6 Sd5+ 59. Tc7 Sxf4 60. f8=L+ Td6
play all play one stop play next play all
James Malcom: A very interesting retro problem, with long-winded bishops promotions to cover its en passant tracks. (2021-01-24)
Hans-Jürgen Manthey: damit der Player klappt, muß 38. La7 in 38. Lba7 geändert werden. (2021-01-25)
James Malcom: Fixed. (2021-01-25)
comment
Keywords: Constrained problem, En passant, Non-standard material, Promotion, Non-Unique Proof Game
Genre: Retro
FEN: 1b3B1b/B1R3b1/1NBr3b/B1k1Kpb1/1R3n1b/B5b1/1P3q1B/b5B1
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-01-25 more...
3 - P0003090
Raymond Smullyan
37 The Chess Mysteries of the Arabian Knights 1981
P0003090
(13+15)
Die wD wurde auf c6 geschlagen, Weiß darf noch rochieren. Ergänze einen weißen Läufer. Ergänze, wenn möglich, den schwarzen Königsläufer oder zeige, auf welchem Feld dieser geschlagen worden ist!
Henrik Juel: Add wLh8, [Lf8] was captured on h6
[Lc1] was captured on c1, so the added wL must be [Pg2] playing g5xLh6-h8=L
. (2019-01-29)
Hans-Jürgen Manthey: R: 1. Td7 h8=L 2. The8 h7 3. O-O-O gxLh6 4. Df6 Lg2 5. Sg6 g5 6. Lh6 b3 7. La2 g4 8. Le6 g3 9. dxDc6 Dc6 10. Se7 De6+ 11. Sa6 Dc4 12. Lg5 Da6 13. h5 Dc6 14. Le7 Da4 15. e5 Da2 16. Sb4 Db1 17. Sa2 a5 18. SxLc1 Sf3 19. Sb3 Sc3 20. Sa5 a4 21. Sc6 a3 (2021-07-27)
comment
Keywords: Constrained problem
Genre: Retro
FEN: 2k1r3/pppr1pp1/n1p2qn1/P3p2p/8/1PN2N2/b1PPPPBP/R3K2R
Reprints: 37 Die Schachgeheimnisse des Kalifen 1984
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2018-03-07 more...
4 - P0005354
anonymous
52 Die Welt 05/04/1947
P0005354
(16+16)
Weiß beginnt mit 1. f3 2. Kf2 3. Kg3 4. Kh4.
Wie muß Schwarz spielen, um darauf im vierten Zug mattzusetzen?
1. f3 e5 2. Kf2 Df6 3. Kg3 Dxf3+ 4. Kh4 Le7#
play all play one stop play next play all
'Schachmatnaja Mosaika' gibt als Autor an: "A. Aljechin", ebenso 'Svobodnaia Gruzia'
Henrik Juel: If Mario is right, the author is no longer unknown (2021-10-31)
Henrik Juel: Black could also play 1... e6 (2021-10-31)
comment
Keywords: Constrained problem, Non-Unique Proof Game, Construction task
Genre: Retro
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Reprints: Der Schachzug (FV Schach Berlin) 18 01-02/1977
390 Schachmatnaja Mosaika , p. 128, 1984
1 Svobodnaia Gruzia 13/01/2005
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-08-10 more...
5 - P0006428
Andrey Frolkin
9133v Die Schwalbe 157, p. 283, 02/1996
P0006428
(13+12) cooked
BP in 20.0
May White castle?
Weiß darf noch rochieren: 1. h4 Sa6 2. h5 Sc5 3. h6 a6 4. hxg7 h5 5. f4 h4 6. f5 h3 7. f6 h2 8. fxe7 Sf6 9. g8=L Lh6 10. Lxf7 Kxf7 11. e8=T Kg6 12. Te6 Dg8 13. Tc6 Db3 14. axb3 dxc6 15. Ta4 Lh3 16. Te4 Tag8 17. Te8 hxg1=D 18. e4 Dd4 19. Lc4 Tg7 20. Lg8 Dd8
Ohne die Bedingung, daß Weiß rochieren darf, gibt es andere BPs: 1. f4 Sa6 2. f5 Sc5 3. f6 a6 4. fxe7 f5 5. e4 f4 6. Lb5 f3 7. Lc6 f2+ 8. Ke2 fxg1=D 9. h4 dxc6 10. h5 Lf5 11. h6 Sf6 12. hxg7 Dd5 13. g8=L Db3 14. Lxh7 Kf7 15. axb3 Lh6 16. Ta4 Tag8 17. Td4 Dxd4 18. e8=T Tg7 19. Lg8+ Kg6 20. Ke1 Dd8
play all play one stop play next play all
SH: Cooked?:
1. e4 f5 2. Ne2 f4 3. h4 f3 4. h5 fxe2 5. f4 exf1B 6. f5 Bc4 7. f6 Bb3 8. axb3 Na6 9. fxe7 Nc5 10. Ra6 Nf6 11. Rc6 dxc6 12. h6 Bh3 13. hxg7 Qd7 14. g8B Bh6 15. Bxh7 Kf7 16. Qf3 Rag8 17. Qe2 Rg7 18. Bg8+ Kg6 19. Qd1 a6 20. e8R Qd8 (2005-06-20)
Henrik Juel: The stipulation question was probably intended as an implicit condition; when answered affirmatively, there should be a unique solution
But SH's solution shows that the problem is cooked (2018-12-09)
more ...
comment
Keywords: Anti-Pronkin, Constrained problem, Unique Proof Game, Castling
Genre: Retro
FEN: 3qR1Br/1pp3r1/p1p2nkb/2n5/4P3/1P5b/1PPP2P1/1NBQK2R
Input: Gerd Wilts, 1996-06-12
Last update: James Malcom, 2022-02-08 more...
6 - P0008784
Niels Høeg
The Chess Amateur 07/1926
P0008784
(1+1)
Längste BP ohne Schach. Welches war der letzte Zug?
R: 5899. Kg2xTh1
play all play one stop play next play all
The game ends after 50 consecutive moves without captures or pawn moves (loss of castling right is not included here), or when there is not enough material to mate (say K-K or K-KS). There are 30 captures and 96 pawn moves (including 8 pawn captures) available, so the longest game seems to last (30+96-8)x50=5900 moves. This cannot be achieved because of the move-loss when the draw-preventing move shifts between white and black. Niels Høeg believed that 2 moves were lost and stated the solution as 5898.- Kb7xTa8. Karl Fabel later showed that only 1.5 moves need be lost.

Since 1926, there have been some relevant innovations to rules and conventions
(1) 50 move rule applies only to retro compositions, and will trigger automatically (no issue with writing down the move). (Codex 1953?)
(2) Removal of rules about draw by insufficient material (Laws 1997)
(3) Dead position rule introduced (Laws 1997)
(4) 75 move rule introduced (Laws 2014)
(5) Dead position rule applies only to retro compositions (Codex 2015)
(6) Articles 9.2 & 9.3 apply to chess problems - this includes 50 move rule and excludes 75 move rule (Codex 2019)

This is certainly a composition rather than a question about over the board chess. And it is certainly a retro composition. So the 50 move rule will dominate the 75 rule. The standard interpretation of interaction between 50 move rule and Dead Position in compositions is that Dead Position assessment *is* aware of looming automatic draw by 50 moves. (Note there is a similar assessment for interaction between Dead Position and Draw by Repetition.)

So we can argue that the game cannot last to 5898.5 moves, because the final move leads to a mandatory draw: either the king captures the last officer, or the king avoids the capture and the game ends in draw under the 50 move rule. So the position is dead at 5898.0. But even 5898.0 is too long for the diagram position with the kings so far apart. In the alternate reality if the last capture of a rook does not take place, there must be sufficient moves left for the kings to come together so that the rook can deliver checkmate. This will take at least 6.0 moves.

There is also still an ambiguity in the rules as to whether checkmate overrides draw by 50 moves. This is explicitly mentioned in the 75 move rule, but not in 50 move rule. I assume that checkmate *does* take priority.

Or does a valid problem only exist in the context of the rules and conventions that pertained at the time of its composition? The Codex does not opine on this general point.

Compare P1331022

Duplicate Diagram: P1101148, P1189676, P1191185, P1304589

A.Buchanan: There's a question whether DP rule has visibility of 3Rep & 50M state. The current consensus among most of the tiny group who might care is that for retros, it does have visibility, but for purely forward problems, it does not. This align with the idea that by default 50M & DP rules apply only to retros (2023-09-06)
A.Buchanan: The old intended interpretation is protected under the Golden Age principle. Suppose we do apply modern rules, codex & clarifications. If Black just played Kb7xRa8, then the alternative leading to mate takes 7.0 moves, so this must have been Black's 5892nd move at the latest. If White just moved, then it was White's 5892nd at latest. So to maximize the length of the game, Black moved last. (2023-09-06) edit (2023-09-06)
more ...
comment
Keywords: Longest Proof Game, Last Move?, only Kings, Non-Unique Proof Game, Dead Position, 50 move rule, Constrained problem, Type A, Miniature, Golden Age (pre-dead), Aristocrat
Genre: Mathematics, Retro
FEN: k7/8/8/8/8/8/8/7K
Reprints: Schackproblemet 1928
Schach und Zahl 1966
Input: Gerd Wilts, 1997-06-20
Last update: A.Buchanan, 2024-01-18 more...
7 - P0502695
John Niemann
192 FEENSCHACH 3 05/1971
P0502695
(1+0) cooked
Ergänze wB und sK, so daß ein korrektes Hilfsmatt entsteht
+sKd8, +wBg7
1. Ke7 Kg2 2. Kf6 Kf3 3. Kg5 g8=D+ 4. Kh4 Dg4#
play all play one stop play next play all
Cook: +wBg7 +sKe8 = h#4
+wBg7 +sKe7,Ske1,Skf2 = h#3.5

Duplicate Diagram: P0530804

HBae: Lösung in feenschach 11/1971 S.192, Lösungsstellung siehe P0530804 (2021-05-24)
comment
Keywords: Aristocrat, Miniature, Kindergarten Problem, Constrained problem
Genre: Fairies, h#
FEN: 8/8/8/8/8/8/8/7K
Input: hpr, 1996-07-23
Last update: A.Buchanan, 2021-05-24 more...
8 - P1013968
Hans Gruber
Frank Müller

3312c Problemkiste 10/1993
Erich Bartel zum (Schachbrett-1)ten Geburtstag gewidmet
P1013968
(16+16)
KBP zu einem Legal Cluster?
Kamikaze rex inklusiv
0 moves
play all play one stop play next play all
Null Züge! Denn schon die PAS ist ein Legal Cluster!
Die Autoren schreiben:
Correction:
Problemkiste, Issue 91, Februar 1994, Seite 128, Abteilung Bemerkungen & Berichtigungen
"Nr. 3212 [sic!] v.Gruber/Müller: wie HG mitteilt muß die Bedingung "KAMIKAZE REX INCLUSIV" heißen, nicht Kamikaze-Circe rex inclusiv. Letzteres wäre nicht nur ein Zuviel an Bedingungen, sondern auch noch UL, weil da auch 31-steinige Stellungen möglich sind."
Dieser Eintrag wird also mit der Korrektur aktualisiert!
Erich Bartel: Das war ein schöner Tag am 21.8.1993 bei Hans Gruber zusammen mit Frank Müller und ich bedanke mich recht herzlich für die Widmung, die mir ausgezeichnet schmeckt, fast so gut wie der super delikate Original-Zwetschgendatschi von Hansens Mutter!!!(\eb) (2007-03-04)
A.Buchanan: I don't get this: please can someone explain (2020-12-09)
Henrik Juel: Under the Kamikaze-Circe condition, where both captured and capturing man are reborn after a capture, the initial game array is legal, but removing one man gives an illegal position, so the IGA is a legal cluster
I have no info about the super delicate original Zwetschgendatschi from Hansen's mother (2020-12-09)
A.Buchanan: Hi Henrik, please clarify the rules for the middle part. Why is e.g. 1. a4 b5 axb5[sBb7] with the white pawn disappearing because it can't be reborn on occupied b2? Is it because the rebirth square for capturer is taken to be the start square not the end square of the move? That seems unlikely because I read in PDB that promotion counts before capturer's rebirth. The other possibility is that capture is illegal if the capturer cannot be reborn, but that seems inconsistent with the spirit of Kamikaze. Normally in Circe, a captured piece doesn't come back if the rebirth square is occupied. That surely would still apply here. Glossary is silent on all this. Thanks! (2020-12-09)
Henrik Juel: I think you are right, Andrew, so the IGA is not a legal cluster under Kamikaze-Circe, and the problem is cooked
In the more serious matter at hand, Zwetschgendatschi means plum cake... (2020-12-09)
A.Buchanan: I can't believe that serious fairy heavyweights such as were engaged in this problem would have made something cooked. And I still don't know what K-C actually is. I will ask HG. Maybe he has some comments on plum cake too? :-) (2020-12-09)
A.Buchanan: Henrik has also been copied on an email from Hans Gruber which states in part:
"The problem by Frank and me (PDB: P1013968) is cooked - the intended solution does not work, because the initial game array is NOT a legal cluster under Kamikaze-Circe rex inclusive. Of course this was unavoidable, as dedication problems always are unsound. We should have known that before but did not see in-how-far we were wrong. It is good to learn about that a bit more than a quarter of a century after. I dare to state, without consulting Frank, that there will be no correction to the problem. The basic idea "initial game array = legal cluster" just does not work." (2020-12-09)
Arnold Beine: Perhaps the problem can be fixed by changing the fairy condition into "Platzwechselcirce rex inklusiv". Then every position with 31 pieces is illegal and the IGA should be legal. (2020-12-09)
Henrik Juel: The problem can probably be saved just by changing the condition to Strict Kamikaze-Circe RI (2020-12-09)
A.Buchanan: What is the exact rule for Kamikaze-Circe please? And how does strict differ? (2020-12-09)
Henrik Juel: In Kamikaze-Circe, as part of a capture, the capturer is reborn if its rebirth square is empty (otherwise it vanishes), and the captured is reborn if its rebirth square is empty (otherwise it vanishes). The rebirth squares are the standard Circe ones (based on the capture square).
Strict Kamikaze-Circe additionally stipulates that a capture is possible only when both rebirth squares are empty; so the piece count never changes (2020-12-09)
A.Buchanan: Thanks Henrik: that doesn't sound very "strict" Kamikaze, sounds like "no-sharp-edges video-game" Kamikaze, where no-one actually can get hurt :) Regular Kamikaze RI is trivially a legal cluster, no? And more interestingly, ordinary Kamikaze is a legal cluster too, because if just one piece is removed, that must have been captured by the *other* king, who is still stuck in his cage. That would be my suggested fix (2020-12-10)
Henrik Juel: Yes, any condition where 31 men cannot occur in a game would do
I chose strict K-C, because that was the most simple change to K-C (2020-12-10)
A.Buchanan: Strict K-C RI the number of pieces for each side must always be the same. K RX you can certainly have 31 pieces on the board but not in game array position. It’s still very simplistic but it’s a bit more subtle (2020-12-10)
comment
Keywords: Unique Proof Game, Kamikaze (RI), Initial Game Array (2), Homebase (2), Constrained problem, Legal cluster
Genre: Fairies, Retro
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: Hans Gruber, 2004-05-01
Last update: A.Buchanan, 2020-12-22 more...
9 - P1044548
Giambattista Lolli
36 Osservazioni teorico-pratiche sopra il giuoco degli scacchi 1763
P1044548
(7+6)
#4
Weiß muss stets Schachbieten und Schwarz zwingen, dasselbe zu tun
1. Sbxc5+ Kc4+ 2. Lb3+ Txb3+ 3. axb3+ Dxb3+ 4. Dxb3#
play all play one stop play next play all
Frank Müller: Das ist ein Bedigungsproblem. "Ich soll durch stetes Schachbieten den Schwarzen dazu zwingen, dasselbe zu thun".
Heute könnte man Doppelschachzwang sagen. (2013-03-14)
SBD: Dann 1. Sbxc5+ Kc4+ 2. Lb3+ Txb3+ 3. axb3+ Dxb3+ 4.
Dxb3# oder auch 2. b3+ usw da Schwarz Schach geben muss. Oder darf Schwarz nicht Schach geben wenn er es kann? (2013-03-15)
Henrik Juel: I think that White must force Black to check, so 2.b3+ does not work.
The condition says 'I (i.e., White) must by constant checking force Black to do the same (i.e., check)' (2013-03-15)
Frank Müller: Korschelt schreibt hierzu: Lange Handbucg S. 146 druckt die Aufgabe falsch ab, schwarzer Se5 statt d5. Da sie dann an 3. ... Kd5 scheitert, so setzt er die wDame nach b7. (2015-04-05)
milan: +bSc7 M.Frelih (2021-05-25)
more ...
comment
Keywords: Brian Stephenson Collection (29553), Constrained problem
Genre: n#
FEN: 8/8/1Q2N3/2pn4/8/KN1kr1q1/PP1p4/3B4
Reprints: 533 Codex der Schachspielkunst , p. 60, 1834
IV.48 Schachspiel-Probleme [Alexandre] 1846
Input: Brian Stephenson, 2004-08-12
Last update: Arnold Beine, 2020-08-25 more...
10 - P1147347
Josef Kling
13 L'Illustration 05/08/1854
P1147347
(3+4)
#6
bei Weiß darf nur der Läufer ziehen
12 Lösungen, 2x3# (gilt nur, wenn die Zusatzbedingung nicht beachtet wird!)

Die Forderung war zunächst verdruckt (#8 statt #6), deshalb wurde die turnusmäßige Publikation der Al wohl verschoben und dann leider ganz vergessen [habe sie zumindest in den nachfolgenden Ausgaben nicht entdeckt].

Originalforderung: "Les blancs font mat en 6 coups, ne jouant que le Fou."
Henrik Juel: I see no #6, but as #8, this works, with one dual:
1.Lh4+ Kf8 2.Lg5 b5! 3.Le7+ Ke8 4.La3,Lc5+ Kd8 5.Lb4 g5,g6! 6.Le7+ Ke8 7.Lg5+ Kf8 8.Lh6# (2022-05-16)
Henrik Juel: Actually, there is a cook and several duals, as the withdrawal checks could go to d6/f6 also
1.Lf6,Lh4+ etc.
How do you manage a #6? (2022-05-16)
Anton Baumann: 1.Lf6,Lh4+ Kf8 2.Lg5 b5! 3.Le7+ Ke8 4.Lb4+ Kd8 5.Ld2[6.Lg5#] c5,c6 6.La5#
2. ... c5,c6 3.Le7+ Ke8 4.Lc5+ Kd8 5.Lxb6#
Die NL kann mit einem wBh4 verhindert werden. (2022-05-17)
comment
Keywords: Miniature Collection (0321113), Miniature, Constrained problem
Genre: n#
FEN: 4k3/2p1B1p1/1p2Q3/8/8/8/8/4K3
Input: Zuncke/Bruder, 2010-09-11
Last update: Mario Richter, 2022-05-15 more...
11 - P1171167


P1171167
(4+1)
#21
Matt in spätestens 21 Zügen, ohne König und Turm zu bewegen
1. Sc5! Kc8 2. Sa6 3. Sb8 Kc8 4. Sd7 5. Sb6 6, Sc8 Kd8 7. Se7 ... bis 16. Sh7 Kh8 17.-19 g2-g6 20. Sf6+ Kf8 21. g7#
play all play one stop play next play all
Alfred Pfeiffer: Bedingungsaufgabe: "Matt in spätestens 21 Zügen, ohne König und Turm zu bewegen."
Autor: unbekannt; Quelle: Carrera VII c. 13. Ponziani Sc. 37.
(Max Lange "Handbuch der Schachaufgaben" 1862, S. 141) (2010-09-15)
Arnold Beine: Weiß kann das Matt auch schon in 19 Zügen erzwingen:
1.Sf6 Kc8 2.Sd7 Kd8 3.Sb6 Ke8 4.Sc8 Kd8 5.Se7 Ke8 6.Sc6 Kf8 7.Sd8 Ke8 8.Sf7 Kf8 9.Sd6 Kg8 10.Se8 Kf8 11.Sg7 Kg8 12.Se6 Kh8 13.Sf8 Kg8 14.Sh7 Kh8 15.g4 Kg8 16.g5 Kh8 17.g6 Kg8 18.Sf6+ Kf8,Kh8 19.g7#, 13.Sg5 Kg8 14.Sh7 Kh8 usw.,
10...Kh8 11.g4 Kg8 12.g5 Kf8 13.Sg7 Kg8 14.Se6 Kh8 15.Sf8 Kg8 16.Sh7 Kh8 17.g6 Kg8 usw. mit weiteren Dualen. (2020-08-27)
Anton Baumann: Noch kürzer mit Turmopfer: 1.g4 Kc8! 2.g5 Kb8 3.g6 Kxa7 4.g7 Kb6 5.g8=D Kc6 6.Dg5 Kd7 7.Df6 Ke8 8.Dg7 Kd8 9.Df7 Kc8 10.Sc5 nebst 11.Db7/Dd7#
1. .. Ke8? 2.g5 Kf8? 3.g6 Ke8,Kg8 4.g7,Sf6+ 5.g8=D,g7#
2. .. Kd8 3.g6 Kc8 4.g7 Kb8 5.g8=D Kxb7 6.Db3 7.Sd6 nebst 8.Db5/Dd7#
offenbar muss auch noch die Umwandlung ausgeschlossen werden! (2021-09-24)
comment
Keywords: Constrained problem
Genre: n#
FEN: 3k4/R7/8/8/4N3/8/6PK/8
Reprints: 8 Schachzeitung , p. 19, 01/1861
Handbuch der Schachaufgaben [Lange] , p. 141, 1862
Input: Felber, Volker, 2010-09-12
Last update: Felber, Volker, 2018-03-10 more...
12 - P1178622
Joseph Ney Babson
403 American Chess Bulletin 12/1908
Xmas Novelty
P1178622
(8+1)
s=12
On the 12th move white's pieces must be so arranged that no square is unattacked, except that on which Black King is stalemated.
play all play one stop play next play all
James Malcom: Solution? (2021-01-03)
comment
Keywords: Constrained problem
Genre: Fairies
FEN: 8/8/8/8/8/6k1/8/RNBQKBNR
Input: Frank Müller, 2010-10-23
Last update: Frank Müller, 2010-10-23 more...
13 - P1182118
Frank J. Marshall
Marshall's Chess "Swindles" , p. 129, 1914
P1182118
(6+3) C+
#2 mit dem Sg6
1. f7 Kxg6 2. f8=S#
play all play one stop play next play all
Der gerade geschlagene wSg6 kommt durch die UW wieder aufs Brett und setzt matt; deshalb auch der wSb1.
Originalforderung: White mates in two - with the Knight at KKt6.
Motto: "A La Napier"
A.Buchanan: It's easy to make a miniature ideal mate version of this cheeky problem, e.g. 8/6pk/3N1PN1/6P1/7K/8/8/8 but the charm of Marshall's original is the apparently pointless wSb1 (2020-10-21)
comment
Keywords: Joke, Constrained problem, Joke promotion (recycle wS), Phoenix
Genre: 2#
Computer test: Popeye v4.85. The only #2 happens to satisfy the constraint
FEN: 8/6pk/4KPN1/6Pp/7P/8/8/1N6
Input: Mario Richter, 2010-12-11
Last update: Rainer Staudte, 2022-09-01 more...
14 - P1186631
Yuri Lebedev
Andrey Frolkin

574 Shakhmatnaya Kompozitsiya 84, p. 62, 10/2008
P1186631
(16+9)
Retract wPf4 to its starting square in 11 retromoves
R: 1. Lh5xTg4+ Sg7-f5 2. g3xBf4 f5-f4 3. Sc1xLe2 Td4-g4 4. Sg4xSf2+ Se4-f2+ 5. f2xDe3+ Kf4-f3 6. g2-g3+
play all play one stop play next play all
Hans-Jürgen Manthey: R: 1. Lh5xTg4+ Sg7-f5 2. g3xf4 f5-f4 3. Sc1xLe2 Td4-g4 4. Sg4xf2+ Se4-f2+ 5. f2xDe3+ Kf4-f3 6. g2-g3+ Lf3-e2+ 7. Sh2-c1 Te2-d2 8. Lb2-c3 Lc3-e1 9. Sf6-g4 Te1-e2 10. Le2-d1 Sd2-e4 11. La3-b2 c5-c4 12. e5xd6 c7-c5 13. Lb2-a3 Tb1-e1 14. La3-b2 Tb8-b1 15. Ta1-f1 Sb1-d2 16. Sh7-f6 b2-b1=S 17. Dd1-g1 b3-b2 18. Lf1-e2 f7-f5 19. Lg4-h5 Tb4-d4 20. Ld7-g4 Th8-b8 21. Lc6-d7 Da7-e3 22. La8-c6 Db8-a7 23. a7-a8=L Dd8-b8 24. a6-a7 Tb8-b4 25. a5-a6 b4-b3 26. e4-e5 b5-b4 27. a4-a5 b7-b5 28. Ta5-g5 Se6-g7 29. Ta8-a5 Sd4-e6 30. a7-a8=T Sc6-d4 31. e2-e4 h5-h4 32. Kh4-h3 Lg4-f3 33. Sg5-h7 Lc8-g4 34. Sf3-g5 Kf5-f4 35. Lc1-a3 Lg7-c3 36. Sg1-f3 Ke6-f5 37. Kg5-h4 Kd7-e6 38. Kf4-g5 Ke8-d7 39. Ke3-f4 Sg8-h6 40. Kd2-e3 h7-h5 41. Ke1-d2 Lf8-g7 42. d2-d3 g7-g6 43. Sc3-a2 d7-d6 44. a2-a4 Ta8-b8 45. a6-a7 Sb8-c6 46. a5-a6 Ta6-a8 47. b4xa5 Ta8-a6 48. b2-b4 a7-a5 49. Sb1-c3 (2021-07-19)
more ...
comment
Keywords: Non-standard material (TL), Constrained problem
Genre: Retro
FEN: 8/4p3/3P2pn/5nR1/2p2PBp/2BPPk1K/2PrNN1P/3BbRQR
Input: Nikolai Beluhov, 2011-02-05
Last update: A.Buchanan, 2021-07-20 more...
15 - P1186643
Thierry Le Gleuher
34 Problemaz 1 04/2007
P1186643
(14+15)
BP in 10,0 (remis)
1. Sf3 Sc6 2. Se5 Sxe5 3. h3 Sg4 4. Th2 Sxh2 5. a3 Sg4 6. hxg4 h6 7. Sc3 Sf6 8. Sb1 Sg8 9. Sc3 Sf6 10. Sb1 Sg8= draw by 3Rep
play all play one stop play next play all
Henrik Juel: The Draw (by repetition) is part of the stipulation
Without it, there would be cooks, e.g.
1.Sg1-f3 Sb8-a6 2.Sf3-g1 Sa6-b4 3.Sg1-f3 Sb4-d5
4.Sf3-g1 Sd5-e3 5.Bh2-h3 Se3-g4 6.Th1-h2 Sg4xh2
7.Sg1-f3 Sh2-g4 8.Sf3-h2 Sg4xh2 9.Ba2-a3 Sh2-g4
10.Bh3xg4 Bh7-h6
and short solutions, e.g.
1.Sg1-f3 Sb8-a6 2.Sf3-d4 Sa6-c5 3.Sd4-b3 Sc5xb3
4.Sb1-a3 Sb3-d4 5.Ta1-b1 Sd4-f5 6.Tb1-a1 Sf5-g3
7.Sa3-b1 Sg3xh1 8.Ba2-a3 Sh1-g3 9.Bh2xg3 Bh7-h6
10.Bg3-g4 (2018-12-25)
Henrik Juel: The shortest solution (without Draw) is unique
1.Sg1-f3 Sb8-c6 2.Sf3-e5 Sc6xe5 3.Bh2-h3 Se5-g4
4.Th1-h2 Sg4xh2 5.Ba2-a3 Sh2-g4 6.Bh3xg4 Bh7-h6
and identical to the first 6.0 moves of the intended solution
Why could the subsequent triple repetition not be achieved by rook moves?
Probably because the position after 7.Ta2 Tb8 8.Ta1 Ta8 is not considered to be a repeated position, as the castling rights have changed (2018-12-25)
A.Buchanan: Yes that's exactly it Henrik: the surviving rooks are nailed to their start squares (2021-06-05)
comment
Keywords: Unique Proof Game, Draw by repetition, Constrained problem, Castling
Genre: Retro
FEN: r1bqkbnr/ppppppp1/7p/8/6P1/P7/1PPPPPP1/RNBQKB2
Input: Gerd Wilts, 2011-02-05
Last update: Arnold Beine, 2021-06-05 more...
16 - P1216594
Ottó Titusz Bláthy
Deutsche Schachzeitung , p. 371, 12/1914
P1216594
(16+6)
Matt in 335 Königszügen
1. Ke1 Ke4 2. Kd1 Kd5 3. Kc1 Ke4 4. Kb1 Kd5 5. Ka2 Ke4 6. Ka3 Kd5 7. Kb3 Ke4 8. Ka2 Kd5 9. Kb1 Ke4 10. Kc1 Kd5 11. Kd1 Ke4 12. Ke1 Kf3 13. Kf1 hxg6 14. Ke1 Ke4 15. Kd1 Kd5 16. Kc1 Ke4 17. Kb1 Kd5 18. Ka2 Ke4 19. Ka3 Kd5 20. Kb3 Ke4 21. Ka2 Kd5 22. Kb1 Ke4 23. Kc1 Kd5 24. Kd1 Ke4 25. Ke1 Kf3 26. Kf1 c4 27. Ke1 Ke4 28. Kd1 Kd5 29. Kc1 Ke4 30. Kb1 Kd5 31. Ka2 Ke4 32. Ka3 Kd5 33. Kb4 Ke4 34. Ka5 Kd5 35. Ka6 Ke4 36. Ka7 Kd5 37. Kb8 Ke4 38. Kc8 Kd5 39. Kd8 Ke4 40. Ke8 Kd5 41. Kf8 Ke4 42. Kg8 Kd5 43. Kh8 Ke4 44. Kh7 Kd5 45. Kg8 Ke4 46. Kf8 Kd5 47. Ke8 Ke4 48. Kd8 Kd5 49. Kc8 Ke4 50. Kb8 Kd5 51. Ka7 Ke4 52. Ka6 Kd5 53. Ka5 Ke4 54. Kb4 Kd5 55. Ka3 Ke4 56. Ka2 Kd5 57. Kb1 Ke4 58. Kc1 Kd5 59. Kd1 Ke4 60. Ke1 Kf3 61. Kf1 g5 62. Ke1 Ke4 63. Kd1 Kd5 64. Kc1 Ke4 65. Kb1 Kd5 66. Ka2 Ke4 67. Ka3 Kd5 68. Kb4 Ke4 69. Ka5 Kd5 70. Ka6 Ke4 71. Ka7 Kd5 72. Kb8 Ke4 73. Kc8 Kd5 74. Kd8 Ke4 75. Ke8 Kd5 76. Kf8 Ke4 77. Kg8 Kd5 78. Kh8 Ke4 79. Kh7 Kd5 80. Kg8 Ke4 81. Kf8 Kd5 82. Ke8 Ke4 83. Kd8 Kd5 84. Kc8 Ke4 85. Kb8 Kd5 86. Ka7 Ke4 87. Ka6 Kd5 88. Ka5 Ke4 89. Kb4 Kd5 90. Ka3 Ke4 91. Ka2 Kd5 92. Kb1 Ke4 93. Kc1 Kd5 94. Kd1 Ke4 95. Ke1 Kf3 96. Kf1 g4 97. Ke1 Ke4 98. Kd1 Kd5 99. Kc1 Ke4 100. Kb1 Kd5 101. Ka2 Ke4 102. Ka3 Kd5 103. Kb4 Ke4 104. Ka5 Kd5 105. Ka6 Ke4 106. Ka7 Kd5 107. Kb8 Ke4 108. Kc8 Kd5 109. Kd8 Ke4 110. Ke8 Kd5 111. Kf8 Ke4 112. Kg8 Kd5 113. Kh8 Ke4 114. Kh7 Kd5 115. Kg8 Ke4 116. Kf8 Kd5 117. Ke8 Ke4 118. Kd8 Kd5 119. Kc8 Ke4 120. Kb8 Kd5 121. Ka7 Ke4 122. Ka6 Kd5 123. Ka5 Ke4 124. Kb4 Kd5 125. Ka3 Ke4 126. Ka2 Kd5 127. Kb1 Ke4 128. Kc1 Kd5 129. Kd1 Ke4 130. Ke1 Kf3 131. Kf1 g3 132. Ke1 Ke4 133. Kd1 Kd5 134. Kc1 Ke4 135. Kb1 Kd5 136. Ka2 Ke4 137. Ka3 Kd5 138. Kb4 Ke4 139. Ka5 Kd5 140. Ka6 Ke4 141. Ka7 Kd5 142. Kb8 Ke4 143. Kc8 Kd5 144. Kd8 Ke4 145. Ke8 Kd5 146. Kf8 Ke4 147. Kg8 Kd5 148. Kh8 Ke4 149. Kh7 Kd5 150. Kg8 Ke4 151. Kf8 Kd5 152. Ke8 Ke4 153. Kd8 Kd5 154. Kc8 Ke4 155. Kb8 Kd5 156. Ka7 Ke4 157. Ka6 Kd5 158. Ka5 Ke4 159. Kb4 Kd5 160. Ka3 Ke4 161. Ka2 Kd5 162. Kb1 Ke4 163. Kc1 Kd5 164. Kd1 Ke4 165. Ke1 Kf3 166. Kf1 gxh2 167. Ke1 Ke4 168. Kd1 Kd5 169. Kc1 Ke4 170. Kb1 Kd5 171. Ka2 Ke4 172. Ka3 Kd5 173. Kb4 Ke4 174. Ka5 Kd5 175. Ka6 Ke4 176. Ka7 Kd5 177. Kb8 Ke4 178. Kc8 Kd5 179. Kd8 Ke4 180. Ke8 Kd5 181. Kf8 Ke4 182. Kg8 Kd5 183. Kh8 Ke4 184. Kh7 Kd5 185. Kg8 Ke4 186. Kf8 Kd5 187. Ke8 Ke4 188. Kd8 Kd5 189. Kc8 Ke4 190. Kb8 Kd5 191. Ka7 Ke4 192. Ka6 Kd5 193. Ka5 Ke4 194. Kb4 Kd5 195. Ka3 Ke4 196. Ka2 Kd5 197. Kb1 Ke4 198. Kc1 Kd5 199. Kd1 Ke4 200. Ke1 Kf3 201. Kf1 g6 202. Ke1 Ke4 203. Kd1 Kd5 204. Kc1 Ke4 205. Kb1 Kd5 206. Ka2 Ke4 207. Ka3 Kd5 208. Kb4 Ke4 209. Ka5 Kd5 210. Ka6 Ke4 211. Ka7 Kd5 212. Kb8 Ke4 213. Kc8 Kd5 214. Kd8 Ke4 215. Ke8 Kd5 216. Kf8 Ke4 217. Kg8 Kd5 218. Kg7 Ke4 219. Kf8 Kd5 220. Ke8 Ke4 221. Kd8 Kd5 222. Kc8 Ke4 223. Kb8 Kd5 224. Ka7 Ke4 225. Ka6 Kd5 226. Ka5 Ke4 227. Kb4 Kd5 228. Ka3 Ke4 229. Ka2 Kd5 230. Kb1 Ke4 231. Kc1 Kd5 232. Kd1 Ke4 233. Ke1 Kf3 234. Kf1 g5 235. Ke1 Ke4 236. Kd1 Kd5 237. Kc1 Ke4 238. Kb1 Kd5 239. Ka2 Ke4 240. Ka3 Kd5 241. Kb4 Ke4 242. Ka5 Kd5 243. Ka6 Ke4 244. Ka7 Kd5 245. Kb8 Ke4 246. Kc8 Kd5 247. Kd8 Ke4 248. Ke8 Kd5 249. Kf8 Ke4 250. Kg8 Kd5 251. Kg7 Ke4 252. Kf8 Kd5 253. Ke8 Ke4 254. Kd8 Kd5 255. Kc8 Ke4 256. Kb8 Kd5 257. Ka7 Ke4 258. Ka6 Kd5 259. Ka5 Ke4 260. Kb4 Kd5 261. Ka3 Ke4 262. Ka2 Kd5 263. Kb1 Ke4 264. Kc1 Kd5 265. Kd1 Ke4 266. Ke1 Kf3 267. Kf1 g4 268. Ke1 Ke4 269. Kd1 Kd5 270. Kc1 Ke4 271. Kb1 Kd5 272. Ka2 Ke4 273. Ka3 Kd5 274. Kb4 Ke4 275. Ka5 Kd5 276. Ka6 Ke4 277. Ka7 Kd5 278. Kb8 Ke4 279. Kc8 Kd5 280. Kd8 Ke4 281. Ke8 Kd5 282. Kf8 Ke4 283. Kg8 Kd5 284. Kg7 Ke4 285. Kf8 Kd5 286. Ke8 Ke4 287. Kd8 Kd5 288. Kc8 Ke4 289. Kb8 Kd5 290. Ka7 Ke4 291. Ka6 Kd5 292. Ka5 Ke4 293. Kb4 Kd5 294. Ka3 Ke4 295. Ka2 Kd5 296. Kb1 Ke4 297. Kc1 Kd5 298. Kd1 Ke4 299. Ke1 Kf3 300. Kf1 g3 301. Ke1 Ke4 302. Kd1 Kd5 303. Kc1 Ke4 304. Kb1 Kd5 305. Ka2 Ke4 306. Ka3 Kd5 307. Kb4 Ke4 308. Ka5 Kd5 309. Ka6 Ke4 310. Ka7 Kd5 311. Kb8 Ke4 312. Kc8 Kd5 313. Kd8 Ke4 314. Ke8 Kd5 315. Kf8 Ke4 316. Kg8 Kd5 317. Kg7 Ke4 318. Kf8 Kd5 319. Ke8 Ke4 320. Kd8 Kd5 321. Kc8 Ke4 322. Kb8 Kd5 323. Ka7 Ke4 324. Ka6 Kd5 325. Ka5 Ke4 326. Kb4 Kd5 327. Ka3 Ke4 328. Ka2 Kd5 329. Kb1 Ke4 330. Kc1 Kd5 331. Kd1 Ke4 332. Ke1 Kf3 333. Kf1 Ke4 334. Kxg2 Kd5 335. Kxg3#
play all play one stop play next play all
Die ausführliche Lösung kann in 'Deutsche Schachzeitung' 03/1915 S. 86-87 und 'Bohemia' 30.05.1915 nachgelesen werden.
In der 'DSZ' mit der Forderung "Mat in 335 Königszügen" und folgenden Anmerkungen abgedruckt: "Der große Schritt in das Reich des Schrankenlosen. ... Was wir diesmal bringen, zeigt erst so recht, was alles möglich wäre, wenn die Problemkomposition alle Schranken durchbricht. Der w.K allein zieht und er setzt schließlich auch selbst (durch Abzugschach) mat. Er sucht einem der Türme, auf e6 oder g2, beizukommen und Schw. verhindert dies durch Königszüge, solange es geht. Bauernzüge von Schw. verursachen wieder, daß der w.K einmal oben, einmal unten anklopfen muß, bis die Bauern festgerannt sind und dann auch dem schw. K der Atem ausgeht. Die längste Königsfahrt besteht hier aus 35 Zügen. In der Position steckt auch manche Falle, die der Löser vermeiden muß."
Henrik Juel: The stipulation says 355 moves, but the solution says 335.
Could we get some details of the main variation, please. (2011-11-01)
Mario Richter: Correct stipulation is "Matt in 335 Königszügen". (2022-10-07)
Henrik Juel: the last move in this monster problem should be 335.Kxg3 (2022-10-07)
James Malcom: Fixed. (2023-03-07)
Henrik Juel: Thanks, Mario and James (2023-03-07)
more ...
comment
Keywords: Constrained problem, Sea Snake
Genre: n#
FEN: Q7/2BBPNpp/1Pr3R1/1Pp5/N7/2R1Pk2/1R4rP/N4K1Q
Reprints: 1620 Bohemia 30/05/1915
Input: Frank Müller, 2011-11-01
Last update: James Malcom, 2023-03-07 more...
17 - P1227941
Antoine Demonchy
CV 100 Fins de Parties Inverses 1882
P1227941
(7+4) cooked
s#35
par le P damé
Autorlösung unbekannt
play all play one stop play next play all
Cook: Weiß und Schwarz wandeln in die Dame um.
Nebenlösung in 19 Zügen:
1. Tg1 d4 2. Dc5 d3 3. Ta4+ Kf3 4. Te4 d2 5. h5 d1=D+ 6. Txd1 Kg3,Kg2 7. De3+ Kh2 8. h6 Kg2 9. h7 Kh2 10. h8=D+ Kg2 11. Deh3+ Kf2 12. Dh2+ Kf3 13. Tf4+ Ke3 14. Sb5 axb5 15. Dg8 b4 16. Te1+ Kd3 17. De2+ Kc3 18. Da2 b3 19. Dc2+ bxc2#
Olaf Jenkner: Was bedeutet par le P damé ? (2011-12-27)
Frank Müller: Vermutlich mit Damenumwandlung. Offen bleibt, ob weiße oder schwarze. (2011-12-27)
Arnold Beine: "par le P(ion) damé" bedeutet "durch einen umgewandelten Bauern". Die angegebene Kurzlösung in 19 Zügen erfüllt nicht diese Bedingung, denn das Matt gibt ein sB und keine schwarze Umwandlungsfigur. (2022-03-15)
comment
Keywords: Constrained problem
Genre: s#
FEN: 8/6R1/p2N4/P2p4/5k1P/R1Q5/1p6/1K6
Input: Frank Müller, 2011-12-25
Last update: Olaf Jenkner, 2011-12-27 more...
18 - P1227947
Antoine Demonchy
CXI 100 Fins de Parties Inverses 1882
P1227947
(8+3) cooked
s#55
par le P damé
Autorlösung unbekannt
play all play one stop play next play all
Cook: Weiß und Schwarz wandeln in die Dame um.
Nebenlösung in 37 Zügen:
1. Df1 Kd2 2. Kb3 Ke3 3. Kc2 Kd4 4. Ld6 Ke3 5. g8=D Kd4 6. Dgf8 Ke3 7. Sf3 Kxe4 8. Le7 Kd5 9. De8 Ke4 10. Dd3+ Kf4 11. Sh2 Kxe5 12. Df3 Kd4 13. Da4+ Ke5 14. Dfh4+ Kd5 15. Dad4+ Kc6 16. Dfe5 Kb7 17. Dd7+ Ka8 18. Dc8+ Ka7 19. De8 Kb7 20. Lh4 Ka7 21. Dc5+ Kb7 22. Kb1 e5 23. Ka1 e4 24. Lg3 e3 25. Sf3 e2 26. Df2 e1=D+ 27. Dxe1 Ka7 28. Lf2+ Kb7 29. Dd8 Kc6 30. De4+ Kb5 31. Ddd3+ Kxa5 32. Dc6 Kb4 33. Lb6 a5 34. Sd2 a4 35. Sb1 a3 36. Dd4+ Kb3 37. Db2+ axb2#
1. ... Kd4 2. Kb3 Kc5,Ke3 3. Ld6+ Kd4 4. Kc2 Ke3 und weiter wie oben
Arnold Beine: "par le P(ion) damé" bedeutet "durch einen umgewandelten Bauern". Die angegebene Kurzlösung in 37 Zügen erfüllt nicht diese Bedingung, denn das Matt gibt ein sB und keine schwarze Umwandlungsfigur. (2022-03-15)
comment
Keywords: Constrained problem
Genre: s#
FEN: 8/2B3P1/p3pQ2/P3P1N1/K3P3/4k3/8/8
Input: Frank Müller, 2011-12-25
Last update: Olaf Jenkner, 2011-12-27 more...
19 - P1240394
August Oeffner
1442 Schachzeitung , p. 287, 09/1862
P1240394
(13+7) C+
s#5
Weiss soll nur einmal Schach geben
1. Ta6+! Sa5 2. Lc6 cxd6 3. Lb6 d5 4. Ld4 dxc4 5. b4 cxb3ep#
play all play one stop play next play all
more ...
comment
Keywords: Constrained problem, En passant as mating move
Genre: s#
Computer test: A. Baumann: Gustav 4.2a
FEN: 3B4/1Bp5/2RP1P2/1p6/k1P1p3/bnP1p3/1PK1P3/RN1N4
Input: Frank Müller, 2012-06-01
Last update: Dieter Berlin, 2024-02-03 more...
20 - P1265153
Pedro Damiano
48 Libro de imparare giocare a Scacchi 1512
P1265153
(4+1)
#6
1. Se7! Ke8 2. Seg6 Kd8 3. Se6+ Kc8! 4. Se7+ Kb8 5. Sc6+ Kc8 6. Tc7#
play all play one stop play next play all
Cook: Leicht in 4 Zügen zu lösen: 1. Sb6 Ke8 2. Sg6 Kd8 3. Tf7 Ke8 4. Tf8#
Die Forderung ist daher zu ändern in: Matt in genau 6 Zügen.
Felber, Volker: Im Handbuch von Lange steht: 'Ohne die, von den alten Autoren auch wirklich versäumte, Bestimmung des Mattfeldes würde das Ziel schon in vier Zügen (...), welche zugleich eine andere Art der Schlussstellung illustriren, erreicht werden können. Bei Guarinus erscheint übrigens die Aufstellung so gedreht, dass sich der schwarze König auf h5, der weisse Thurm aug g8 etc. präsentirt.'.

Die Forderung beim Diagramm im Handbuch lautet: 'Matt auf c8 in sechs Zügen.'. (2020-04-21)
SCHRECKE: Nach 1.Se7 Ke8 2.Seg6 Kd8 3.Se6+ Ke8! geht zwar 4.Te7#, aber die Forderung 'Matt auf c8 in sechs Zügen' ist nicht zu erreichen. Damit ist das im Sinne der Aufgabenstellung UNLÖSBAR! (2020-04-22)
milan: wTa7-a1 wKh1-d1 wSc8-e7 1.Ta7!M.Frelih (2020-05-02)
Arnold Beine: Die angegebene "Korrektur" von milan funktioniert nicht. Wenn der sK im 4. Zug nach d8 und im 5. Zug nach e8 zieht, kann Weiß die Forderung mit Matt auf c8 nicht erfüllen. (2024-02-18)
comment
Keywords: Constrained problem, Rex solus (s), Aristocrat
Genre: n#
FEN: 2Nk1N2/R7/8/8/8/8/8/7K
Reprints: Osservazioni theoretico-pratiche il sopr , p. 553, 1763
48 Schachpartieen und Endspiele des Damiano 1857
715 Handbuch der Schachaufgaben [Lange] , p. 457, 1862
Input: Frank Müller, 2013-03-22
Last update: Felber, Volker, 2020-04-22 more...
21 - P1265245
Julius Mendheim
II.21 Taschenbuch für Schachspieler 1814
P1265245
(10+13)
#8 von h2
sBh2 ist unverletzlich
1. Df1+ exf1=D 2. Txg1+ Df5 3. Sg2+ Kg4 4. Se1+ hxg1=S 5. Th4+ Kxh4 6. h8=D+ Kg4 7. Sf6+ Kg3 8. Dh2#
play all play one stop play next play all
Arnold Beine: Nach 4...hxg1=D#!? in meinen Augen unlösbar, denn die Umwandlungsdame ist doch unverletzlich. Auch nach 5.Kxg1 Lc5+ gäbe es kein Matt im 8. Zug. Oder bedeutet "unverletzlich", dass auch Schwarz keinen Zug ausführen darf, der Weiß zwingt, diesen Stein zu schlagen? Dann würde 4...hxg1=L!! immer noch für Unlösbarkeit sorgen. (2024-02-18)
comment
Keywords: Constrained problem
Genre: n#
FEN: 8/7P/b3BQpR/rp4P1/1bp1N1RN/3p3k/3Bp2p/1r1n2nK
Input: Frank Müller, 2013-03-23
Last update: Arnold Beine, 2017-06-25 more...
22 - P1266118
Reginald A. Brown
77 Chess Problems [Brown] 1844
P1266118
(8+8) cooked
s#8 mit Sa1
1. Da5+! Kc4 2. T8d4+ Kb3 3. Da4+ Kc3 4. T2d3+ Sxd3 5. Txd3+ Dxd3 6. Se2+ Dxe2 7. Le5+ Kd3 8. Db3+ Sxb3#
play all play one stop play next play all
Cook: NL
Anton Baumann: NL:1.T2d5+ Kb6 2.Tb8+ Db7 3.Txb7+ Kc6 4.Tc5+ Kxc5 5.Da5+ Kc4 6.Dc7+ Kd3 7.Tb3+ Sxb3#
1. ... Kb4 2.Da5+ Kc4 3.Da4+ Kc3 4.Tc8+ Dc7 5.Tc5+ Dxc5 6.Txc5+ Kd3 7.Db3+ Sxb3#
4. ... Lxc8 5.Tc5+ Kd3 6.Db3+ Sxb3#
1. ... Kc4 2.Da4+ Kc3 3.Tc8+ ... (2022-05-31)
comment
Keywords: Constrained problem
Genre: s#
FEN: Q2R4/7q/8/2k5/5Bb1/p3PBN1/p1pR1n2/n1K5
Input: Frank Müller, 2013-04-02
Last update: Marcin Banaszek, 2022-06-02 more...
23 - P1266564
Pedro Damiano
43v Libro de imparare giocare a Scacchi 1512
P1266564
(6+3) cooked
#5
erst Schach dann Matt durch den Bauer
1. Dc7+ Sxc7 2. Ta8+ Sxa8 3. Sd7+ Sb6 4. axb6+ Ka8 5. b7#
play all play one stop play next play all
Korrektur der P1265149
Cook: 1. Sa4+ Sb6 2. Tb7+ Ka8 3. Ta7+ Kxa7 4. axb6+ Ka8 5. b7#
SP: (1) This is not the position given by Lolli (Bc5d4).
(2) The supposed cook (from Franz & Lasa) does not work either in the diagram
or in Lolli's position. (1... Sb6+??)
(3) Like most problems do be found in Damiano this cannot be attributed to
Damiano; the same positions can be found in Lucena, the Göttingen Ms, etc.
See Murray for details.
(4) Lolli's stipulation is simply a pawn mate; no extra condition (2022-09-09)
milan: wTb8-b6+bLb8 sTc7 sSd7 wSf8 1.D×c7 L×c7 2.S×d7 Sd6 3.T×d6 Lb6 4.a×b6+ Ka8 5.b7# by stockfish
M.Frelih (2022-12-11)
comment
Keywords: Constrained problem
Genre: n#
FEN: 1RK5/k7/pN6/PnB1Q3/8/8/8/8
Reprints: 40 Osservazioni teorico-pratiche sopra il giuoco degli scacchi 1767
Input: Frank Müller, 2013-04-06
Last update: Frank Müller, 2013-04-06 more...
24 - P1266821
Hirsch Silberschmidt
80 Die neu entdeckten Geheimnisse im Gebiet 1826
P1266821
(8+8)
#5 mit Le1
b) schwarzes #2 mit Sb5
a) 1. Sb6+ Kc5 2. Txc3+ bxc3 3. Dxb5+ Kd6 4. Lg3+ Te5 5. Lxe5#
b) 1. ... Dg7+ 2. Ke6 Sd4#
play all play one stop play next play all
Arnold Beine: Nebenlösung in b):
1. ... Le7+! 2. Ke6 Sd4# (2020-07-30)
Vendelbo: Diese Aufgabe ist kaputt: 1. a8b6 c4d4! und es gibt keine Lösung. Schwarz kann auch 1. a8b6 c4c5 2. c1c3 b4c3 3. a5b5 c5d4! und wieder mal keine Lösung. (2024-02-15)
comment
Keywords: Constrained problem
Genre: n#
FEN: N4b2/7q/5K2/Qn1r4/1pk3PP/2np4/8/1RR1B3
Input: Frank Müller, 2013-04-11
Last update: Arnold Beine, 2020-07-30 more...
25 - P1269503

914 Leipziger Illustrirte Zeitung 05/07/1862
P1269503
(8+10) cooked
a) +wKa7: #7
b) +wKh8: s#14
c) +wKd1: #16 mit Be3
a) 1. Ka8 Kc4! 2. Tc2+ Sxc2 3. Sxe5+ Sxe5 4. Dxe6+
b) 1. S8a7+ Kc4 2. Tc3+ Kd5 3. Sxb4+ Kd6 4. Sb5+ Ke7 5. Lh4+ Ke8 6. Tg8+ Lf8 7. Dh5+ Tf7 8. Lc6+ Tdd7 9. Txd3 exd3 10. Sxd3 e4 11. Sb4 e5 12. Le7 Kxe7 13. Dxe5+ Kd8 14. Txf8+ Txf8#
c) 1. S8a7+ Kc4 2. Tc3+ Kd5 3. Sxb4+ Kd6 4. Sb5+ Ke7 5. Lh4+ Ke8 6. Tg8+ Lf8 7. Dh5+ (Dg6+) Tf7 8. Lc6+ Tdd7 9. Txd3 exd3 10. e4 d2 11. Sd4 b5! 12. Lxb5 exd4 13. e5 d3 14. Sd5 exd5 15. e6 d4 16. exd7,exf7#
play all play one stop play next play all
Cook: a) NL in 5 Zügen: 3. Dxe6+ Td5 4. La6+ b5 5. Sa5#

Duplicate Diagram: P1378117, P1378117, P1378117

In der 'LIZ' ohne Autor/Quellenangabe zusammen mit folgendem Text veröffentlicht:
"Ein türkischer Sultan, der ohne leibliche Thronerben war, ließ vor seinem Sterben nachfolgende Schachposition mit der Bestimmung ausstellen, daß der glückliche Entzifferer zum Throne gelangen solle. Es war nämlich weder eine Forderung noch der weiße König angegeben. Von zwei Bewerbern placirte der eine die fehlende Figur auf H8 für ein Selbstmatt in 14, der andere auf D1 für ein Bauermatt in 16 Zügen. Die Prüfungscommission enschied sich für keinen, verlangte vielmehr ein einfaches Matt beim Stande auf A7 und beschloß die Krönung desjenigen, welcher in dieser Gestalt die Aufgabe zuerst in der kürzesten Zügenzahl lösen würde. Ein persischer Prinz, der zur Concurrenz zugelassen, das Matt in sieben Zügen herausfand, erlangte den Preis: die Krone des türkischen Reichs. Wir aber gedenken zur Belohnung der glücklichen Löser ihre Namen bekannt zu machen und wollen hiermit unsere Schachfreunde ausdrücklich zum Studium des zusammengesetzten Problems auffordern."

Forderungen:
1) Weißer König auf A7: Weiß zieht an und setzt mit dem siebenten Zuge Matt.
2) Weißer König auf H8: Weiß zieht an und zwingt Schwarz in 14 Zügen Matt zu setzen.
3) Weißer König auf D1: Weiß zieht an und soll mit seinem Bauer im 16. Zuge Matt setzen.
milan: [bRd4-d5+bBe8] 1.wKa8! M.Frelih (2022-03-05)
comment
Keywords: Stipulation change, Constrained problem
Genre: n#, s#
FEN: 2N5/1B1r4/1pN1p3/1kb1pQ2/1n1rp3/R2nP3/6R1/4B3
Input: Frank Müller, 2013-05-05
Last update: Marcin Banaszek, 2022-03-10 more...
26 - P1275766
William Bone
226 Chess Player's Chronicle 12/1844
P1275766
(4+4) C+
#5
Weiß darf nicht schlagen
1. f4+ Kh5 2. Se6 g5 3. f5 g4 4. Sg7+ Kg5 5. h4#
play all play one stop play next play all
herbert bastian: Die angegebene Quelle ist zwar richtig, aber das ist nicht die Erstveröffentlichung. Erstmals wurde sie von Saint Amant in seiner Schachspalte in der französischen Zeitschrift L'Illustration vom 1.7.1843 als Aufgabe Nr. 4 veröffentlicht. Sie stammt wie andere von Saint Amant veröffentlichte Aufgaben aus einem Originalmanuskript von Montigny, das man digital in der Bibliothek in Cleveland findet und downloaden kann. Einzelheiten dazu in meinem Artikel in Heft 8 "Schach" von 2020, S. 39, wo auch der Link nach Cleveland angegeben ist. Montigny starb 1840, somit ist die Aufgabe noch älter. Montigny war ein Sammler, insofern könnte es einen anderen Autor geben. (2021-04-11)
more ...
comment
Keywords: Constrained problem
Genre: n#
Computer test: popeye 4.85
FEN: 8/8/5ppp/6k1/3N4/5PKP/8/8
Reprints: 52 Leipziger Illustrirte Zeitung 25/01/1845
V.128 Schachspiel-Probleme [Alexandre] 1846
Input: Frank Müller, 2013-08-03
Last update: Arnold Beine, 2024-02-18 more...
27 - P1276269
Trevangadacharya Shastree
76 Essays on Chess [Shastree] 1814
P1276269
(6+3) cooked
#11 mit Bc3
1. Tf8+ Ka7 2. Lc5+ Tb6 3. Sb4 e6 4. Kd4 e5+ 5. Kc4 e4 6. Kb3 e3 7. Ka3 e2 8. Le3 e1=D 9. Sbc6+ Ka6 10. Ta8+ Kb5 11. c4#
play all play one stop play next play all
Cook: Abkürzungen:
6. Sd5 e3,Ka6 7. Lxb6 Ka6,e3 8. Kb3 e2,Kb5 9. Ta8 Kb5,e2 10. c4#
4. Ke4 e5 5. Sd5 Ka6 6. Sxb6 Ka7 7. Th8,T~8 Ka6 8. Ta8+ Kb5 9. Kd5 e4 10. c4#
(Anton Baumann)
Anton Baumann: cook: mit Bedingung 'Matt auf Feld b5' findet Gustav Abkürzungen:
6.Sd5 e3,Ka6 7.Lxb6 Ka6,e3 8.Kb3 e2,Kb5 9.Ta8 Kb5,e2 10.c4#
oder 4.Ke4 e5 5.Sd5 Ka6 6.Sxb6 Ka7 7.T.8 Ka6 8.Ta8+ Kb5 9.Kd5 e4 10.c4# (2019-04-16)
Alfred Pfeiffer: Oskar Korschelt schreibt in 'Der gereinigte Alexander' zu dieser Aufgabe: "Sh 76 hat wSd2 statt c2, Druckfehler." (2019-04-16)
Anton Baumann: Mit wSd2 geht es aber noch schneller: 3.Sdc4 Ka6 4.Sxb6 Ka7! 5.Ke6 Ka6 6.Ta8+ Kb5 7.Kd5 e6+ 8.Kd6 e5 9.c4# (2019-04-17)
SP: The supposed cooks are no cooks as they all involve capture of the bR contrary
to the "Burd" rule. As Shastree puts it on p. 29: “The following Positions are
agreeably to the Hindostanee Play where the losing party cannot be checkmated
unless he has a piece remaining on the Board.”
The misprint occurs in the original, and is evident by the solution given on
p. 156 (3.Sb4). There are other choices, but all quotes follow Lewis in placing
the Kt on c2. (2022-05-23)
SP: That should be "all quotes except for Mazel". (2022-05-23)
comment
Keywords: Constrained problem, Pawn mate
Genre: n#
FEN: k7/4pR2/r7/N3K3/8/B1P5/2N5/8
Reprints: 97E Oriental Chess [Lewis] 1817
X.67 Schachspiel-Probleme [Alexandre] 1846
2293v Wiener Schachzeitung 07-08/1908
Input: Frank Müller, 2013-08-11
Last update: Alfred Pfeiffer, 2019-04-17 more...
28 - P1289139
Joseph Ney Babson
1400 American Chess Bulletin 02/1920
P1289139
(5+5)
s#104
at the latter's QKt square
Challenge!! Motto: Will-o'-the Wisp
Später nochmals mit weiteren Forderungen veröffentlicht.
vgl. mit Petroff 1857
Arnold Beine: s. P1318429 bzw. P1201772 (Petrov) (2017-04-24)
James Malcom: This is surely cooked, somehow. (2020-11-19)
comment
Keywords: Constrained problem
Genre: s#
FEN: 7k/8/6Q1/5p2/5p2/4pK2/4P2p/5R1B
Input: Frank Müller, 2014-10-13
Last update: Frank Müller, 2016-03-19 more...
29 - 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...
30 - P1335845
Otto von Oppen
83 Schachzeitung , p. 271, 08/1847
P1335845
(4+3)
#12
Springermatt ohne zu schlagen
1. Lg3! Kg1 2. Ke1 h4 3. Lb8 h3 4. Le4 Kh1 5. Sb6 Kg1 6. Sd5 Kh1 7. Se3 Kg1 8. Sg4 Kh1 9. La7 h2 10. Se3 Kg1 11. Sf1+ Kh1 12. Sg3#
play all play one stop play next play all
Anton Baumann: kürzer: .. 7.Se3 Kg1 8.Sf5! Kh1 9.Kf2 h2 10.Sg3#
8. ... h2 9.La7+ Kh1 10.Sg3#
7. ... h2 8.Sf5 Kg1 9.La7+ Kh1 10.Sg3# (2023-03-07)
comment
Keywords: Constrained problem
Genre: n#
FEN: N7/8/8/7p/8/5B2/6p1/3KB2k
Input: Felber, Volker, 2017-07-12
Last update: Felber, Volker, 2017-07-12 more...
31 - P1340437
F. Dunne
Liverpool Albion
P1340437
(5+2)
#10 durch Bf5
1. f8=L Kg8 2. Le7 Kh8 3. f6 Kg8 4. h7+ Kh8 5. Kf7 Kxh7 6. Lf8 Kh8 7. Kg6 Kg8 8. Lh6 Kh8 9. Lg7+ Kg8 10. f7#
play all play one stop play next play all
im 'Recreationist' Erwähnung alternativer Lösungen von C. Durward and H.J.C. Andrews mit 1. f8=S sowie Hinweis, daß mit 1. f8=D ein Matt in 9 Zügen erreicht werden kann (ohne Angabe der konkreten Zugfolgen).
Originalforderung: White gives mate with Pawn now on K B 5, in ten moves.

12/1873 S.157 Mr. Dunne's Pawn-mate, page 137, was inserted for a particular reason we may explain in giving its solution. It is meant that White should queen a pawn.
01/1874: Das 'Recreationist'-Team wußte schon beim Nachdruck um die Inkorrektheit des Problems - der Hintergrund ist eine für die damalige Zeit nicht ganz untypische Angelegenheit ('Recreationist' 01/1874, S.173): "We feel a word of apology is due to our readers for the publication of this Pawn-mate, for two reasons: firstly, we knew the problem to be defective; and secondly, we knew it had been published in the 'Liverpool Albion´s' chess column. We had occasion to request the favour of a notice of our Magazine in the chess column of the above-named paper. The editor assured us such notice could only be inserted as an advertisement. To this of course we could offer no objection, but the editor of the Chess Column oblingly added that the Chess matter of the 'R.' was of too poor a standard to be recommended to his readers. We determined to test the high standard of the chess supplied to the readers of the 'L.A.', and feeling instinctively that the position in question had a suspicious look about it, tried it, and found it wanting. This, therefore, we considered a point in our favour, for the position had passed publication by the Chess editor of the 'L.A.' without any remark as to its unsoundness. Then, further to test the ability of our readers, we determined to publish it as we did without comment, and to keep the 'rotten egg' [P1305834] suitable companionship. It is but fair to add that even Mr. Dunne himself did not know his problem was unsound till some time after its publicatuion in the 'R.'. Before long at least twenty of our solvers discovered the flaw, which had passed unnoticed by the editor and subscribers of the high standard Chess in the 'L.A.', and we rest quite satisfied. ..."
SP: Not 1. f8=D as this would mate with the wrong man. What was intended,
presumably, was to give the bK some leeway and then promote. Singularly weak
as polemics: if the worst sin that can be adduced against an 1870s editor is
letting a cook in a 10-move problem get through, he would be doing very well!
J. White was much wiser in his later editorial work. (2022-06-17)
comment
Keywords: Constrained problem, under-promotion (L), Kindergarten Problem
Genre: n#
FEN: 7k/5P2/6KP/5Pp1/6P1/8/8/8
Reprints: Recreationist 25/11/1873
Input: Mario Richter, 2017-09-30
Last update: Mario Richter, 2017-09-30 more...
32 - P1340676
L. Smigielski
618 Schachzeitung , p. 334, 08/1854
P1340676
(6+5)
#5
Springermatt
1) 1. Kb7! a6 2. Se6 dxe6 3. d7 e5 4. d8=S e4 5. Sc6#
oder auch
2) 1. Sf3! a6 2. Sd4 cxd4 3. Sxd7 d3 4. Sb8 d2 5. Sc6#
play all play one stop play next play all
Beim Diagramm fehlt ein Hinweis auf eine 2. Lösung die mit 'oder auch' in der LB angegeben ist.
Anton Baumann: weitere NL: 1.Se2 a6 2.Sd4 ...
oder 1.Se6 dxe6 2.Kb7 ... 1. ... a6 2.Se2,Sf3 dxe6 d7 ...
Für die Autorabsicht wird der wSg1 nicht benötigt!
Vermutlich wurde dieser Springer nur eingefügt, um die Löser zu verwirren.
Korrektur: ohne wSg1 geht nur die Autorabsicht 1.Kb7! (2023-05-06)
comment
Keywords: Constrained problem
Genre: n#, Fairies
FEN: 5N2/p1Kp4/3P4/k1p5/p1P5/P7/8/6N1
Input: Felber, Volker, 2017-10-03
Last update: Alfred Pfeiffer, 2017-10-04 more...
33 - P1349244
Otto Wülfing
1422 Schachzeitung , p. 283, 09/1862
P1349244
(5+5)
s#5
durch Turm und ohne Springer zu schlagen
Anton Baumann: unlösbar (2023-11-25)
comment
Keywords: Constrained problem
Genre: s#
FEN: 1k6/r7/p7/P2Q4/p2R4/R7/K7/1n6
Input: Felber, Volker, 2018-04-19
Last update: Felber, Volker, 2018-04-19 more...
34 - P1389104
Finneon
Discord Chess Problems & Studies Server 4/4/2021
13. Platz
CP&S Discord Server Composition Tourney 0 {}
P1389104
(7+2)
White to mate Black in 8, with the final mate given by an en passant capture
Henrik Juel: A miniature with many men (2021-04-30)
A.Buchanan: These days "Miniature" is added automatically when the diagram is created. But I hadn't added all the pieces yet, and when I did add them, the program didn't recheck apparently. (2021-04-30)
Henrik Juel: OK, Andrew, but I hope you will supply the intended solutions to all these weird problems (2021-05-01)
A.Buchanan: Will do (2021-05-02)
comment
Keywords: Constrained problem, Joke, Minimal (b)
Genre: n#
FEN: 2NK4/3p1k2/8/4PP2/1B6/8/2B5/7R
Reprints: MatPlus.net Forum 4/4/2021
Input: A.Buchanan, 2021-04-30
Last update: Alfred Pfeiffer, 2021-07-28 more...
35 - P1398780
Andrew Buchanan
AA007 The Hopper Magazine I01 24/12/2021
P1398780
(16+15)
Drawn game. PG in 9.0
Moldenhauer: Computerprüfung: C- oder Cooked Stelvio 1.4 in 1 Sekunde.
Die Diagrammstellung ist verschieden erreichbar.
Hier 2 Möglickeiten:
1.Sa3 d5 2.Sb1 Kd7 3.f3 De8 4.Kf2 e5 5.Kg3 De6 6.Kh4 Dg6
7.Kh3 e4 8.fxe4 f6 9.e5 Ke8+
1.Sa3 d5 2.Sb1 Dd6 3.f3 Da6 4.Kf2 Sd7 5.Kg3 Dg6+ 6.Kh4 e5
7.Kh3 e4 8.fxe4 f6 9.e5 Sb8+ (2023-07-29)
A.Buchanan: Hi Moldenhauer this problem is sound. You are forgetting the constraint: "Drawn game". (2023-07-29)
Moldenhauer: Hi Andrew, ich habe deshalb nur die Diagrammstellung angegeben.
Ich hoffe Andrew ist erlaubt! Mein Vorname ist Klaus. (2023-07-29)
A.Buchanan: Hi Klaus nice to meet you, thanks for all your work. The point of HC+ is that it combines human & computer logic intelligently. It makes zero sense to solve robotically for 9.0. It is definitely valuable to solve this kind of stipulation (1) as PG in 5.0, and then (2) as PG A-B for 2.0. Apparent duals can then be resolved through reasoning about castling, e.p. and premature draw. (2023-07-29)
comment
Keywords: Draw by repetition, Unique Proof Game, Constrained problem, Castling, En passant (not!)
Genre: Retro
FEN: rnb1kbnr/ppp3pp/5pq1/3pP3/8/7K/PPPPP1PP/RNBQ1BNR
Input: A.Buchanan, 2022-01-31
Last update: Miguel Ambrona, 2023-07-26 more...
36 - P1401711
Andrew Buchanan
OP008 The Hopper Magazine I01 24/12/2021
"Ak ja, retten, retten; hvad hjælper det, at du har retten, når du ikke har nogen magt?"
P1401711
(3+3) C+
h#2.5 (2 solutions)
Position after Black's 5,696th move
1. ... Se6 2. Th2 Ta8 3. Th7 Txe8#
1. ... Txa2 2. Lg6 Tg2 3. Lh7 Sf7#
not 1. ... 0-0-0 2. Lg6 Tg1 3. Lh7 Sf7#?? (game just ended by 50M)
play all play one stop play next play all
A full solution is really too big for PDB, but it is available at https://www.thehoppermagazine.com/AA084
Henrik Juel: solutions
1...Sg5-e6 2.Ra2-h2 Ra1-a8 3.Rh2-h7 Ra8*e8 #
1...Ra1*a2 2.Be8-g6 Ra2-g2 3.Bg6-h7 Sg5-f7 #
not 1...0-0-0? 2.Be8-g6 Rd1-g1 3.Bg6-h7 Sg5-f7 #
because White has lost his right to castle, as Andrew will explain, I hope... (2023-07-29)
A.Buchanan: White has not necessarily lost the right to castle, but if he can castle then the 50M rule triggers before the mate can be executed. It is pretty complicated, sorry. (2023-07-29)
A.Buchanan: We don't do mottoes much these days. Mrs Baird was a big fan. On page 301 of the July 1916 Chess Amateur P.H. Williams wrote:
"I think it was Mrs Baird who did more to search Shakespeare for accidental (or deliberate) reference to chess, since all her retractors had Shakespearean mottoes, and her knowledge of the Avonian bard was obviously extensive. Her example was followed by other composers of retractors, who considered such positions would be incomplete without some reference to the poet."
But there are other great playwrights, and Asteroid 5696 was named after one such, whose bitter quotation accurately states this problem's theme. (2023-07-31)
Henrik Juel: The citation is by danish philosopher Søren Kierkegaard (1813-1855) and means something like
Oh yes, to be right, to be right; it does not help much that you are right, when you have no power. (2023-07-31)
Henrik Juel: Asteroid 5696 is named Ibsen after the norwegian playwright Henrik Ibsen (1828-1906), who must have borrowed the citation from Søren Kierkegaard (2023-07-31)
A.Buchanan: Thanks for this Henrik - I resolved the question here: https://philosophy.stackexchange.com/questions/101229/did-ibsen-originate-this-statement
Beyond the castling try, the other innovation here is in the solution beginning 1...Se6. Castling convention and 50M I think are trying to combine by Retro Strategy (being optimistic about 50M rights like we are with castling, rather than pessimistic like we are with ep). (2023-08-04)
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comment
Keywords: Aristocrat, Miniature, 50 move rule, Castling, Exchange of roles (T/S, Guard/Mate), Chumakov theme (l/t, simplified), Retro Strategy (RS), Model mate (2), Constrained problem
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 and analysis
FEN: 4b2k/8/8/6N1/8/8/r7/R3K3
Reprints: AA084 The Hopper Magazine I04 13/07/2023
Input: A.Buchanan, 2022-06-09
Last update: A.Buchanan, 2023-08-27 more...
37 - P1413923
Andrew Buchanan
R611 The Problemist 07/2023
P1413923
(15+11)
Proof Game in 12.5
Game over!

A long-standing question (since a problem by Michel Caillaud P1004030, and also one by Guus Rul) asks how the Draw by Triple Repetition (3Rep) convention interacts with the Dead Position rule (DP). This is not an issue over the board, since 3Rep must be claimed. Miguel Ambrona proposed that for retro problems, DP rule has visibility of any looming 3Rep or 50M draw, while for purely forward problems, e.g. help dead position (HDP), it does not. This distinction works extremely well: it enables problems such as Michel's and this new one, while allowing us to eliminate a class of cooks which would be ubiquitous in longer HDP problems. So the HDP design space does not hit an effective ceiling at 4.0 moves. The concept of "retro problem" is already used in Codex Articles 17 & 17A for 50 move rule & DP, so is not new.
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Keywords: Unique Proof Game, Constrained problem, Draw by repetition, Castling, En passant
Genre: Retro
FEN: 2Q1kb1r/1pp1ppp1/3q3p/2Pp1R2/8/1P6/1P1P1PPP/1NB1KBNR
Reprints: www.chess.com 18/12/2023
Input: A.Buchanan, 2023-12-05
Last update: A.Buchanan, 2023-12-18 more...
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