No. 715 HN
Mario Richter: Stellung ist inkorrekt (8 wBB + wUW-L), und die angegebene Lösung paßt auch nicht zu Stellung+Forderung. (2010-01-05)
A.Buchanan: OK there were numerous typos in both the diagram and Henri's solution - which is visible only in edit mode, but it did show the intended plan, and I think I've succeeded in correcting them both. If someone can find Funkschach to check, I would appreciate. This is a nice problem for its time and well worth salvaging. (2020-09-19)
Henrik Juel: Nice that 4.d2xSc3 and 13... b3xDa2 are the only possible uncaptures (2020-09-19)
Mario Richter: Sorry, Henrik, but unless the position is still incorrect, the problem is cooked: there's no need for an unpromotion on a1 and the earlier uncapture of Rook h1 (which would be necessary to provide a tempo move after 11. Sa1-b3):
wCaps: d2xBc3 g6xSh7 f3xSg4
sCaps: a6xSb5xSc4, c7xDd6 and (e.g.) Db6xTg1
R: 1. Td1-a1 f3-f2 2. Ta1-d1 f4-f3 3. Lc1-e3 f5-f4 4. d2xBc3! c4-c3 5. a2-a3 Kb4-a4 6. Tb1-a1 b5xSc4 and so on (2020-09-20)
Henrik Juel: You are right, Mario
The problem is cooked (2020-09-20)
A.Buchanan: Can we have wBc3 to e3, removing wLe3 and sBf2. Add sTe1. Yes there is non-standard material but can work on that. Is it sound? (2020-09-20)
Mario Richter: Yes Andrew, your Suggestion (wBc3 to e3, removing wLe3 and sBf2. Add sTe1) makes the Problem correct. With a white double pawn on the e-file the uncapture of a black pawn is impossible (d2xBauer_e3??), so only the intended solution works! (2020-09-20)
Mario Richter: Re-checking Andrew's correction, I have to correct myself: the problem is still cooked:
wPawnCaps: g6xSh7 d2xLe3 f3xSg4
sPawnCaps: b3xDa2 a5xLb4 f2xTg1=L c7xSd6
sProms: a2-a1=T f2xTg1=L
R: 1. Td1-a1 Tc4xSc7! 2. Td2-d1 Ta1-e1 3. Td1-d2 a2-a1=T 4. Ta1-d1 b3xDa2 5. Db1-a2 b4-b3 6. Dd1-b1 a5xLb4 7. Ld2-b4 Tf4-c4 8. a2-a3 Kb4-a4 9. Lc1-d2 Kc4-b4 10. Tb1-a1 a6-a5 11. Ka5-b6 Da7-a8 12. Ta1-b1 Dd4-a7 13. Ka4-a5 Dh8-d4 14. Sb5-c7 Lb6-d8 15. Sc3-b5 Dd8-h8 16. Sb1-c3 Tf8-f4 17. d2xLe3 Kd5-c4 18. Kb3-a4 Lg1-e3 19. Kc3-b3 f2xTg1=L! (2020-09-23)
A.Buchanan: Well done Mario! Have to say I didn't like my repair so much because of the retro dual R: 1. Ta1-a1 fxLe1=T 2. Le2-e1,Ta1-d1. (2020-09-23)
Frank Müller: Vielleicht besser: dann s#1 (2010-01-06, hidden)
Frank Müller: Diagrammfehler: wLf1 (statt g1) (2010-08-18, hidden)
A.Buchanan: fixed the diagram and stipulation following suggestions here, but the solution is still bogus (2020-09-18, hidden) more ...
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Henrik Juel: C+ Popeye 4.61 (2022-01-26)
Henrik Juel: 1.f7 thr. 2.f8=D hxg6 3.Db5#
1... Kh5/h5/hxg6 2.Df5+/f8=L/f8=S (2022-01-26)
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e018
Sally: Doppelsetzung (Echo) ohne Nebenspiele; das entsprechende Modellmatt lässt sich im vier-Steiner, nicht als reines Echo darstellen. Die Aufgabe ist an sich nur eine von 2 Varianten eines 4- Steiners von Knotek - die Abkürzung strafft aber den Inhalt und ermöglichte es, einen guten Schlüsselzug vorzubauen.
Nr.575 Mattaufg. mit 3 und 4 Steinen I. Teil(Speckmann 1976) (2012-04-11)
SCHRECKE: C+, popeye 4.87 (2022-04-12)
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SCHRECKE: Ohne sBf6 geht nur die 1. Lösung! (2022-03-09)
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Anton Baumann: DL (Zugumstellung) 3. ... f4 4.Sc6 Kd3 5.Le1 f3 ... (2023-05-06)
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SCHRECKE: C+, popeye 4.87 (2022-12-14)
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Böttcher = P1290388
SCHRECKE: C+, popeye 4.87 (2022-04-12)
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Quelle schreibt in der LB vom 05.02.1928: 'Ein Zwilling zu Berhausens s3#.' (P1291078)
SCHRECKE: C+, popeye 4.87 (2022-04-12)
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A.Buchanan: Corrected typo: Tf6g6 are really pawns in the skakbladet diagram (2020-09-18)
Mario Richter: also C+ with rawbats (2020-09-20)
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T.R. Dawson zu dieser Aufgabe und P1375497: "Two interesting combinations of retraction play with helpmating; the elementary exhibition of the PxPe.p. retraction, as in these problems, must soon be abandoned however, I think in favour of greater elaborations."

Originalforderung: Hvidhar trukket sidst. 3 Træk tidligere var det muligt, at Hvid ved Sorts Hjælp kunde gøre Mat i 3 Træk. Hvorledes?
Klarstellung Heft 18 (December 1923) S.211: "For ikke at misforstaans, burde Fordringen lyde '3 hvide Træk tidligere' etc. -"
Walter Lindenthal: Wie wäre es mit einer wDa7:
R 1.Ke3xBd3 Be4xd3ep 2.Bd2-d4 Dd7xDa7+ 3.Da~-a7
& vor 1.wDa7-a8 Ke7-f6 2.Da8-g8 Dd7-e7 3.Dg8-g6# (2020-09-15)
Walter Lindenthal: Korrektur Flüchtigkeitsfehler in der NL:
... & vor 1.wDa~-a8 ... (2020-09-16)
A.Buchanan: Yes Walter something not quite right. Maybe wP missing from diagram. But even then g7-g5 is not forced so how is the White e.p. ok? (2020-09-17)
Walter Lindenthal: It's not ok with today's rules! (That's why my solution avoids it). As far as I understand, the ep-rules were handled differently in that time (1923!), ep had not to be 'forced'. As Dawson (above) seems to hint to that fact ("... must soon be abandoned...") (2020-09-17)
Walter Lindenthal: And I do not think a wP (or any other w piece) is missing mistakenly! The idea seems to be that the wK MUST start to retract into fields EACH threatened twice, so only the (first) ep can R-save the wK. And if there is NO additional wP (or a wQ like mine) then again ONLY the wK can retract. And such he is "forced" into the next threat, where the second ep would R-save him again and finally lead to mate. If there is no second ep (old style 1923!) then there's no mate in h#2.5. So, any (rectractable) additonal w piece would destroy this tactic (like my wQa7 does by eliminating the need for wK's move e3-f4 into the second threat from bPg5). The only point is if this additional w piece can establish a different mate in h#2.5 (as my wQ can). (2020-09-17)
A.Buchanan: Hi Walter, thanks for your responses. I checked skakbladet: the diagram, stipulation and intended solution are accurate. However by 1915 (Dawson & Hunsdorfer "Retrograde Analysis" page 111), the notion of analytic en passant was already established. Moreover, the other problem (P1375497) that Dawson apparently mentioned does *not* contain a forward en passant, so I think he was referring to the simplistic retro logic present in both. Maybe Dawson was being polite in his criticism of the unproven forward e.p. here.
So how about the chess? Is Dd7xDa7+ & Kf4-e3 the only retraction which leads to any #2.5? It seems that other retractions might at least be possible: e.g. S~-b3, T~-c4, D~-a7, and any of them could be any uncapture. Maybe I will look at this later, but I must go now. (2020-09-18)
Walter Lindenthal: Hi Andrew, thanks for the clarifying info! So, the "unproven ep"-issue was decided even before 1923 (I just assumed because of Dawson's comment...), but then the problem would have been broken from start! Of course, some other retractions might be possible, too, but I have not analysed any other one if it could lead to a h#2.5 solution. My only intention with wQa7 was to show that the intended solution was not the only one possible, so it's cooked. (2020-09-18)
Mario Richter: The intended solution is the only possible one (checked with 'rawbats'), if we accept that g7-g5 preceeded the other retractions (perhaps the idea was help-play in the retractions).
The suggested R 1.Ke3xBd3 Be4xd3ep 2.Bd2-d4 Dd7xDa7+?? doesn't work (illegal black pawn structure)! (2020-09-20)
Walter Lindenthal: Hi Mario - GREAT, yes, you're right! The bBe1 is the 'killjoy'... 5+7+4=16, there is no room for ANY additional w stone! thanks! (2020-09-21)
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