"das zeigt die Paradoxie der AP-Bedingung!" (MS). Gleich 1. bxc3 ep.? und dann 2. Kd4 geht natürlich nicht, weil zuvor außer K,T-Zug und c2-c4 auch S-Züge und c3-c4 möglich waren. "Interessanter und genau abgestimmter Lösungsverlauf" (GW) "Sehr schöne und ökonomische Verbindung zwischen konsequentem sh# und AP!" (TB) Mit 'Typ Tauber/Caillaud' ist gemeint, daß die einzelnen Stellungen doch als retroanlytischer Gesamtkomplex betrachtet werden dürfen/sollen (sonst wäre keine AP-Legalisierung möglich), obwohl sie ja aufgrund der retroanalytischen Neubewertung nach jedem Rückzug voneinander unabhängig sind. 6L./5+5P.
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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James Malcom: Solution? (2021-01-18, hidden)
SCHRECKE: 1.Lf1 2.Kh3 3.Kh4 4.Lh3 5.f:g3 e.p. 6.Kh5 7.Kg6 8.Kf7 9.Ke8 10.0-0-0 11.Td7 a8D# (2021-01-18, hidden) more ...
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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)
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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)
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Michel Caillaud: cooked : 2.Kc3 3.Db2 4.d2 5.Ld3 6.d4 Lxb4# (2021-09-14, hidden) more ...
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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)
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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)
A.Buchanan: One question was not raised all those years ago. How does the "series mover" condition affect AP? How does flipping the player change our knowledge of the game history? If it's *consequent* series mover, then the flip erases all history: we use retro analysis to derive any information of the history of the game. But for regular series movers, the history is in some sense retained. Here, White's castling rights in the diagram position are the same as those after 6 single moves by Black. Is this correct? (2022-03-20, hidden) more ...
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James Malcom: Where is the cook? (2021-09-14)
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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)
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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)
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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)
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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)
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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)
Erich Bartel: Lösung: 1.-8.Kf1 9.h1D 10.Dh8 18.Kb1 19.Da1 20.b2 Sd4= (2008-01-25, hidden) more ...
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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)
Joshua Green: This problem is unsound. Here's one cook:
1. Lf4 2. Lxh2 3. Ke4 4. Kf3 5. Kxf2 6. Tg1 7. Ke1 8. g5 9. g4 10. g3 11. Tf4 12. fTf1 13. Sg4 14. Sf2 15. Sxh1 16. f5 17. f4 18. f3 19. f2 20. h5 21. h4 22. d5 23. d4 24. d3 25. d2 26. d1L 27. Le2 28. Ld7 29. Le8 hTxe8= (2022-09-13, hidden)
Arnold Beine: @Joshua: Your given "cook" must be wrong. Perhaps you have tested another position.
... 6.Tg1?? (or 6.g1=T?) ... 11.Tf4???
What about the bPc5 that never moved in your cook? (2022-09-13, hidden)
Joshua Green: @Arnold, sorry. It looks like I was accidentally starting with the position after Black's seventh move in the intended solution. I'm not immediately seeing any way to make the same idea work against the actual position. (2022-09-14, hidden) more ...
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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)
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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)
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Kr4Bzzz: NL: 51. Kf5 56. e1=T 57. Te8 fxe8=D= (2010-04-15)
Mario Richter: Wenn die Forderung lautete "ser-h=57", dann gäbe es genau eine Lösung. (2010-04-28)
Anton Baumann: als ser-h=57 C+ (Gustav 4.2a) (2023-10-30)
SCHRECKE: Ohne den wBh3 geht nur die AL! (2023-10-30)
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Joshua Green: I've been testing some modifications to Popeye with the goal of improving its performance against seriesmovers. Unfortunately, my improved -- assuming there are no bugs -- Popeye cooked this problem, offering:
1. Txd3 2. Th3 3. Kd5 4. Kc4 5. Le5 6. Lf5 7. Lxb1 8. La2 9. b1L 10. Da1 11. Kd3 12. Tc2 13. Lb2 14. La3 15. Tb2 16. Kc2 17. Th2 18. Te2+ Kxe2= (2022-09-13)
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paul: a) 1.Rh2 2.Bh1 3.Rfg2 4.Bf2 5.Rg1 6.Bg2 7.Rgh1 8.Kg1 9.Bf1 10.Rg2 11.Kh2 12.Bg1 13.Rf2 14.Qg2 15.Kh3 16.Rh2 17.Qh1 18.g2 Sf5=
b) 1.Bh3 2.Qg2 3.Rh2 4.Qh1 5.Bg2 6.Rh3 7.Bh2 8.Kg1 9.Bf1 10.Qg2 11.Kh1 12.Bg1 13.Kh2 14.Qh1 15.g2 16.Kg3 17.Rh2 18.Kh3 Sf5=
The judge was H.P.Rehm. (2013-05-06)
Joshua Green: This is C+ according to a special version of Popeye that I've modified to improve performance on ser-h= and ser-h# problems (hopefully without introducing any bugs). (2022-09-15)
A.Buchanan: Hi Joshua: great stuff, thanks! I guess you’ve been working with a test set of previously validated compositions? Do you plan to merge your branch with the main Popeye version? (2022-09-16)
Ian Shanahan: In FIDE Album 1992-1994 (2014-01-20, hidden) more ...
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SCHRECKE: C+, popeye 4.87 (2023-03-23)
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Erich Bartel: im Problemist ist hierzu vermerkt:
Jonathan's Artikel has already apeared, in translation,in the
March 1995 issue of Rochade-Europa, and these originals
should compete in the informal tourney of that magazine.
--------
Die Frage ist nun welches Magazin als Erstqueelle gelten soll. (2011-11-28)
Anton Baumann: NL: 1.d6 2.dxe5 .. 6.e1=D .. 8.Dg8 .. 13.e1=L 14.Lxh4 15.Lg5 16.Lxh6 .. 23.Kh8 24.Lg7 .. 29.h1=T 30.Th7 Sg6# (2020-11-21)
Alfred Pfeiffer: Lösung: 1.Kb2 2.Kc3 3.Kd4 4.K:e5 5.Kf6 6.e5 7.e4 8.e3 9.e2 10.e1L 11.L:h4 12.Lg5 13.L:h6 14.Kg7 15.Kh8 16.Lg7 17.h5 18.h4 19.h3 20.h2 21.h1L 22.L:d5 23.Lg8 24.d5 25.d4 26.d3 27.d2 28.d1T 29.Th1 30.Th7 Sg6#
Eck-zu-Eck-Marsch, drei Exzelsiors und Unterverwandlungen. (2011-11-28, hidden) more ...
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Urversion NL mit wKe4!
Kevin Begley: Correction of 'P0577029' (was this actually published)? (2011-05-18)
paul: Cook: 1.Q×e7(Sd8) 2.K×d8(Sc7) 3.Qf8 4.S×c7(Sa8) 5.Se8 6.Q×g7(Qf8) 7.Qh8 8.Q×f6(Qh8) 9.Qe7 Qf6= (2024-02-28)
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Alfred Pfeiffer: Was ist "Iceberg"? Ist das eine Art Augsburger Schach? (2010-09-29)
Alfred Pfeiffer: "Iceberg" ist Peter Fayers' Version des Augsburger Schachs, bei folgenden drei Unterschiede gelten (Quelle: "http://www.bcvs.ukf.net/augsb.htm"):
E1: Pieces may combine, but not kings, and NOT PAWNS either (not even when moving to promote).
E2: The game-array queen is not separable, and Qs are distinct from (R+B)s.
E3: A pawn may promote to any combination seen in the diagram, or generated during play. (2010-10-11)
A.Buchanan: Link is now at https://www.mayhematics.com/s/augsb.htm. Piece images to make up the diagrams have been lost, but you can get a clue what the diagrams were by clicking in the html. (2019-02-04)
paul: c) 1.Lb2 (2022-08-17, hidden) more ...
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Henrik Juel: After 5.gxf3ep last move could be Ka2-a1 (2021-09-15)
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Henrik Juel: 1.d5-d4-d3-d2-d1B-b3xa2-c4-f1 10.Bg2-h1 12.fxg3ep+?? fxg3# seems to fail, because after 11.Bh1 last move could be g2xh3. So what is the solution? (2003-04-28)
Thierry LE GLEUHER: After 11.Bh1, the last move can't be g2xh3 because of the promoted Bishop h1 imcompatible with the bP structure! (2005-09-25)
Henrik Juel: Why couldn't Black have a promoted bishop at this point in time (where wPa2 and bPd7 are no longer present)? 12 captures by black pawns seem to match the 12 missing white men. (2005-09-26)
Joost de Heer: Besides, after g2-g4, the black bishop must've been promoted too. (2005-09-30)
James Malcom: Henrik, gxh3 is indeed impossible. The last move sequence must go something like 1. Kfg8 Se2+ 2. Kh2 Sg1. There cannot be a knight on h3, cutting off g8, or a queen/rook, cutting off h2. Thus, gxh3 has already been played and g2-g4 is the last move. (2021-09-14)
Mario Richter: Position after 11. Lh1 is 8/5p2/8/5ppp/5pPk/5p1P/5P1K/6nb.
It can be easily shown, that in this position g2xLh3! is indeed a legal retraction.
So the ep-capture is not justified and the problem is cooked!
(Perhaps Thierry forgot, that black pawn d7 "disappeared"? But Joost's argument raises [correctly] the question, why a promoted Bishop on h1 is ok for g2-g4, but not for g2xLh3.) (2021-09-15)
Henrik Juel: So the good guys (Mario, Joost, and me) win over the bad ones (Thierry and James) by 3-2... (2021-09-15)
Henrik Juel: or, less insultingly:
By a majority of 3 to 2 this problem has been deemed cooked (2021-09-15)
James Malcom: Make it 4 to 1: 1. Nc3 Nf6 2. Nd5 Rg8 3. Nxf6+ exf6 4. d4 Ke7 5. Bg5 fxg5 6. Nf3 Kf6 7. Ne5 Bd6 8. Nc6 Bg3 9. hxg3 bxc6 10. Rh4 gxh4 11. b4 Ba6 12. a4 Bd3 13. b5 Bf5 14. b6 h3 15. bxa7 h2 16. a5 Bh3 17. g4 Re8 18. a6 Re5 19. dxe5+ Kg5 20. Qd6 Qe7 21. c4 Qe6 22. Ra3 Qd5 23. cxd5 cxd5 24. Rf3 Nc6 25. e4 Re8 26. a8=Q Nd4 27. a7 cxd6 28. Qxe8 dxe4 29. Qe6 dxe6 30. a8=Q exf3 31. Bd3 Kh4 32. Bf5 exf5 33. Qa5 g5 34. Qd5 h5 35. Qe4 dxe5 36. Qf4 exf4 37. Kf1 h1=B 38. Kg1 Ne2+ 39. Kh2 Ng1 40. gxh3 (2021-09-16)
A.Buchanan: Is it also possible that in the critical position, the last moves were R: 1.gxSh3 exXf345 2.X~? I think I can fix the cook R: 1.gxLh3 nicely. (2021-09-17)
A.Buchanan: Can it be fixed? What’s the point of wPa2 elimination anyway? Move it to d5. Shift bPf3d7 to f6d6. Add bPb7. Then the cook becomes thematic try as we have 5.b1=L 7.Lxd6 8. Lh1 13.d1=L 14.Ldf3 15.e.p. fxg3# The exact capture count in the critical position prevents bPa from having promoted to L on b1. Does this work? (2021-09-16, hidden)
A.Buchanan: 8/1p1p2pp/8/3P2pp/P4pPk/1p5P/Pn3P1K/6n1 shc#13 fixes both gxLh3 & gxSh3 using an additional sS, which I've tried to make as active as possible. Please break it if you can :) (2021-09-17, hidden) more ...
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SCHRECKE: C+, Jacobi v0.7.5 (2020-12-07)
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SCHRECKE: C+, Jacobi v0.7.5 (2020-12-07)
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SCHRECKE: C+, Jacobi v0.7.5 (2020-12-07)
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SCHRECKE: C+, Jacobi v0.7.5 (2020-12-07)
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Version M. Caillaud
SCHRECKE: C+, Jacobi v0.7.5 (2020-12-07)
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wDU, sDU =Grashüpfer
SCHRECKE: C+, popeye 4.61 (2021-11-08)
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paul: Jacobi+
For example, the input for d is:
stip ~1
stip ser-h#5
pieces white Ka8 Pf7 black Ke4 Qf4 Bd3e7 Sd5h5 Pc3d4e5d6g7 (2022-08-13)
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SCHRECKE: C+, popeye 4.87 (2023-03-23)
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Die obige Kurzlösung wurde innerhalb der Lösungsbesprechung in Heft 201, 06/2003, S. 154 angegeben.
paul: Cook: 1.g4 2.Rd3 3.f4 4.Kb3 5.f5 6.gxh5 7.f6 8.h6 9.h7 10.h8=Q 11.Qxh4 12.Qxa4 13.h4 14.Kc3 15.Qxc2 16.a4 17.a5 18.h5 19.a6 20.h6 21.a7 22.f7 23.h7 24.a8=S 25.f8=B 26.h8=R 27.Sb6 28.Sc4 29.Sd2 30.Rh4 31.Rc4 32.Bb4 33.Qb2+ Bxb2# (2022-06-20, hidden) more ...
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No. 11661 HN
A.Buchanan: “Obvious promotion” is intended to refer to something visible in the initial diagram position, not what is reached during series play (2022-09-14)
A.Buchanan: Similarly “non-standard material (2022-09-14)
paul: La solution commence avec 1.d4xc3 e.p.! La source est diagrammes 148/2004. (2011-07-03, hidden)
A.Buchanan: Similarly “non-standard material”, though this is not present in the problem at any point. (2022-09-14, hidden) more ...
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No. 1304 HN
A.Buchanan: Made changes, and checked that all moves comply with the condition. Following the e.p., b4-b5 is legitimate last move. (2021-01-19)
A.Buchanan: I think this is missing Grid Chess fairy condition, and the final mating move Lxb3#. Cute mate. (2021-01-18, hidden) more ...
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No. 1857 HN

Original ohne wBb2, mit wKb1 und sLa3 veröffentlicht, dann aber NL 1. bxc3ep, weil danach W als letzten Zug Kc2-b1 hatte.
In der korrigierten Stellung (03/1988, S.400) hat W nach bxc3ep nie einen letzten Zug, so daß dieser ep-Schlag erst am Ende der Lösung gespielt werden darf.

Außerdem bietet der Autor noch einen Zwilling an: Ks a5/a4, zusätzlich Monchromes Schach, shc#6
('The Problemist' (03/1988, S.400): "Composer's Mono-chromatic setting (1.bc ep.) saves another unit, but has less interest."
paul: Or directly 1.bxc3ep etc. (2011-10-24)
James Malcom: Is this cooked, as paul's comment seems to indicate? (2021-09-14)
Henrik Juel: No, I don't think so, James
After 1.bxc3ep White has no previous move (2021-09-15)
Mario Richter: Paul's comment was obviously referring to the origial position, not to the one now presented in the diagram.
(The germsn comment above reads: Originally published without white pawn b2, with wKb1 and black Bishop a3. But then cooked by 1. bxc3ep, because W has Kc2-b1 as last move [preceeded by c4xb3+]) (2021-09-16)
TBr: Korrekte Forderung: shc#9, Gitterschach (2011-10-21, hidden) more ...
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No. 2142 HN
James Malcom: Due to the implementation of odds castling, the castling animation is off here. Is there any possible fix or circumvention? (2021-09-14)
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{ DL}
Arnold Beine: Bei der Forderung "genau" ist eine Kurzlösung kein Dual. (2022-08-15)
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SCHRECKE: C+, popeye 4.87 (2023-03-23)
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wKR=kgl.Prinzessin sKL=kgl.Bauer
SCHRECKE: a)+b): C+, popeye 4.87 (2022-04-12)
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A.Buchanan: Baffled - what’s going on here? (2023-04-07)
shankar ram: Series Draw in 4 moves (Draw by repetition rule not applicable)! (2023-04-08)
A.Buchanan: Thanks Shankar, I could translate the stipulation. But now the penny has dropped that it means reach a drawing position, not a draw. But normally after ser-goal in n moves, it is formally Black to move. E.g. ser-# in n, or ser-= in n. But here, the rule is broken: it's White to move to sneak a 5th move in. After 4. c6 with Black to move as usual, there is 4. ... Qh3+/Qh7+. So I think it should be ser-draw in 5. I will mark it as a study as well as fairies. (2023-04-08)
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Anton Baumann: C+ Gustav 4.2a (2023-04-21)
Ladislav Packa: Ka7-a1 and ser-=12 C+ by Popeye v4.63 (2023-04-21)
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Anton Baumann: C+ Gustav 4.2a (2023-04-21)
Ladislav Packa: More interesting would be a switchback after Kc7-g3 and ser-=10. (2023-04-21)
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Anton Baumann: C+ Alybadix (2020-11-11)
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Anton Baumann: C+ Alybadix (2023-04-24)
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SCHRECKE: C+, popeye 4.87
1.Kf2 2.Ke3 3.Kd4 4.Kc4 5.Kb5 6.Kb6 7.Ka7 Db7# (2022-05-25)
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Anton Baumann: C+ Alybadix (2020-12-16)
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wKU sKU=kgl. Grashüpfer
Moldenhauer: Computerprüfung: C+ Popeye 4.88 (2022-08-05)
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Erich Bartel: Figuren-Parallelogramm z1b1c3b3 (2021-04-05)
Erich Bartel: Quelle: A6 Problemkiste (106) VIII 1996; (2007-02-09, hidden) more ...
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Anton Baumann: C+ Alybadix (2022-04-02)
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Cf: P1347359
paul: Checked with stip ser-dia 31 forsyt rsbqkbsr/pppppppp/1RR5/BPPP1S2/1QB4K/5PP1/P3P2P/3S4 by Jacobi. (2021-09-27)
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Erich Bartel: Nachdruck:
StrateGems (26) IV-VI 2004 (2012-06-03)
Joshua Green: See P1242245 for a later rendering of this idea. (2022-05-20)
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The white king walks both main diagonals there and back. Fairy
stalemate (Autor)
Anton Baumann: C+ Alybadix (2022-03-25)
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Erich Bartel: Bei der angegebenen Quelle dürfte es sich um einen Nachdruck handeln.
Lt. dem mir vorliegenden Original-Heft ist die Erstquelle:
1894 Europe Echecs (162) VII 1972.---
weiterer Nachdruck:
2) 81 feenschach (16) VIII 1973 (2009-07-27)
SBD: The 1978 second edition of Rice and Dickins The Serieshelpmate gives the BCM source, so perhaps the error crept in by the citation of the secondary source through the book? (2009-08-01)
Anton Baumann: C+ Alybadix (2020-11-17)
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Pawns reborn on their 1st rank have no legal move in this condition (unlike Circe-Parrain).
Henrik Juel: I am not a co-author of this problem, which appeared in The Problemist under Kevin's name only
In the solution section there was no mention of a cook (2021-05-17)
Henrik Juel: Kevin has pointed out a memory lapse on my part
He found a cooked problem by me from 1993 with the same idea and suggested that his 2006 problem should include me as co-author, which I accepted (2022-03-31)
Anton Baumann: "cooked" ist zu entfernen! Was soll denn hier cooked sein? (2022-04-06)
Henrik Juel: I believe that Kevin has found a cook (2022-04-06)
Kevin Begley: Norbert Geissler found this impressive cook (which Popeye does not find).
We have long known of a simple correction (I believe it's fully sound starting with wSf7, as ser-=18, but the circuit would be lost). I had hoped to reveal details of the cook (and the solver) in the publication of a correction, but time slipped away, then I stopped composing for some years. Truth be told, prospects to fully restore the intent were never likely. (2022-04-10)
Kevin Begley: Henrik,
I have an email dated (Mon, Apr 25, 2011, 7:18 AM) from Norbert Geissler to you, Stephen Emmerson, Hans Gruber and myself, confirming that I had proposed "Henrik Juel & Kevin Begley" (and that you had agreed with this arrangement).
In that email, we were both notified of the cook.
You then replied (Apr 25, 2011, 9:29 AM): "I stopped composing for real some time ago, so I am not very disappointed. As I recall, Kevin bears the main responsibility for the problem."
I will forward the entire chain to you, if you wish.
I do bear responsibility for this, and I am sorry I have not yet provided a correction (I stopped composing for a long while myself). I will do my best to make that happen this year. (2022-03-31, hidden) more ...
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Anton Baumann: C+ Alybadix (2020-11-27)
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paul: Cook: 1.a1=B 2.Ka2 3.Ka3 4.b1=S 5.Sd2 6.h3 7.Sf1 8.h1=R 9.Rh2 10.Ra2 11.h2 12.h1=R 13.Rh4 14.Ra4 15.h4 16.h3 17.h2 18.h1=R 19.Rh5 20.Re5 21.h5 22.h4 23.h3 24.h2 25.h1=Q 26.Qh6 27.Sd2 28.Sb3 29.Kb2 30.R4a3 31.a4 32.Qc1+ (2022-06-22, hidden) more ...
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Frank Müller: Korrektur in Springaren 08/1997, Seite 161: ser-Beweispartie in 53 mit wDa8 (2009-11-12)
paul: Author's solution: 1. b4 2. b5 3. b6 4. ba7 5. ab8=B 6. Ba7 7. Be3 8. Bg5 9. f4 10. Kf2 11. Kg3 12. Kh4 13. g3 14. Bh3 15. Bf5 16. h3 17. Rh2 18. Rf2 19. Rf3 20. Ra3 21. Rxa8 22. Ra4 23. Re4 24. a4 25. a5 26. a6 27. a7 28. a8=N 29. Nb6 30. Nxc8 31. Na7 32. Nc6 33. Nxd8 34. Ne6 35. Nxf8 36. Ng6 37. Nxh8 38. Ng6 39. Ne5 40. Ng4 41. Nh6 42. Nxg8 43. Nh6 44. Ng4 45. Ne3 46. Ng2 47. e3 48. Ne2 49. Nec3 50. Na2 51. c3 52. Qa4 53. Qa8# (2010-05-29)
paul: See P1007350 (2010-05-29)
Henrik Juel: I believe that the year should be 1997 (2022-09-30)
Mario Richter: Klären Stellung: In einer ser-BP sollte Weiß 16 Steine haben. (2009-11-12, hidden)
hpr: Hier schient sowieso was schiefgegangen zu sein, denn die Quelle ist auch nicht vollständig. Kann mal jemand bitte nachschauen?! (2009-11-12, hidden)
Frank Müller: Stellung berichtigt und Datum ergänzt. (2009-11-12, hidden) more ...
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Anton Baumann: Lösung gemäss YACPDB (ID=364241):
1.Sf2 2.Sd3 3.f2 4.Lc6 5.Kc5 6.Kc4 7.Ld4 8.La1 9.b2 10.b3 11.Da3 12.b4 13.Txe5 14.Te6 Dg8= (2021-12-29)
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Joost de Heer: 1. gxf8=R 2. Rxf7 3. Rb7 5. f8=R 7. Rfb6 10. f8=R 12. Rxa5 16. f8=R 18. Rc3 23. f8=R 24. Rf2 Rxa5#

5 rook-promotions on the same square, zugzwang-selfmate. (2010-04-05)
SBD: Nice direct return of the f2 pawn as a rook to f2. (2010-04-06)
Joost de Heer: Cooked by Wilfried Seehofer, correction will appear in next SG. (2010-04-25)
Joost de Heer: Correction never appeared, as far as I know:

bPc4 to d4, -bSb3, +bBc3, solution unchanged (apart from 18. Rd3) (2012-07-31)
Joost de Heer: The correction is also cooked with a mate with the white king on a1. (2023-01-10)
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Richter:Frank Müller:
Die Plazierung dieser Aufgabe, die letztendlich ein innerer Kompromiß
ist, bereitete mir die größten Schwierigkeiten. Ich konnte nie das
Gefühl loswerden, daß die zweite Lösung wegen ihrer Ähnlichkeit (aber
ohne vollständige Analogie) lediglich die Kultivierung einer
schemabedingten Nebenlösung ist. Ich gestehe, daß ich bei nur einer
Lösung diese Aufgabe höher eingestuft hätte. Andererseits wollte ich
dieses Problem nicht zu weit rückstufen.
Die Darstellung der dreifachen Bahnung auf der Grundreihe für den
schwarzen König macht den besonderen Reiz der Aufgabe. Hinzu kommt
Schachschutz auf g2 für 10.Kf2 einmal durch den Lf1, zum anderen durch
die Umwandlungsfigur. Hingegen erfolgt der Schachschutz für den wKa8
einmal direkt durch einen Auswahlzug der Umwandlungsfigur (2.Sf3), zum
anderen indirekt durch die Wahl der Umwandlungsfigur (1.g1=D? 3.Dg2+).
Anton Baumann: Bei der 2. Lösung muss es heissen: ... 15.Tg1 16.Lf1 Dxg1= (2021-10-25, hidden) more ...
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Richter:Frank Müller:
Dies ist die Abkürzung der leider nebenlösigen \Nr90 wohl vom gleichen
Autor.
Anton Baumann: Dual: 2.Sd2 3.Kb1 4.Kc2 5.Lb1 6.Ta2 7.Lb2 8.La1 9.b2 10.b3 11.Ta3 12.La2 13.Kb1 Kxd2=
mit 'ZU': 3.Lb1 4.Ta2 5.Kb2 6.Kc2 7.Lb2 8.La1 ... (2021-10-26)
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Anton Baumann: C+ Alybadix (2020-11-25)
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Preisbericht in "http://problem64.beda.cz/silo/bcf_awards_1970_71.pdf"
Anton Baumann: C+ Alybadix (2020-12-02)
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Anton Baumann: Vergl. yacpdb (ID = 360227)
Korrektur Diagramm: sBc4 nach c5 !
Ergänzung Auszeichnung: 1. Preis
Lösung: 1.Lh8 2.Sg7 3.Kf7 4.e5 5.e4 6.exd3 7.d2 8.d1=L 9.Lf3 10.La8 11.Sb7 12.d5 13.d4 14.dxe3 15.e2 16.e1=L 17.Lc3 18.La1 19.b2 20.Ld5 21.Lh1 22.g2 Sxc5 =
1.La1 2.b2 3.Ld5 4.La8! 5.Sb7 ... 8.dxe3 9.e2 10.e1=L 11.Lc3 12.Lh8 13.Sg7 14.Kf7 ... 17.exd3 18.d2 19.d1=L 20.Lf3 21.Lh1 22.g2 Sxc5 = (2022-04-19)
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Korrektur s. P1192112
Anton Baumann: 1.Kg1 2.Kf2 3.Ke1 4.Kd1 5.Kc1 6.Kb2 7.Ka3 8.Kxb4 9.Ka5 10.Kb6 11.Kxc7 12.Kb6 13.Kc5 14.Kxd4 15.Kc5 16.Kb4 17.Ka3 18.Kb2 19.Kc1 20.Kd1 21.Kxe2 22.Kf2 23.Kxg2 24.Kf3 25.Ke4 26.Kd5 27.Kc6 28.Ld5 29.Lg8 30.Kd5 31.Ke6 32.Kf7 33.Kxg7 34.Kf7 35.Ke6 36.Kd5 37.Ke4 38.Kf3 39.Kg2 40.Kh1 Txg8=
Soll inkorrekt sein (??) vergl. P1192112 (2021-01-15)
paul: Cooked by R. McCracken: 8.Kxb4 11.Kxd4 14.Kxd4 19.Kxc2 22.c1=Q 23.Qxf4 24.Qxh4 25.Qf4 26.Qxe5 27.Qe8 28.Kxe2 31.c1=R 32.Rxf1 33.Rg1 36.hxg2 39.f2 40.Kf1 dxe8=R= (2022-03-31)
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Erich Bartel: ND 1) G27 Fide-Album 1989-1991
2) StrateGems (5) I-III 1999
3) A2180 Phenix (75) VI 1999 (2012-06-18)
paul: Obviously: 18.c1=R 19.Rc3 20.Rcg3 (because 20.Sg3 is with check) (2022-06-06, hidden) more ...
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Erich Bartel: zum Vergleich Vorläufer P1110583 (2010-07-15)
Anton Baumann: C+ Alybadix (2020-11-28)
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paul: Pawn h5 is not necessary, I believe. (2022-06-19)
paul: Just have seen, without Ph5 we have an alternative promotion: 1.f8=B 2.Bxd6 3.Ba3 4.d6 5.d7 6.d8=B 7.Bxc7 8.Bxe5 9.c7 10.c8=Q 11.Qxg4 12.Qgg3 13.Qxb3 14.Ka2 15.Ba1 etc (2022-06-19)
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Erich Bartel: weiterer Nachdruck:
2) Umwandlungsrekorde im Serienzüger XII 1989, (Ergänzugen und Tabellen) (2010-07-16)
paul: Compare with P1110715, but here is one more promotion. (2013-11-16)
Anton Baumann: Vorgänger vergl. P1112565 (2022-01-08)
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Erich Bartel: weitere Nachdrucke:
2) 868 Fide-Album 1980-82,
3) Umwandlungsrekorde im Serienzüger XII 1989, (Ergänzugen und Tabellen)
4) 65 Anthologie des bulgarischen Problemschachs 2003 (2010-07-16)
Anton Baumann: C+ Gustav 4.2a (2021-11-19)
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Erich Bartel: Nachdruck:
Die Schwalbe (169) II 1998 (2013-02-26, hidden)
paul: Cooked by 17.Sxa4 18.Sc3 19.Sxe2 20.b7 21.b8=Q 22.Qxb4 23.Qc5 24.b4 25.bxa5 26.a6 27.a7 28.Sf4 29.a8=R 30.Ra3 31.Rd3 exf5# (2022-06-08, hidden)
paul: Take the alternative as a try, because Black has Kf2 (instead exf5#). Sorry. (2022-06-08, hidden)
paul: The problem is however cooked by 17.Scd3 18.b7 19.b8=Q 20.Qxb4 21.Qc5 22.b4 23.bxa5 24.a6 25.a7 26.a8=S 27.Sb6 28.Sxa4 29.Sc3 30.Sxe2 31.Sef4 exf5# (or gxf5#) (2022-06-08, hidden) more ...
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Erich Bartel: Ein Wahnsinnsproblem!!! (2012-06-13)
Arnold Beine: Leider kaputt, denn am Ende kann man das Selbstmatt auch mit einem Batterieschach erzwingen. Die thematisch vergleichbare P1278403 hat aber gute Chancen, korrekt zu bleiben. (2022-10-22)
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paul: Cook: 1.b8=B 2.c8=B 3.Bf4 4.Bh6 5.c7 6.d8=R 7.Kh7 8.d5 9.d6 10.d7 11.Rg8 12.d8=R 13.Be6 14.Rg6 15.d4 16.d5 17.d6 18.d7 19.Rh8 20.d8=S 21.Bg8 22.Sf7 23.c8=Q 24.Qc4 25.Qf1+ KxQ= (2022-07-05, hidden) more ...
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SCHRECKE: a)+b): C+, Gustav 4.2a, Brute Force (2021-12-27)
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paul: Checked by Jacobi with this code:
forsyth 4r3/1PR2P2/3k1B2/2R1p1p1/1Q2p1sp/4p2B/1RppK3/2S1S3
option InitialMoveNumber 0.5
stip ~1
stip ser-h=6

In about 30 min. (2023-06-30)
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Anton Baumann: C+ Alybadix (2021-11-13)
Henrik Juel: C+ Popeye 4.61 also (instantly) (2021-11-13)
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paul: Cook: 1.b1=B 2.Ba2 3.Sdc5 4.d3 5.d2 6.d1=B 7.Bxf3 8.Bxg2 9.f3 10.f2 11.f1=B 12.Bfa6 13.b5 14.Rcb6 15.Ba8 16.g2 17.gxh1=B 18.Bhb7 19.h1=B 20.Bhd5 21.Se4 22.c5 23.Bdc6 24.Bad5 25.c4 26.Kg6 (2022-06-03)
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Erich Bartel: Nachdrucke:
1) A34 Problemkiste (97) II 1995
2) 50500) 2001 Series XII 2001
3) Problemkiste (141) VI 2002 (2010-09-06)
HBae: NL (Kurzlösung) 1.a8=S 4.Sg5 8.Ke7 Kg7,Kh8 sy. Symmetrieachse = a1-h8.
siehe BUB zu JS und PK (2023-01-19)
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paul: Le problème série noire dualistique a été tranché depuis 1989 dans feenschach:
Manfred Seidel & Hans-Peter Reich, feenschach 1989
11 Serienzüge, dann 2 Lösungen (en allemand):
1) 1. a5 2. d5 6. Ra4 7. Fg4 8. Fxe2 9. Fh5 10. Dd7 11. Dg4, puis 1. c3+ Dxd1#
2) 1. d5 2. e5 3. Fh3 4. Dg5 5. Dxg2 9. Rc4 10. Fb4 11. c5, puis 1. e3+ Dxf1#
La version non-dualistique a été résolue par Nicolas Dupont:
1.a5 3.Td6 4.b6 6.Fxe2 7.Ca6 8.Da8 14.Ra4 15.Fg4 16.Df3, puis 1.c3+ Dxd1# (2021-08-25)
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Mario Richter: Illegale Stellung - nicht-erklärbares Schach durch sTh8. (2010-09-25)
Anton Baumann: mögliche Korrektur: sDf6 nach f8, wBd6 nach e5, ser-s#12
1.b8=L 4.e8=S 7.Sxh8 8.Sg6 9.h8=L 11.La7 12.Df3+ Dxf3#
Diese Version ist C+, Gustav 4.1d (2021-08-21)
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paul: Cook: 17.c3 18.c4 19.Sc5 20.Sb3 21.Sa1 22.R4b3 23.Bb4 24.Ba3 25.Sb4 26.Sa2 27.b4 28.Bb5 29.Ba4 30.b5 Kf2= (2023-05-17)
hans: 1.h4 2.h3 3.h2 4.h1=L 5.Le4 6.f3 7.Kf4 8.Ke3 9.Kd2 10.Kd1 11.Dc1 12.Th2 13.Thb2 14.d2 15.Lb1 16.c2 17.Sc3 18.Sa2 19.c3 20.c4 21.Sc5 22.Sb3 23.Sa1 24.T4b3 25.Lb4 26.La3 27.b4 28.Lb5 29.La4 30.b5 Kf2= (2012-05-12, hidden) more ...
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8-mal passive Kontragrashüpferumwandlung (\eb).
HBae: NL (Kurzlösung) 1.E×a3 [Ba2=KG] 2.Eb2 3.Ec2 4.E×b3[Bc2=KG] 5.Ec4 6.E×d3 [Bc4] 7.Ee2 8.Ef2 9.E×e3[Bf2=KG] 10.Ef4 11.E×g3 [Bf4] 12.Eh2 13.E×h3[Bh2=KG]=. [HG]
siehe BuB zu JS und PK (2023-01-19)
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