2287 problem(s) found in 4171 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT K='logisches Problem' AND G='Fairies'] [download as LaTeX]
1 - P0000250
Nikita M. Plaksin
Valery Liskovets
7577v Die Schwalbe 132 12/1991
(5+14)
ser-h#5 (AP)
Typ Tauber/Caillaud
Nikita M. Plaksin
Valery Liskovets
7577v Die Schwalbe 132 12/1991
(5+14)
ser-h#5 (AP)
Typ Tauber/Caillaud
1. Kd4 2. bxc3ep (zuletzt nur K,T-Zug oder c2-c4) 3. e5! (nicht 4. Tb4? weil zuletzt K,T-Zug und 0-0-0 nicht mehr möglich) 4. Tb4 (zuletzt e7xLd8=S möglich) 5. Tc4 & 1. 0-0-0# nicht 1. Td1#? wegen fehlender AP-Legalisierung
Keywords: Castling (wl), a posteriori (AP), Promotion (S), Seriesmover, Consequent, En passant
Genre: Retro, Fairies
FEN: 1N6/1ppppppn/n7/1Pq1k3/1pP1r3/4p3/1r6/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-29 more...
Genre: Retro, Fairies
FEN: 1N6/1ppppppn/n7/1Pq1k3/1pP1r3/4p3/1r6/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-29 more...
2 - P0000598
Nikita M. Plaksin
Andrej N. Kornilow
3876 Die Schwalbe 74 04/1982
(4+3)
h=2 (AP)
Circe, Monochromes Schach
Nikita M. Plaksin
Andrej N. Kornilow
3876 Die Schwalbe 74 04/1982
(4+3)
h=2 (AP)
Circe, Monochromes Schach
Intended solution:
1. dxc3ep[+wBc2] 0-0 2. axb1=S Kh2=
2. ... Kf2? 3. Kf3!
1. ... Tf1? 2. axb1=S Kf2? 3. Kf3!
1. dxc3ep[+wBc2] 0-0 2. axb1=S Kh2=
2. ... Kf2? 3. Kf3!
1. ... Tf1? 2. axb1=S Kf2? 3. Kf3!
Erich Bartel: vom Dual 2.-- Kf2/Kh2 abgesehen C+ PY V4.41.--- (2008-11-07)
A.Buchanan: This is cooked in Popeye because White can't castle as long as bK controls f1! This is quite reasonable behaviour in monochrome: compare to en passant! So no solution (2022-02-18)
comment
A.Buchanan: This is cooked in Popeye because White can't castle as long as bK controls f1! This is quite reasonable behaviour in monochrome: compare to en passant! So no solution (2022-02-18)
comment
Keywords: a posteriori (AP) (Type Petrovic), Circe, En passant as key, Monochromatic Chess, Castling (wk), Miniature, Golden Age (Monochrome castling)
Genre: Retro, Fairies
FEN: 8/8/8/8/2Pp4/8/p3k3/1N2K2R
Reprints: (IV) Quartz 4 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-18 more...
Genre: Retro, Fairies
FEN: 8/8/8/8/2Pp4/8/p3k3/1N2K2R
Reprints: (IV) Quartz 4 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-18 more...
3 - P0000792
Klaus Wenda
1557 Die Schwalbe 33 06/1975
2. Preis
(13+12)
#2 Längstzüger
b) sTa7 nach d7
Klaus Wenda
1557 Die Schwalbe 33 06/1975
2. Preis
(13+12)
#2 Längstzüger
b) sTa7 nach d7
Anton Baumann: Autorabsicht: Die weiss-schwarzen Rochaden schliessen sich gegenseitig aus.
a) 1.O-O? Tf8! daher: 1.Tf1! O-O 2.Sxe7#
b) 1.Tf1? O-O! daher: 1.O-O! Tf8 2.Sxg7#
Aber in der Urfassung (= nebenstehendes Diagramm) geht in a) und b) die NL:
1.Tg1 O-O 2.Txg7,Sf5xh6#
Korrektur in 'Schwalbe' 04/1976 S.464: sLb7 nach g6, sBc5 nach b7
Ausgezeichnet wurde die korrigierte Fassung 1557v (vergl. 'Die Schwalbe' 06/1977 S.82) (2022-12-09)
comment
a) 1.O-O? Tf8! daher: 1.Tf1! O-O 2.Sxe7#
b) 1.Tf1? O-O! daher: 1.O-O! Tf8 2.Sxg7#
Aber in der Urfassung (= nebenstehendes Diagramm) geht in a) und b) die NL:
1.Tg1 O-O 2.Txg7,Sf5xh6#
Korrektur in 'Schwalbe' 04/1976 S.464: sLb7 nach g6, sBc5 nach b7
Ausgezeichnet wurde die korrigierte Fassung 1557v (vergl. 'Die Schwalbe' 06/1977 S.82) (2022-12-09)
comment
Keywords: Maximummer, Castling (wksk)
Genre: Retro, Fairies
FEN: 4k2r/rb2pNbp/1P5p/p1pppN2/8/8/PPPPP2P/2BQK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
Genre: Retro, Fairies
FEN: 4k2r/rb2pNbp/1P5p/p1pppN2/8/8/PPPPP2P/2BQK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
1. ... hxg6ep[+sBg7] 2. a1=T+ Txa1[+sTh8] 3. 0-0 Ta3 4. bxa3[+wTa1] 0-0-0 5. Td8 Txd8[+sTh8]#
Wenn die beiden Könige noch nicht gezogen haben (was a.p. durch Ausführung beider Rochaden bewiesen wird!), kann der letzte Zug nur g7-g5 gewesen sein, Sonst ginge z.B. auch R: 1. Kd8-e8 c2xDb3
Wenn die beiden Könige noch nicht gezogen haben (was a.p. durch Ausführung beider Rochaden bewiesen wird!), kann der letzte Zug nur g7-g5 gewesen sein, Sonst ginge z.B. auch R: 1. Kd8-e8 c2xDb3
Henrik Juel: A possible retroplay is -1... g7 -2.f3xP(-Pg7) b5 -3.f2 c6xP(-Pb2) -4.b2 a6 -5.Rc3 a7 -6.Rc1 b6xB -7.Bf8 b7 -8.Bg7xB etc. (2003-04-22)
Michel Caillaud: The problem is not cooked as indicated, as under A Posteriori convention both castles must be played in the solution to justify the en passant key. (2022-02-04)
A.Buchanan: If wK never moved, then Black has not just moved another non-capturing pawn, as White has no prior move. The only possible exception is R: 1. g7-g5 f/h3xPg4[+Pg7]. (2022-02-04)
comment
Michel Caillaud: The problem is not cooked as indicated, as under A Posteriori convention both castles must be played in the solution to justify the en passant key. (2022-02-04)
A.Buchanan: If wK never moved, then Black has not just moved another non-capturing pawn, as White has no prior move. The only possible exception is R: 1. g7-g5 f/h3xPg4[+Pg7]. (2022-02-04)
comment
Keywords: a posteriori (AP), Circe, En passant as key, Castling (wgsk), Promotion in forward play, Valladao Task
Genre: Retro, Fairies, h#
Computer test: HC+ all solutions begin with ep, but only one includes both white & black castling
FEN: 4k3/8/7p/p1p3pP/Pp4Pp/RP4p1/pP6/4K3
Reprints: feenschach 61, p. 534, 08/1982
RA152 diagrammes 72 07-08/1985
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-28 more...
Genre: Retro, Fairies, h#
Computer test: HC+ all solutions begin with ep, but only one includes both white & black castling
FEN: 4k3/8/7p/p1p3pP/Pp4Pp/RP4p1/pP6/4K3
Reprints: feenschach 61, p. 534, 08/1982
RA152 diagrammes 72 07-08/1985
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-28 more...
Gerald Ettl: 1.h8D# (2023-04-03)
Gerald Ettl: die Stellung loest sich auf indem der sK nach b1 - g8 wandert und dann Sg5 weg zieht. Der wK kommt ueber g5 raus. (2023-04-03)
Michel Caillaud: The original stipulation is #1 (durch wen?).
As wPe2 cannot be captured on its file, the 4 white captures for 4 Pawns to promote to Knights are (a2)xb3, (e2)xf3 and (f2)xg3 2 times, and wPb2 was captured on its file by (Dd8)xb6.
As b3-b2 (before (a2xb3)) and c7-c6 (before (Dd8)xb6) cannot be immediately retracted, only bK and wSs can play the last moves.
As indicated by Gerald, bK has to go to g8 to unlock the position, freeing bSg5.
When Ka3-a2 is retracted, previous white move places the 6 white Knights on black squares; the resulting Retro-Opposition implies that black is to play in the diagram position.
1.b1S#! (1g8D#?) (2023-04-04)
Gerald Ettl: Danke Michel fuer Dein Erklärung.
Ich löse so auf, dass Schwarz am Rückzug ist:
R: 1.Kc1b1 Se4d6 2.Bb2b3 Sc5d3 3.Kb1a2 Sd3c1 4.Ka2b1 Sh1f2 5.Kb1a2 Sa4c5 6.Bb3b4 Sf5e3 7.Ka2a3 La1d4 8.Ka3a4 Sf2h1 9.Ka4a5 Sh1f2 10.Ka5b6 Sf2h1 11.Kb6c7 Sh1f2 12.Kc7d8 Sf2h1 13.Kd8e8 Sh1f2 14.Ke8f8 Sf2h1 15.Kf8g8 Sh1f2 16.Sg5f3 Kh6g5 17.Sb8a6 Sc5a4 18.Sa6c5 Sd6b5 19.Sc5b3 Sb5c7 20.Sb3a1 Sc7a6 21.Sa1b3 Sa6b8 22.Sb3a1 Sa4b6 23.Sa1b3 Sb6c8 24.Sb3a1 Sb8b7[+wBb7] 25.Sa1b3 Bb7a6[+sLb7] 26.Sb3a1 Sc8c7[+wBc7] 27.Lb7c8 Kg5h4 28.Sa1b3 Tg6g5 29.Sb3a1 Tg5f5 30.Sa1b3 Tf5f4 31.Tg7g5 Sf2h1 32.Tg5b5 Sh1f2 33.Tb5b8 Sf2h1 34.Bb4b5 Sh1f2 35.Bb5b7 Bc7b6[+sLc7] 36.Tb8a8 Tf4e4 37.Lc7f4 Sg3f5 38.Kg8f8 Sf2h3 39.Lf4h6 Sc1e2 40.Kf8e8 Se2g3 41.Lh6f8 Sf5h6 42.Sb3a1 Sh6g8 43.Sa1b3 Sg8g7[+wBg7] 44.Sb3a1 Bg7g6 45.Sa1b3 Bg6f5[+sTg6] 46.Tg6g8 Bf5f4 47.Tg8h8 Sg3f5 48.Sb3a1 Sf5h6 49.Sa1b3 Sh6g8 50.Sb3a1 Sg8g7[+wBg7] 51.Bg4g5 Bg7g6 52.Bf6g7[+wDf6] Lh5g4 53.Sa1b3 Bh7h6 54.Sb3a1 Bh6h5 55.Bg5h6[+wTg5]
warum geht das nicht? Den Zug b1S# habe ich vorher ueberhaupt nicht gesehen. (2023-04-04)
Gerald Ettl: Jetzt habe ich es gesehen: der wBa2 musste ja von a2 geschlagen haben. (2023-04-04)
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Gerald Ettl: die Stellung loest sich auf indem der sK nach b1 - g8 wandert und dann Sg5 weg zieht. Der wK kommt ueber g5 raus. (2023-04-03)
Michel Caillaud: The original stipulation is #1 (durch wen?).
As wPe2 cannot be captured on its file, the 4 white captures for 4 Pawns to promote to Knights are (a2)xb3, (e2)xf3 and (f2)xg3 2 times, and wPb2 was captured on its file by (Dd8)xb6.
As b3-b2 (before (a2xb3)) and c7-c6 (before (Dd8)xb6) cannot be immediately retracted, only bK and wSs can play the last moves.
As indicated by Gerald, bK has to go to g8 to unlock the position, freeing bSg5.
When Ka3-a2 is retracted, previous white move places the 6 white Knights on black squares; the resulting Retro-Opposition implies that black is to play in the diagram position.
1.b1S#! (1g8D#?) (2023-04-04)
Gerald Ettl: Danke Michel fuer Dein Erklärung.
Ich löse so auf, dass Schwarz am Rückzug ist:
R: 1.Kc1b1 Se4d6 2.Bb2b3 Sc5d3 3.Kb1a2 Sd3c1 4.Ka2b1 Sh1f2 5.Kb1a2 Sa4c5 6.Bb3b4 Sf5e3 7.Ka2a3 La1d4 8.Ka3a4 Sf2h1 9.Ka4a5 Sh1f2 10.Ka5b6 Sf2h1 11.Kb6c7 Sh1f2 12.Kc7d8 Sf2h1 13.Kd8e8 Sh1f2 14.Ke8f8 Sf2h1 15.Kf8g8 Sh1f2 16.Sg5f3 Kh6g5 17.Sb8a6 Sc5a4 18.Sa6c5 Sd6b5 19.Sc5b3 Sb5c7 20.Sb3a1 Sc7a6 21.Sa1b3 Sa6b8 22.Sb3a1 Sa4b6 23.Sa1b3 Sb6c8 24.Sb3a1 Sb8b7[+wBb7] 25.Sa1b3 Bb7a6[+sLb7] 26.Sb3a1 Sc8c7[+wBc7] 27.Lb7c8 Kg5h4 28.Sa1b3 Tg6g5 29.Sb3a1 Tg5f5 30.Sa1b3 Tf5f4 31.Tg7g5 Sf2h1 32.Tg5b5 Sh1f2 33.Tb5b8 Sf2h1 34.Bb4b5 Sh1f2 35.Bb5b7 Bc7b6[+sLc7] 36.Tb8a8 Tf4e4 37.Lc7f4 Sg3f5 38.Kg8f8 Sf2h3 39.Lf4h6 Sc1e2 40.Kf8e8 Se2g3 41.Lh6f8 Sf5h6 42.Sb3a1 Sh6g8 43.Sa1b3 Sg8g7[+wBg7] 44.Sb3a1 Bg7g6 45.Sa1b3 Bg6f5[+sTg6] 46.Tg6g8 Bf5f4 47.Tg8h8 Sg3f5 48.Sb3a1 Sf5h6 49.Sa1b3 Sh6g8 50.Sb3a1 Sg8g7[+wBg7] 51.Bg4g5 Bg7g6 52.Bf6g7[+wDf6] Lh5g4 53.Sa1b3 Bh7h6 54.Sb3a1 Bh6h5 55.Bg5h6[+wTg5]
warum geht das nicht? Den Zug b1S# habe ich vorher ueberhaupt nicht gesehen. (2023-04-04)
Gerald Ettl: Jetzt habe ich es gesehen: der wBa2 musste ja von a2 geschlagen haben. (2023-04-04)
comment
Keywords: Circe (Mars), Non-standard material, Promotion
Genre: Retro, Fairies
FEN: 1n6/p2ppprP/2p2pRK/2N2NnB/N3N1p1/6N1/1pPP4/B1k4N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2019-02-03 more...
Genre: Retro, Fairies
FEN: 1n6/p2ppprP/2p2pRK/2N2NnB/N3N1p1/6N1/1pPP4/B1k4N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2019-02-03 more...
6 - P0001273
Luigi Ceriani
145 Europe Echecs 84 01/1965
(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
Luigi Ceriani
145 Europe Echecs 84 01/1965
(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
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...
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...
1. Lf1 2. Kh3 3. Kh4 4. Lh3 5. fxg3ep 6. Kh5 7. Kg6 8. Kf7 9. Ke8 10. 0-0-0 11. Td7 a8=D#
The cluster of tries: Kh2-g1-f1-e1-d2-...-e8 0-0-0 Td7 is 12 moves: just too long.
In the diagram, e.p. is not permitted because last move might be g3-g4. So first shift sK to h4. Following the e.p., check that wK does now have prior move: Kf4-f3. Now sK scampers to e8 where through consequent forgetfulness it recovers castling rights.
The cluster of tries: Kh2-g1-f1-e1-d2-...-e8 0-0-0 Td7 is 12 moves: just too long.
In the diagram, e.p. is not permitted because last move might be g3-g4. So first shift sK to h4. Following the e.p., check that wK does now have prior move: Kf4-f3. Now sK scampers to e8 where through consequent forgetfulness it recovers castling rights.
Keywords: En passant, Castling (sg), Seriesmover, Consequent, Valladao Task, Promotion in the mating move (D), Switchback (l), Promotion (D), Königswanderung
Genre: Retro, Fairies
FEN: r7/P1p5/pp6/5pp1/5pP1/5K1b/4rp1k/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
Genre: Retro, Fairies
FEN: r7/P1p5/pp6/5pp1/5pP1/5K1b/4rp1k/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
1. dxc3ep 2. Kxb5 3. Kxc6 4. Kd7 5. Ke8 6. 0-0 Th8#
Aber es geht auch R: 1. Kb2-a1!?
Aber es geht auch R: 1. Kb2-a1!?
Henrik Juel: The intention may be 1.dxc3ep 2.Kxb5xc6-d7-e8 6.0-0 Rh8#, but -1.Kb2 seems possible. Should bBc1 be moved to a3? (2003-04-28)
GW: Yes, that's the solution, and the problem seems indeed to be cooked. I don't know if it has been corrected. (2003-04-28)
James Malcom: I believe that the intent was that the sLc1 must be the orginal one, and thus an impediment to wK foreplay, as the wBc6 "must" have captured the e7 Black pawn on its way from f2. The problem is that it can take the alternative path fxsLf3xDe3xTc5-c6, and there are enough White pieces to capture for the Black b pawn to make it d2 and for the d pawn to make it to c2. (2021-09-14)
James Malcom: It turns out this problem actually was corrected a long time later: P1012052 (2021-09-14)
comment
GW: Yes, that's the solution, and the problem seems indeed to be cooked. I don't know if it has been corrected. (2003-04-28)
James Malcom: I believe that the intent was that the sLc1 must be the orginal one, and thus an impediment to wK foreplay, as the wBc6 "must" have captured the e7 Black pawn on its way from f2. The problem is that it can take the alternative path fxsLf3xDe3xTc5-c6, and there are enough White pieces to capture for the Black b pawn to make it d2 and for the d pawn to make it to c2. (2021-09-14)
James Malcom: It turns out this problem actually was corrected a long time later: P1012052 (2021-09-14)
comment
Keywords: En passant, Castling (sk), Seriesmover, Consequent, Non-standard material (sLb1), Valladao Task, Promotion in the retro play (sLb1), Obvious promotion ((sLb1)), Superseded by (P1012052)
Genre: Retro, Fairies
FEN: 7r/5pnR/2P3pR/pPpn3p/1kPp4/3P4/P2pP3/Kbb5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-09-14 more...
Genre: Retro, Fairies
FEN: 7r/5pnR/2P3pR/pPpn3p/1kPp4/3P4/P2pP3/Kbb5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-09-14 more...
1) 1. gxh8=G# justified by history R: 1. Kg6xGh5 & e.g. Gf5-h5+
2) 1. g8=CM# justified by history R: 1. Kg6xCMh5 & e.g. CMe4-h5+
Im orthodoxen Schach hat Schwarz keinen letzten Zug. Schwarz muß deshalb im letzten Zug eine Märchenfigur entschlagen haben. Und sehen P1390927
2) 1. g8=CM# justified by history R: 1. Kg6xCMh5 & e.g. CMe4-h5+
Im orthodoxen Schach hat Schwarz keinen letzten Zug. Schwarz muß deshalb im letzten Zug eine Märchenfigur entschlagen haben. Und sehen P1390927
Henrik Juel: something is wrong here: the claimed mates are not even checks
also, the stipulation should probably be something like
Black and White retract, then #1 (2021-04-30)
A.Buchanan: Yes the problem was the word "dann" which indicates help retractor. The retro is just justification for the forward moves, showing that the relevant fairy piece existed earlier in the game so can be promoted to. What's the name of that principle? (2021-05-01)
Henrik Juel: Thanks for the clarification, Andrew
I think that this principle should remain unnamed and just stay in the joke realm (2021-05-01)
A.Buchanan: Sorry not to be clear: I am talking about the standard fairy principle that if you want to promote to a Princess, there should be one on the problem diagram. Not talking about the joke extension which riffs off this. I thought this was called "Bartel Theme", but it turns out that is something different. (2021-05-01)
Henrik Juel: I see; that principle probably was established soon after the invention of grasshoppers and nightriders
I do not think it has a name, but it is so well established that it should be in the Codex (2021-05-01)
more ...
comment
also, the stipulation should probably be something like
Black and White retract, then #1 (2021-04-30)
A.Buchanan: Yes the problem was the word "dann" which indicates help retractor. The retro is just justification for the forward moves, showing that the relevant fairy piece existed earlier in the game so can be promoted to. What's the name of that principle? (2021-05-01)
Henrik Juel: Thanks for the clarification, Andrew
I think that this principle should remain unnamed and just stay in the joke realm (2021-05-01)
A.Buchanan: Sorry not to be clear: I am talking about the standard fairy principle that if you want to promote to a Princess, there should be one on the problem diagram. Not talking about the joke extension which riffs off this. I thought this was called "Bartel Theme", but it turns out that is something different. (2021-05-01)
Henrik Juel: I see; that principle probably was established soon after the invention of grasshoppers and nightriders
I do not think it has a name, but it is so well established that it should be in the Codex (2021-05-01)
more ...
comment
Keywords: Joke, Last Moves?, Minimal
Pieces: = Grasshopper (G), = Camel (CM)
Genre: Retro, Fairies
FEN: 7b/6P1/7B/6Pk/2P5/3B3K/2P1P3/8
Reprints: (17) diagrammes 97 04-06/1991
MatPlus.net Forum 6/4/2021
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-19 more...
Pieces: = Grasshopper (G), = Camel (CM)
Genre: Retro, Fairies
FEN: 7b/6P1/7B/6Pk/2P5/3B3K/2P1P3/8
Reprints: (17) diagrammes 97 04-06/1991
MatPlus.net Forum 6/4/2021
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-19 more...
1. dxe3ep 2. fxg1=L 3. 0-0 Tg4#
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
Henrik Juel: If Black may castle, his latest move must have been e3xf2, so last move was e2-e4. (Orig. wPg2 promoted on g8, so g7xh6 happened early). 1.dxe3 e.p. 2.fxg1=B 3.0-0 Tg4# (2003-03-21)
James Malcom: Is this the only know one-sided Valladao in a regular series-mover helpmate? Also, can wPh2 be shifted to h4, and the White king moved as well to h2, so that the solution must have 2. fxg1=S? (2020-12-08)
James Malcom: Well, the only known one that is in minimum form, i.e. 3 moves, I mean. (2020-12-08)
A.Buchanan: This is nice. I don't know of any other AP ser-h#3 with Valladao, but there could be. Just been cleansing PDB data today. Yes I think wKh3 and wBh2 can be transposed, and then promote to S instead of L. (2020-12-08)
A.Buchanan: The legality of intermediate positions in seriesmovers is of course irrelevant. However, each proof game that leads to the diagram position with BTM can be appended with a black move and then a "flip" that changes the player to move. We can repeat this process, and then the fact that Black retains castling rights implies that they had castling rights in the diagram position. I think this is the right formal way to handle AP with seriesmovers. (2022-03-21)
comment
James Malcom: Is this the only know one-sided Valladao in a regular series-mover helpmate? Also, can wPh2 be shifted to h4, and the White king moved as well to h2, so that the solution must have 2. fxg1=S? (2020-12-08)
James Malcom: Well, the only known one that is in minimum form, i.e. 3 moves, I mean. (2020-12-08)
A.Buchanan: This is nice. I don't know of any other AP ser-h#3 with Valladao, but there could be. Just been cleansing PDB data today. Yes I think wKh3 and wBh2 can be transposed, and then promote to S instead of L. (2020-12-08)
A.Buchanan: The legality of intermediate positions in seriesmovers is of course irrelevant. However, each proof game that leads to the diagram position with BTM can be appended with a black move and then a "flip" that changes the player to move. We can repeat this process, and then the fact that Black retains castling rights implies that they had castling rights in the diagram position. I think this is the right formal way to handle AP with seriesmovers. (2022-03-21)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sk), Seriesmover, Promotion, Valladao Task
Genre: Retro, Fairies
Computer test: C+ Popeye v4.85 + thinking
FEN: 4k2r/1pp2p1p/1N5p/3P4/2RpP2N/P2P1P1K/p1BP1p1P/B3b1R1
Reprints: Die Schwalbe 143 10/1993
Rex Multiplex 45-46 05/1994
(XII) Quartz 5 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
Genre: Retro, Fairies
Computer test: C+ Popeye v4.85 + thinking
FEN: 4k2r/1pp2p1p/1N5p/3P4/2RpP2N/P2P1P1K/p1BP1p1P/B3b1R1
Reprints: Die Schwalbe 143 10/1993
Rex Multiplex 45-46 05/1994
(XII) Quartz 5 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
11 - P0002853
Marcel Eugene Nordlohne
Umwandlungen in Märchenfiguren 2 1993
(3+3) C+
s#4
Längstzüger
wDU,sDU=Grashüpfer
Marcel Eugene Nordlohne
Umwandlungen in Märchenfiguren 2 1993
(3+3) C+
s#4
Längstzüger
wDU,sDU=Grashüpfer
1. fxe6ep! Gf6 2. e7 Gd8 3. e8=G Gf8 4. Gg8 Gh8#
Der letzte Zug kann unter der Längstzügerbedingung nur Be7-e5 gewesen sein. Also ist der En-passant-Schlüssel erlaubt. Erstaunlicherweise gelingt der En-passant-Nachweis in Miniaturfassung.
Der letzte Zug kann unter der Längstzügerbedingung nur Be7-e5 gewesen sein. Also ist der En-passant-Schlüssel erlaubt. Erstaunlicherweise gelingt der En-passant-Nachweis in Miniaturfassung.
Henrik Juel: C+ Popeye 4.61 after analysis
Last move obviously was not made by Ga6
neither by Kf1 from e1, e2, f2, because K-f1 would not be a longest move
neither by Pe5 from d6, e6, f6, by a similar reason
So last move was e7-e5, and White may capture ep (2021-05-02)
comment
Last move obviously was not made by Ga6
neither by Kf1 from e1, e2, f2, because K-f1 would not be a longest move
neither by Pe5 from d6, e6, f6, by a similar reason
So last move was e7-e5, and White may capture ep (2021-05-02)
comment
Keywords: En passant, Maximummer, Model mate
Pieces: = Grasshopper (G)
Genre: Retro, Fairies
Computer test: HC+ Popeye 4.61 after analysis
FEN: 8/8/*2q7/4pP2/8/8/7*2Q/5k1K
Reprints: Probleemblad 05-06/196?
Input: Gerd Wilts, 1995-06-03
Last update: Marcin Banaszek, 2021-05-10 more...
Pieces: = Grasshopper (G)
Genre: Retro, Fairies
Computer test: HC+ Popeye 4.61 after analysis
FEN: 8/8/*2q7/4pP2/8/8/7*2Q/5k1K
Reprints: Probleemblad 05-06/196?
Input: Gerd Wilts, 1995-06-03
Last update: Marcin Banaszek, 2021-05-10 more...
1. Kf8 2. Ke8 3. 0-0 4. Kh8 5. Tg8 Sf7#
Version Bernd Schwarzkopf
Henrik Juel: Not 1.Kf8 2.Tg8 3.Tg7 4.Kg8 5.Kh8 6.Tg8, one move too many (2021-01-18)
comment
Henrik Juel: Not 1.Kf8 2.Tg8 3.Tg7 4.Kg8 5.Kh8 6.Tg8, one move too many (2021-01-18)
comment
Keywords: Seriesmover, Consequent, Castling
Genre: Retro, Fairies
FEN: 6kr/7p/7K/6N1/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
Genre: Retro, Fairies
FEN: 6kr/7p/7K/6N1/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
1. Df2 2. Db6 3. Kd4 4. bxc3ep 5. Dg6 6. De4 Lb6
Cook: 1. Kd4 2. Kc3 3. Db2 4. d2 5. Ld3 6. d4 Lxb4#
Cook: 1. Kd4 2. Kc3 3. Db2 4. d2 5. Ld3 6. d4 Lxb4#
Keywords: Seriesmover, Consequent, En passant
Genre: Retro, Fairies
FEN: 8/8/p7/B2pp3/KpP1k3/Np1p4/q7/1b6
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
Genre: Retro, Fairies
FEN: 8/8/p7/B2pp3/KpP1k3/Np1p4/q7/1b6
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
14 - P0003925
Ivan Skoba
1009 diagrammes 47 09-10/1980
(6+7) C+
a) ser-h#6 (AP)
b) nach dem Schlüssel von a): ser-h#5
Ivan Skoba
1009 diagrammes 47 09-10/1980
(6+7) C+
a) ser-h#6 (AP)
b) nach dem Schlüssel von a): ser-h#5
a)
1. dxc3ep 2. Kc4? 3. Tb5 4. Td5 5. Dc5 6. Kd4 Th4#
2. Tb2! 3. Kb3 4. Kc2 5. Kc1 6. c2 0-0#
b)
1. Tb2? 2. Kb3 3. Kc2 4. Kc1 5. c2 0-0#
1. Kc4! 2. Tb5 3. Td5 4. Dc5 5. Kd4 Th4#
There are basically two candidate solutions ending Th4# and 0-0#. In (a), the initial ep can only be justified by 0-0#. On the other hand in the diagram position in (b), White's last move must have been with wK or wR, so 0-0# is disrupted. Nice and paradoxical.
1. dxc3ep 2. Kc4? 3. Tb5 4. Td5 5. Dc5 6. Kd4 Th4#
2. Tb2! 3. Kb3 4. Kc2 5. Kc1 6. c2 0-0#
b)
1. Tb2? 2. Kb3 3. Kc2 4. Kc1 5. c2 0-0#
1. Kc4! 2. Tb5 3. Td5 4. Dc5 5. Kd4 Th4#
There are basically two candidate solutions ending Th4# and 0-0#. In (a), the initial ep can only be justified by 0-0#. On the other hand in the diagram position in (b), White's last move must have been with wK or wR, so 0-0# is disrupted. Nice and paradoxical.
Keywords: Seriesmover, Castling (wk), Cant Castler, a posteriori (AP) (Type Petrovic), En passant as key
Genre: Retro, Fairies
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 8/8/8/8/bkPp4/pr1pq3/3PRP2/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-29 more...
Genre: Retro, Fairies
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 8/8/8/8/bkPp4/pr1pq3/3PRP2/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-29 more...
1. a1=S 2. fxg3ep 3. Kxh5 4. Kxg6 5. Kf7 6. Ke8 7. 0-0-0 a8=D#
8. TT
Henrik Juel: Solution: 1.a1S 2.fxg3ep 3.Kxh5xg6-f7-e8 7.0-0-0 a8Q#. Nice problem with open and hidden ep capture (exf3ep). (2003-04-28)
James Malcom: How is the key here justified? (2021-01-18)
James Malcom: I am still pondering. (2021-09-14)
Henrik Juel: James, I have forgotten all about this weird stipulation during the time since my last comment, and I don't know what I meant so say...
The definition of shc in the PDB is not complete; here is the Schwalbe definition:
Konsequenter Serienzüger: Ein Serienzüger, bei dem nach jedem Zug die Retroanalyse der Stellung neu durchgeführt wird (also ohne Kenntnis früherer Züge und Analysen). Beispielsweise wird eine Rochade wieder möglich, wenn König und Rochadeturm auf ihre entsprechenden Felder ziehen (weil in der neuen Analyse "vergessen" ist, dass beide bereits gezogen haben).
In english, something like
After each move the analysis of the position is done afresh, without knowledge of previous moves;
for instance, castling is possible when king and rook reach their original squares (e8 and a8), because it is 'forgotten' that they have already moved (2021-09-14)
Henrik Juel: You ask for a justification of the key, James; there is no need for justification, Black can do whatever he likes
In the position after 1.a1=S last move must have been g2-g4, so the ep capture 2.fxg3ep is legitimized
In the position after 2.fxg3ep last move must have been Ke3xPf3, with the double check being explained by exf3ep++; this is the 'hidden ep capture' I mentioned in my youthful comment (2021-09-14)
James Malcom: Thanks Henri! While the definition is incomplete, the reset each time is deducible from looking over shc problems. The justification here, however, was a level above me somehow. (2021-09-14)
comment
Henrik Juel: Solution: 1.a1S 2.fxg3ep 3.Kxh5xg6-f7-e8 7.0-0-0 a8Q#. Nice problem with open and hidden ep capture (exf3ep). (2003-04-28)
James Malcom: How is the key here justified? (2021-01-18)
James Malcom: I am still pondering. (2021-09-14)
Henrik Juel: James, I have forgotten all about this weird stipulation during the time since my last comment, and I don't know what I meant so say...
The definition of shc in the PDB is not complete; here is the Schwalbe definition:
Konsequenter Serienzüger: Ein Serienzüger, bei dem nach jedem Zug die Retroanalyse der Stellung neu durchgeführt wird (also ohne Kenntnis früherer Züge und Analysen). Beispielsweise wird eine Rochade wieder möglich, wenn König und Rochadeturm auf ihre entsprechenden Felder ziehen (weil in der neuen Analyse "vergessen" ist, dass beide bereits gezogen haben).
In english, something like
After each move the analysis of the position is done afresh, without knowledge of previous moves;
for instance, castling is possible when king and rook reach their original squares (e8 and a8), because it is 'forgotten' that they have already moved (2021-09-14)
Henrik Juel: You ask for a justification of the key, James; there is no need for justification, Black can do whatever he likes
In the position after 1.a1=S last move must have been g2-g4, so the ep capture 2.fxg3ep is legitimized
In the position after 2.fxg3ep last move must have been Ke3xPf3, with the double check being explained by exf3ep++; this is the 'hidden ep capture' I mentioned in my youthful comment (2021-09-14)
James Malcom: Thanks Henri! While the definition is incomplete, the reset each time is deducible from looking over shc problems. The justification here, however, was a level above me somehow. (2021-09-14)
comment
Keywords: Seriesmover, Consequent, En passant, Promotion (sD x2), Valladao Task, Castling, Promotion in the mating move
Genre: Retro, Fairies
FEN: r7/P1pp4/pp3pP1/4rpbP/5pPk/3n1K1b/p6q/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
Genre: Retro, Fairies
FEN: r7/P1pp4/pp3pP1/4rpbP/5pPk/3n1K1b/p6q/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
1. exf6ep d5,g5 2. Db7 g5,d5 3. Dg7#
Henrik Juel: The longest among the possible last moves is f7-f5, so 1.exf6ep d5 2.f7 g5 3.f8Q!?, but 3... Kh7!, so something is wrong. (2003-11-11)
HBae: Habe korrigiert. Es fehlte der wKa1 und die wDb2. (2019-10-11)
Henrik Juel: C+ by Popeye 4.61 (assuming that the condition was in effect also in last move)
Pd7 and Pg4 can be removed without affecting correctness or symmetry (2019-10-11)
Bernd Schwarzkopf: Pd7 and Pg4 are necessary. Without them last move could have been: 1.Pg6xXf5. (2021-02-06)
Mario Richter: Interestingly, the Editors of 'TfS' too thought that Pd7 and Pg4 can be omitted (s. 'TfS' 06/1938, p.124: "Går det inte lika bra utan bönderna på d7 och g4? (BL). Jo, det går visst lika bra. Men förf. har väl haft någon mening med dem också (Red.)."
I think, pawn d7 serves to prevent retractions like R: 1.Pg6xXf5, but pawn d4 is only there to complete the symmetry (i.e. Pd7 is necessary, Pg4 is not). (2021-02-06)
A.Buchanan: Doesn’t bPg4 stop 2.Dg2 as a dual? (2021-02-07)
comment
HBae: Habe korrigiert. Es fehlte der wKa1 und die wDb2. (2019-10-11)
Henrik Juel: C+ by Popeye 4.61 (assuming that the condition was in effect also in last move)
Pd7 and Pg4 can be removed without affecting correctness or symmetry (2019-10-11)
Bernd Schwarzkopf: Pd7 and Pg4 are necessary. Without them last move could have been: 1.Pg6xXf5. (2021-02-06)
Mario Richter: Interestingly, the Editors of 'TfS' too thought that Pd7 and Pg4 can be omitted (s. 'TfS' 06/1938, p.124: "Går det inte lika bra utan bönderna på d7 och g4? (BL). Jo, det går visst lika bra. Men förf. har väl haft någon mening med dem också (Red.)."
I think, pawn d7 serves to prevent retractions like R: 1.Pg6xXf5, but pawn d4 is only there to complete the symmetry (i.e. Pd7 is necessary, Pg4 is not). (2021-02-06)
A.Buchanan: Doesn’t bPg4 stop 2.Dg2 as a dual? (2021-02-07)
comment
Keywords: Maximummer, En passant as key, Asymmetrical solution, Symmetrical position
Genre: Retro, 3#, Fairies
FEN: 7k/3p2p1/4p3/4Pp2/6p1/8/1Q6/K7
Reprints: (II) Problem 65-68 01/1960
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
Genre: Retro, 3#, Fairies
FEN: 7k/3p2p1/4p3/4Pp2/6p1/8/1Q6/K7
Reprints: (II) Problem 65-68 01/1960
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
* 1. ... Lxd2#
1. dxe3ep 2. Txf2 3. Txg2 4. Th2 5. Th3 0-0#
1. dxe3ep 2. Txf2 3. Txg2 4. Th2 5. Th3 0-0#
Keywords: Castling (wk), Seriesmover, En passant as key, a posteriori (AP) (Type Petrovic)
Genre: Retro, Fairies
Computer test: HC+ Popeye 4.61 after analysis
FEN: 8/8/8/4pPp1/1p1pPkpb/1P1P2p1/pPpr1PP1/rbB1K2R
Reprints: 103 Bilten 1970 1971
(70) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-20 more...
Genre: Retro, Fairies
Computer test: HC+ Popeye 4.61 after analysis
FEN: 8/8/8/4pPp1/1p1pPkpb/1P1P2p1/pPpr1PP1/rbB1K2R
Reprints: 103 Bilten 1970 1971
(70) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-20 more...
18 - P0004481
Matjaz Zigman
Länderkampf Mazedonien-Slowenien 1969
7. Platz
(13+10) cooked
ser-h#6** AP
Matjaz Zigman
Länderkampf Mazedonien-Slowenien 1969
7. Platz
(13+10) cooked
ser-h#6** AP
* 1. ... Sb5#
* 1. ... dxc3#
1. fxe3ep 2. e2 3. exf1=S 4. Se3? 5. Sd5 Sb5# (must castle for AP)
1. fxe3ep 2. e2 3. exf1=S 4. Sxd2 5. Sxb1 6. Sxa3 0-0-0!#
Luckily, illegal diagram can be fixed by removal of bPa4.
Then if White castling rights remain, ep is on. One ser-h#5 & 18 ser-h#6 tries exist, in which White does not bother to castle.
We don't know if this was a transcription error or a counting mistake by the composer.
Cook: Too many pawn captures required. For retro problem, illegal position implies unsound
* 1. ... dxc3#
1. fxe3ep 2. e2 3. exf1=S 4. Se3? 5. Sd5 Sb5# (must castle for AP)
1. fxe3ep 2. e2 3. exf1=S 4. Sxd2 5. Sxb1 6. Sxa3 0-0-0!#
Luckily, illegal diagram can be fixed by removal of bPa4.
Then if White castling rights remain, ep is on. One ser-h#5 & 18 ser-h#6 tries exist, in which White does not bother to castle.
We don't know if this was a transcription error or a counting mistake by the composer.
Cook: Too many pawn captures required. For retro problem, illegal position implies unsound
Mike Neumeier: The solution appears to be 1.fxe4 e.p. 2.e2 3.exf1=L 4.Ld3 5.Le4 6.Ld5 dxc3#. Was that the intention? (2013-02-18)
Arno Tüngler: There would even be a solution in 5 moves by 3.exf1=S 5.Sd5 Sb5#
However, in order to justify the e.p.-key there must be an AP prove that e2-e4 (and not a move by the wK or wRa1) was played as White's last move. Thus the only solution giving this in 6 moves is 1.fxe3 e.p. 2.e2 3.exf1=S 4.Sxd2 5.Sxb1 6.Sxa3 0-0-0!# (2013-02-18)
Mike Neumeier: It struck me as odd there was no solution listed. (2013-02-18)
Mike Neumeier: And, if we let black be the idle side, there is 1.Txg1 2.Lxh1 3.Kd1 4.Kc1 hxg1T#. What does the asterisk(*) mean? (2013-02-18)
Henrik Juel: The asterisk means that there is also a white mate if he had the move, in this case two mates, so maybe the stipulation should be ser-h#6**:
1.dxc3,Sb5# (2013-02-18)
Ladislav Packa: It is clear that e.p. is possible only when the wK and wRa1 done neither move.
What convention is used here for the right to castling? (2013-02-18)
Henrik Juel: White may castle, unless you can show that he has lost the right to castle. In this problem there is a major difficulty, I think: the position seems illegal.
White pawns captured b2xc3xd4xe5xf6, g2xf3, and axb, promoting on b8; Black captured bxc, dxc, and g3xh2. We cannot explain the destiny of [Ph7].
Reversing wPh3 and bPh2 seems to handle the illegality, but then the problem can be solved in 5 moves (2013-02-18)
Mike Neumeier: Thanks, Henrik. Maybe the stipulation was a typo. Considering all comments together, perhaps the stip should have been ser-h#5**, with the one solution being the 5-mover Arno gave. I do not think it can be proven, except by a stipulation of ser-h#5 that en passant is possible. Which leads to the question of convention. Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible? Whether there is castling or not here seems immaterial. It is just another 6-move solution. There are 19 6-movers (Popeye) with one queenside castling among them. But only the one 5-mover. (2013-02-18)
Henrik Juel: By convention, an en passant capture as first move is not permitted, unless the pawn double step can be shown by some kind of retro analysis.
(Conversely, a castling is permitted, unless it can be shown by retro analysis that the right to castle has been lost, i.e., that king and/or rook has moved) (2013-02-18)
A.Buchanan: Suggested repair: Remove bPa4. Add AP to stipulation.
As Henrik points out, the diagram as it stands is illegal. We can't swap wPh3 & bPh2, because that allows 2 h#1 cooks.
Suppose we remove bPa4 instead. Then we still at least 5 White pawn captures: bxcxdxexf6 & gxf3. But wPh3 did still move from h2, so there are 3 Black pawn captures bxc, dxc & gxh2. So [bhP] died without capturing or promoting. [waP] must have promoted - so either [baP] was captured to clear the way, or [waP] captured to promote, and later [baP] promoted. Either way, all the numbers add up, and the position is legal. wPe4 cannot have just come from d3. So AP is triggered. Forward logic works just as before without bPa4. There is 1 5-move try, and 18 6-move tries, but AP eliminates them all because of need for castling to retrospectively justify the ep. (2013-02-23)
A.Buchanan: Mike asked: "Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible?"
Answer: No. The stipulation does imply a default player to move, but otherwise cannot be used as a premise to determine game state (castling, en passant). (2022-03-21)
A.Buchanan: WinChloe & Yacpdb have the same cooked diagram as PDB, so I think this cook was in the design. I would like to post a sound version with bPa4 removed, and I think it would just be “Zigman correction” with no one else’s name mentioned in the credits. Is everyone ok with this? (2022-05-30)
Mario Richter: 'Bilten 1970' (printed 1971) contains on pp. 62-81 a complete report on the "Problem Duel Slovenia - Macedonia 1969", the problem here can be found on page 79 and is printed there exactly as given here.
Andrew's handling of the suggested correction (remove black pawn a4) is o.k. for me. (2022-05-31)
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Arno Tüngler: There would even be a solution in 5 moves by 3.exf1=S 5.Sd5 Sb5#
However, in order to justify the e.p.-key there must be an AP prove that e2-e4 (and not a move by the wK or wRa1) was played as White's last move. Thus the only solution giving this in 6 moves is 1.fxe3 e.p. 2.e2 3.exf1=S 4.Sxd2 5.Sxb1 6.Sxa3 0-0-0!# (2013-02-18)
Mike Neumeier: It struck me as odd there was no solution listed. (2013-02-18)
Mike Neumeier: And, if we let black be the idle side, there is 1.Txg1 2.Lxh1 3.Kd1 4.Kc1 hxg1T#. What does the asterisk(*) mean? (2013-02-18)
Henrik Juel: The asterisk means that there is also a white mate if he had the move, in this case two mates, so maybe the stipulation should be ser-h#6**:
1.dxc3,Sb5# (2013-02-18)
Ladislav Packa: It is clear that e.p. is possible only when the wK and wRa1 done neither move.
What convention is used here for the right to castling? (2013-02-18)
Henrik Juel: White may castle, unless you can show that he has lost the right to castle. In this problem there is a major difficulty, I think: the position seems illegal.
White pawns captured b2xc3xd4xe5xf6, g2xf3, and axb, promoting on b8; Black captured bxc, dxc, and g3xh2. We cannot explain the destiny of [Ph7].
Reversing wPh3 and bPh2 seems to handle the illegality, but then the problem can be solved in 5 moves (2013-02-18)
Mike Neumeier: Thanks, Henrik. Maybe the stipulation was a typo. Considering all comments together, perhaps the stip should have been ser-h#5**, with the one solution being the 5-mover Arno gave. I do not think it can be proven, except by a stipulation of ser-h#5 that en passant is possible. Which leads to the question of convention. Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible? Whether there is castling or not here seems immaterial. It is just another 6-move solution. There are 19 6-movers (Popeye) with one queenside castling among them. But only the one 5-mover. (2013-02-18)
Henrik Juel: By convention, an en passant capture as first move is not permitted, unless the pawn double step can be shown by some kind of retro analysis.
(Conversely, a castling is permitted, unless it can be shown by retro analysis that the right to castle has been lost, i.e., that king and/or rook has moved) (2013-02-18)
A.Buchanan: Suggested repair: Remove bPa4. Add AP to stipulation.
As Henrik points out, the diagram as it stands is illegal. We can't swap wPh3 & bPh2, because that allows 2 h#1 cooks.
Suppose we remove bPa4 instead. Then we still at least 5 White pawn captures: bxcxdxexf6 & gxf3. But wPh3 did still move from h2, so there are 3 Black pawn captures bxc, dxc & gxh2. So [bhP] died without capturing or promoting. [waP] must have promoted - so either [baP] was captured to clear the way, or [waP] captured to promote, and later [baP] promoted. Either way, all the numbers add up, and the position is legal. wPe4 cannot have just come from d3. So AP is triggered. Forward logic works just as before without bPa4. There is 1 5-move try, and 18 6-move tries, but AP eliminates them all because of need for castling to retrospectively justify the ep. (2013-02-23)
A.Buchanan: Mike asked: "Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible?"
Answer: No. The stipulation does imply a default player to move, but otherwise cannot be used as a premise to determine game state (castling, en passant). (2022-03-21)
A.Buchanan: WinChloe & Yacpdb have the same cooked diagram as PDB, so I think this cook was in the design. I would like to post a sound version with bPa4 removed, and I think it would just be “Zigman correction” with no one else’s name mentioned in the credits. Is everyone ok with this? (2022-05-30)
Mario Richter: 'Bilten 1970' (printed 1971) contains on pp. 62-81 a complete report on the "Problem Duel Slovenia - Macedonia 1969", the problem here can be found on page 79 and is printed there exactly as given here.
Andrew's handling of the suggested correction (remove black pawn a4) is o.k. for me. (2022-05-31)
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Keywords: Castling (wg), Seriesmover, a posteriori (AP) (Type Petrovic), Illegal position, En passant as key, Promotion (s), Valladao Task, Superseded by (P1401546)
Genre: Retro, Fairies
Computer test: Popeye v4.87 but simple retro-logic shows illegality
FEN: 8/8/5P2/2p1p3/p1pkPp2/N1p2P1P/2PP1PBp/RN2KRbr
Reprints: 104 Bilten 1970 1971
(72) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-01 more...
Genre: Retro, Fairies
Computer test: Popeye v4.87 but simple retro-logic shows illegality
FEN: 8/8/5P2/2p1p3/p1pkPp2/N1p2P1P/2PP1PBp/RN2KRbr
Reprints: 104 Bilten 1970 1971
(72) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-01 more...
19 - P0005297
Les Marvin
Journal of Recreational Mathematics
(3+3)
Welches war der erste Zug und welches waren die letzten Züge?
Tic-Tac-Toe
"Experten" haben gespielt.
Les Marvin
Journal of Recreational Mathematics
(3+3)
Welches war der erste Zug und welches waren die letzten Züge?
Tic-Tac-Toe
"Experten" haben gespielt.
Alain Brobecker: Full analysis of such Tic-Tac-Toe games with expert players can be found here: http://abrobecker.free.fr/text/tictactoe.pdf (2022-01-01)
comment
comment
Keywords: First Move?, Last Moves?, Tic-Tac-Toe, no 8x8 board
Genre: Retro, 2#, Fairies
FEN: iIIqqqqq/2iqqqqq/1iIqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2022-03-08 more...
Genre: Retro, 2#, Fairies
FEN: iIIqqqqq/2iqqqqq/1iIqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2022-03-08 more...
20 - P0006067
Nikita M. Plaksin
Andrej N. Kornilow
(2) diagrammes 15 07-09/1994
(2+7)
Ist die Stellung legal?
Monochromes Schach
Nikita M. Plaksin
Andrej N. Kornilow
(2) diagrammes 15 07-09/1994
(2+7)
Ist die Stellung legal?
Monochromes Schach
R: 1. ... 0-0 2. Kf8-e7 Lg8-h7 3. Kg7-f8 Lh7-g8 4. Le7-d8 Lg8-h7 5. La3-e7 Lh7-g8 6. Kh6-g7 e7xTf6 7. Kg5-h6
Kees: R: -1. … 0-0 -2. Kf8-e7 Lg8-h7 -3. Kg7-f8 Lh7-g8 -4. Le7-d8 Lg8-h7 -5.Le7-a3 Lh7-g8 -6. Kh6-g7 e7xTf6 Kg5-h6 (2022-02-16)
A.Buchanan: The final position in Kees' solution has wTf6 which must be promoted. This could have been wBb, which captured e.g. bBa5, bBb6 e.p., bLa7, bSb8.
wTf7 is also promoted, and might be sBh having capturing wBg5, wBh6 e.p., wLg2, wTh1. sBfxLg6 completes the picture. I don't see any difficult captures e.g. of S remaining, so the position looks to be legal. (2022-02-16)
comment
A.Buchanan: The final position in Kees' solution has wTf6 which must be promoted. This could have been wBb, which captured e.g. bBa5, bBb6 e.p., bLa7, bSb8.
wTf7 is also promoted, and might be sBh having capturing wBg5, wBh6 e.p., wLg2, wTh1. sBfxLg6 completes the picture. I don't see any difficult captures e.g. of S remaining, so the position looks to be legal. (2022-02-16)
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Keywords: Monochromatic Chess, Minimal
Genre: Retro, Fairies
FEN: 3B1rk1/2p1Kr1b/5pp1/8/8/8/8/8
Input: Gerd Wilts, 1995-07-16
Last update: A.Buchanan, 2022-02-16 more...
Genre: Retro, Fairies
FEN: 3B1rk1/2p1Kr1b/5pp1/8/8/8/8/8
Input: Gerd Wilts, 1995-07-16
Last update: A.Buchanan, 2022-02-16 more...
21 - P0006406
Nikita M. Plaksin
4.B Caissas Schloßbewohner 1, p. 73, 1983
(4+2)
Welches waren die letzten 3 Einzelzüge?
Alle Bauern sind Berolinabauern
Nikita M. Plaksin
4.B Caissas Schloßbewohner 1, p. 73, 1983
(4+2)
Welches waren die letzten 3 Einzelzüge?
Alle Bauern sind Berolinabauern
Teilaufgabe B von P1393956
Henrik Juel: R: 1.BEb7-a8+ Ka8-a7 2.BEc7-b8+, not 1... Ka8xYa7? (2021-09-23)
comment
Henrik Juel: R: 1.BEb7-a8+ Ka8-a7 2.BEc7-b8+, not 1... Ka8xYa7? (2021-09-23)
comment
Keywords: Last Moves?, Berolina chess
Pieces: = Berolina Pawn (BE)
Genre: Retro, Fairies
FEN: RRK5/k7/*2p*2P6/8/8/8/8/8
Reprints: (26) Die Schwalbe 156, p. 227, 12/1995
Input: Gerd Wilts, 1996-06-12
Last update: A.Buchanan, 2022-09-13 more...
Pieces: = Berolina Pawn (BE)
Genre: Retro, Fairies
FEN: RRK5/k7/*2p*2P6/8/8/8/8/8
Reprints: (26) Die Schwalbe 156, p. 227, 12/1995
Input: Gerd Wilts, 1996-06-12
Last update: A.Buchanan, 2022-09-13 more...
22 - P0006467
Michel Caillaud
Jacques Rotenberg
3927 Problemkiste 102, p. 168, 12/1995
(14+15)
BP in 6.0
N statt S in der PAS
wSU,sSU=Nachtreiter
Michel Caillaud
Jacques Rotenberg
3927 Problemkiste 102, p. 168, 12/1995
(14+15)
BP in 6.0
N statt S in der PAS
wSU,sSU=Nachtreiter
a) 1. Nxe7 Nxe2 2. Nb1 Nxc1 3. Df3 Nca5 4. Ke2 Ngc6+ 5. Kd3 Nb8 6. Ke4 Ng8+
paul: Verified by Jacobi with this input:
stip dia 6
forsyth rnbqkbnr/pppp1ppp/8/8/4K3/5Q2/PPPP1PPP/RN3BNR
cond cavaliermaj (2023-06-16)
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stip dia 6
forsyth rnbqkbnr/pppp1ppp/8/8/4K3/5Q2/PPPP1PPP/RN3BNR
cond cavaliermaj (2023-06-16)
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Keywords: Unique Proof Game, Cavalier majeur
Pieces: = Nightrider (N)
Genre: Retro, Fairies
FEN: r*2nbqkb*2nr/pppp1ppp/8/8/4K3/5Q2/PPPP1PPP/R*2N3B*2NR
Input: Gerd Wilts, 1996-06-16
Last update: A.Buchanan, 2021-02-19 more...
Pieces: = Nightrider (N)
Genre: Retro, Fairies
FEN: r*2nbqkb*2nr/pppp1ppp/8/8/4K3/5Q2/PPPP1PPP/R*2N3B*2NR
Input: Gerd Wilts, 1996-06-16
Last update: A.Buchanan, 2021-02-19 more...
1. fxg6ep+ Kg5 2. Lh4#
R: 1. g7-g5
R: 1. g7-g5
Keywords: En passant as key
Pieces: = Grasshopper (G)
Genre: Retro, 2#, Fairies
Computer test: HC+ Popeye 4.61 after very simple analysis
FEN: 8/5*2Q1K/5P1P/5Ppk/6p1/6B1/6N1/8
Reprints: 115 Caissa's Wild Roses 1935
Input: Gerd Wilts, 1996-07-16
Last update: A.Buchanan, 2021-05-03 more...
Pieces: = Grasshopper (G)
Genre: Retro, 2#, Fairies
Computer test: HC+ Popeye 4.61 after very simple analysis
FEN: 8/5*2Q1K/5P1P/5Ppk/6p1/6B1/6N1/8
Reprints: 115 Caissa's Wild Roses 1935
Input: Gerd Wilts, 1996-07-16
Last update: A.Buchanan, 2021-05-03 more...
a) +wKc8?: Schwarz hat keinen letzten Zug
b) +wKg7: 1. ... Gc8 2. Gf4 Gh8
b) +wKg7: 1. ... Gc8 2. Gf4 Gh8
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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h#2 +wGb8 wGe4 1.bGa5 Ke7 2.Gf5 Gf8# M.Frelih (2021-09-05)
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Keywords: Add pieces
Pieces: = 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...
Pieces: = 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...
25 - P0006953
Josef Haas
feenschach 21, p. 258, 04/1974
(6+1) cooked
Welches war der letzte Zug?
Circe
Josef Haas
feenschach 21, p. 258, 04/1974
(6+1) cooked
Welches war der letzte Zug?
Circe
R: 1. 0-0
Cook: R: 1. Kf2-g1 Kh2xBh3[+wBh2]
Cook: R: 1. Kf2-g1 Kh2xBh3[+wBh2]
HBae: cook siehe feenschach-25, 10/1974, S. 372 (2015-12-02)
A.Buchanan: Bh2->L (2015-12-03)
A.Buchanan: A number of these problems by Haas were cooked. HBae gives a teasing link to an article which possibly includes some corrections. Has anyone got access to this article, please? (2021-05-13)
HBae: Hello Andrew, link to feenschach 10/1974 (website Kotesovec):
http://problem64.beda.cz/silo/feenschach25_1974.pdf (2021-05-13)
A.Buchanan: Thanks HBae! (2021-05-13)
A.Buchanan: In this one B-=D seems to be Type C, not type A?
In WinChloe there are a batch more from a few years later 34 p.143, feenschach 47 (juil. 79). If we're uploading anything, should maybe skip to the end? (2021-05-13)
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A.Buchanan: Bh2->L (2015-12-03)
A.Buchanan: A number of these problems by Haas were cooked. HBae gives a teasing link to an article which possibly includes some corrections. Has anyone got access to this article, please? (2021-05-13)
HBae: Hello Andrew, link to feenschach 10/1974 (website Kotesovec):
http://problem64.beda.cz/silo/feenschach25_1974.pdf (2021-05-13)
A.Buchanan: Thanks HBae! (2021-05-13)
A.Buchanan: In this one B-=D seems to be Type C, not type A?
In WinChloe there are a batch more from a few years later 34 p.143, feenschach 47 (juil. 79). If we're uploading anything, should maybe skip to the end? (2021-05-13)
comment
SCHRECKE: In a) geschah zuletzt 0. ... Td8-a8, also ist die Rochade nicht mehr möglich.
In b) kam zuletzt die sD von h1, schwarze Rochade also noch spielbar.
Lösung:
a) 1. gxh5 Td8 2. Df7+ Kxf7#
b) 1. Dg6+ Dxg6 2. Sf5 0-0-0# (2023-11-07)
Joost de Heer: Slight error in previous comment: Last move was Td8xT/Da8, not Td8-a8 (as Td8-d1 would be longer) (2023-11-07)
comment
In b) kam zuletzt die sD von h1, schwarze Rochade also noch spielbar.
Lösung:
a) 1. gxh5 Td8 2. Df7+ Kxf7#
b) 1. Dg6+ Dxg6 2. Sf5 0-0-0# (2023-11-07)
Joost de Heer: Slight error in previous comment: Last move was Td8xT/Da8, not Td8-a8 (as Td8-d1 would be longer) (2023-11-07)
comment
Keywords: Maximummer
Genre: Retro, s#, Fairies
FEN: r3k2K/6QP/R1P5/4P1pq/6PN/8/8/8
Input: Gerd Wilts, 1996-08-13
Last update: A.Buchanan, 2023-06-03 more...
Genre: Retro, s#, Fairies
FEN: r3k2K/6QP/R1P5/4P1pq/6PN/8/8/8
Input: Gerd Wilts, 1996-08-13
Last update: A.Buchanan, 2023-06-03 more...
R: 1. Kg1xSh2#
Keywords: Last Move? (KxS), Checkless, Miniature
Genre: Retro, Fairies
FEN: 8/8/8/8/8/7P/2Q3PK/4k2R
Input: Gerd Wilts, 1996-08-18
Last update: A.Buchanan, 2023-05-15 more...
Genre: Retro, Fairies
FEN: 8/8/8/8/8/7P/2Q3PK/4k2R
Input: Gerd Wilts, 1996-08-18
Last update: A.Buchanan, 2023-05-15 more...
28 - P0007435
Adamas
(47) feenschach 40 11-12/1977
(1+3) cooked
Welches war der letzte Zug?
Längstzüger
Adamas
(47) feenschach 40 11-12/1977
(1+3) cooked
Welches war der letzte Zug?
Längstzüger
R: 1. Bd2xDe1=L+
Cook: R: 1. Bd2xDc1=L+,d2xT/L/Sc1=L+
Cook: R: 1. Bd2xDc1=L+,d2xT/L/Sc1=L+
Mario Richter: Position is as given in 'feenschach'. It seems that R: 1. d3xDc1+,d3xT/L/Sc1+ also works and that adding a black pawn on the 7th rank (e.g. h7) would prevent the cook.
Can somebody confirm my observation? (2022-01-10)
Mario Richter: small typo: It seems that R: 1. d2xDc1+,d2xT/L/Sc1+ also works (2022-01-10)
Henrik Juel: Your observation, with last moves d2xDTLDc1=L+, seems correct (2022-01-10)
comment
Can somebody confirm my observation? (2022-01-10)
Mario Richter: small typo: It seems that R: 1. d2xDc1+,d2xT/L/Sc1+ also works (2022-01-10)
Henrik Juel: Your observation, with last moves d2xDTLDc1=L+, seems correct (2022-01-10)
comment
Keywords: Maximummer, Last Move? (BxD=L), Aristocrat, Miniature
Genre: Retro, Fairies
FEN: 8/8/8/8/8/4K1k1/8/2b1b3
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-10 more...
Genre: Retro, Fairies
FEN: 8/8/8/8/8/4K1k1/8/2b1b3
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-10 more...
29 - P0007436
Adamas
(48) feenschach 40 11-12/1977
(1+3) cooked
Welches war der letzte Zug?
Längstzüger
Adamas
(48) feenschach 40 11-12/1977
(1+3) cooked
Welches war der letzte Zug?
Längstzüger
R: 1. Bg2xTh1=L+
Cook: R: 1. Bg2xDf1=S+,Bg2xTf1=S+,Bg2xLf1=S+,Bg2xSf1=S+,
Cook: R: 1. Bg2xDf1=S+,Bg2xTf1=S+,Bg2xLf1=S+,Bg2xSf1=S+,
Henrik Juel: What is the other last move, Andrew? (2022-01-08)
A.Buchanan: I didn’t analyse these problems, just corrected the genre when I saw that it was incomplete. Here how about e.g. R: 1. Bc6-h8+ Kb7-a8 2. Bh8xPc6 (2022-01-09)
Mario Richter: R: 1. ... Bc6-h1+? is illegal - it leaves the wK in check!
The cooks are the discovered checks by 1. ... g2xYf1+ (2022-01-09)
comment
A.Buchanan: I didn’t analyse these problems, just corrected the genre when I saw that it was incomplete. Here how about e.g. R: 1. Bc6-h8+ Kb7-a8 2. Bh8xPc6 (2022-01-09)
Mario Richter: R: 1. ... Bc6-h1+? is illegal - it leaves the wK in check!
The cooks are the discovered checks by 1. ... g2xYf1+ (2022-01-09)
comment
Keywords: Maximummer, Last Move? (BxT=L), Aristocrat, Miniature
Genre: Retro, Fairies
FEN: K7/8/8/8/8/8/7k/5n1b
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-09 more...
Genre: Retro, Fairies
FEN: K7/8/8/8/8/8/7k/5n1b
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-09 more...
30 - P0007437
Ewald Reichel
(49) feenschach 40 11-12/1977
(4+8) cooked
Welches war der letzte Zug?
Längstzüger
Ewald Reichel
(49) feenschach 40 11-12/1977
(4+8) cooked
Welches war der letzte Zug?
Längstzüger
R: 1. Ba2xLb1=L+
Cook: R: 1. a2xSb1=L+
Cook: R: 1. a2xSb1=L+
Henrik Juel: What is the other last move, Andrew? (2022-01-08)
Henrik Juel: Well, maybe 1.Pa2xSb1=L+ also works (2022-01-08)
comment
Henrik Juel: Well, maybe 1.Pa2xSb1=L+ also works (2022-01-08)
comment
Keywords: Maximummer, Last Move? (BxL=L)
Genre: Retro, Fairies
FEN: 8/8/8/3p4/p2p4/R1pK4/1k1PP3/rbb5
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-09 more...
Genre: Retro, Fairies
FEN: 8/8/8/3p4/p2p4/R1pK4/1k1PP3/rbb5
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-09 more...
31 - P0007676
Wolfgang Dittmann
(B) feenschach 50 04-06/1980
(13+1)
Welches war der letzte Zug?
Kamikaze
Wolfgang Dittmann
(B) feenschach 50 04-06/1980
(13+1)
Welches war der letzte Zug?
Kamikaze
R: 1. Dd1-f1 ... 2. Se1-d3+
A.Buchanan: Last move sure, but prior to that could be Kg1-h2, Kh1-h2 or even Kg1xTh2. wSa1 needs to be present so it wasn’t uncaptured on h2, but I think it can replace wLb2, which cannot I think be uncaptured on h2. Am I missing sth? (2022-07-15)
Mario Richter: Does the Kamikaze-rule only apply to Black's moves? Otherwise, how can the diagram position be reached from the initial game array?
If Kamikaze applies to White too, I think, wSa1 also prevents Tg1xSh1[-wTh1] as an alternative last move ... (2022-07-15)
A.Buchanan: Hi Mario: thanks for this. Kamikaze applies to both sides. By default it’s Rex exclusive so kings can go round mopping up odd pieces. Black used three of Pacg BQSS to capture wBRR inside the cage. I now see wSSBQ must remain on the board of else there may just have been a capture outside the cage. If there was just a Kamikaze capture inside the cage, it can only have been Q&R or S&R (or vice versa). e.g. bSf3xRg1(-bSg1). Since a1 is occupied, can’t get bK past wR to retract via b2. (2022-07-15)
more ...
comment
Mario Richter: Does the Kamikaze-rule only apply to Black's moves? Otherwise, how can the diagram position be reached from the initial game array?
If Kamikaze applies to White too, I think, wSa1 also prevents Tg1xSh1[-wTh1] as an alternative last move ... (2022-07-15)
A.Buchanan: Hi Mario: thanks for this. Kamikaze applies to both sides. By default it’s Rex exclusive so kings can go round mopping up odd pieces. Black used three of Pacg BQSS to capture wBRR inside the cage. I now see wSSBQ must remain on the board of else there may just have been a capture outside the cage. If there was just a Kamikaze capture inside the cage, it can only have been Q&R or S&R (or vice versa). e.g. bSf3xRg1(-bSg1). Since a1 is occupied, can’t get bK past wR to retract via b2. (2022-07-15)
more ...
comment
Keywords: Kamikaze, Last Move? (D), Type A
Genre: Retro, Fairies
FEN: 8/8/8/8/8/1P1N3P/PBPPPPPk/N1K2Q2
Input: Gerd Wilts, 1996-09-02
Last update: A.Buchanan, 2022-07-15 more...
Genre: Retro, Fairies
FEN: 8/8/8/8/8/1P1N3P/PBPPPPPk/N1K2Q2
Input: Gerd Wilts, 1996-09-02
Last update: A.Buchanan, 2022-07-15 more...
A.Buchanan: I guess that the two castlings are intended to be mutually exclusive, but I can't see why that would be. (2022-05-27)
comment
comment
Keywords: Twinning by board reflection
Pieces: = Grasshopper (G)
Genre: Retro, Fairies
FEN: bb2k2r/1rp1*2q*2Q*2q1/n6p/8/n3p1p*2Q/4p3/1PP1P1PP/R3KR2
Input: Gerd Wilts, 1996-10-03
Last update: A.Buchanan, 2023-06-02 more...
Pieces: = Grasshopper (G)
Genre: Retro, Fairies
FEN: bb2k2r/1rp1*2q*2Q*2q1/n6p/8/n3p1p*2Q/4p3/1PP1P1PP/R3KR2
Input: Gerd Wilts, 1996-10-03
Last update: A.Buchanan, 2023-06-02 more...
33 - P0008449
Jean-Michel Trillon
3574 diagrammes 118 07/1996
(6+4)
Orthorekonstruktion in 12.5
Echecs sentinelles
Weiß am Zug
Jean-Michel Trillon
3574 diagrammes 118 07/1996
(6+4)
Orthorekonstruktion in 12.5
Echecs sentinelles
Weiß am Zug
1. e5 Kb8 2. e6 Ka8 3. e7 Kb8 4. e8=T Ka8 5. Th8 Kb8 6. Th1 Ka8 7. Te1 Kb8 8. Te2 Ka8 9. Te1[+wBe2] Kb8 10. Tb1+ Ka8 11. Tb8+ Kxb8 12. e4 Ka8 13. e3
Frank Müller: Welche Bauern sind Berolinabauern? (2010-08-15)
Joost de Heer: No berolina pawns.
Solution: 1.é5 Rb8 2.é6 Ra8 3.é7 Rb8 4.é8=T Ra8 5.Th8 Rb8 6.Th1 Ra8 7.Té1 Rb8 8.Té2 Ra8 9.Té1(+é2) Rb8 10.Tb1+ Ra8 11.Tb8+ R×b8 12.é4 Ra8 13.é3 (2023-08-27)
A.Buchanan: Quite delightful. Can it be validated in Jacobi as an A-B PG? (2023-08-28)
comment
Joost de Heer: No berolina pawns.
Solution: 1.é5 Rb8 2.é6 Ra8 3.é7 Rb8 4.é8=T Ra8 5.Th8 Rb8 6.Th1 Ra8 7.Té1 Rb8 8.Té2 Ra8 9.Té1(+é2) Rb8 10.Tb1+ Ra8 11.Tb8+ R×b8 12.é4 Ra8 13.é3 (2023-08-27)
A.Buchanan: Quite delightful. Can it be validated in Jacobi as an A-B PG? (2023-08-28)
comment
34 - P0008667
Maryan Kerhuel
Didier Innocenti
2524 Phénix 51 04/1997
(9+15) cooked
BP in 15,0
Circe Assassin
Maryan Kerhuel
Didier Innocenti
2524 Phénix 51 04/1997
(9+15) cooked
BP in 15,0
Circe Assassin
paul: Cooked by: 1.h3 d5 2.Rh2 Bxh3(ph2) 3.Sxh3(Bc8) Bxh3 4.g4 a5 5.Bg2 Bxg4(pg2) 6.c4 a4 7.b4 e6 8.Bb2 Bxb4(pb2) 9.Sa3 Ra5 10.Qxa4(pa7) Qc8 11.QxKe8(Ke8) (suicide of wQ)b5 12.Sc2 dxc4(pc2) 13.a3 Bxe2(pe2) 14.Ra2 Bxa3(pa2) 15.Kd1 b4. (2010-06-19)
Arnold Beine: Here is something wrong. Either Paul's cook does not work, because 10.Qxa4(Pa7)+ is a check (10...Qc8??) or in the given condition "Rex inclusive" must be added. (2023-07-26)
paul: Right, Arnold. Anyway, the proof game is cooked: 1.a4 e6 2.Sa3 B×a3 3.b4 a5 4.Bb2 a×b4(b2) 5.Ra2 R×a4(a2) 6.e4 Ra5 etc, as in author's solution: 7.Ba6 b×a6(Bf1) 8.B×a6(a7) d5 9.B×c8(bBc8) Bb7 10.c4 Qc8 11.Qc2 d×c4(c2) 12.Se2 B×e4(e2) 13.h3 Bf5 14.Rh2 B×h3(h2) 15.Kd1 B×g2(g2). (2024-03-13)
comment
Arnold Beine: Here is something wrong. Either Paul's cook does not work, because 10.Qxa4(Pa7)+ is a check (10...Qc8??) or in the given condition "Rex inclusive" must be added. (2023-07-26)
paul: Right, Arnold. Anyway, the proof game is cooked: 1.a4 e6 2.Sa3 B×a3 3.b4 a5 4.Bb2 a×b4(b2) 5.Ra2 R×a4(a2) 6.e4 Ra5 etc, as in author's solution: 7.Ba6 b×a6(Bf1) 8.B×a6(a7) d5 9.B×c8(bBc8) Bb7 10.c4 Qc8 11.Qc2 d×c4(c2) 12.Se2 B×e4(e2) 13.h3 Bf5 14.Rh2 B×h3(h2) 15.Kd1 B×g2(g2). (2024-03-13)
comment
Keywords: Circe (Assassin), Unique Proof Game
Genre: Retro, Fairies
FEN: 1nq1k1nr/p1p2ppp/4p3/r7/1pp5/b7/PPPPPPPP/3K4
Input: Gerd Wilts, 1997-05-28
Last update: Mario Richter, 2013-09-01 more...
Genre: Retro, Fairies
FEN: 1nq1k1nr/p1p2ppp/4p3/r7/1pp5/b7/PPPPPPPP/3K4
Input: Gerd Wilts, 1997-05-28
Last update: Mario Richter, 2013-09-01 more...
VL: Solution:
1.Rg2=wR Rb6=wR!! 2.Re8=bR! 0-0#! (2... Rf1#??)
2.Ra(c)8=wR??
Here only one w rook is volage since they both stand on light squares
of ODD ranks: any promotee had to visit a square of the opposite
colour at least once in order to arrive from the 8th rank! Similarly,
only one bl rook is volage (EVEN ranks). By the AP logic, we have a
right to justify the 1st W's move by subsequent castling, which shows
that it is Rh1 that is volage. Bl rooks play by the post factum
(subordination) logic: the first one turns out volage.
Cooked (feenschach, 126): 1.Ra2(Rc2,Rb3...)=wR Rh7+ 2.Kf8 Ra8#
Correction: Rh8- e8. (2002-04-05)
VL: Correction (H.130): Rh8 to d8 (rather than to the light square e8). (2021-02-12)
A.Buchanan: shifted sT to d8 as per request (2021-02-12)
comment
1.Rg2=wR Rb6=wR!! 2.Re8=bR! 0-0#! (2... Rf1#??)
2.Ra(c)8=wR??
Here only one w rook is volage since they both stand on light squares
of ODD ranks: any promotee had to visit a square of the opposite
colour at least once in order to arrive from the 8th rank! Similarly,
only one bl rook is volage (EVEN ranks). By the AP logic, we have a
right to justify the 1st W's move by subsequent castling, which shows
that it is Rh1 that is volage. Bl rooks play by the post factum
(subordination) logic: the first one turns out volage.
Cooked (feenschach, 126): 1.Ra2(Rc2,Rb3...)=wR Rh7+ 2.Kf8 Ra8#
Correction: Rh8- e8. (2002-04-05)
VL: Correction (H.130): Rh8 to d8 (rather than to the light square e8). (2021-02-12)
A.Buchanan: shifted sT to d8 as per request (2021-02-12)
comment
Keywords: a posteriori (AP), Retro-volages, Post Factum (PF), Miniature, Castling, Castling as mating move
Genre: Retro, Fairies, h#
FEN: 3r4/4pk2/8/8/8/8/1r6/1R2K2R
Input: Gerd Wilts, 1997-06-16
Last update: James Malcom, 2021-02-12 more...
Genre: Retro, Fairies, h#
FEN: 3r4/4pk2/8/8/8/8/1r6/1R2K2R
Input: Gerd Wilts, 1997-06-16
Last update: James Malcom, 2021-02-12 more...
1. Sc8 2. Sxb6 3. Sxa4 4. Sb2 5. Sd3 6. Se1 7. Sf3 8. exd3ep 9. Kd4 10. Ke4 11. Sd4 12. gxf3ep 13. Kf4 14. Kg4 15. f4 16. Sf5 exf3#
NL. Verbesserung 1998 erschienen: P1000933
Cook: 1. Lf7 2. Txe5 3. Kd5 4. Ke6 5. Kf6 6. Sg7 7. Se6 8. Tg6 9. Dh7 10. Dg7 dxe5#
außerdem Dual in der AL
1. Sc8 2. Sxb6 3. Sxa4 4. Sc5
NL. Verbesserung 1998 erschienen: P1000933
Cook: 1. Lf7 2. Txe5 3. Kd5 4. Ke6 5. Kf6 6. Sg7 7. Se6 8. Tg6 9. Dh7 10. Dg7 dxe5#
außerdem Dual in der AL
1. Sc8 2. Sxb6 3. Sxa4 4. Sc5
Keywords: Seriesmover, Consequent, En passant (x2)
Genre: Retro, Fairies
FEN: 8/4n3/1P2b3/3rPprn/P1kPpPpq/2p1p1p1/4P1Pb/7K
Input: Gerd Wilts, 1998-06-26
Last update: Mario Richter, 2021-09-16 more...
Genre: Retro, Fairies
FEN: 8/4n3/1P2b3/3rPprn/P1kPpPpq/2p1p1p1/4P1Pb/7K
Input: Gerd Wilts, 1998-06-26
Last update: Mario Richter, 2021-09-16 more...
1. gxf3ep 2. Dg4 3. 0-0-0 4. Te8 5. Kd8 6. Dxh5 0-0-0#
Henrik Juel: 1.gxf3ep 2.Qg4 3.000 4.Re8 5.Kd8 6.Qxh5 000#. Both castlings are necessary to legitimize the en passant capture. (2003-09-29)
A.Buchanan: I don't think this is consequent series mover, under which every position would considered afresh from the perspective of legality i.e. possible histories. Instead, this is regular series mover where one player is simply saying "pass". It's hard to imagine how AP might operate in a consequent context. I will change the keyword suitably. (2021-11-08)
comment
A.Buchanan: I don't think this is consequent series mover, under which every position would considered afresh from the perspective of legality i.e. possible histories. Instead, this is regular series mover where one player is simply saying "pass". It's hard to imagine how AP might operate in a consequent context. I will change the keyword suitably. (2021-11-08)
comment
Keywords: a posteriori (AP), Seriesmover, Castling, En passant as key
Genre: Retro, Fairies
FEN: r3k3/ppp1p3/2p5/6PP/5Ppq/1P4nB/PrP1P1p1/R3K1nb
Input: Gerd Wilts, 1998-06-26
Last update: A.Buchanan, 2021-11-08 more...
Genre: Retro, Fairies
FEN: r3k3/ppp1p3/2p5/6PP/5Ppq/1P4nB/PrP1P1p1/R3K1nb
Input: Gerd Wilts, 1998-06-26
Last update: A.Buchanan, 2021-11-08 more...
1. c5 2. c4 3. c3 4. c2 5. c1=S 6. Kf7 7. Kg6 8. Kh5 9. Kxh4 10. h5 g3#
Vor Kf7-e8 geschah Bg6xh7, deshalb muß zuerst einer der sB durch Umwandlung eliminiert werden.
In der Diagrammstellung haben die Bauern jeweils alle fehlenden gegnerischen Steine geschlagen:
W: Bdxexfxgxh, Baxbxa (oder Bcxbxa)
S: Bfxg, Bdxcxb, Bexdxcxb, Bbxa
Nach 1. Kf7? oder 1. Kf8 2. Kg7 3. Kg6 hätte W jeweils keinen letzten Zug. (R: h3-h4 sperrt den Th1 aus, der aber auch zu den Schlagopfern der sBB gehörte).
Also muß im Vorwärtsspiel zunächst eine Stellung mit geänderter Schlagbilanz erreicht werden.
Der Versuch 1. Kf8 2.Kg8 3. Kxh7 4. Kg6 5. Kh5 6. Kxh4? - aber nun hat W keinen letzten Zug - zeigt, daß der wBh7 erhalten bleiben muß, also muß die virtuelle Rücknahme h2,h3-h4 ermöglicht werden. Dies geschieht durch "Elimination" des Bc7, indem dieser auf c1 umwandelt. Bei einem sUW-Lc1 würden weiterhin 7 sBB-Schläge erforderlich sein, sUW-Dc1 oder sUW-Tc1 böten Schach, sodaß nur ein sUW-Sc1 übrig bleibt.
Nach 1. c5 2. c4 3. c3 4. c2 5. c1=S 6. Kf7 hat W nun h2,h3-h4 als Rücknahmemöglichkeit, weil nun der wTh1 nicht mehr notwendig als Schlagobjekt benötigt wird.
Vor Kf7-e8 geschah Bg6xh7, deshalb muß zuerst einer der sB durch Umwandlung eliminiert werden.
In der Diagrammstellung haben die Bauern jeweils alle fehlenden gegnerischen Steine geschlagen:
W: Bdxexfxgxh, Baxbxa (oder Bcxbxa)
S: Bfxg, Bdxcxb, Bexdxcxb, Bbxa
Nach 1. Kf7? oder 1. Kf8 2. Kg7 3. Kg6 hätte W jeweils keinen letzten Zug. (R: h3-h4 sperrt den Th1 aus, der aber auch zu den Schlagopfern der sBB gehörte).
Also muß im Vorwärtsspiel zunächst eine Stellung mit geänderter Schlagbilanz erreicht werden.
Der Versuch 1. Kf8 2.Kg8 3. Kxh7 4. Kg6 5. Kh5 6. Kxh4? - aber nun hat W keinen letzten Zug - zeigt, daß der wBh7 erhalten bleiben muß, also muß die virtuelle Rücknahme h2,h3-h4 ermöglicht werden. Dies geschieht durch "Elimination" des Bc7, indem dieser auf c1 umwandelt. Bei einem sUW-Lc1 würden weiterhin 7 sBB-Schläge erforderlich sein, sUW-Dc1 oder sUW-Tc1 böten Schach, sodaß nur ein sUW-Sc1 übrig bleibt.
Nach 1. c5 2. c4 3. c3 4. c2 5. c1=S 6. Kf7 hat W nun h2,h3-h4 als Rücknahmemöglichkeit, weil nun der wTh1 nicht mehr notwendig als Schlagobjekt benötigt wird.
Henrik Juel (PDB 2021-09-15): This is one of the few shc#'s without ep capture and/or castling
Henrik Juel: I don't agree entirely with you comment, James; this seems clearer:
1.Kf7? does not work, because now White has no last move ([Th1] was captured by a pawn)
1.Kf8 2.Kg7 3.Kg6? fails for the same reason
After 5.c1=S 6.Kf7 last move was h3-h4 or h2-h4
This is one of the few shc#'s without ep capture and/or castling (2021-09-15)
James Malcom: Henrik, what comment of mine? I merely edit swapped some text. (2021-09-15)
Henrik Juel: OK, so it was probably Gerd who wrote it in 1998; the early updates are no longer available, when you click on 'more...'
The german text roughly translates to:
Pg6xh7 happened before Kf7-e8, so first one of the black pawns must be eliminated by promotion.
I still like my comment better... (2021-09-15)
comment
Henrik Juel: I don't agree entirely with you comment, James; this seems clearer:
1.Kf7? does not work, because now White has no last move ([Th1] was captured by a pawn)
1.Kf8 2.Kg7 3.Kg6? fails for the same reason
After 5.c1=S 6.Kf7 last move was h3-h4 or h2-h4
This is one of the few shc#'s without ep capture and/or castling (2021-09-15)
James Malcom: Henrik, what comment of mine? I merely edit swapped some text. (2021-09-15)
Henrik Juel: OK, so it was probably Gerd who wrote it in 1998; the early updates are no longer available, when you click on 'more...'
The german text roughly translates to:
Pg6xh7 happened before Kf7-e8, so first one of the black pawns must be eliminated by promotion.
I still like my comment better... (2021-09-15)
comment
Keywords: Seriesmover, Consequent, Promotion (s), Excelsior
Genre: Retro, Fairies
FEN: 4k3/2p4P/7p/P5p1/pp4pP/bp6/pP2PPP1/K4B2
Input: Gerd Wilts, 1998-06-26
Last update: Mario Richter, 2021-09-16 more...
Genre: Retro, Fairies
FEN: 4k3/2p4P/7p/P5p1/pp4pP/bp6/pP2PPP1/K4B2
Input: Gerd Wilts, 1998-06-26
Last update: Mario Richter, 2021-09-16 more...
1. a8=T+ Kd7 2. e8=L+ Ke6 3. f8=S+ Kf5 4. c8=D+ Se6 5. 0-0-0 Ke4 6. Dxc6+ Kf5 7. g4+ fxg3 e.p. 8. Df3+ Sf4 9. Lb5 axb5 10. Ta1 bxc4 11. Tab1 cxb3 12. Dd3+ Sxd3#
paul: Cooked by 1.f8=Q Kd7 2.c8=B Kd6 3.e8=S Ke5 4.Ra5+ Kd4 5.Qc5+ Ke4 6.Re3+ fxe3 7.Qxc6+ Kd3 8.Bf5+ Se4 9.Kd1 exd2 10.Qf6 Kxc4 11.Be6+ Kd3 12.Qc3+ Sxc3#
2... Kc7 3.a8=S+ Kd6 4.e8=S++ Ke5 5.Re3+ Se4 6.d4+ Kxd4 7.Rd1+ Sd2 8.Rf3 Ke5 9.Qf5+ Kd4 10.g3 fxg3 11.Re3 Kxe3 12.Qf2+ gxf2# (Jacobi)
See P1287941 as correction. (2022-04-06)
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2... Kc7 3.a8=S+ Kd6 4.e8=S++ Ke5 5.Re3+ Se4 6.d4+ Kxd4 7.Rd1+ Sd2 8.Rf3 Ke5 9.Qf5+ Kd4 10.g3 fxg3 11.Re3 Kxe3 12.Qf2+ gxf2# (Jacobi)
See P1287941 as correction. (2022-04-06)
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Keywords: Valladao Task, Allumwandlung
Genre: Fairies
FEN: 2k5/P1P1PP2/p1p5/6n1/2P2p2/1R5P/3P2P1/R3K3
Input: Gerd Wilts, 1999-02-27
Last update: Olaf Jenkner, 2013-09-22 more...
Genre: Fairies
FEN: 2k5/P1P1PP2/p1p5/6n1/2P2p2/1R5P/3P2P1/R3K3
Input: Gerd Wilts, 1999-02-27
Last update: Olaf Jenkner, 2013-09-22 more...
40 - P0501007
Thomas Brand
Jörg A. R. Kuhlmann
Bo Lindgren
Arno Tüngler
Kjell Widlert
3707v feenschach 63 12/1982
(1+8) cooked
h#8
Circe (nur Offiziere) Rex inklusive
Thomas Brand
Jörg A. R. Kuhlmann
Bo Lindgren
Arno Tüngler
Kjell Widlert
3707v feenschach 63 12/1982
(1+8) cooked
h#8
Circe (nur Offiziere) Rex inklusive
1. b1=S Kb2 2. a1=T Kxa1[+sTh8] 3. Te8 Kxb1[+sSg8] 4. c1=L Kxc1[+sLf8] 5. Ke1 Kc2 6. Se7 Kd3 7. f1=D+ Ke3 8. De2+ Kxe2[+sDd8]#
Korrektur in feenschach 1988
Cook: Michel Caillaud (2022-07-19): cooked using Jacobi 0.7.5 :
1. a1=L Kb3 2. b1=T Ka2 3. c1=D Kxa1[+sLf8] 4. Sb6 Kxb1[+sTa8] 5. Te8 Kxc1[+sDd8]+ 6. Ke1 Kc2 7. Sd5 Kd3 8. Se7+ Ke2#
Korrektur in feenschach 1988
Cook: Michel Caillaud (2022-07-19): cooked using Jacobi 0.7.5 :
1. a1=L Kb3 2. b1=T Ka2 3. c1=D Kxa1[+sLf8] 4. Sb6 Kxb1[+sTa8] 5. Te8 Kxc1[+sDd8]+ 6. Ke1 Kc2 7. Sd5 Kd3 8. Se7+ Ke2#
Michel Caillaud: cooked using Jacobi 0.7.5 :
1.a1=L Kb3 2.b1=T Ka2 3.c1=D Kxa1(Lf8) 4.Sb6 Kxb1(Ta8) 5.Te8 Kxc1(Dd8)+ 6.Ke1 Kc2 7.Sd5 Kd3 8.Se7+ Ke2# (2022-07-19)
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1.a1=L Kb3 2.b1=T Ka2 3.c1=D Kxa1(Lf8) 4.Sb6 Kxb1(Ta8) 5.Te8 Kxc1(Dd8)+ 6.Ke1 Kc2 7.Sd5 Kd3 8.Se7+ Ke2# (2022-07-19)
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Keywords: Circe (Rex inklusive), Allumwandlung
Genre: Fairies
FEN: n7/8/4pp2/8/8/2K5/pppk1p2/8
Reprints: The Problemist , p. 91, 09/1985
Rex Multiplex 12/1988
969 Minimalkunst im Schach 2006
My problems [Arno Tüngler] , p. 82, 06/2009
Input: Gerd Wilts, 1996-06-06
Last update: Mario Richter, 2022-07-19 more...
Genre: Fairies
FEN: n7/8/4pp2/8/8/2K5/pppk1p2/8
Reprints: The Problemist , p. 91, 09/1985
Rex Multiplex 12/1988
969 Minimalkunst im Schach 2006
My problems [Arno Tüngler] , p. 82, 06/2009
Input: Gerd Wilts, 1996-06-06
Last update: Mario Richter, 2022-07-19 more...
41 - P0501008
Karl Pohlheim
5883 Schach 12/1968
Ein Sylversterscherz
(1+8) cooked
ser-h=16
b) sGh8 nach h2
sDU=Grashüpfer
Karl Pohlheim
5883 Schach 12/1968
Ein Sylversterscherz
(1+8) cooked
ser-h=16
b) sGh8 nach h2
sDU=Grashüpfer
a) 1. Kg3 2. Kh2 3. Gh1 4. Kg2 5. Gg1 6. Kf2 7. Gf1 8. Ke2 9. Ge1 10. Kd2 11. Gd1 12. Kc2 13. Gc1 14. Kb2 15. Gb1 16. Ka1 Ka3=
b) 1. Kg5 2. Kh6 3. Gh7 4. Kg7 5. Gg6 6. Kf6 7. Gf5 8. Ke5 9. Ke4 10. Kd4 11. Gd3 12. Kc3 13. Gc2 14. Kb2 15. Gb1 16. Ka1 Ka3=
Cook: in b) 1. Ke5 2. Ke6 3. Gd5 4. Kf5 5. Gg4 6. Ke4 7. Ge3 8. Kd3 9. Kc2 10. Kb2 11. Ga8 12. Gb1 13. Ka1 14. Ga3 Kxa3=
b) 1. Kg5 2. Kh6 3. Gh7 4. Kg7 5. Gg6 6. Kf6 7. Gf5 8. Ke5 9. Ke4 10. Kd4 11. Gd3 12. Kc3 13. Gc2 14. Kb2 15. Gb1 16. Ka1 Ka3=
Cook: in b) 1. Ke5 2. Ke6 3. Gd5 4. Kf5 5. Gg4 6. Ke4 7. Ge3 8. Kd3 9. Kc2 10. Kb2 11. Ga8 12. Gb1 13. Ka1 14. Ga3 Kxa3=
HBae: Korrekturvorschlag b) sKf4 nach h7
1.Kh6 2.Gh5 3.Gh7 4.Kg7 ... usw. Eine Prüfung mit Popeye habe ich nach nach 13 Std. abgebrochen. (2021-06-06)
HBae: Mit sKh7 auch NL: 1.Ga8 2.Ga3 3.Kg6 4.Gg5 5.Kf5 6.Ke4 7.Gh4 8.Gh3 9.Gd4 10.Ke3 11.Ge2 12.Kd3 13.Kc3 14.Kb2 15.Gb1 16.Ka1 Kxa3= (2021-06-06)
comment
1.Kh6 2.Gh5 3.Gh7 4.Kg7 ... usw. Eine Prüfung mit Popeye habe ich nach nach 13 Std. abgebrochen. (2021-06-06)
HBae: Mit sKh7 auch NL: 1.Ga8 2.Ga3 3.Kg6 4.Gg5 5.Kf5 6.Ke4 7.Gh4 8.Gh3 9.Gd4 10.Ke3 11.Ge2 12.Kd3 13.Kc3 14.Kb2 15.Gb1 16.Ka1 Kxa3= (2021-06-06)
comment
Keywords: Seriesmover
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 1*2q*2q*2q*2q*2q*2q*2q/8/8/8/K4k2/8/8/8
Input: Gerd Wilts, 1996-06-06
Last update: Vaclav Kotesovec, 2020-07-01 more...
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 1*2q*2q*2q*2q*2q*2q*2q/8/8/8/K4k2/8/8/8
Input: Gerd Wilts, 1996-06-06
Last update: Vaclav Kotesovec, 2020-07-01 more...
1. Da8 2. Kb7 3. Kc6 4. Kd5 5. Ke4 6. Kf3 7. Kg2 8. Kg1 9. Dh1 10. g2 11. Lh2 12. g3 13. Lh3 14. g4 Ke2=
paul: This problem could be done as an ser-auto-stalemate (with inverted colors). (2013-04-30)
Olaf Jenkner: It's not a good idea to add another fairy condition. (2021-03-26)
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Olaf Jenkner: It's not a good idea to add another fairy condition. (2021-03-26)
more ...
comment
Keywords: Seriesmover, Homebase
Genre: Fairies
Computer test: Gustav 4.1d
FEN: 2k5/2bb4/8/6p1/6p1/5qp1/8/4K3
Input: Gerd Wilts, 1996-06-06
Last update: Arnold Beine, 2022-09-14 more...
Genre: Fairies
Computer test: Gustav 4.1d
FEN: 2k5/2bb4/8/6p1/6p1/5qp1/8/4K3
Input: Gerd Wilts, 1996-06-06
Last update: Arnold Beine, 2022-09-14 more...
1. b1=D 2. Db4 3. De1 4. b4 5. b3 6. b2 7. b1=D 8. Db6 9. Dbg1 10. b5 11. b4 12. b3 13. b2 14. b1=D 15. Db8 16. c1=D 17. Dcf4 18. Dff1 19. Tf2 20. f5 21. f4 22. f3 23. Sh1 24. Dbh2 25. Tg3 26. Lh3 27. Dfg2 28. Def1 29. Ke1 Kc2 =
Cook: NL (s.PK137):
von George P.~Sphicas wird folgende von
Mike Neumeier gefundene NL mitgeteilt: 1.Tf2 2.b1\prom{S} 3.Sd2
4.Sf3 5.Sg1 6.b4 9.b1\prom{L} 10.La2 11.Lad5 12.Lf1 13.L5g2 14.b5
18.b1\prom{D} 19.Db4 20.De1 21.f5 23.f3 24.Sh1 25.Tg3 26.Lh3
27.L1g2 28.Df1 29.Ke1 K×c2=
Cook: NL (s.PK137):
von George P.~Sphicas wird folgende von
Mike Neumeier gefundene NL mitgeteilt: 1.Tf2 2.b1\prom{S} 3.Sd2
4.Sf3 5.Sg1 6.b4 9.b1\prom{L} 10.La2 11.Lad5 12.Lf1 13.L5g2 14.b5
18.b1\prom{D} 19.Db4 20.De1 21.f5 23.f3 24.Sh1 25.Tg3 26.Lh3
27.L1g2 28.Df1 29.Ke1 K×c2=
Erich Bartel: Nachdrucke:
1) G29 Fide-Album 1989-1991
2) 975 Minimalkunst im Schach 2006 (2013-03-07)
paul: See P1330761 as correction. (2022-07-10)
comment
1) G29 Fide-Album 1989-1991
2) 975 Minimalkunst im Schach 2006 (2013-03-07)
paul: See P1330761 as correction. (2022-07-10)
comment
Keywords: Promotion (dddd), Rex solus (w), Seriesmover
Genre: Fairies
FEN: 8/1p3p2/5r2/1p6/6p1/3K2nr/1pp1p1b1/3k4
Input: Gerd Wilts, 1996-06-06
Last update: Alfred Pfeiffer, 2013-07-17 more...
Genre: Fairies
FEN: 8/1p3p2/5r2/1p6/6p1/3K2nr/1pp1p1b1/3k4
Input: Gerd Wilts, 1996-06-06
Last update: Alfred Pfeiffer, 2013-07-17 more...
1. f3 Kb8 2. f2 Ka8 3. f1=L Kb8 4. Ld3 Ka8 5. Lb1 Kb8 6. La2 Ka8 7. Db1 Kb8 8. f5 Ka8 9. f4 Kb8 10. f3 Ka8 11. f2 Kb8 12. f1=L Ka8 13. Tf2 Kb8 14. Kf7 Ka8 15. Ke6 Kb8 16. Kd5 Ka8 17. Kc4 Kb8 18. Kc3 Ka8 19. Kb2 Kb8 20. Ka1 Ka8 21. Tb2 Kb8 22. T8f2 Ka8 23. Lf5 Kb8 24. Lc2 Ka8 25. d3 Kb8 26. Le3 Kc7 27. e5 Kxd6 28. Lc1 Ke6 29. Td2 Kf5 30. e4 Kg4 31. Le2+ Kxh3 32. e3 Kg2 33. Led1+ Kf1 34. e2+ Ke1=
Keywords: Rex solus (w), Superseded by (P0501023), Move Length Record
Genre: Fairies
FEN: K1b2rk1/4pr2/1b1n1p2/5q2/3p1p2/1p5n/8/8
Reprints: The Problemist 07/2008
Input: Gerd Wilts, 1996-06-06
Last update: James Malcom, 2022-04-27 more...
Genre: Fairies
FEN: K1b2rk1/4pr2/1b1n1p2/5q2/3p1p2/1p5n/8/8
Reprints: The Problemist 07/2008
Input: Gerd Wilts, 1996-06-06
Last update: James Malcom, 2022-04-27 more...
45 - P0501023
Attila Koranyi
Zdravko Maslar
Sakkelet 1991
Maslar, Version Koranyi
(1+13)
h=36
(h=35.5)
Attila Koranyi
Zdravko Maslar
Sakkelet 1991
Maslar, Version Koranyi
(1+13)
h=36
(h=35.5)
1. ... Kb8 2. Df5 Ka8 3. f3 Kb8 4. f2 Ka8 5. f1=L Kb8 6. Ld3 Ka8 7. Lb1 Kb8 8. La2 Ka8 9. Db1 Kb8 10. f5 Ka8 11. f4 Kb8 12. f3 Ka8 13. f2 Kb8 14. f1=L Ka8 15. Tf2 Kb8 16. Kf7 Ka8 17. Ke6 Kb8 18. Kd5 Ka8 19. Kc4 Kb8 20. Kc3 Ka8 21. Kb2 Kb8 22. Ka1 Ka8 23. Tb2 Kb8 24. T8f2 Ka8 25. Lf5 Kb8 26. Lc2 Ka8 27. d3 Kb8 28. Le3 Kc7 29. e5 Kxd6 30. Lc1 Ke6 31. Td2 Kf5 32. e4 Kg4 33. Le2+ Kxh3 34. e3 Kg2 35. Led1+ Kf1 36. e2+ Ke1=
1.5 moves more then the famous P0501022
1.5 moves more then the famous P0501022
Michel Caillaud: Cooked
In Winchloé, the h=35 is given as Zdravko Maslar, Problem 1958, with a cook :
1.Dh5 Ka8 2.f3 Kb8 3.f2 Ka8 4.f1=L Kb8 5.Ld3 Ka8 6.Lb1 Kb8 7.La2 Ka8 8.f5 Kb8 9.f4 Ka8 10.f3 Kb8 11.f2 Ka8 12.f1=L Kb8 13.Ld3 Ka8 14.Ldb1 Kb8 15.Tf2 Ka8 16.Tb2 Kb8 17.Tff2 Ka8 18.Lçf5 Kb8 19.Lfç2 Ka8 20.d3 Kb8 21.Lé3 Kç7 22.é5 K×d6 23.Lç1 Kd5 24.Td2 Ké4 25.Dd1 Kf5 26.Kh7 Ké6 27.Kh6 Ké7 28.Kh5 Kf7 29.Kg4 Kf6 30.Kf3 Kf5 31.Ké2 Kg4 32.Ké1+ K×h3 33.é4 Kg3 34.é3 Kh3 35.é2 Kg2=
and P0501022 as the correction
It seems Zdravko analyzed the whole thing in due time... (2022-07-26)
more ...
comment
In Winchloé, the h=35 is given as Zdravko Maslar, Problem 1958, with a cook :
1.Dh5 Ka8 2.f3 Kb8 3.f2 Ka8 4.f1=L Kb8 5.Ld3 Ka8 6.Lb1 Kb8 7.La2 Ka8 8.f5 Kb8 9.f4 Ka8 10.f3 Kb8 11.f2 Ka8 12.f1=L Kb8 13.Ld3 Ka8 14.Ldb1 Kb8 15.Tf2 Ka8 16.Tb2 Kb8 17.Tff2 Ka8 18.Lçf5 Kb8 19.Lfç2 Ka8 20.d3 Kb8 21.Lé3 Kç7 22.é5 K×d6 23.Lç1 Kd5 24.Td2 Ké4 25.Dd1 Kf5 26.Kh7 Ké6 27.Kh6 Ké7 28.Kh5 Kf7 29.Kg4 Kf6 30.Kf3 Kf5 31.Ké2 Kg4 32.Ké1+ K×h3 33.é4 Kg3 34.é3 Kh3 35.é2 Kg2=
and P0501022 as the correction
It seems Zdravko analyzed the whole thing in due time... (2022-07-26)
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46 - P0501043
Theodor Steudel
AHK-Memorial 1983-1985
2. Preis
(1+4) C+
ser-h#5
magischer Kb8
b) wKb8 nach g8
Theodor Steudel
AHK-Memorial 1983-1985
2. Preis
(1+4) C+
ser-h#5
magischer Kb8
b) wKb8 nach g8
a) 1. Th7 2. Kh2 3. La8(w) 4. Tb7(w) 5. Kh1 Th7# b) 1. Le4 2. Kg2 3. Th8(w) 4. Lh7(w) 5. Kh1 Le4#
Keywords: Seriesmover, Aristocrat, Miniature, Aristocrat
Genre: Fairies
Computer test: popeye 4.87
FEN: 1K6/8/8/8/8/8/6br/6nk
Input: Gerd Wilts, 1996-06-06
Last update: Arnold Beine, 2022-09-14 more...
Genre: Fairies
Computer test: popeye 4.87
FEN: 1K6/8/8/8/8/8/6br/6nk
Input: Gerd Wilts, 1996-06-06
Last update: Arnold Beine, 2022-09-14 more...
* 1. ... Tc4 2. Ka1 Tc1#
1. Lc3 Ta2 2. Lb2 Ta4 3. La3 Td4 4. Lb4 Td1#
1. Lc3 Ta2 2. Lb2 Ta4 3. La3 Td4 4. Lb4 Td1#
Keywords: Maximummer, no 8x8 board, Aristocrat, Pure Round Trip (L)
Genre: Fairies
FEN: qqqqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq/1B2qqqq/1k2qqqq/2r1qqqq/1K2qqqq
Input: hpr, 1996-07-23
Last update: Alfred Pfeiffer, 2018-12-11 more...
Genre: Fairies
FEN: qqqqqqqq/qqqqqqqq/qqqqqqqq/qqqqqqqq/1B2qqqq/1k2qqqq/2r1qqqq/1K2qqqq
Input: hpr, 1996-07-23
Last update: Alfred Pfeiffer, 2018-12-11 more...
48 - P0502695
John Niemann
192 FEENSCHACH 3 05/1971
(1+0) cooked
Ergänze wB und sK, so daß ein korrektes Hilfsmatt entsteht
John Niemann
192 FEENSCHACH 3 05/1971
(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#
Cook: +wBg7 +sKe8 = h#4
+wBg7 +sKe7,Ske1,Skf2 = h#3.5
1. Ke7 Kg2 2. Kf6 Kf3 3. Kg5 g8=D+ 4. Kh4 Dg4#
Cook: +wBg7 +sKe8 = h#4
+wBg7 +sKe7,Ske1,Skf2 = h#3.5
Duplicate Diagram: P0530804
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...
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...
1) 1. Da7 Ge1 2. Tfb6 Gxc4#
2) 1. Nc8 Gxd8 2. Dd6 Ge1#
3) 1. Tfc6 Gxc4 2. Nd6 Gxd8#
2) 1. Nc8 Gxd8 2. Dd6 Ge1#
3) 1. Tfc6 Gxc4 2. Nd6 Gxd8#
Keywords: Cycle
Pieces: = Grasshopper (G), = Nightrider (N)
Genre: Fairies
Computer test: Alybadix
FEN: 3b*2Q3/2p5/r4rp1/*2Q5p1/2pq2pk/3p2p1/4*2Q*2nK1/8
Input: Torsten Linss, 1996-10-09
Last update: Arnold Beine, 2022-12-30 more...
Pieces: = Grasshopper (G), = Nightrider (N)
Genre: Fairies
Computer test: Alybadix
FEN: 3b*2Q3/2p5/r4rp1/*2Q5p1/2pq2pk/3p2p1/4*2Q*2nK1/8
Input: Torsten Linss, 1996-10-09
Last update: Arnold Beine, 2022-12-30 more...
1) 1. Se4 Ge5 2. Gxe5 Gb5#
2) 1. Ge5 Gc6 2. Txc6 Gd8#
3) 1. Tc6 Ge4 2. Sxe4 Gxd3#
2) 1. Ge5 Gc6 2. Txc6 Gd8#
3) 1. Tc6 Ge4 2. Sxe4 Gxd3#
Keywords: Cycle
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 1*2qr5/4pq2/*2Qp1p3K/2bk1P*2Q1/1prp4/2pn4/5n2/1*2Q6
Input: Torsten Linss, 1996-10-09
Last update: Vaclav Kotesovec, 2023-01-11 more...
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 1*2qr5/4pq2/*2Qp1p3K/2bk1P*2Q1/1prp4/2pn4/5n2/1*2Q6
Input: Torsten Linss, 1996-10-09
Last update: Vaclav Kotesovec, 2023-01-11 more...
51 - P0504716
Miroslav Henrych
6070 Die Schwalbe 108 12/1987
(5+7)
h#3
b) wKd2 nach e2
c) wKd2 nach d7
Miroslav Henrych
6070 Die Schwalbe 108 12/1987
(5+7)
h#3
b) wKd2 nach e2
c) wKd2 nach d7
a) 1. Gc3 Sd6 2. Kd4 dxc4 3. Ge5 Se6#
b) 1. Ge3 dxc4 2. Ke4 Se6 3. Ge5 Sd6#
c) 1. Kd5 Se6 2. Ge7 Sd6 3. Ge5 dxc4#
b) 1. Ge3 dxc4 2. Ke4 Se6 3. Ge5 Sd6#
c) 1. Kd5 Se6 2. Ge7 Sd6 3. Ge5 dxc4#
WB- WS- WS
klären Popeye-Anbindung
Anton Baumann: C+ Alybadix
Informalturnier 1987: Lob (2023-07-15)
comment
klären Popeye-Anbindung
Anton Baumann: C+ Alybadix
Informalturnier 1987: Lob (2023-07-15)
comment
Keywords: Cycle
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 3N4/1p5p/7P/4kN2/2p4p/3P2p1/3K4/4*2q3
Input: Torsten Linss, 1996-10-09
Last update: hpr, 1999-07-10 more...
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 3N4/1p5p/7P/4kN2/2p4p/3P2p1/3K4/4*2q3
Input: Torsten Linss, 1996-10-09
Last update: hpr, 1999-07-10 more...
52 - P0504717
Markus Manhart
54 harmonie 33 31/07/1991
1. Preis
(4+11) C+
h#3
3.1...
wDU,sDU=Grashüpfer
Markus Manhart
54 harmonie 33 31/07/1991
1. Preis
(4+11) C+
h#3
3.1...
wDU,sDU=Grashüpfer
1) 1. Lc4 Ga5 2. Kd4 Gg6 3. Ge4 c3#
2) 1. c4 c3 2. Kd3 Ga5 3. e4 Gg6#
3) 1. Tc4 Gg6 2. Kd5 c3 3. Se4 Ga5#
2) 1. c4 c3 2. Kd3 Ga5 3. e4 Gg6#
3) 1. Tc4 Gg6 2. Kd5 c3 3. Se4 Ga5#
Keywords: Cycle
Pieces: = Grasshopper (G)
Genre: Fairies
Computer test: Popeye 4.85
FEN: 6*2Q1/*2Q5p1/p1p1*2q3/2p1p3/1r2k3/4p3/2P1bn2/2K5
Reprints: B) Rochade Europa , p. 58, 08/1996
Input: Torsten Linss, 1996-10-09
Last update: Vaclav Kotesovec, 2020-09-30 more...
Pieces: = Grasshopper (G)
Genre: Fairies
Computer test: Popeye 4.85
FEN: 6*2Q1/*2Q5p1/p1p1*2q3/2p1p3/1r2k3/4p3/2P1bn2/2K5
Reprints: B) Rochade Europa , p. 58, 08/1996
Input: Torsten Linss, 1996-10-09
Last update: Vaclav Kotesovec, 2020-09-30 more...
1. ... Ge3 2. Ke2 Gg1 3. Kf3 Lh5#
1. Ge2 Ge3 2. Gd2 Gc1 3. Gc2 Lb5#
1. Ge2 Ge3 2. Gd2 Gc1 3. Gc2 Lb5#
Keywords: Echo
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 4B3/8/8/4K3/8/*2Q2k4/5*2q*2q1/8
Input: Markus Manhart, 1998-02-16
Last update: hpr, 1999-11-04 more...
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 4B3/8/8/4K3/8/*2Q2k4/5*2q*2q1/8
Input: Markus Manhart, 1998-02-16
Last update: hpr, 1999-11-04 more...
1. Kh4 nCHh2(=nS) 2. nCHf1(=nL) nCHa6(=nT) 3. e7
nCHg6(=nD) 4. e8=S nCHb1(=nS) break 5. nCHa3(=nL)
nCHf8(=nT) 6. Sg7 nCHa8(=nD) 7. nCHa1(=nS)
nCHc2(=nL) break 8. Kh5 nCHh7(=nT)#
nCHg6(=nD) 4. e8=S nCHb1(=nS) break 5. nCHa3(=nL)
nCHf8(=nT) 6. Sg7 nCHa8(=nD) 7. nCHa1(=nS)
nCHc2(=nL) break 8. Kh5 nCHh7(=nT)#
{ nDR} = Chamäleon
SCHRECKE: Aus der Lösung geht hervor, dass ein wBe6 fehlt, aber das scheint mir hoffnungslos kaputt zu sein. (2023-07-09)
Anton Baumann: mögliche Korrektur: wKh4 +wBe2 +sBc3 nCH-D nach b6 (2+4+1) s#9 Längstzüger
mit der Lösung: 1.e4 nCHg1(=nS)+ 2.nCHe2(=nL) nCHa6(=nT) 3.e5 nCHg6(=nD) 4.e6 nCHb1(=nS) 5.e7 nCHa3(=nL) 6.e8=S nCHf8(=nT) 7.Sg7 nCHa8(=nD) 8.nCHa1(=nS) nCHc2(=nL) 9.Kh5 nCHh7(=nT)#
5. ... nCHd2(=nL)? geht nicht wegen Selbstschach
C+ Alybadix (2023-07-12)
comment
SCHRECKE: Aus der Lösung geht hervor, dass ein wBe6 fehlt, aber das scheint mir hoffnungslos kaputt zu sein. (2023-07-09)
Anton Baumann: mögliche Korrektur: wKh4 +wBe2 +sBc3 nCH-D nach b6 (2+4+1) s#9 Längstzüger
mit der Lösung: 1.e4 nCHg1(=nS)+ 2.nCHe2(=nL) nCHa6(=nT) 3.e5 nCHg6(=nD) 4.e6 nCHb1(=nS) 5.e7 nCHa3(=nL) 6.e8=S nCHf8(=nT) 7.Sg7 nCHa8(=nD) 8.nCHa1(=nS) nCHc2(=nL) 9.Kh5 nCHh7(=nT)#
5. ... nCHd2(=nL)? geht nicht wegen Selbstschach
C+ Alybadix (2023-07-12)
comment
Keywords: Maximummer
Genre: Fairies
FEN: 8/5p2/8/8/5k2/1p5K/*1Q7/8
Input: Ralf Krätschmer, 1998-03-30
Last update: hpr, 1999-11-07 more...
Genre: Fairies
FEN: 8/5p2/8/8/5k2/1p5K/*1Q7/8
Input: Ralf Krätschmer, 1998-03-30
Last update: hpr, 1999-11-07 more...
1. Kb2 Ga8 2. Td7 Gc6 3. Kc3 Gf3 4. Kd4 Ga8 5. Ke5 Gc6 6. Kf6 Gg2 7. Kg7 Ga8 8. Kh8 Gc6 9. Tg7 Gh1#
Anton Baumann: NL: 1.Ta4+ Kb8! 2.h8=D+ Kc7 3.De8 Ga8 4.Dh5 Ga3 5.Tb4 Ga8 6.Tb2 Lh1 7.Dh7+ Kd6,Kd8 8.Dd3+ Ld5 9.Db1 Ga2# (2022-02-02)
HBae: Korrekturvorschlag (Neufassung)
weiß: Ka1 Gd7f4h7 (4) schwarz: Kb7 Gd5e4 (3)
1.Kb2 Ga8 2.Kc3 Gc6 3.Kd4 Gf3 4.Ke5 Ga8 5.Kf6 Gc6 6.Gf7 Gg2 7.Kg7 Ga8 8.Kh8 Gc6 9.Gg7 Gh1 #
C+ Popeye v4.37 (2022-02-03)
comment
HBae: Korrekturvorschlag (Neufassung)
weiß: Ka1 Gd7f4h7 (4) schwarz: Kb7 Gd5e4 (3)
1.Kb2 Ga8 2.Kc3 Gc6 3.Kd4 Gf3 4.Ke5 Ga8 5.Kf6 Gc6 6.Gf7 Gg2 7.Kg7 Ga8 8.Kh8 Gc6 9.Gg7 Gh1 #
C+ Popeye v4.37 (2022-02-03)
comment
Keywords: Maximummer
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 8/kb5P/8/3*2q4/3R*2q3/8/8/K7
Input: Ralf Krätschmer, 1998-03-30
Last update: hpr, 1999-11-07 more...
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 8/kb5P/8/3*2q4/3R*2q3/8/8/K7
Input: Ralf Krätschmer, 1998-03-30
Last update: hpr, 1999-11-07 more...
56 - P0537646
Evgeny A. Vaulin
2092 Probleemblad 1992
(5+14)
h#2
b) CIRCE
c) Madrasi
d) Patrouilleschach
Evgeny A. Vaulin
2092 Probleemblad 1992
(5+14)
h#2
b) CIRCE
c) Madrasi
d) Patrouilleschach
a) 1. Sd6 Lc3 2. Dd5 Le1#
b) 1. Dxe5[+wLc1] Lxe3 2. Tf4 Lf2#
b) 1. Dxe5[+wLc1] Lxe3 2. Tf4 Lf2#
SCHRECKE: a)-d): C+, popeye 4.87
a) 1. Tf4+ Sxg5 2. Te4 Sxf3# (korrigiert)
b) 1. Dxe5[+wLc1] Lxe3 2. Tf4 Lf2#
c) 1. Sd6 Lc3 2. Dd5 Le1#
d) 1. Tf8+ Lf6 2. Lg3 Lxg5# (2023-03-06)
comment
a) 1. Tf4+ Sxg5 2. Te4 Sxf3# (korrigiert)
b) 1. Dxe5[+wLc1] Lxe3 2. Tf4 Lf2#
c) 1. Sd6 Lc3 2. Dd5 Le1#
d) 1. Tf8+ Lf6 2. Lg3 Lxg5# (2023-03-06)
comment
Keywords: Circe, Madrasi, Patrol Chess
Genre: Fairies, h#
FEN: bb6/4K2n/4Brpp/2p1B1qp/2n4k/r3pp1N/4P3/8
Input: Michal Dragoun, 1998-04-07
Last update: HBae, 2017-09-15 more...
Genre: Fairies, h#
FEN: bb6/4K2n/4Brpp/2p1B1qp/2n4k/r3pp1N/4P3/8
Input: Michal Dragoun, 1998-04-07
Last update: HBae, 2017-09-15 more...
a) 1. Dh8 Sf4 2. Dh5 Sxg2[+sBg7]#
b) 1. Lf4 Sxd4[+sTh8] 2. Th5 Sxf3[+sBf7]#
c) 1. Dh8 Sf4 2. Dh5 Sxg2#
d)
b) 1. Lf4 Sxd4[+sTh8] 2. Th5 Sxf3[+sBf7]#
c) 1. Dh8 Sf4 2. Dh5 Sxg2#
d)
klären : Lösung d?
SCHRECKE: a)-d): C+, popeye 4.87
a) 1. Th6+ Sxg5 2. Th5 Sxf3# (korrigiert)
b) 1. Lf4 Sxd4[+sTh8] 2. Th5 Sxf3[+sBf7]#
c) 1. Dh8 Sf4 2. Dh5 Sxg2#
d) 1. Lf4 Kxf6 2. Kh5 Kg5# (2023-03-06)
comment
SCHRECKE: a)-d): C+, popeye 4.87
a) 1. Th6+ Sxg5 2. Th5 Sxf3# (korrigiert)
b) 1. Lf4 Sxd4[+sTh8] 2. Th5 Sxf3[+sBf7]#
c) 1. Dh8 Sf4 2. Dh5 Sxg2#
d) 1. Lf4 Kxf6 2. Kh5 Kg5# (2023-03-06)
comment
Keywords: Circe, Madrasi, Patrol Chess
Genre: Fairies, h#
FEN: q7/3BK3/3PNr2/6b1/3rP2k/4pp2/4P1pP/8
Input: Michal Dragoun, 1998-04-07
Last update: HBae, 2017-09-15 more...
Genre: Fairies, h#
FEN: q7/3BK3/3PNr2/6b1/3rP2k/4pp2/4P1pP/8
Input: Michal Dragoun, 1998-04-07
Last update: HBae, 2017-09-15 more...
a) 1. Db3 cxb3 2. Ld2 Dc2#
b)
b)
Genre: Fairies, h#
FEN: 8/3p4/3Kn3/p1QNP2p/Rb1rk3/p1q2n2/2P1r3/5b2
Input: Michal Dragoun, 1998-04-07
Last update: hpr, 1999-11-09 more...
a) 1. Lc4 Ke3 2. Ke6 Kf4#
b)
b)
Genre: Fairies, h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 3q4/1n3p2/b2r1N2/5kB1/7P/1B3p2/1n3K2/4R3
Input: Michal Dragoun, 1998-04-07
Last update: hpr, 1999-01-03 more...
60 - P0538311
Bernd Ellinghoven
Klaus Wenda
Kjell Widlert
FIDE Tagung 1996
Spezialpreis
(8+10)
h#2
2.1...
b) Pao b3, Vao c2
Bernd Ellinghoven
Klaus Wenda
Kjell Widlert
FIDE Tagung 1996
Spezialpreis
(8+10)
h#2
2.1...
b) Pao b3, Vao c2
a1) 1. Ld3 hxg4 2. Tc4 Sf3#
a2) 1. Td3 b4 2. fxe4 Txe4#
b)
a2) 1. Td3 b4 2. fxe4 Txe4#
b)
Genre: Fairies, h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/4p3/2p1K1p1/2P1RpN1/3kP1p1/1r4pP/1PbP4/2r5
Input: Michal Dragoun, 1998-04-07
Last update: hpr, 1999-11-09 more...
61 - P0539968
Unto Heinonen
470 Suomen Tehtäväniekat 1994
3. ehrende Erwähnung
(5+9) C+
h#2
b) Circe
c) Platzwechselcirce
d) Madrasi
Unto Heinonen
470 Suomen Tehtäväniekat 1994
3. ehrende Erwähnung
(5+9) C+
h#2
b) Circe
c) Platzwechselcirce
d) Madrasi
a) 1. Kd1 Le4 2. Le1 Lxf3#
b) 1. Txg6[+wLf1] d4 2. Txc6[+wTh1] Ld3#
c) 1. fxg6[+wLf7] Ta6 2. Txa6[+wTa1] Lxb3[+sSf7]#
d) 1. Ld1 Tc4 2. Le1 d4#
b) 1. Txg6[+wLf1] d4 2. Txc6[+wTh1] Ld3#
c) 1. fxg6[+wLf7] Ta6 2. Txa6[+wTa1] Lxb3[+sSf7]#
d) 1. Ld1 Tc4 2. Le1 d4#
a) 1. Db1 Dd3 2. Kd2 Sc2#
b) 1. Df4 Dd2 2. Ld3 Se3#
b) 1. Df4 Dd2 2. Ld3 Se3#
63 - P0557149
Franz Pachl
Markus Manhart
Torsten Linss
Andernach TT 05/1996
1. ehrende Erwähnung
(5+2)
h#2
b) sTd8
Antiandernachschach
Franz Pachl
Markus Manhart
Torsten Linss
Andernach TT 05/1996
1. ehrende Erwähnung
(5+2)
h#2
b) sTd8
Antiandernachschach
a) 1. Lxh4 Tf6 (=s) 2. Lg5 (=w) Lxf6#
b) 1. Txd6 Lf8 (=s) 2. Td7 (=w) Txd8#
b) 1. Txd6 Lf8 (=s) 2. Td7 (=w) Txd8#
SCHRECKE: a)+b): C+, popeye 4.87
Lösung b) exakt: 1. Txd6 Ld8(=s) 2. Td7(=w) Txd8# (2023-08-24)
comment
Lösung b) exakt: 1. Txd6 Ld8(=s) 2. Td7(=w) Txd8# (2023-08-24)
comment
Keywords: Anti-Andernach Chess
Genre: Fairies
FEN: 3b3k/8/3R2Q1/3P3K/7B/8/8/8
Input: Franz Pachl, 1998-09-16
Last update: A.Buchanan, 2019-02-04 more...
Genre: Fairies
FEN: 3b3k/8/3R2Q1/3P3K/7B/8/8/8
Input: Franz Pachl, 1998-09-16
Last update: A.Buchanan, 2019-02-04 more...
Anton Baumann: 1. ... Lc4 2.Sb5 Ge6 3.Sa7 La6#
1.Sc6 Ld7 2.Da7 Ge6 3.Sb8 Lc8#
C+ Alybadix (2023-08-22)
comment
1.Sc6 Ld7 2.Da7 Ge6 3.Sb8 Lc8#
C+ Alybadix (2023-08-22)
comment
Keywords: Axis Echo
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: *2Qq*2Q5/nk5K/*2Q2pB3/P7/8/6*2Q1/8/8
Input: Markus Manhart, 1999-01-26
Last update: hpr, 1999-12-04 more...
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: *2Qq*2Q5/nk5K/*2Q2pB3/P7/8/6*2Q1/8/8
Input: Markus Manhart, 1999-01-26
Last update: hpr, 1999-12-04 more...
65 - P0563366
Joel Fridlizius
7558 Tidskrift för Schack 09/1932
F. Hansson gewidmet
(5+6)
h#3
b) wKe3 nach b4
Joel Fridlizius
7558 Tidskrift för Schack 09/1932
F. Hansson gewidmet
(5+6)
h#3
b) wKe3 nach b4
Anton Baumann: a) 1.Ne1 Ng3 2.Nh7 Tg1 3.Kh8 Ne2#
b) 1.Nxa5 Nxc7 2.Ng8 Ta7 3.Kh8 Nb5#
C+ Alybadix (2023-09-04)
comment
b) 1.Nxa5 Nxc7 2.Ng8 Ta7 3.Kh8 Nb5#
C+ Alybadix (2023-09-04)
comment
Keywords: Axis Echo
Pieces: = Nightrider (N)
Genre: Fairies
FEN: 8/1*2nn3k1/8/P7/4p3/1p2K3/5p1P/R4*2N2
Input: Markus Manhart, 1999-01-30
Last update: hpr, 1999-12-04 more...
Pieces: = Nightrider (N)
Genre: Fairies
FEN: 8/1*2nn3k1/8/P7/4p3/1p2K3/5p1P/R4*2N2
Input: Markus Manhart, 1999-01-30
Last update: hpr, 1999-12-04 more...
SCHRECKE: C+, popeye 4.87
1. ... Sb6 2. Gca8 Sd7+ 3. Ka7 Tc7#
1. Ka7 Sc7 2. Gaa8 Sb5+ 3. Kb8 Tb6# (2022-11-30)
comment
1. ... Sb6 2. Gca8 Sd7+ 3. Ka7 Tc7#
1. Ka7 Sc7 2. Gaa8 Sb5+ 3. Kb8 Tb6# (2022-11-30)
comment
Keywords: Axis Echo, Aristocrat
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: Nk*2q5/8/*2q1R5/8/4K3/8/8/8
Input: Markus Manhart, 1999-01-31
Last update: Gunter Jordan, 2022-12-01 more...
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: Nk*2q5/8/*2q1R5/8/4K3/8/8/8
Input: Markus Manhart, 1999-01-31
Last update: Gunter Jordan, 2022-12-01 more...
a) 1. b1=G Ka5 2. Gb5 Sd5 3. Gb8 Sb6#
b) 1. Ga1 Kb6 2. Ga5 Gb5 3. Ga3 b3#
b) 1. Ga1 Kb6 2. Ga5 Gb5 3. Ga3 b3#
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 8/k*2q6/8/8/KN1*2Q4/8/1p6/8
Input: Torsten Linss, 1999-03-02
Last update: hpr, 1999-12-11 more...
a) 1. g5 Sxa3 2. Kc3 Tg4 3. Kb3 Sb5=
b) 1. Tf4 g7 2. Tf8 gxf8=D 3. Sxa3 Dxa3=
b) 1. Tf4 g7 2. Tf8 gxf8=D 3. Sxa3 Dxa3=
Keywords: Model stalemate, Miniature
Genre: Fairies
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/6p1/1N6/1R6/p2k4/8/1K6
Reprints: 1376 Problemista 145-147 10-12/1973
Input: Torsten Linss, 1999-03-02
Last update: Marcin Banaszek, 2021-03-29 more...
Genre: Fairies
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/6p1/1N6/1R6/p2k4/8/1K6
Reprints: 1376 Problemista 145-147 10-12/1973
Input: Torsten Linss, 1999-03-02
Last update: Marcin Banaszek, 2021-03-29 more...
*) 1. ... Te7 2. Ne2 Ne1 3. Ng6 Th7#
1) 1. Na2 Nf2 2. Kh5 Te6 3. Ng5 Th6#
1) 1. Na2 Nf2 2. Kh5 Te6 3. Ng5 Th6#
SCHRECKE: Nebenlösung
1. Na2 Te5 2. Nb4 Tg5 3. Nh7 Nf7# (2023-01-12)
HBae: Correction +wPawn d5 (5+2) C+ Popeye v4.37 (2023-01-12)
comment
1. Na2 Te5 2. Nb4 Tg5 3. Nh7 Nf7# (2023-01-12)
HBae: Correction +wPawn d5 (5+2) C+ Popeye v4.37 (2023-01-12)
comment
Keywords: Reihen-Echo
Pieces: = Nightrider (N)
Genre: Fairies
FEN: 8/8/1K5k/8/8/3*2N4/4R1P1/2*2n5
Input: Ralf Binnewirtz, 1998-10-22
Last update: hpr, 1999-12-12 more...
Pieces: = Nightrider (N)
Genre: Fairies
FEN: 8/8/1K5k/8/8/3*2N4/4R1P1/2*2n5
Input: Ralf Binnewirtz, 1998-10-22
Last update: hpr, 1999-12-12 more...
s) 1. La3 2. Sb5 3. Tc3 4. Kb4 Sd3#
w) 1. ... Tg5 2. Se6 3. Lf4 4. Kf6 Sd5#
w) 1. ... Tg5 2. Se6 3. Lf4 4. Kf6 Sd5#
ANDROMEDA FAIRY CHESS MATCH GREAT BRITAIN v HUNGARY 1993-95
SECTION 1 (Serieshelpmates in 4-8 moves)
more ...
comment
SECTION 1 (Serieshelpmates in 4-8 moves)
more ...
comment
Keywords: Seriesmover
Genre: Fairies
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/6K1/6Pn/p7/k4NR1/2n2r2/1bPB4/8
Input: Torsten Linss, 1999-04-07
Last update: Arnold Beine, 2022-09-14 more...
Genre: Fairies
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/6K1/6Pn/p7/k4NR1/2n2r2/1bPB4/8
Input: Torsten Linss, 1999-04-07
Last update: Arnold Beine, 2022-09-14 more...
1. Sa6 BEe8=S 2. Sc5 BEb8=S#
1. BEe8=T Sa6 2. BEb8=T Sc7#
1. BEe8=T Sa6 2. BEb8=T Sc7#
Pieces: = Berolina Pawn (BE)
Genre: Fairies
FEN: Knn5/*3P2*3P4/1pk5/1p1p4/8/8/8/8
Input: Torsten Linss, 1999-04-07
Last update: Vaclav Kotesovec, 2023-09-14 more...
72 - P0570721
Edgar D. Holladay
7655 Ideal-Mate Review 61 01-03/1996
Ehrende Erwähnung
(2+2) C+
s=5
Längstzüger
Duplex
b) Verschiebung a1 nach c1
Edgar D. Holladay
7655 Ideal-Mate Review 61 01-03/1996
Ehrende Erwähnung
(2+2) C+
s=5
Längstzüger
Duplex
b) Verschiebung a1 nach c1
a) 1. Kc2 Dg1 2. Kb3 Da7 3. De5+ Kc6 4. Dc3+ Dc5 5. Ka4 Dxc3=
a) 1. ... Kc6 Dh8 2. Kb5 Da1 3. De3+ Kc2 4. Dc5+ Dc3 5. Ka4 Dxc5=
b) 1. Dd3+ Ke6 2. Kg4 Da1 3. Dd6+ Kf7 4. Df4+ Df6 5. Kh5 Dxf4=
b) 1. ... Dd5+ Ke2 2. Kg4 Da7 3. Dd2+ Kf1 4. Df4+ Df2 5. Kh3 Dxf4=
a) 1. ... Kc6 Dh8 2. Kb5 Da1 3. De3+ Kc2 4. Dc5+ Dc3 5. Ka4 Dxc5=
b) 1. Dd3+ Ke6 2. Kg4 Da1 3. Dd6+ Kf7 4. Df4+ Df6 5. Kh5 Dxf4=
b) 1. ... Dd5+ Ke2 2. Kg4 Da7 3. Dd2+ Kf1 4. Df4+ Df2 5. Kh3 Dxf4=
Keywords: Aristocrat, Minimal, Miniature, Maximummer
Genre: Fairies
Computer test: Alybadix
FEN: 8/8/8/2qk4/8/2QK4/8/8
Input: Torsten Linss, 1999-04-07
Last update: Alfred Pfeiffer, 2022-01-31 more...
Genre: Fairies
Computer test: Alybadix
FEN: 8/8/8/2qk4/8/2QK4/8/8
Input: Torsten Linss, 1999-04-07
Last update: Alfred Pfeiffer, 2022-01-31 more...
73 - P0574239
Hans-Peter Reich
5365 feenschach 88 07-09/1988
3. ehrende Erwähnung
(10+3) C+
s#6
Circe
Hans-Peter Reich
5365 feenschach 88 07-09/1988
3. ehrende Erwähnung
(10+3) C+
s#6
Circe
1. gxf3[+sBf7] a4 2. Sde6 fxe6[+wSb1] 3. Lg5 e5 4. Lf4 exf4[+wLc1] 5. De3 fxe3[+wDd1] 6. b4 axb3ep[+wBb2]#
Preisbericht feenschach 10/1990 S. 372, Preisrichter Thomas Kolkmeyer
Preisbericht feenschach 10/1990 S. 372, Preisrichter Thomas Kolkmeyer
74 - P0574240
Markus Johannes Ott
Hans-Peter Reich
7434 Die Schwalbe 1991
8. Lob
(5+13) C+
s#2
königliche Grashüpfer a4/d7
Madrasi Rex inclusive
Markus Johannes Ott
Hans-Peter Reich
7434 Die Schwalbe 1991
8. Lob
(5+13) C+
s#2
königliche Grashüpfer a4/d7
Madrasi Rex inclusive
1. Ta5! droht 2. kGd4+ d5#
1. ... d5 2. Tc6+ Lb4#
1. ... kGa7 2. Te5+ a5#
1. ... Gb4 2. Gc6+ Gb3#
1. ... d5 2. Tc6+ Lb4#
1. ... kGa7 2. Te5+ a5#
1. ... Gb4 2. Gc6+ Gb3#
Keywords: Madrasi (Rex inklusive)
Pieces: = Grasshopper (G)
Genre: Fairies
Computer test: SCHRECKE: popeye 4.89
FEN: 4*2Qb2/1*2q1k4/p2p1rp1/4Rp2/K1R4*2q/2*2Qp4/3pp3/3*2q4
Input: hpr, 1999-10-21
Last update: Dieter Berlin, 2024-03-29 more...
Pieces: = Grasshopper (G)
Genre: Fairies
Computer test: SCHRECKE: popeye 4.89
FEN: 4*2Qb2/1*2q1k4/p2p1rp1/4Rp2/K1R4*2q/2*2Qp4/3pp3/3*2q4
Input: hpr, 1999-10-21
Last update: Dieter Berlin, 2024-03-29 more...
1. Sce6 Txg5 2. Dh7 Taf5 3. Txg5[+wTb5] Tbe5#
76 - P0574256
Hans-Peter Reich
Achim Schöneberg
7992 Die Schwalbe 1993
(5+6)
h#2
b) sKd5 nach f8
Strict-CIRCE
Hans-Peter Reich
Achim Schöneberg
7992 Die Schwalbe 1993
(5+6)
h#2
b) sKd5 nach f8
Strict-CIRCE
a) 1. a1=D Lxa1[sDd8] 2. Dd6+ Te5#
b) 1. c1=T Txc1[sTh8] 2. Tg8+ Lg7#
b) 1. c1=T Txc1[sTh8] 2. Tg8+ Lg7#
Anton Baumann: vergl. 'Die Schwalbe' 02/1993: es fehlt sBd3 (sonst viele NL in b))
Steinkontrolle (5+7)
mit sBd3 C+ Alybadix (2023-06-17)
comment
Steinkontrolle (5+7)
mit sBd3 C+ Alybadix (2023-06-17)
comment
*) 1. ... Lg3+ 2. Kg1#
1) 1. Df3+ Lg3? Kg1#
1. ... Lf4 2. Kg1+ Kh2 3. Dh3#
(1. Df1+? Lg3/Lg1, 1. Dg2+? Lg3!)
1) 1. Df3+ Lg3? Kg1#
1. ... Lf4 2. Kg1+ Kh2 3. Dh3#
(1. Df1+? Lg3/Lg1, 1. Dg2+? Lg3!)
Keywords: Aristocrat, Minimal, Miniature
Genre: Fairies
Computer test: SCHRECKE: popeye 4.89
FEN: 8/8/8/8/8/7Q/5K1b/7k
Input: hpr, 1999-10-22
Last update: Dieter Berlin, 2024-03-13 more...
Genre: Fairies
Computer test: SCHRECKE: popeye 4.89
FEN: 8/8/8/8/8/7Q/5K1b/7k
Input: hpr, 1999-10-22
Last update: Dieter Berlin, 2024-03-13 more...
1) 1. Ka1 Le5 2. Da2 Dxd4[wDd1][-wDd4]#
2) 1. Db4 Da8 2. Ka3 Lxa5[wLc1][-wLa5]#
2) 1. Db4 Da8 2. Ka3 Lxa5[wLc1][-wLa5]#
1. Kd5 Kb1 2. Kc4 Lc6 3. Kb3 Kc1 4. Ka2 Lxa8 5. Ka1 Ld5=
Genre: Fairies
FEN: q7/8/4k3/8/B7/8/8/K7
Input: Felber, Volker, 1999-12-20
Last update: hpr, 2000-01-28 more...
*) 1. ... Sf7=
1) 1. Kxh8 Ka4 2. Kg8 Ka3 3. Kh8 Kb2 4. Kg8 Kc1 5. Kh8 Kd1 6. Kg8 Ke1 7. Kh8 Kf2 8. Kg8 Kg3 9. Kh8 Kf4 10. Kg8 Kxe4 11. Kh8 Kd5 12. Kg8 Kxc4 13. Kh8 Kxd3 14. Kg8 Kc4 15. Kh8 d4 16. Kg8 d5 17. Kh8 d6 18. Kg8 d7 19. Kh8 d8=S 20. Kg8 Sf7=
1) 1. Kxh8 Ka4 2. Kg8 Ka3 3. Kh8 Kb2 4. Kg8 Kc1 5. Kh8 Kd1 6. Kg8 Ke1 7. Kh8 Kf2 8. Kg8 Kg3 9. Kh8 Kf4 10. Kg8 Kxe4 11. Kh8 Kd5 12. Kg8 Kxc4 13. Kh8 Kxd3 14. Kg8 Kc4 15. Kh8 d4 16. Kg8 d5 17. Kh8 d6 18. Kg8 d7 19. Kh8 d8=S 20. Kg8 Sf7=
Genre: Fairies
FEN: 5bkN/1p2p1p1/1P2P1P1/KPp1P3/2p1p3/2PpP3/3P4/8
Input: Felber, Volker, 1999-12-20
Last update: hpr, 2000-01-30 more...
81 - P0574405
Tivadar Kardos
Pavlos N. Moutecidis
Feladvanykedvelök Lapja 1970
5. ehrende Erwähnung
(2+5) cooked
ser-h=20
Tivadar Kardos
Pavlos N. Moutecidis
Feladvanykedvelök Lapja 1970
5. ehrende Erwähnung
(2+5) cooked
ser-h=20
Erich Bartel (2008-01-25):
1. Kb2 2. Ka3 3. Kb4 4. Kc4 5. Kd3 6. Ke3 7. Kf2 8. Kf1 9. h1=D 10. Dh8 11. Kf2 12. Ke3 13. Kd3 14. Kc4 15. Kb4 16. Ka3 17. Kb2 18. Kb1 19. Da1 20. b2 Sd4=
Cook: Anton Baumann (2020-11-20):
1. a1=L 2. Ld4 3. Lg1 4. h1=T 5. Th8 6. Le3 7. Lc1 8. Kb2 9. Ta8 10. Ta1 11. Tb1 12. Ka1 13. b2 14. d4 15. d3 16. d2 Sc3=
1. Kb2 2. Ka3 3. Kb4 4. Kc4 5. Kd3 6. Ke3 7. Kf2 8. Kf1 9. h1=D 10. Dh8 11. Kf2 12. Ke3 13. Kd3 14. Kc4 15. Kb4 16. Ka3 17. Kb2 18. Kb1 19. Da1 20. b2 Sd4=
Cook: Anton Baumann (2020-11-20):
1. a1=L 2. Ld4 3. Lg1 4. h1=T 5. Th8 6. Le3 7. Lc1 8. Kb2 9. Ta8 10. Ta1 11. Tb1 12. Ka1 13. b2 14. d4 15. d3 16. d2 Sc3=
Anton Baumann: kürzer, wenn eine Umwandlungsfigur rechts abschirmt:
1.a1=L 3.Lg1 4.h1=T 6.Ta8 8.Lc1 9.Kb2 11.Tb1 12.Ka1 13.b2 16.d2 Sc3= (2020-11-20)
more ...
comment
1.a1=L 3.Lg1 4.h1=T 6.Ta8 8.Lc1 9.Kb2 11.Tb1 12.Ka1 13.b2 16.d2 Sc3= (2020-11-20)
more ...
comment
Keywords: Seriesmover
Genre: Fairies
FEN: 8/8/8/3p4/8/1p6/p3N2p/1k1K4
Input: Felber, Volker, 1999-12-20
Last update: Mario Richter, 2020-11-20 more...
Genre: Fairies
FEN: 8/8/8/3p4/8/1p6/p3N2p/1k1K4
Input: Felber, Volker, 1999-12-20
Last update: Mario Richter, 2020-11-20 more...
*) 1. ... Dh8=
1) 1. Lxh7 Kc1 2. Lg8 Kd1 3. Lh7 Ke1 4. Lg8 Kf2 5. Lh7 Ke3 6. Lg8 Kf4 7. Lh7 Kxg4 8. Lg8 Kf4 9. Lh7 g4 10. Lg8 Lf2 11. Lh7 Lxa7 12. Lg8 Lb8 13. Lh7 a7 14. Lg8 a8=D 15. Lh7 Dh1 16. Lg8 Dh8=
1) 1. Lxh7 Kc1 2. Lg8 Kd1 3. Lh7 Ke1 4. Lg8 Kf2 5. Lh7 Ke3 6. Lg8 Kf4 7. Lh7 Kxg4 8. Lg8 Kf4 9. Lh7 g4 10. Lg8 Lf2 11. Lh7 Lxa7 12. Lg8 Lb8 13. Lh7 a7 14. Lg8 a8=D 15. Lh7 Dh1 16. Lg8 Dh8=
Genre: Fairies
FEN: 4k1b1/p2pPp1Q/P2P1Pp1/6P1/6pB/p2p2P1/P2P4/1K6
Input: Felber, Volker, 1999-12-20
Last update: hpr, 2000-01-30 more...
1. c1=T 2. Tc4 3. Th4 4. c4 5. c3 6. c2 7. c1=T 8. Tcc4 9. Tcf4 10. Ke4 11. d5 12. d4 13. d3 14. d2 15. d1=T 16. Td3 17. Tdf3 18. g1=T 19. Tg5 20. Tgh5 21. g5 22. g4 23. g3 24. g2 25. g1=T 26. T1g5 27. Lg4 28. Tgf5 29. Lg5 30. h6 The8=
Erstdarstellung der fünffachen Turmumwandlung im Ser.H=, vollständig
schlagfrei, auch im weißen Schlußzug. -eb-
Erich Bartel: t/GUW, u/ttttt.---
Nachdrucke:
1) 1064 Ajedrez Magico (28) VII 1971.---
2) feenschach (150) I-III 2003,S.44.--- (2008-07-06)
more ...
comment
schlagfrei, auch im weißen Schlußzug. -eb-
Erich Bartel: t/GUW, u/ttttt.---
Nachdrucke:
1) 1064 Ajedrez Magico (28) VII 1971.---
2) feenschach (150) I-III 2003,S.44.--- (2008-07-06)
more ...
comment
Keywords: Seriesmover, Promotion (konsekutiv ttttt)
Genre: Fairies
FEN: 1KbR3R/3p3p/5ppb/2pkn3/8/7p/2p2PpP/7B
Input: Felber, Volker, 1999-12-20
Last update: Arnold Beine, 2022-09-14 more...
Genre: Fairies
FEN: 1KbR3R/3p3p/5ppb/2pkn3/8/7p/2p2PpP/7B
Input: Felber, Volker, 1999-12-20
Last update: Arnold Beine, 2022-09-14 more...
84 - P0574436
Endre Szentai
Feladvanykedvelök Lapja 1970
1.-2. Preis
(5+1) C+
ser-h=32
b) sKf1 nach h6, +wTg1, ser-h=36
Endre Szentai
Feladvanykedvelök Lapja 1970
1.-2. Preis
(5+1) C+
ser-h=32
b) sKf1 nach h6, +wTg1, ser-h=36
a) 1. Kg1 2. Kh2 3. Kh3 4. Kh4 5. Kg5 6. Kf6 7. Kf7 8. Ke8 9. Kd8 10. Kc7 11. Kb8 12. Ka7 13. Ka6 14. Kb5 15. Kxb4 16. Kb5 17. Ka6 18. Ka7 19. Kb8 20. Kc7 21. Kd8 22. Ke7 23. Kf6 24. Kg5 25. Kh4 26. Kh3 27. Kh2 28. Kg1 29. Kf1 30. Ke1 31. Kd1 32. Kc1 Ke2=
b) 1. Kh5 2. Kh4 3. Kh3 4. Kh2 5. Kxg1 6. Kh2 7. Kh3 8. Kh4 9. Kg5 10. Kf6 11. Kf7 12. Ke8 13. Kd8 14. Kc7 15. Kb8 16. Ka7 17. Ka6 18. Kb5 19. Kxb4 20. Kb5 21. Ka6 22. Ka7 23. Kb8 24. Kc7 25. Kd8 26. Ke7 27. Kf6 28. Kg5 29. Kh4 30. Kh3 31. Kh2 32. Kg1 33. Kf1 34. Ke1 35. Kd1 36. Kc1 Ke2=
b) 1. Kh5 2. Kh4 3. Kh3 4. Kh2 5. Kxg1 6. Kh2 7. Kh3 8. Kh4 9. Kg5 10. Kf6 11. Kf7 12. Ke8 13. Kd8 14. Kc7 15. Kb8 16. Ka7 17. Ka6 18. Kb5 19. Kxb4 20. Kb5 21. Ka6 22. Ka7 23. Kb8 24. Kc7 25. Kd8 26. Ke7 27. Kf6 28. Kg5 29. Kh4 30. Kh3 31. Kh2 32. Kg1 33. Kf1 34. Ke1 35. Kd1 36. Kc1 Ke2=
Keywords: Seriesmover, Aristocrat, Miniature, Aristocrat
Genre: Fairies
Computer test: Gustav 4.1d
FEN: 8/8/1N6/8/1BN1B3/5K2/8/5k2
Input: Felber, Volker, 1999-12-20
Last update: Arnold Beine, 2022-09-15 more...
Genre: Fairies
Computer test: Gustav 4.1d
FEN: 8/8/1N6/8/1BN1B3/5K2/8/5k2
Input: Felber, Volker, 1999-12-20
Last update: Arnold Beine, 2022-09-15 more...
1) 1. De5 Sf4 2. Kf8 Sg6#
2) 1. De3 Sg3 2. Kd6 Sf5#
2) 1. De3 Sg3 2. Kd6 Sf5#
a) 1. Th7 ... 2. Kg7 ... 3. Kh8 ... 4. T1g7 Le5=
b) 1. ... Txh4 h5 2. Kxh3 Th6 3. Txh5 T6h7 4. Th6 Txh6#
b) 1. ... Txh4 h5 2. Kxh3 Th6 3. Txh5 T6h7 4. Th6 Txh6#
Keywords: Twinning by continuation, Stipulation change, Seriesmover
Genre: Fairies, s#
Computer test: Gustav 4.2a
FEN: 1B3kB1/4N1r1/7p/7R/7p/7p/7K/6r1
Input: Felber, Volker, 1999-12-26
Last update: Arnold Beine, 2022-08-17 more...
Genre: Fairies, s#
Computer test: Gustav 4.2a
FEN: 1B3kB1/4N1r1/7p/7R/7p/7p/7K/6r1
Input: Felber, Volker, 1999-12-26
Last update: Arnold Beine, 2022-08-17 more...
a) 1. Sc3 Le4 2. Dd8 Lxd3 3. Se4 Lxd8 4. Kf5 Lg5=
b) 1. Se3 Lf3 2. e4 La5 3. e5 Ld2 4. Dg4+ Lxg4=
b) 1. Se3 Lf3 2. e4 La5 3. e5 Ld2 4. Dg4+ Lxg4=
Genre: Fairies
FEN: B7/3q4/1B2p3/3np2K/5k2/3p4/8/8
Input: Felber, Volker, 1999-12-26
Last update: hpr, 2000-02-13 more...
1. Kf5 2. Sd3 3. Sf4 4. Kg4 5. Kf3 6. Sd5 7. Se3 8. Ke2 9. Kd3 10. Sf5 11. Sd4 12. Kxc4 13. Kd5 14. Sf3 15. Se5 16. Ke6 Se3=
Keywords: Seriesmover, Pure Round Trip (k)
Genre: Fairies
Computer test: Alybadix
FEN: 4N3/4p3/p1P1k1P1/rp2n3/pPP1R3/K7/6N1/8
Reprints: FIDE Album 1965-1967
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-15 more...
Genre: Fairies
Computer test: Alybadix
FEN: 4N3/4p3/p1P1k1P1/rp2n3/pPP1R3/K7/6N1/8
Reprints: FIDE Album 1965-1967
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-15 more...
89 - P0575605
Norbert Ringeltaube
3843 Deutsche Schachzeitung 03/1978
Herbert Grasemann zum 60. Geburtstag gewidmet
(3+7)
h#2
2.1...
Doppel-Längstüger
Norbert Ringeltaube
3843 Deutsche Schachzeitung 03/1978
Herbert Grasemann zum 60. Geburtstag gewidmet
(3+7)
h#2
2.1...
Doppel-Längstüger
1) 1. Db2 e8=L 2. Dh2 Lxc6#
2) 1. Dh8 c8=S 2. Lb8 Sb6#
2) 1. Dh8 c8=S 2. Lb8 Sb6#
Keywords: Maximummer
Genre: Fairies
FEN: k7/2P1P3/K1p3p1/8/8/2p3b1/p6q/8
Input: Felber, Volker, 1999-12-27
Last update: Felber, Volker, 2021-05-19 more...
Genre: Fairies
FEN: k7/2P1P3/K1p3p1/8/8/2p3b1/p6q/8
Input: Felber, Volker, 1999-12-27
Last update: Felber, Volker, 2021-05-19 more...
1. d2 2. d1=D 3. Dxa4 4. Dc4 5. a4 6. a3 7. a2 8. a1=D 9. Dg1 10. Dxg6 11. De8 12. g5 13. g4 14. g3 15. gxh2 16. h1=D 17. Dh3 18. Dhe6 19. h3 20. h2 21. h1=D 22. Dh4 23. Dhd8 24. h4 25. h3 26. h2 27. h1=D 28. Dhh4 29. Dhe7 30. Dc6+ bxc6#
Keywords: Seriesmover, Promotions to queens, konsekutive Umwandlungen 5, Kindergarten Problem
Genre: Fairies
Computer test: Alybadix
FEN: 8/1K1k2p1/3p2Pp/pP5p/P3P2p/2Pp4/7P/8
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-15 more...
Genre: Fairies
Computer test: Alybadix
FEN: 8/1K1k2p1/3p2Pp/pP5p/P3P2p/2Pp4/7P/8
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-15 more...
1. f1=S 2. Sxh2 3. Sg4 4. h2 5. h1=T 6. Th5 7. Tg5 8. h5 9. h4 10. h3 11. h2 12. h1=T 13. Th6 14. Thg6 15. h5 16. h4 17. h3 18. h2 19. h1=T 20. Th8 21. Tb8 22. Txb7 23. Kb8 24. Kc8 25. Kd8 26. Ke7 27. Kf6 28. Kxf5 29. Df6 30. c3 31. c2 32. c1=L 33. Lxb2 34. Le5 35. b2 36. b1=L 37. La2 38. Le6 39. dxc5 40. c4 41. c3 42. c2 43. c1=L 44. Lcf4 e4#
Erich Bartel: Nachdruck:
1) Umwandlungsrekorde I 1985.---:w (2008-01-27)
Anton Baumann: C+ Gustav 4.2a (2023-09-16)
Vaclav Kotesovec: Anton, this is an incredible result. Was your test "brute force" ? Can you also add an elapsed time ? (2023-09-16)
Vaclav Kotesovec: 4 hours later... Since I couldn't believe this fantastic result at first, I tested the problem myself and Gustav tested it in less than 4 hours. Thank you Olaf for the great improvement! (2023-09-16)
comment
1) Umwandlungsrekorde I 1985.---:w (2008-01-27)
Anton Baumann: C+ Gustav 4.2a (2023-09-16)
Vaclav Kotesovec: Anton, this is an incredible result. Was your test "brute force" ? Can you also add an elapsed time ? (2023-09-16)
Vaclav Kotesovec: 4 hours later... Since I couldn't believe this fantastic result at first, I tested the problem myself and Gustav tested it in less than 4 hours. Thank you Olaf for the great improvement! (2023-09-16)
comment
Keywords: Seriesmover
Genre: Fairies
Computer test: Anton Baumann: Gustav 4.2a, V. Kotesovec: Gustav 4.2e (Lösezeit: 03 : 55 : 44,97 h, Hashtables: 100000 MB)
FEN: b7/kNp4p/2Kp3p/1PP2P2/2p5/PpqP3p/1P2Pp1P/8
Reprints: 632 FIDE Album 1977-1979 , p. 126,
Input: Felber, Volker, 1999-12-27
Last update: Vaclav Kotesovec, 2023-09-16 more...
Genre: Fairies
Computer test: Anton Baumann: Gustav 4.2a, V. Kotesovec: Gustav 4.2e (Lösezeit: 03 : 55 : 44,97 h, Hashtables: 100000 MB)
FEN: b7/kNp4p/2Kp3p/1PP2P2/2p5/PpqP3p/1P2Pp1P/8
Reprints: 632 FIDE Album 1977-1979 , p. 126,
Input: Felber, Volker, 1999-12-27
Last update: Vaclav Kotesovec, 2023-09-16 more...
*) 1. d3 2. Lg1 3. Df2 4. Da7 5. Lb6 6. Tc1 7. Tc6 bxc6#
1) 1. ... b6, dann Tc8 2. Kb8 3. Lg2 4. Df3 5. Da8 6. Lb7 7. Tc7+ bxc7#
1) 1. ... b6, dann Tc8 2. Kb8 3. Lg2 4. Df3 5. Da8 6. Lb7 7. Tc7+ bxc7#
Keywords: Seriesmover, Minimal, Turton, Räumung and Bahnung
Genre: Fairies
Computer test: Jacobi
FEN: br6/bk1K4/p3p3/1P6/3p4/8/8/5q1r
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-17 more...
Genre: Fairies
Computer test: Jacobi
FEN: br6/bk1K4/p3p3/1P6/3p4/8/8/5q1r
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-17 more...
93 - P0575717
Emiliano F. Ruth
harmonie 46 06/1996
2. Preis, Abt. A, 8. harmonie-TT 1995/96
(7+3) cooked
1w->ser-h=34
Circe
Kürzestzüger
Emiliano F. Ruth
harmonie 46 06/1996
2. Preis, Abt. A, 8. harmonie-TT 1995/96
(7+3) cooked
1w->ser-h=34
Circe
Kürzestzüger
1. ... Lg1, dann Kg5 2. Kg6 3. Kxf6[+wBf2] 4. Ke6 5. Kd6 6. Kc6 7. Kc5 8. Kb5 9. Kb4 10. Kb3 11. Kb2 12. Kb1 13. Kc1 14. Kd1 15. Ke1 16. Kf1 17. Kxg1[+wLc1] 18. Kf1 19. Ke1 20. Kd1 21. Kxc1 22. Kb1 23. Kb2 24. Kb3 25. Kb4 26. Kb5 27. Kxb6[+wSg1] 28. Kc6 29. Kd6 30. Ke6 31. Kf6 32. Kg6 33. Kg5 34. Kg4 h4=
Cook: 1. ... Sa8, dann g1=T 2. Th1 3. Txh2 4. Th3 5. Th4 6. Kxg3[+wBg2] 7. Kg4 8. Kg5 9. Tg4 10. Tf4 11. Txe4[+wBe2] 12. Tf4 13. Kg6 14. Kxf6[+wBf2] 15. Ke6 16. Ke5 17. Kd5 18. Kxc5[+wLc1] 19. Kd5 20. Ke5 21. Kf5 22. Kg5 23. Kg4 24. Kh4 25. Tf3+ exf3=
Cook: 1. ... Sa8, dann g1=T 2. Th1 3. Txh2 4. Th3 5. Th4 6. Kxg3[+wBg2] 7. Kg4 8. Kg5 9. Tg4 10. Tf4 11. Txe4[+wBe2] 12. Tf4 13. Kg6 14. Kxf6[+wBf2] 15. Ke6 16. Ke5 17. Kd5 18. Kxc5[+wLc1] 19. Kd5 20. Ke5 21. Kf5 22. Kg5 23. Kg4 24. Kh4 25. Tf3+ exf3=
Arnold Beine: Kurzlösung in 25 Zügen:
1.Sa8! (besetzt das Wiedergeburtsfeld des sT, s. letzten Zug), dann g1=T 2.Th1 3.Txh2 4.Th3 5.Th4 6.Kxg3[+wBg2] 7.Kg4 8.Kg5 9.Tg4 10.Tf4 11.Txe4[+wBe2] 12.Tf4 13.Kg6 14.Kxf6[+wBf2] 15.Ke6 16.Ke5 17.Kd5 18.Kxc5[+wLc1] 19.Kd5 20.Ke5 21.Kf5 22.Kg5 23.Kg4 24.Kh4 25.Tf3+ exf3=
Mit dem Zusatz "exakt" in der Forderung wäre die Aufgabe korrekt (C+, Jacobi). (2022-09-20)
comment
1.Sa8! (besetzt das Wiedergeburtsfeld des sT, s. letzten Zug), dann g1=T 2.Th1 3.Txh2 4.Th3 5.Th4 6.Kxg3[+wBg2] 7.Kg4 8.Kg5 9.Tg4 10.Tf4 11.Txe4[+wBe2] 12.Tf4 13.Kg6 14.Kxf6[+wBf2] 15.Ke6 16.Ke5 17.Kd5 18.Kxc5[+wLc1] 19.Kd5 20.Ke5 21.Kf5 22.Kg5 23.Kg4 24.Kh4 25.Tf3+ exf3=
Mit dem Zusatz "exakt" in der Forderung wäre die Aufgabe korrekt (C+, Jacobi). (2022-09-20)
comment
Keywords: Seriesmover, Circe, Minimummer
Genre: Fairies
FEN: 8/8/1N3P2/2B4p/4P1k1/3K2P1/6pP/8
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-20 more...
Genre: Fairies
FEN: 8/8/1N3P2/2B4p/4P1k1/3K2P1/6pP/8
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-20 more...
94 - P0575718
Hans Ott
Hans Peter Rehm
harmonie 46 06/1996
Lob, Abt. A, 8. harmonie-TT 1995/96
(4+8) C+
1w->ser-h#7
2 Lösungen
Hans Ott
Hans Peter Rehm
harmonie 46 06/1996
Lob, Abt. A, 8. harmonie-TT 1995/96
(4+8) C+
1w->ser-h#7
2 Lösungen
1) 1. ... Se8, dann Sxd6 2. Se4 3. Ke3 4. Kf4 5. Kf5 6. Sg5 7. Df4 Sg7#
2) 1. ... Sh5, dann Lf7 2. Kd4 3. Kc5 4. Kxd6 5. Ke6 6. De7 7. Sd6 Sg7#
2) 1. ... Sh5, dann Lf7 2. Kd4 3. Kc5 4. Kxd6 5. Ke6 6. De7 7. Sd6 Sg7#
Keywords: Seriesmover
Genre: Fairies
Computer test: Popeye 4.61
FEN: 8/3p2N1/3P1pb1/4p3/1qn5/3k1B2/1p6/1K6
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-15 more...
Genre: Fairies
Computer test: Popeye 4.61
FEN: 8/3p2N1/3P1pb1/4p3/1qn5/3k1B2/1p6/1K6
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-15 more...
95 - P0575724
Theodor Steudel
harmonie 46 06/1996
3. ehrende Erwähnung, Abt. B, 8. harmonie-TT 1995/96
(6+13)
1w->ser-h=13*
Theodor Steudel
harmonie 46 06/1996
3. ehrende Erwähnung, Abt. B, 8. harmonie-TT 1995/96
(6+13)
1w->ser-h=13*
*) 1. f2 Ta1=
1) 1. ... Ta1, dann Tg2 2. Tde2 3. Kd2 4. Sa2 5. Sb4 6. Sxc6 7. Sb4 8. Sa2 9. Sc1 10. Kd1 11. Td2 12. Tge2 13. f2 c6=
1) 1. ... Ta1, dann Tg2 2. Tde2 3. Kd2 4. Sa2 5. Sb4 6. Sxc6 7. Sb4 8. Sa2 9. Sc1 10. Kd1 11. Td2 12. Tge2 13. f2 c6=
PR Torsten Linß zur 2. ehrenden Erwähnung (P0575722) und dieser Aufgabe: "Im Satz wird ein schwarzes `Eigenpatt' mit einem weißen Tempozug beantwortet. In der Lösung ist Weiß genötigt, sich im Vorschaltzug dieses Tempos zu berauben. Mit langzügigen Manövern stellt Schwarz einen anderen Tempozug bereit. Dabei kommt es zu Rückkehren von 2 bzw. 4 schwarzen Steinen. Die doppelte Rückkehr ist trotz `halben' Inhaltes wegen ihrer Eleganz und Subtilität besser. Der Unterschied, den ein Tempo ausmachen kann, ist gut herausgearbeitet, allerdings hätte mir ein Wechsel im Spiel (z. B. anderes Pattbild) noch mehr zugesagt."
paul: Probably a white Pawn b2 is missing. (2013-05-06)
Anton Baumann: mit wBb2 C+ Gustav 4.2a (Test für alle Züge 1w) (2022-05-30)
Mario Richter: in der mir vorliegenden Version des Preisberichtes (31. Januar 1997) steht im Diagramm kein wB auf b2 (2022-05-31)
comment
paul: Probably a white Pawn b2 is missing. (2013-05-06)
Anton Baumann: mit wBb2 C+ Gustav 4.2a (Test für alle Züge 1w) (2022-05-30)
Mario Richter: in der mir vorliegenden Version des Preisberichtes (31. Januar 1997) steht im Diagramm kein wB auf b2 (2022-05-31)
comment
Keywords: Seriesmover
Genre: Fairies
FEN: 8/2p5/2P5/2P5/6p1/1p1pppPp/2prr2P/1RnkbK2
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-15 more...
Genre: Fairies
FEN: 8/2p5/2P5/2P5/6p1/1p1pppPp/2prr2P/1RnkbK2
Input: Felber, Volker, 1999-12-27
Last update: Arnold Beine, 2022-09-15 more...
96 - P0575920
Unto Heinonen
106 harmonie 39 30/11/1992
1. Preis
(3+9) C+
h#2
b) Madrasi
c) Circe
d) PlatzwechselCIRCE
Unto Heinonen
106 harmonie 39 30/11/1992
1. Preis
(3+9) C+
h#2
b) Madrasi
c) Circe
d) PlatzwechselCIRCE
a) 1. Kb5 Ke2 2. Lc5 Lxd7#
b) 1. Kd3 Ta7 2. Sc4 Txd7#
c) 1. Kb3 Ta2 2. Sxf5[+wLf1] Lc4#
d) 1. Kd5 Lxd7[+sLf5] 2. Lc5 Td4#
b) 1. Kd3 Ta7 2. Sc4 Txd7#
c) 1. Kb3 Ta2 2. Sxf5[+wLf1] Lc4#
d) 1. Kd5 Lxd7[+sLf5] 2. Lc5 Td4#
97 - P0576057
Thomas Maeder
Manfred Rittirsch
harmonie 59 09/1999
3. Platz
Andernach 1999
PR Theodor Tauber + Bernd Ellinghoven + Hans Gruber
(6+11)
h#2
2.1...
Bi-Woozles
Thomas Maeder
Manfred Rittirsch
harmonie 59 09/1999
3. Platz
Andernach 1999
PR Theodor Tauber + Bernd Ellinghoven + Hans Gruber
(6+11)
h#2
2.1...
Bi-Woozles
1) 1. Lh2 Kg3 2. Tg2 Sb2#
2) 1. Tg2 Kf2 2. Lg3 Sd6#
2) 1. Tg2 Kf2 2. Lg3 Sd6#
Thomas Maeder: After the meeting, we found a better setting
(first published on p. 1801 in idee & form
63, July 1989):
Kd5 Qf6 Sf7g6
Kh6 Rg4 Bf8 Sh8g7 Ph7e6g5h5c3g3
1.Tc4 Kd4 2.Lc5 Se7#
1.Lb4 Kc5 2.Tc4 Sf4# (2000-12-27)
Henrik Juel: C+ Popeye 4.61 (2023-02-02)
comment
(first published on p. 1801 in idee & form
63, July 1989):
Kd5 Qf6 Sf7g6
Kh6 Rg4 Bf8 Sh8g7 Ph7e6g5h5c3g3
1.Tc4 Kd4 2.Lc5 Se7#
1.Lb4 Kc5 2.Tc4 Sf4# (2000-12-27)
Henrik Juel: C+ Popeye 4.61 (2023-02-02)
comment
98 - P0576058
Markus Manhart
Thomas Maeder
Andernach TT 05/1999
6. Platz
PR Theodor Tauber + Bernd Ellinghoven + Hans Gruber
(6+6)
h#2
2.1...
Bi-Woozles
Markus Manhart
Thomas Maeder
Andernach TT 05/1999
6. Platz
PR Theodor Tauber + Bernd Ellinghoven + Hans Gruber
(6+6)
h#2
2.1...
Bi-Woozles
1) 1. Dh6 Lg7 2. Kd4 Tg4#
2) 1. Dh5 Tag5 2. Kd5 Lg2#
2) 1. Dh5 Tag5 2. Kd5 Lg2#
99 - P0576059
Colin Peter Sydenham
Andernach TT 05/1999
13. Platz
PR Theodor Tauber + Bernd Ellinghoven + Hans Gruber
(2+2)
h#2
2.1...
Duplex
Mono-Heffalumps
Colin Peter Sydenham
Andernach TT 05/1999
13. Platz
PR Theodor Tauber + Bernd Ellinghoven + Hans Gruber
(2+2)
h#2
2.1...
Duplex
Mono-Heffalumps
s) 1. Kc4 Db1 2. Dc3 Kc5#
w) 1. ... Kd6 Kdd4 2. Dd5 Dd7#
w) 1. ... Kd6 Kdd4 2. Dd5 Dd7#
Keywords: Heffalumps (Mono-), Aristocrat, Minimal (D/d), Miniature
Genre: Fairies
FEN: 8/1Q6/2K5/8/8/3k3q/8/8
Reprints: 15 feenschach 132, p. 64, 06-08/1999
harmonie 59, p. 216-217, 09/1999
Input: Felber, Volker, 1999-12-30
Last update: Gunter Jordan, 2023-02-02 more...
Genre: Fairies
FEN: 8/1Q6/2K5/8/8/3k3q/8/8
Reprints: 15 feenschach 132, p. 64, 06-08/1999
harmonie 59, p. 216-217, 09/1999
Input: Felber, Volker, 1999-12-30
Last update: Gunter Jordan, 2023-02-02 more...
1. Dg8 f4 2. Se5 f5 3. Sg6 fxg6 4. Dh7 gxh7[+sDd8] 5. Da8 h8=D#
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The problems of this query have been registered by the following contributors:
Gerd Wilts (46)hpr (8)
Torsten Linss (9)
Markus Manhart (4)
Ralf Krätschmer (2)
Michal Dragoun (6)
Franz Pachl (2)
Ralf Binnewirtz (1)
Felber, Volker (22)
A.Buchanan: "By 'type Tauber/Caillaud' it is meant that the individual positions may/should be regarded as a retroanalytic overall complex (otherwise no AP legalization would be possible), although they are independent of each other due to the retroanalytic reassessment after each retraction." (2023-06-29)
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