Ryan McCracken: Cooked...Nf6/Nh6 is one of many move defects. (2001-09-13)
Moldenhauer: Computerprüfung: Cooked Stevio 1.2 1 Sekunde.
Keine Lösung: BP 16.0, BP 16.5.
Beispiel Notation:1.f4 Sc6 2.Sf3 Sf6 3.f5 Tg8 4.g4 Sxg4 5.f6 a5 6.fxe7 a4 7.exd8=S a3
8.Sxb7 axb2 9.Sa5 bxa1=S 10.Lb2 Lb7 11.Lxg7 Lc5 12.Sd4 Se3 13.c3 Sc4 14.Sc2 Lg1
15.Ld4 Tc8 16.Lf2 Sd4 17.Sb3 Lf3.
Die Idee mit den Buchstaben SL im Diagramm finde ich super! (2023-05-20)
A.Buchanan: Yes I like the letter idea, which can only work in German. Do you think this was intentionally a non-unique PG? Given the triumphs of the unique PG genre, such problems do not impress, but it wouldn't be "cooked". Or maybe we should classify all non-unique PGs here as cooked? What do you think? (2023-05-21)
Henrik Juel: Testing gives the result 'cooked'
This problem is correct in the sense that all shortest proof games last 17.0 moves
Still the current C+ is rather misleading
I suggest no label, neither C+ nor cooked (2023-05-21)
Moldenhauer: Ich nehme an, dass die Absicht es Sam Loyd zu widmen, das Ziel war.
Deshalb für diese Ausführung C+. Wenn der Autor selbst angibt das es nur
einen Lösungsweg gibt ist die Aufgabe cooked. NUPG haben andere
Schachproblemfreunde festgestellt und als Schlüsselwort angegeben. Oder irre ich? (2023-05-21)
Olaf Jenkner: Machine translation is very poor as far as problem chess is concerned. (2023-05-23)
A.Buchanan: Hi Moldenhauser. Non-unique PGs are uncooked if 1) the move total is correct, (2) the intended theme appears in every shortest solution. For a problem like this where the theme is the final diagram, the second is vacuous. In 1986 some unique PGs were being already published, and I t can be difficult to discern the intention of the composer but I think this one was intended to be non-unique. I don’t think any non-unique PG can be classed as C+. (2023-05-23)
A.Buchanan: Hallo Moldenhauser, Ich stimme mit Ihnen ein. Dieses alte Problem ist für sich genommen C+: (1) die Anzahl der Züge stimmt, (2) das Thema wird in jeder Partie erreicht. Nicht jedes nicht eindeutige Beweisspartie wird diese beiden Tests bestehen, obwohl der zweite Test in diesem Beispiel inhaltslos ist. Im Jahr 1986 wurden sicherlich einzigartige Beweisspartie entwickelt, und es ist nicht immer klar, ob die Absicht Einzigartigkeit war. Drei neuere Probleme, die Eindeutigkeit erreichen, sind k='einzigartiger Beweisspartie' und K='Buchstabenproblem'. (2023-05-22, hidden) more ...
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Anton Baumann: NL: R: 1.Ke8-d8 g7-g6 V: 1.Th1..6 Le7#
Korrektur in 'Die Schwalbe' 06/1989 S.78: +sBh7 (2022-12-30)
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Michel Caillaud: cooked by Stelvio 0.93 :
1.b4 c5 2.b5 c4 3.b6 c3 4.bxa7 d5 5.e4 d4 6.f4 d3 7.f5 dxc2 8.d4 g5 9.Lf4 c1=L 10.d5 g4 11.f6 g3 12.fxe7 gxh2 13.g4 c2 14.Lg3 Lf4 15.g5 c1=T 16.g6 Tc6 17.Dd3 f5 18.Sd2 h5 19.g7 Tg6 20.0-0-0 Sf6 21.Te1 h4 22.d6 h3 23.g8=L Th4 24.Lb3 L8h6 25.Ld1 Kf7 26.d7 Dh8 27.d8=T Le6 28.Td4 Sbd7 29.e5 Tf8 30.e8=T Kg8 31.Tc8 Lf7 32.e6 b6 33.e7 Lb8 34.e8=S f4 35.Te7 f3 36.Se2 f2 37.Tg1 h1=S 38.Da6 h2 39.Lh3 f1=S 40.Ta4 Sf2 41.Tc2 h1=L... (2022-12-20)
Paulo Roque: obs: 28...S8d7 (2008-11-22, hidden) more ...
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"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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Anton Baumann: Neufassung vergl. P0006288 (2023-01-06)
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Reto: This is now the new record for the longest computer tested SPG. Strategy seeking takes up all the time, playing the 800 strategies on the other hand is done within a few seconds. (2023-08-11)
Reto: With the upcoming v1.6, this is solved by Stelvio in around 20min. The improvement is due to the fact that the cage in the south-west is now detected (either the wRa1 is captured inside the cage or the wBc1 is captured at home) (2023-09-02)
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 29:20:32 Stunden. (hh:mm:ss)
Da NUPG C+
Notation: 1.b4 e5 2.d4 Le7 3.d5 f5 4.d6 h5 5.dxe7 d5 6.b5 Dd6 7.b6 Kd7 8.bxa7 b5
9.c4 b4 10.c5 d4 11.cxd6 c5 12.e8=T Sc6 13.Tf8 c4 14.e4 d3 15.Dg4 d2+ 16.Ke2 c3
17.Kd3 d1=L 18.Sd2 fxg4 19.Tf4 exf4 20.e5 Sd4 21.e6+ Kc6 22.e7 b3 23.e8=T b2
24.Te6 Lb3 25.Th6 Lf7 26.d7+ gxh6 27.d8=T Lce6 28.Td5 Te8 29.Kc4 b1=L 30.a8=T Lh7
31.Ta3 c2 32.Tg3 Se7 33.a4 Thf8 34.a5 Sg6 35.a6 Sh8 36.a7 fxg3 37.a8=T gxh2
38.T8a3 Lhg8 39.Th3 gxh3 40.Se4 Lxd5+
3 schwarze Läufer stehen zum Schluss auf den weißen Felder d5, f7, g8. (2023-09-21)
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Henrik Juel: It is illegal for Black to supplement anything on b6, because [Ta8] was captured in its corner and the other missing black men were captured by white pawns (2016-03-28)
Henrik Juel: ... as wLb3 is a pawn promoted on e8 or g8
Nice type Høeg defensive retractor
Here are some other explanatory comments
In retraction 1 White chooses to move his king back to h3; Black could choose to supplement a black man on g3 (or nothing), but supplementing a pawn is the only way to maintain legality (Kh3 stands in double check from Lc8 and Dh8); again moving Pg3 back to h3 and White supplementing a pawn on g4 is forced (this e.p. case is the only one where the supplementing does no happen on the abandoned square)
In retraction 2 the white retraction is forced, and then moving Kd6 back to d7 to uncheck is illegal because of the double check from Sb6 and Pc6, so Black must uncheck by moving Kd6 back to e6 and White choose to supplement a pawn on the abandoned square
In retraction 3 White chooses to move Pd6 back to e5, forcing another e.p. situation (2023-04-08)
Henrik Juel: The Proca type is easy to define: White and Black alternate retractions, until White can mate with a forward move
The Høeg type is usually defined the same way, except that the other side decides which man (if any) was captured; but this can be detailed as follows:
1. White chooses a man and 'moves it back'
2. Black chooses which man (if any) to 'supplement' on the abandoned square
(only now is the white retraction complete)
3. Black chooses a man and 'moves it back'
4. White chooses which man (if any) to 'supplement' on the abandoned square
(only now is the black retraction complete)
etc. etc. until, following a white retraction, White can mate with a forward move
In tries, Black can ruin the white plan by mating White with a forward move after a black retraction
It goes without saying that the resulting retractions must be legal
'supplement' is my (poor) translation of the danish term 'supplere'; maybe 'add' would be better
'the abandoned square' needs a special interpretation in the e.p. case, which happens twice in this problem
These details may be the cause why new type Høeg defensive retractors are rarely seen, as type Proca is more natural and straightforward (2023-04-08)
A.Buchanan: Thanks Henrik. Yesterday, I went through all the defensive retractors to clear up keywords & genres. There were a very few where the stip did not specify the VRZ Type, and others where Anticirce did not specify Calvet vs Cheylan. The answers are probably obvious to you, and if you want to comment on those, then I will update the stips & keywords.
A more general question: Typ Friedlich appears to be the German for Type Pacific: can we standardize on one? (2023-04-08)
Henrik Juel: Thanks Andrew for enabling me to post my type Høeg spiel once again
Anticirce without specification usually means that both Calvet and Cheylan work
Friedlich is indeed german for Pacific, and as the PDB is a german product, I guess we must live with the present conditions (2023-04-08)
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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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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 63 Sekunden.
Keine Lösung: BP 29.5, BP 30.0. (2023-04-15, hidden) more ...
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Moldenhauer: Ergänzung Stelvio 1.2 C+: Keine Lösung: BP 8.5, 9.0.
10. dxc6ep++ Doppelschach. (2023-05-24, hidden) more ...
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Sally: Eine eindeutige BwP., die lange Zeit den Längenreord hielt und, wenn auch mit langer Verzögerung, wesentlich zum Boom dieser Gattung in den letzten 15 Jahren beigetragen hat. Im-
mer noch eine faszinierende Aufgabe. Nr. KF-13 Schwalbe Okt. 2005. (2011-12-05)
Moldenhauer: Ergänzung: Keine Lösung: BP 40.5, BP 41.0. (2023-05-20)
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Silvio Baier: Der wesentliche thematische Inhalt ist bereits nach 31,5 Zügen erreicht. Bis dahin ist es C+ (Euclide 0.98). (2010-08-04)
James Malcom: Is this fully C+ then? (2021-01-27)
Henrik Juel: No, only the first 31.5 moves are tested OK (2021-01-27)
James Malcom: No Henrik, as in is the entire problem testable. (2021-01-27)
Henrik Juel: I guess that testing the entire problem would take an unreasonably long time (2021-04-06)
A.Buchanan: It might be possible these days: the motivation for stopping at 31.5 is that the promotion theme had been demonstrated by then. But there’s still e.g. bSb8 as random impostor (2021-04-07)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 50:24:34 Stunden.(hh:mm:ss)
Keine Lösung: BP 34.5, BP 35.0. (2023-05-08)
Henrik Juel: Thanks for your patience, Moldenhauer; more than two days... (2023-05-08)
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 26.0, BP 26.5. (2023-05-08, hidden) more ...
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Paulo Roque: obs: Cook: 26...D6b6 (2008-12-24)
paul: See P0001634 as correction. (2010-09-25)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 in 31:54:04 Stunden. (hh:mm:ss).
1 Lösung mit 20 cooks. (2024-01-07)
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Henrik Juel: Dual: ... 4.d4 La6 5.d5 Tc8 6.d6 Tc7 7.dxc7 Db8 8.c8L d6 9.Lf5 ... (2004-02-12)
Moldenhauer: Ergänzung Stelvio 1.2. Keine Lösung BP 19.0, BP 19.5. (2023-05-20, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 25.5, BP 26.0. (2023-04-02)
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paul: Correction of P0001580 (2010-09-25)
Silvio Baier: Mit 30....dc+ ist das C+ (Euclide 0.98, Retractor für den letzten Zug) (2011-07-08)
James Malcom: So this is fully C+? (2021-01-27)
Henrik Juel: No, another problem in 30.0 moves ending with 30. Tc2 dxc4+ is tested OK (2021-01-27)
Silvio Baier: Mit 30....dc+ ist das C+ (Euclide 0.98) (2011-07-08, hidden)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 65:30:51 Stunden. (hh:mm:ss)
Keine Lösung: BP 30.0, BP 30.5. (2023-05-08, hidden) more ...
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A.Buchanan: The unsoundness in this non-unique PG is that 6 promotions (apparently intended) are not forced (2023-05-11)
Moldenhauer: Computerprüfung: C+ Cooked Stelvio 1.11 48 Sekunden.
Keine Lösung: BP 36.0, BP 36.5. (2023-05-10, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stelvio 1.2 18 Sekunden.
Keine Lösung: BP 23.5, BP 24.0. (2023-05-12)
Silvio Baier: C+ (Euclide 0.93) (2009-05-12, hidden) more ...
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Kostas Prentos: A correction was published in Phenix, 2009 (No.186/Pg.7979)
Solution:
1. Sf3 f5 2. Sd4 Kf7 3. Sb5 Kg6 4. Sd6 exd6 5. e4 Le7 6. e5 Lh4 7. e6 Dg5 8. e7 Dg3 9. e8=T Se7 10. Lc4 Tf8 11. 0-0 Tf6 12. f3 Kh5 13. Te1 Th6 14. Lf7+ Sg6 15. T1e6 dxe6 16. Txc8 Sd7 17. Th8 Txh8 18. Le8 (2022-12-08)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.2 00:33:56 Minuten. (hh:mm:ss)
Keine Lösung: BP 17.5, BP 18.0.
Beispiel: 1.Sf3 f5 2.d4 Kf7 3.e4 g5 4.Lc4+ Kg6 5.e5 Lg7 6.e6 Lxd4
7.0–0 Lxb2 8.Dd6 exd6 9.e7 Kh5 10.e8S Se7 11.Te1 Tf8 12.Sxg5 Tf6
13.f3 Th6 14.Lf7+ Sg6 15.Te7 Lf6 16.Se6 dxe6 17.Td7 Lh4 18.Lg5 Dxg5
19.Sg7# (2023-05-30)
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Renny Bosch: This problem together with P0002238 and P0002243 form a lovely triad in each of which the rightmost three white pawns promote to a common target (Q's, R's, or B's) which subsequently get captured. Studying their similarities and differences, and what it took to make each one work, gives one a good appreciation of the art of SPG composition.

To see them all at once, enter
PROBID='P0002238' OR PROBID='P0002241' OR PROBID='P0002243'
into the search window on the home page.  (2008-03-23)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00;56:06 Minuten. (hh:mm:ss)
Keine Lösung: BP 33.0, BP 33.5. (2023-04-19)
Reto: With Stelvio 1.33, solving time drops to 20sec (2023-06-10)
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paul: Correction of P0000067. (2018-03-03)
Reto: The new length record for a fully tested SPG. (2023-09-18)
Olaf Jenkner: How long does it take with one core and how many cores did you use? (2023-09-18)
Reto: I used 15 seekers and 15 players, but not all of them are active all the time. My guess is that it takes about 7-10h single-threaded. (2023-09-19)
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Henrik Juel: So Michel's proof game is not cooked; yet... (2019-09-10)
Reto: This is now the length record for a fully tested SPG. (2023-09-17)
Moldenhauer: Keine Lösung: BP 46.5. (2024-01-05)
paul: Cooked: 1.a3 g5 2.Sh3 a5 3.Sf4 a4 4.Sg6 hxg6 5.Ra2 Rh4 6.h3 Rb4 7.h4 Ra5 8.h5 Bg7 9.Rh4 Be5 10.Rd4 Bh2 11.g3 Kf8 12.Bg2 Kg7 13.Bc6 Kf6 14.Kf1 Qf8 15.Kg2 Qh6 16.Kf3 g4+ 17.Ke4 Qf4+ 18.Kd3 Kg5 19.gxf4+ Kh4 20.Qe1 Kh3 21.Qd1 g3 22.Ra1 g2 23.Qh1 g1=S 24.Qd5 Sf3 25.Qc5 Kg2 26.Qc4 Kf1 27.Qc5 Ke1 28.Qc4 Kd1 29.Qd5 Kxc1 30.Qb3 axb3 31.Ra2 bxa2 32.h6 axb1=S 33.h7 Sc3 34.h8=R Sa4 35.Rh5 Sb6 36.Rb5 Sa8 37.Rb6 Sg5 38.Ra6 Sh7 39.Ra7 b6 40.Rb7 Ra7 41.Bd5 Sa6 42.Rb8 Bb7 43.Ba2 Bd5 44.Bb1 Ba2 45.b3 Kb2 46.Rf8 Ka1 47.Kc3 Bg1.[N. Plaksin, Die Schwalbe 130/1991] (2019-09-09, hidden)
paul: Comparing the obtained positions, it was observed that the pawn in the g6 must be at f6. A mistake by the publishers, but mine! (2019-09-10, hidden) more ...
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Henrik Juel: Cook: -1.a7xFb8=L Kb7xTa8, 1.Ka6 axb8DT#. (2004-04-01)
Mario Richter: vgl. mit P0006207 (2009-12-24)
Adrian Storisteanu: Possible (expensive, 50% more units) fix: Ka8 Bd8 / Kc8 (2+1) -- R: 1.e7xRd8=B Rd7-d8 & 1.Rc7 e8Q#. On the plus side, there is a wK in the picture now. (2016-02-10)
Adrian Storisteanu: See P1389595 (not clear which was first). (2024-03-19)
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Paulo Roque: obs: 23...Kd7 24.T8h3 f6 (2008-12-29)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 184:20:33 Stunden. (hh:mm:ss)
1.d4 Sa6 2.Ld2 Sb4 3.e3 Sf6 4.Ld3 c5 5.Ke2 c4 6.h4 c3 7.h5 cxd2 8.c4 Sc2 9.c5 Sg4
10.c6 e5 11.c7 Lc5 12.dxc5 d6 13.c6 Le6 14.c8=S Dg5 15.Se7 Sh2 16.Sf5 Sf1 17.c7 a5
18.c8=S a4 19.Sce7 a3 20.Sg6 axb2 21.a4 hxg6 22.a5 Th6 23.a6 gxf5 24.a7 Tg6
25.h6 Kd7 26.h7 f6 27.h8=T Sa3 28.T8h3 Th8 29.a8=T Thh6 30.Ta4 Lg8 31.Te4 fxe4
32.f4 Dg4+ 33.Tf3 exf3+
Keine Lösung: BP 32.0 wegen der Retraktion. Da NUPG C+. (2023-09-28)
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Branko Koludrovic: P.S.
The black pawn a4(on the diagramm)came from c7 after capturing the white bishop on b6, then moving to b5 and capturing a white officer on a4. (2010-09-28)
A.Buchanan: I think sBf must have captured on g2 & then promoted on f1 to S, otherwise it’s impossible to unlock the southeast cage (2023-07-11)
A.Buchanan: So that means it’s sound. However I don’t see that it needs to have been sL shielding on f1 (2023-07-11)
A.Buchanan: Can sBf capture to e-file to be captured directly by wBf rather than capture to g2 to promote? If so, then sBf could capture L on a dark square, meaning that last move could have been c6xb5 (2023-07-11, hidden) more ...
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Retro
b) Weiß am Zug
Adrian Storisteanu: Two knight promotions differently motivated.
The technical challenge was in averting the short second solution in a) as set play, in order to keep the two phases nicely apart, and also, of course, preventing black tempi and potential white retro moves. (2000-11-20)
A.Buchanan: Could wBc3 simply be removed for b)? I guess author's preference is to just slide the pawn over, to keep the twins as similar as possible (2023-12-06, hidden) more ...
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Konstruktionsausschreiben 06/1973
Moldenhauer: Computerprüfung: C+ NUPG BP 28.5 cooked in 1 Sekunde.
Keine Lösung: BP 26.5 bis BP 28.0.
Beispiel: 1.Sc3 Sa6 2.Sd5 Tb8 3.b4 Ta8 4.La3 Tb8 5.Db1 Ta8 6.Kd1 Tb8 7.Kc1 Ta8
8.Kb2 Tb8 9.Kc3 Ta8 10.Kd3 Tb8 11.Ke3 Ta8 12.Kf3 Tb8 13.Kg3 Ta8
14.Kh3 Tb8 15.b5 Ta8 16.b6 Sb8 17.Db5 Sc6 18.Da6 bxa6 19.b7 Sh6
20.b8D Tg8 21.Db1 Lb7 22.Dd1 Dc8 23.Sf6+ Kd8 24.Sxg8 Sb8 25.Sf6 Sg8
26.Sd5 Ke8 27.Lc1 Dd8 28.Sc3 Lc8 29.Sb1 (2023-04-02)
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Juha Saukkola:: Solution not unique! (2000-09-29)
Moldenhauer: Computerprüfung: C+ NUPG Stelvio 1.11 00:02:12 Minuten. (hh:mm:ss)
Keine Lösung: BP 29.5, BP 30.0. (2023-04-24)
Moldenhauer: Beispiel Stelvio 1.11: 1.Sc3 Sc6 2.b3 Sf6 3.La3 a5 4.g4 a4 5.Lg2 Ta5 6.h4 Tf5
7.Lc5 axb3 8.Ld5 b6 9.a4 bxc5 10.gxf5 g5 11.hxg5 e5 12.Th4 h5 13.Tc4 h4
14.a5 h3 15.a6 h2 16.a7 h1D 17.Ta6 De4 18.Sf3 Le7 19.Sd4 exd4 20.a8D Se5
21.Te6 dxc3 22.Dc6 dxc6 23.d4 fxe6 24.dxe5 exd5 25.gxf6 Le6
26.fxe6 dxc4 27.fxe7 Ddd3 28.cxd3 Th3 29.Dc2 Te3 30.dxe4 bxc2 31.fxe3 (2023-04-24)
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Moldenhauer: Ergänzung: C+ Stelvio 8 Sekunden.
Keine Lösung: BP 25.5. (2023-06-24)
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Moldenhauer: Computerprüfung: C+ Stevio 1.11 00:49:45 Minuten.(hh:mm:ss)
Keine Lösung: BP20.0, BP 20.5. (2023-03-16)
Henrik Juel: Hello Moldenhauer
Your tests with Stelvio are much appreciated
Did you notice any improvement by version 1.11 over version 1.0? (2023-03-16)
Moldenhauer: Hallo Henrik Juel
Danke! Seit der neuen Version Stelvio 1.11 hat sich die Anzahl der Strategien reduziert.
Beispiel: P1000406 von 6775 auf 4994, was natürlich das Tempo erhöht. Habe ich zufällig bemerkt.
Dieses Beispiel habe ich auch gespeichert. Herr Reto Aschwanden schrieb glaube ich auch darüber.
Bei mir laufen immer mehrere Stelvios gleichzeitig. Vielleicht ist mein Rechner deshalb
gegenüber anderen langsamer. AMD Ryzen 3950X 16 Kerne 3 Jahre alt. (2023-03-16)
Henrik Juel: Thanks, Moldenhauer (2023-03-20)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 5 Sekunden.
Keine Lösung: BP 26.0, BP 26.5. (2023-03-30)
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Henrik Juel: 0... fxe3ep 1.d8S Kxd5 2.0-0-0#. Not -1.f2? and Lg1 is caught. (2004-09-16)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
Henrik Juel: In principle I agree, Vaclav
But PDB does not (yet) allow HC+, so I find it acceptable to use the C+ label, when you also tell the whole story after 'Computer test:' below (2023-08-03)
A.Buchanan: Hopefully Gerd will have more time at some point, and can expand the functionality in this and other areas. In the meantime, engine solving of conventional retros including AP, is in its infancy. Retractor 2 has some effectiveness, but is still basic. And there is nothing that yet grasps the intricacies of castling/ep etc. However AP problems do often contain considerable forward chess, and the C+ tag is very useful to filter out those that have already been solved forwardly, without pretending that these are in any sense fully solved (2023-08-04)
Ladislav Packa: I know the definition of AP, but I don't understand the logic behind it. The move 1.fxe3 e.p. proves that White CAN castling. But the solution (2. ...Rd1#?) claims that castling is MANDATORY. From my point of view, AP is correct when only castling is necessary for the solution and the Rook move would be a dual. (2023-08-04)
A.Buchanan: @Ladislav: I am not sure how to help you. Maybe you can read this page from Retro Corner: https://www.janko.at/Retros/Glossary/APosteriori.htm (2023-08-05)
Ladislav Packa: Andrew, what should the article help me with? I quote the final sentence:
Some people still oppose this rule and argue that it should certainly not be the default convention. (2023-08-05)
Henrik Juel: You could view it this way, Ladislav
h#2 means that it is Black to move, so White made last move
What was last move? A little analysis shows just three possibilities: f2-f4, move by Ta1, or move by Ke1
So normally we cannot assume that last move was f2-f4
But if White can castle, then the last move was f2-f4
So if we could start with 0... 0-0-0, then 1.fxe3ep would be legitimate
AP says that you are allowed to reverse the sequence of events; first do the ep capture, then later legitimize it by castling
Was this helpful? (2023-08-05)
Ladislav Packa: Henrik, you don't have to explain that to me. I've done a few AP issues myself, like P1348357. But that doesn't mean I agree with AP's logic. I already wrote it - the term "you can castle" is applied as "you must castle". But these are only problems where, in addition to castling, the Rook move can also be used, I consider that a dual.
From that point of view, the P1000662 issue is perfectly fine for me. (2023-08-05)
Joost de Heer: AP: By castling, you prove a posteriori that the ep-capture was not just a try but the actual solution. Without castling, the ep-solution just is that: a try.
So: Try 1. fe3 ep e8=S 2. Kd5 Rd1 - but ep capture not allowed, as there is no proof that f2-f4 must've been the last move.
Solution 1. fe3 ep e8=S 2. Kd5 OOO - Now the ep capture was justified because white castled, thereby proving that the last move before the diagram position indeed was f2-f4. (2023-08-06)
Joost de Heer: See e.g. P1052919 : The try is an ep capture which is unjustified. (2023-08-06)
A.Buchanan: OK Ladislav: I think I get your point. If one solution with castling justifies the e.p., then based on that certainty, why shouldn't an alternative solution with no castling *then* be allowed as well? There are problems in which one twin shows 0-0-0 and the other shows 0-0. Each is based on the other in a similar way, so the idea of dependency is not new. Why are we not allowed to add other "parasitic" solutions as well? Why can we only have the "paying" solutions? We can't say that we are restricted to one solution: that's not the way chess problems operate! And this is just in the help world - in the adversarial world it might get even more complicated. Is this your issue, Ladislav? (2023-08-07)
Ladislav Packa: I don't want to unnecessarily prolong this discussion. However, I will add one more note: in this position, white castling is also possible without e.p. in Black's 1st move. If B1 were an indifferent move, then white can 1...0-0-0! The Codex of Chess Composition writes about it in Article 16 (1):
Castling convention. Casting is permitted unless it can be proven that it is not permissible.
In our case, 0-0-0 is possible because White's last move exists - e2-e4! It does not matter if it is this move or some a2-a3, both moves are equivalent. EP does not prove the possibility of casting, it would be legal even without it. (2023-08-07)
A.Buchanan: Ladislav was what I wrote your issue pls? Y/N :-) (2023-08-07)
Ladislav Packa: I have no problem, I'm just expressing my own opinion about the AP convention. (2023-08-07)
A.Buchanan: OK cos I think the point I raised is a real one that should be addressed by theory some day. Clearly from the nice problem that you composed Ladislav you understand the mechanics very well. From a justification perspective it's all a bit iffy, but that's why it's controversial. Under RS it's really the only way one can end up actually eping, and it's proved compositionally fertile. So that's enough to justify (2023-08-08)
Ladislav Packa: No need to apologize. This is a normal discussion with different views on the issue. Maybe it will come to some conclusion.
I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p. (2023-08-08)
Joost de Heer: "I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p."
You misinterpret AP. The e.p. capture does not authorize castling, castling provides a justification later on (hence the 'a posteriori') for the legality of ep.
Usually, for ep justification you need to examine all game trees that lead to the diagram, and only if all game trees end with the double-step, then ep is allowed.
With AP, you examine all the game trees including the actual play. If all those game trees have as last move before the diagram position the double step, then ep is possible.
In this case, if white doesn't castle, then there are game trees which don't have as last move the double step, and therefore AP logic dictates that the ep capture was illegal. However, all game trees which lead to the diagram and which have castling in the actual play have as last move before the diagram position the double step, hence AP dictates that the ep capture is legal. (2023-08-09)
Ladislav Packa: Joost: A simple question - is white allowed to castle after any 1st move by black (except e.p.)? (2023-08-09)
Joost de Heer: Of course he is. AP only is used to combine the ep justification with castling, not the castling right per se. (2023-08-09)
A.Buchanan: I agree with both. Also "cooked" is closer to "unsound". And we choose not to have "C+ cooked", which might legitimately mean "C-" (2023-08-04, hidden) more ...
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Henrik Juel: Natch 3.1 made no progress after half an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:03:45 Minuten. (hh:mm:ss)
Keine Lösung: BP 23.0, BP 23.5. (2023-04-01, hidden) more ...
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Henrik Juel: Natch 3.1 made no progress after half an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 27:15:46 Stunden.
Keine Lösung: BP 21.5, BP 22.0. (2023-02-18)
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Retro!-patt, Aufspaltung Doppelschritt!, Excelsior, Umwandlung
Erich Bartel: Nachdruck:
1) 365 Minimalkunst im Schach 2006.--- (2008-03-08, hidden)
A.Buchanan: Currently, the stip is "h#5 0.2.1..." Really not a fan of the "0." usage. Normally, I would prefer "h#4.5 2.1..." but here it should be "h#5 2.1...". (2023-12-07, hidden) more ...
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Achim Schöneberg: Vergleiche P1414844 (2024-02-07)
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t/Allumwandlung, u/61112, u/SDlt
Vorläufer: A.C.White,1930, s.Nr.1056
s. A. C. White 364v L'Echiquier 07-08/1930
Anton Baumann: Die Aufgabe stammt gem. 'Die Schwalbe' 05/1963 von André Simonet, Brüssel
Vorgänger vergl. P0542142 (2022-11-16)
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t/Allumwandlung, u/61113, u/DTsl
Yuri Bilokin: you can see 1...g8=Q 2.cxb1=B Qb3#
1...g8=Q 2.cxb1=S Qb3#
1...Bxc2 2.Ka2 Ra4# (2024-02-29)
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Yuri Bilokin: correction wKa7-b1, +bNd1 7k/5p1b/3p4/8/8/8/1Pn5/1K1n4 (2+6) h#7

1.f5 b4 2.f4 b5 3.f3 b6 4.f2 b7 5.f1=B b8=B 6.Bc4 Bxd6 7.Bcg8 Be5# (MM)

Excelsior (black)
Excelsior (white)
Promotion (Bb, 1, 1)
Model mate × 1 (2023-05-10)
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Verst-4
klären Stellung
SCHRECKE: Unlösbar - wLb1 gehört vielleicht nach c2!? (2023-03-10)
HBae: habe korrigiert, auch das Datum war falsch (2023-03-10)
SCHRECKE: C+, popeye 4.87 (2023-03-10)
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Yuri Bilokin: correction wKg4-f4, -bBa6, +bRe7, then a1=c1 6N1/5Pr1/4k3/8/2P4K/8/8/8 (4+2) h#2*

1...f8=Q 2.Rd7 Qf6#
1.Rg5 f8=Q 2.Rf5 Qe7#

Echo mates (mirrored diagonally, 0, 0)
Promotion (QQ, 2)
Echo (2023-06-04)
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Yuri Bilokin: correction bQa1-b7, -bPd6 3K4/1q2p3/8/4n3/8/8/3P4/7k (2+4) h#3
1.Sc4 d3 2.e5 dxc4 3.e4 c5 4.e3 c6 5.e2 c7 6.e1=R c8=Q 7.Rg1 Qh3# (MM)

Slow Excelsior
Active sacrifice (black, delayed)
Excelsior (black)
Excelsior (white)
Promotion (Qr, 1, 1)
Model mate × 1 (2023-05-10)
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klären AL ?
Yuri Bilokin: correction +bQh4(h3) 8/7k/1p3p1p/8/2n4q/p1pp2n1/p1p4P/1bK5 (2+13) (2023-05-08)
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Yuri Bilokin: possibly wKg8-h8 7K/4p3/7k/6p1/8/5p2/6P1/8 (2+4) (2023-02-01)
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Erich Bartel: Nachdruck:
Die Schwalbe (2) XII 1969 (2013-02-22)
Yuri Bilokin: possibly wKb3-a2 7k/1p4pn/8/8/8/8/K4P2/8 (2+4) h#7
1.b5 f4 2.b4 f5 3.b3+ Ka3 4.b2 f6 5.b1=B f7 6.Ba2 f8=S 7.Bg8 Sg6# (MM) (2023-02-01)
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Jakob Leck: Compare to P0559688 (2023-01-03)
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Branko Koludrovic: This task has been improved, and the new setting has been printed in"Die Schwalbe"Feb 2002:
White: Ke8,Bc7d6e7f6g7(6)
Black:Ke6,Sd5e5 Bc2d7f4f5f7(8) H#3
1.c1D c8S 2.Dc7 g8S 3.Dd8+ exd8S#
Task Type dSSS (2010-04-05)
Anton Baumann: auch oben genannte Version ist C+ (Gustav 4.2a) (2023-02-18)
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James Malcom: I don’t believe that the great Fabel, with his keen eye, would have not seen such a simply cook as the taking of Black’s c5 pawn. I suspect that a Black bishop was on d4 but was forgotten placing the problem in the database, or it was a misprint. I am changing this problem to not cooked for now. However, someone should check the original source in the event I am wrong, which is always a possibility. (2020-03-15)
Mario Richter: Hi, James, there's indeed a black bishop on d4. Original stipluation: "h#48 (nicht exakt!)"
A scan for some issues of 'Schachmatt' is available in the archive at stere.ro, see
www.stere.ro/wp-content/uploads/2014/06/stere_sah_istoria_sahului-1947-1949-Schachmatt1.pdf (2020-03-26)
Joost de Heer: Stupid question, but won't moving Bh3 to g4 add another halfmove? (2023-01-30)
Henrik Juel: Yes, Joost, but this would make the solution even less exact
0... Lf3,Lh3,Lh5 1.g4 Lxg4 etc. (2023-01-30)
Joost de Heer: @Henrik: So? This is an attempt at a non-unique (exact is something else) h# length record, the level of non-uniqueness is irrelevant. And as James pointed out, adding Kg4-f4 will add yet another move.
Only possible cooks are when white starts with 1... Bh5 2. g4 Be8 (so not capturing Pg4) but I don't see how that could lead to a quicker mate. (2023-01-31)
James Malcom: I double-checked that myself, Joost, and found that going around for wPg2 actually takes far longer, which would be required any PxB action to take place, but that takes far longer than the normal solution. I know Olaf's Gustav would check, since it C+ the h#111. (2023-01-31)
James Malcom: Is this the record for the longest legal, but dualed, helpmate? It is as far as I know. Is it possible to correct this problem with a black bishop placed on d4 (this is legally possible as Black has a pawn to spare), which would uphold the original solution? (2020-01-27, 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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Henrik Juel: Natch 3.1 made no progress in half an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:15:12 Minuten. (hh:mm:ss)
Keine Lösung: BP 22.0, BP 22.5. (2023-03-27)
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Reto: This is a case where strategy analysis in Stelvio is in a certain sense optimal: Only 1 strategy remains after analysis, namely the one that belongs to the solution. (2023-04-25)
Henrik Juel: Natch 3.1 did not even find the solution in 3 hours (2018-12-14, hidden)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 29.5, BP 30.0. (2023-04-24, hidden) more ...
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Henrik Juel: Natch 3.1 produced nothing in more than 3 hours (2018-12-12)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 01:02:15 Stunden.
Keine Lösung: BP 15.0, BP 15.5. Notation OK. (2023-02-18)
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Henrik Juel: Natch 3.1 did nothing in almost an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 84:43:01 Stunden. (hh:mm:ss)
Keine Lösung: BP 22.0, BP 22.5. (2023-04-01, hidden) more ...
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Kommentare:
Als einziger letzter schwarzer Zug, der die weiße Rochade
erhält, kommt nur Bh7-h5 infrage, was dem Weißen den e.p.-Schlag
ermöglicht, um nicht die Rochade oder das Mattnetz zerstören zu
müssen (H.P.Suwe)
Gianni Donati: This intends to show the Valladao theme in the minimum number of moves. (2004-03-19)
VL: The waiting ep capture. Cf. also P1000260
(by T.Petrovic, 2000) somewhat enriched
thematically with the illegal try 2... Rd1#??. (2004-06-03)
A.Buchanan: The “illegal try” 2. Rd1+ is not actually mate because c2 is unprotected. This is kind of “logic dual” spoils the A.P. motivation for the castling (2020-05-20)
Henrik Juel: White captured h6xg7 and once more, e.g. axPb-b8=Y
Black captured cxdxe, dxexf, fxg, and once more, e.g. a2xb1=Y
Possible retroplay 1... h7-h5 2.h6xSg7 Df8-f4 3.h5-h6 Sf4-h3 4.Sh3-g1 etc., preserving the castling right
Any other black last move would force White to retract Ta1, as d2xc3, d2xe3, and h6xh7 would be illegal retractions
I agree with Andrew that the double motivation of 2... 000# is a weakness:
a. legitimizing h7-h5
b. mating (2020-05-21)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
A.Buchanan: See discussion at P0009121 (2023-08-04)
A.Buchanan: Maybe can replace wPf2 with wR, wRf1 with wB & shift bPg3 to g2? (2020-05-20, hidden) more ...
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Yuri Bilokin: you can see
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.d5 Qb8#
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.g5 Qb8#
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.Sg4 Qb8#
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.Sf5 Qb8# (2023-12-28)
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Henrik Juel: Natch 3.1 found the solution but did not finish testing in more than an hour (2018-12-18)
Henrik Juel: Stelvio 2.0 is slightly faster than Stelvio 1.14: 55 sec. (2024-02-04)
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Anton Baumann: Informalturnier 1999: 2.Preis (2023-02-27)
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SCHRECKE: C+, popeye 4.87 (2022-12-14)
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Erich Bartel: Nachdruck:
1) 62701 s\#L.wuw IX 2004.--- (2008-03-09)
Henrik Juel: C+ Popeye 4.61 (2023-01-17)
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Henrik Juel: an important keyword is missing: Donati theme
same in (some of) P1011889-P1011929 (2018-12-18)
Henrik Juel: Natch 3.1 found nothing in half an hour (2018-12-18)
Michel Caillaud: Reprint : H71 p.585, FIDE Album 2001-03

Cooked using Jacobi 0.7.5 with constraints :
1.c4 e5 2.c5 Ld6 3.cxd6 g5 4.d3 g4 5.e3 g3 6.Dg4 c5 7.Sc3 c4 8.Sa4 c3 9.Ld2 Sh6 10.0-0-0 c2 11.b4 Tg8 12.Kb2 gxf2 13.g3 Tg6 14.Lg2 f1=L 15.Ka3 c1=L+ 16.Sb2 Le2 17.Le1 Ld2 18.Tc1 Sg8 19.Txc8 Lc1 20.Tc3 Lf3 21.Se2 Dh4 22.Tf1 Sf6 23.Lh1 Lg2 24.Tf3 Lf1 25.Dh3

Correction : sTg6 to g4 (21...Tg4 instead of 21...Tg6 in the solution)
With the set of constraints used to uncork the cook, Jacobi 0.7.5 finds only the intended solution... (2021-09-20)
Michel Caillaud: The correction sTg6 to g4 was published in StrateGems 99 (p.138, sept. 2022)
It is C+ by Stelvio 0.93 (2022-12-18)
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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. 1449 HN
AV: Je crois que c'est un Fou blanc en h8 et non un Cavalier, et les derniers coups sont gxFh8=F+ Rf6-e7 fxDe8=D+ (au lieu de fxCg8=C#). Avec l'antérieur cxTb8=T, cela donne une "Allumwandlung" avec une pièce noire prise de la même nature. (2005-06-09)
Henrik Juel: You must be right, AV, because with wSh8 the retroplay might be the bland -1.f7xDe8=D Kf6 etc. without unpromotion on h8. (2005-06-13)
Michel Caillaud: Indeed the diagram is wrong : there should be a white Bishop on h8
AUW spread over retro play (-1.fxDe8=D+ Kf6-e7 -2.gxLh8=L+ ...-n.cxTb8=T) and forward play 1.fxSg8=S#
Matching nature of promoted piece and piece captured by promotion. (2023-03-25)
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2044
SCHRECKE: a)-e): C+, popeye 4.89 (2023-10-22)
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