Die Schwalbe

3359 problem(s) found in 1272 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20220117] [download as LaTeX]

1 - P0000498
Thomas Volet
3097 Die Schwalbe 62 04/1980
3. ehrende Erwähnung
P0000498
(10+13)
Wie weit muß der sK maximal nach links gelaufen sein?
Beispielauflösung mri:
R: 1. ... Ke1xSf1 2. Se3-f1 Kf1-e1 3. Sf5-e3 Ke1-f1 4. Sg7-f5 Kf1-e1 5. Sf5xTg7 Ke1-f1 6. Se3-f5 Tg8-g7 7. Sf1-e3 Tg7-g8 8. Tg1-g2 Tg8-g7 9. Lf3-h1 Tg7-g8 10. Tg2-g1 Tg8-g7 11. Se3-f1 Kf1-e1 12. Sc4-e3 Ke1-f1 13. Se5-c4 Kf1-e1 14. Sd3-e5 Tg7-g8 15. Kd1-c1 Tg8-g7 16. Sc1-d3 Tb1-b2 17. Le4-f3 Lb2-a1 18. Sd3-c1 Lc1-b2 19. Sb4-d3 Ta1-b1 20. Sd5-b4 Tb1-a1 21. Sc3-d5 Ta1-b1 22. Sb1-c3 Lb2-c1 23. Kc1-d1 Le5-b2 24. Kb2-c1 Ld6-e5 25. Sc3-b1 Ke1-f1 26. Sa4-c3 Kd1-e1 27. Sc5-a4 Tf8-g8 28. Sd3-c5 Tg8-f8 29. Se1-d3 Tc1-a1 30. Tg1-g2 Ta1-c1 31. Lg2-e4 Tc1-a1 32. Lf1-g2 Ta1-c1 33. Sf3-e1 Tc1-a1 34. Th1-g1 Ta1-c1 35. Sg1-f3 Tc1-a1 36. Lg2-f1 Ta1-c1 37. Ld5-g2 Ke1-d1 38. Lc4-d5 Kf1-e1 39. Ld3-c4 Kg2-f1 40. Sf3-g1 Kh3-g2 41. Sd4-f3 Kg4-h3 42. Sb5-d4 Kh5-g4 43. Sc3-b5 Kh6-h5 44. Sb1-c3 Kg7-h6 45. Kc1-b2 Kf8-g7 46. Kd1-c1 Ke8-f8 47. Ke1-d1 Le5-d6 48. Le4-d3 Lg7-e5 49. Lg2-e4 Lf8-g7 50. Lf1-g2 g7-g6 51. g2-g3 g3xBh2 52. Tg1-h1 f4xDg3 53. Dc3-g3 e5xTf4 54. Db2-c3 d6xLe5 55. Dc1-b2 Sg6-h8 56. Dd1-c1 Th8-g8 57. Lb2-e5 Se5-g6 58. Lc1-b2 Sc6-e5 59. b2xDa3 Da5-a3 60. Te4-f4 Dd8-a5 61. Td4-e4 a3-a2 62. Tc4-d4 a4-a3 63. Td4-c4 a5-a4 64. Tc4-d4 Ta4-a1 65. Td4-c4 Tb4-a4 66. Tc4-d4 Tb6-b4 67. Ta4-c4 Ta6-b6 68. Ta1-a4 Ta8-a6 69. a2xSb3 Sd4-b3 70. Th1-g1 c7xSd6 71. Sf5-d6+ Sb8-c6 72. Sh4-f5 Sf5-d4 73. Sf3-h4 Sh6-f5 74. Sg1-f3 Sg8-h6 75. Sh3-g1 a7-a5 76. Sg1-h3
play all play one stop play next play all
Henrik Juel: -1... Ke1:S, wS uncaptures a bR in the NE corner, screens on f1 to let wB out, then on c1 to let bBa1 out; retract this bB to f8 and g7-g6, use 2 S's to extract bK via h3 and c7 to e8, etc. (2003-11-27)
hans: .....Rb1+ Nc1 Ba1 ???? Solution don't work! (2010-05-23)
Henrik Juel: I think the solution I gave in 2003 works fine. Here it is spelled out in more detail.
Following -1... Ke1xSf1 -2.Se3 Kf1 -3.Sf5 Ke1 -4.Sg7 Kf1 -5.Sf5xRg7 Ke1, Black can make tempo retractions with Rg7 allowing -6.Se3, -7.Sf1, -8.Rg1, -9.Bf3, -10.Rg2, -11.Se3 Kf1 -12.Sc4, -13.Se5, -14.Sd3, -15.Kd1 Rg8 -16.Sc1 Rb1 -17.Be4 Bg7 -18.Sd3 Rb2 -19.Kc1 Bf8 -20.Sf4 g7 etc. (2010-05-24)
Mario Richter: Henrik, may I ask you to look at this problem again?
First, the intended solution:
The black King needs only to go to the d-file, not further to the left (mainly by avoiding an early retraction of g7-g6). This can indeed be done by using the two Bishops as shields:
1. Ke1xSf1
2. wSxTg7
3. wS shields on f1
4. now white Bishop get out, providing tempo moves for White
5. wSc1
6. black rook a1, black Bishop b2-c1
7. wS shields on b1
8. now black Bishop and wK can get out
9. sKd1,wSe1
10. wLf1, wTh1
11. wSg1
12. now white Bishop and sK can get out
13. sK back home via g7

So far, so good. But now my question: What happened to the white Pawn h2? (2022-03-21)
Mario Richter: I first thought that the provable retraction "black Pawn g3 x white Pawn h2" will introduce some difficulties in resolving the position. But as my sample resolution given above shows, this is not the case ... (2022-03-21)
Henrik Juel: I see your point, Mario (2022-03-22)
comment

Genre: Retro
FEN: 2b4n/1p1ppp1p/6p1/8/8/PP4P1/prPPPPRp/b1K2k1B
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-03-21 more...
2 - P0000574
Thomas Volet
The Problemist 11/1980
P0000574
(13+14)
Woher kamen die sTTa8 und h8?
Henrik Juel: The rooks have switched place. Following -1... Rb8:S -2.Sb6, bRb8 retracts further to g8, bBd8 retracts to f8 and bPe6 to e7, to allow wK to exit via e6; retract bRh8 to h6 and h7xRg6 etc. wPg4 cannot retract until orig. wRh1 is home. This is the only Volet retro with a K in check, composed in Tom's youth, before he saw the light! (2003-11-25)
Thomas Volet: The exception in this case to the otherwise observed constraint of not having a check in the diagram or stipulation as to who is on the move relates to a collegial colloquy at the time with with the composer J.G. Mauldon, who was active in this task. (2022-03-24)
A.Buchanan: Terrific composition! Often in a Type C composition, one might retract the check to give Type B. But the thematic Ta8 here could not exist without a check in diagram. There are different areas of design space are accessible depending upon level of self-imposed constraints. Some retro composers like Baibakov systematically explore Types B&C as well as Type A. I don't see this as any kind of defect.
I particularly like that the lock on h-file only allows one sT to squeeze itself out. (2022-03-24)
comment
Keywords: Impostor (tt), Which are original men? (tt), Interchange (tt)
Genre: Retro
FEN: r1bbn2r/Kpp3p1/3pppp1/7B/5kP1/1P6/pPPPPP1P/NQB5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-24 more...
3 - P0000583
Andrey Frolkin
Günter Lauinger

3468v Die Schwalbe 72 12/1981
P0000583
(10+12) C+
h#2 (AP)
Satzspiel:
*1. ... 0-0! 2. cxd4 Tc1#

Verführung:
1. exd3ep? 0-0 2. cxb4 Tf4# aber zuletzt R: 1. d2-d4?? ist illegal, da es den wLc1 als schwarzes Bauernschlagobjekt ausschließt

Lösung:
1. ... 0-0! (AP-Legalierung des weissen Anzugrechtes) 2. cxd4 Tc1#
play all play one stop play next play all
Alle Versuche, das Satzspiel 1. ... 0-0! 2. cxd4 Tc1# durch einen schwarzen Vorschaltzug aufrecht zu erhalten, scheitern.
Mario Richter: 1. exd3ep 0-0 2. cxb4 Tf4 ist nur Verfuehrung (scheitert an schwarzer Bauern-Schlagbilanz).
Loesung: 1. ... w0-0! (AP-Legalierung des weissen Anzugrechtes) 2.cxd4 Tc1# (2009-02-10)
A.Buchanan: Supersedes P0000553 (2022-01-08)
Mario Richter: Hi Andrew, I only quoted from the "official solution" ('Die Schwalbe' Heft 72, 12/1981, p.399).
Perhaps a better way to get anwers to your interesting questions is to ask the authors directly ... (2022-01-08)
A.Buchanan: AP Type Petrovic is a try, because wLc1 was captured at home, yet Black still made 6 pawn captures.
AP Type Keym is the actual solution, because if it was really BTM, then White would have lost castling rights.
So there are two kinds of AP here. (2022-03-21)
more ...
comment
Keywords: Castling (wk), a posteriori (AP) (Type Keym), En passant as key, a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: BC+ Popeye v4.87
FEN: 8/2p5/1pPp4/bRpP4/BPkPp3/qp2p2p/rP2P3/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
4 - P0000598
Nikita M. Plaksin
Andrej N. Kornilow

3876 Die Schwalbe 74 04/1982
P0000598
(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!
play all play one stop play next play all
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
Keywords: a posteriori (AP) (Type Petrovic), Circe, En passant as key, Monochrome, 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...
5 - P0000615
Andrey Lobusov
4088 Die Schwalbe 77 10/1982
14.Lob
P0000615
(15+6) cooked
h#2* (AP)
1. ... Kg5 2. 0-0?? Se7#
1. ... Txh7 2. Kf8 Txh8#

1. cxb3ep Txh7 2. Kf8?? Txh8#
1. cxb3ep Kg5 2. 0-0! Se7#
play all play one stop play next play all
Wh captured 10 by pawns (including hxPg-g8) but not on the last move. wPa3 came from a2: wPa5 captured bPa.
Bl took only b3xa2.
In set play, Bl has no last move except for K or R.
For the main play, the intention is that if Bl retains castling rights, last moves were R: 1. b2-b4 b3xa2 e.g. 2. Bg1-d4 b4-b3 3. d4xc5 etc. However there are retro cooks e.g. R: 1. Bg1-d4 b3xa2 2. d4xc5 etc, with wPa5 the original wPb. wPb4 & wPb5 come from further East.
Cook: Retro logic for main play is unsound.
A.Buchanan: Not trivial to fix as shifting wPa5 to b6 leads to promotion cooks. Adding black material must be carefully done to avoid disrupting uniqueness of e.p. as tempo. (2022-03-11)
A.Buchanan: OK just to say I have found a form of this idea that works. This one was the very devil to fix - the final form includes different mates and retro logic, so becomes an "after" rather than a "correction", and I will find somewhere else to publish first. (2022-03-19)
more ...
comment
Keywords: Castling (sk), a posteriori (AP) (Type Petrovic), En passant as key, Tempo Move
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/5p1p/5K1R/PPP2N2/1PpB4/P1P5/p1RNB1P1/Q7
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-11 more...
6 - P0000649
André Hazebrouck
2261 Die Schwalbe 47 10/1977
4. Preis
P0000649
(8+14)
h#2.5
b) AP (2 Lösungen)
a) 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#
b) also 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#
is clearly the intention
play all play one stop play next play all
Can count 1+7 visible pawn captures, leaving 1+1 unexplained. Need to resolve wPfgh & bPg. Suppose bK never moved, then one of:
1) wPf waylaid, wPhxPg=, wPg=. Here w00 right can remain.
2) Or similarly, wPh waylaid, wPfxPg=, wPg=. Again w00 right can remain.
3) bPgxPh=, wPfxg= (or wPfxe then captured by bPf), wPg= w00 right lost.
1+1 captures explained in either case.
On the other hand if bK moved, then maybe wPf=, and only requires one more capture to resolve g&h files. The w00 right can remain. So there is a captured unit unaccounted for, and we can't validate the ep.
On the other hand if we are in case 3 above, then we can't know that wPbxa captured dark bishop, so White might retract a6-a7.
We also need to know that w00 rights remain. s000 rights (i.e. prior movement of sTa8) are not relevant.
If we know that w00, w000 & b000 remain, then we are in case 1 or 2 above with bPg captured unpromoted. Therefore wPb6xLa7 due to bishop shade. Therefore R: 1. b7-b5 b6xLa7 to unblock.

Only one candidate solution does not begin with ep: 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#. So this is the solution for a).

Let ????? denote validity of w00,w000,b00,b000,ep. Possibilities are: YYY?Y and all of YYN??, YNY??, NYY??, YNN??, NYN??, NNY??, NNN??. So there are 30 possibilities.
Under PRA, the solution parts would be YYYYY, YYNYN, YNYYN & NYYYN. For the first there are 20 solutions, while the second and third have 0 solutions. So this is not the right paradigm.
Under SPRA, there would be a single solution part YYYYY with 20 solutions. So this is not the right paradigm either.

Under RS with AP, the solution of a) still works. This comes from ?YY??. On the other hand, if the first move is ep, then we are in YYYYY. So all castlings are valid. But when we get to perform the mate, we need to know that based on castlings actually performed, the ep is valid. So the solution must include w00. So is 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#.

By combining the information of both of these solutions, we know we are in YYY?? so YYY?Y is the only possibility, and ep is legal. However, the combination of these two solutions would also validate any of the other 18 ep solutions that do not include 2. ... 0-0. They contribute no new evidence, but how to exclude them?

This is one of the issues with "AP Consolidation". Normally the number of solutions is not a constraint. But here perhaps we should insist that *only* two solutions are allowed?
Cook: 1. ... cxb6ep 2. La6 Tb1,~ 3. 0-0-0 a8=D#
18 different possibilities for W2 which are validated by the same logic that validates 2. ... 0-0.
Henrik Juel: [I don't follow the silly convention of writing black moves first]. 1.0-0-0 0-0 2.dTg1+ Kh8 3.Txh6#. b) If Ke1,e8 and Th1 never moved, White captured f/hxPg and b6xLa7; if Ta1 also never moved, last white move was b6xLa7. 1.cxb6ep La6 2.0-0/0-0-0 0-0-0 3.a8Q#. The a) solution also works in b). (2003-12-18)
A.Buchanan: Where a single solution must justify itself, then the AP logic seems clear. But where evidence from different solutions must be Consolidated to justify themselves or other solutions, there are a number of unresolved issues. A general charitable approach: “this looks worthy enough, let’s say that it’s sound” is really counter-productive (2022-03-23)
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comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (wkwlsk), En passant as key
Genre: h#, Retro
Computer test: Forward logic Popeye v4.87 & basic retro-logic works, but then AP Consolidated protocol seems to cook the problem.
FEN: r1b1k2r/P2p4/2p1n2p/ppPq4/1np5/p7/P2PP3/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
7 - P0000760
André Hazebrouck
1120 Die Schwalbe 23 10/1973
P0000760
(9+10) C+
h#2 (AP)
1. bxc3ep Sa6 2. 0-0-0 Tc4#
play all play one stop play next play all
Henrik Juel: -1.c2 c5xLb4 -2.Lc3 c6 -3.Lb2 c7 -4.Lc1 a6! -5.b2 b2xDSa2 etc. White captured sDTSS by fxexd, gxfxe, allowing Black to capture f7xPe6xPd5 etc. (2003-12-18)
A.Buchanan: Why not e.g. wSb1? (2022-03-04)
Henrik Juel: That also seems to work (2022-03-04)
A.Buchanan: I really like the motivation for ep. I guess the motivation for Rb1 is to be inside the cage that forms around it, but it’s still a bit loose. Note there is no retro try. (2022-03-05)
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comment
Keywords: En passant as key, Castling (sg), a posteriori (AP) (Type Petrovic), Volet Pawn
Genre: h#, Retro
Computer test: HC+ Popeye 4.87 + simple retro reflection
FEN: rN2k3/1p1pp1pp/8/p7/RpP5/PP6/p2PP3/KR6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-05 more...
8 - P0000899
Giuseppe Brogi
743 Die Schwalbe 06/1972
P0000899
(8+15) cooked
h#2
b) wSa1
a) 1. T2xh3 Txh3 2. 0-0 Th8#
b) 1. La4 0-0 2. Tf8 Te1#
play all play one stop play next play all
a) White cannot cross-capture, so because of sTh2, w0-0 impossible.
wBe captured sD, either to be captured on f-file or to promote without disrupting sKe8.
b) sTh2 can return to a8 only via e8 as Black cannot cross-capture, so s0-0 is impossible.
wBe was waylaid or promoted after disrupting sKe8.
Cook: 1. T2xh3 Lb4 2. Txg3 Txh8#
1. Sd2 Lf6 2. Tf8 Sd6#
(cooked by MR)
See P0003736 a companion problem.
more ...
comment
Keywords: Cant Castler, Castling (wksk), Cross-capture (s,w), Superseded by (P1399805)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: B3k2r/pN1p1p2/1pp3p1/bb3p2/8/p1B3PP/5P1r/nn2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-16 more...
9 - P0000933
Luigi Ceriani
Sahovski vjesnik 12/1951
1. Preis
P0000933
(2+11)
Letzter Zug?
R: 1. Kg8xDh8
play all play one stop play next play all
A.Buchanan: Solved in YouTube: https://www.youtube.com/watch?v=Do_rkLQnrpk (2022-02-17)
more ...
comment
Keywords: Type A, Last Move? (KxD), Economy record (Last Move? Type A eq), Type B (a fortiori), Economy record (Last Move? Type B eq)
Genre: Retro
FEN: 3bkN1K/pppprp1p/4p1p1/8/8/8/8/8
Reprints: 1372 FEENSCHACH 08-09/1952
Problem 10-12 12/1952
1.2A Eigenartige Schachprobleme , p. 172, 2010
1.2B Eigenartige Schachprobleme , p. 172, 2010
YouTube 13/02/2022
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-30 more...
10 - P0000957
Theophilus Harding Willcocks
Die Schwalbe 1959
P0000957
(4+8)
Letzter Zug?
R: 1. a7xTb8=L
play all play one stop play next play all
Yoav Ben-Zvi: WBd8 can be replaced with a BN after moving BPg7 to f7 and BPh5 to h6. (2018-08-08)
A.Buchanan: Yes after all these years, I found this a few months ago, but when I told Thomas Brand, he said that Werner Keym had found it - one of a series of modifications through seeking to avoid non-standard material. (2018-08-12)
Henrik Juel: Werner's improvement can be found in his 'Eigenartige Schachprobleme' from 2010, p.196, dia 1.68 (2018-08-12)
A.Buchanan: I thought it was more recent (2018-08-12)
A.Buchanan: In fact Werner was not trying to avoid non-standard material, but to prefer "cheaper" knights over bishops. But this is not the canonical ordering, which regards knights and bishops as equivalent. So the older record will win out in classical terms. (2019-10-04)
A.Buchanan: And non-standard material is no defect, according to the 1977 grading criteria, so old Theophilus reigns supreme (2019-10-04)
Henrik Juel: The further retroplay is
-1... h6 -2.a6 and e.g. -2... Ka7 -3.a5 Ta8 -4.Tb8 Ka6 -5.Kc8
and wK out via g6 (2019-10-04)
A.Buchanan: I don't know if a bishop is more expensive than a knight for these economy records. If modifying the criteria, I personally would prefer to minimize non-standard material. But more generally, how to handle a later version of economy record, with similar force? In some cases, the change is trivial, just transforming one unit which might as well be a dummy. At the other extreme, the matrix is completely different. In "Eigenartige Schachprobleme", Werner Keym stoically prefers the earlier record. Even the original composer(s) of a record problem are forbidden from making any artistic improvement to a published economy record! Over the years, the only change to the grading criteria is that Type C problem now needs to contain a check. Any old check-free Type C would a fortiori have been a Type B economy record at least. Here in PDB, I've tagged some cases of "ex-aequo" (search k='economy:eq') but I fear that Werner would not approve of this. What should we do? (2022-03-31)
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comment
Keywords: Type A, Last Move? (BxT=L), Promotion (L), Economy record (Last move? Type A), Non-standard material (L)
Genre: Retro
FEN: kBRB4/1pKpp1p1/1pp5/7p/8/8/8/8
Reprints: 1.49A Eigenartige Schachprobleme , p. 190, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-30 more...
11 - P0001086
Peter Kahl
Jan R. Mortensen
Sveto Stambuk

Problem 5-6, p. 92, 12/1951
6. ehrende Erwähnung
P0001086
(4+1)
Welches war der letzte Zug?
R: 1. Dh3xSh2+
play all play one stop play next play all
im Original ist als Dritter Autor D. Suboticanec angegeben, habe entsprechend geändert.
Zu P. Kahl heißt es ergänzend: [aus] Osterholz-Scharmbeck
HBae: Ist "Klaus Peter Kahl" identisch zu "Peter Kahl" ? (2010-09-22)
Henrik Juel: Yes, probably
I know that 'Jan Robert Mortensen' is identical to 'Jan Mortensen'; he never used his middle name (2017-03-08)
A.Buchanan: It’s possible to put the middle name into the “addition” field in the author table, rather than have two names in the “first name” field. Then in overview list the middle name won’t appear, but it will still appear above each composition as here. Curiously the addition field is common across all four languages (German, English, French, Unicode) supported by the author table. There is no comment field to allow for random extra info to be stored, unlike the source table. So if we drop Robert, we lose it completely. (2022-03-31)
Henrik Juel: Thanks for the info, Andrew
I guess we will have to live with it
For Jan, there is really no risk of confusion
But for Knud Hannemann (son of the less illustrious composer Harald Hannemann) it is annoying to see him cast as Knud Harald Hannemann, just because it was a custom in his family that the sons were given the father's first name
They still use this custom in Russia, but with 'son of' added, e.g., Vladimir Vladimirovitch Putin (2022-03-31)
Henrik Juel: A chess problem example is Nikita Michaelovitch Plaksin
But the rendering of his name in the PBD also gives an obvious partial solution to my problem
Change the middle name of the two danish composers to the corresponding initial:
Jan R. Mortensen and Knud H. Hannemann (2022-03-31)
Henrik Juel: Thanks for the 'initialization', Andrew (2022-03-31)
A.Buchanan: OK fixed those, Henrik. HBae's original question concerned "Klaus Peter Kahl" vs "Peter Kahl". The former only has 2 problems here, and "both" collaborated with Jan Mortensen, so surely this is one person? (2022-03-31)
A.Buchanan: Kahls unified (2022-04-01)
Henrik Juel: I hear no protests from Osterholz-Scharmbeck, so that seems fine (2022-04-01)
comment
Keywords: Type C, Last Move? (DxS), Economy record (Last Move? Type C)
Genre: Retro
FEN: 8/8/8/8/7P/6P1/5K1Q/7k
Reprints: 1375 FEENSCHACH 08-09/1952
(C10) feenschach 160 07-08/2005
1.11C Eigenartige Schachprobleme , p. 176, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-01 more...
12 - P0001107
Michel Caillaud
3332 Die Schwalbe 66 12/1980
2. Preis
P0001107
(7+9)
h#4.5 (AP)
Circe
1. ... hxg6ep[+sBg7] 2. a1=T+ Txa1[+sTh8] 3. 0-0 Ta3 4. bxa3[+wTa1] 0-0-0 5. Td8 Txd8[+sTh8]#
play all play one stop play next play all
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
Keywords: a posteriori (AP), Circe, En passant as key, Castling (wgsk), Promotion in forward play, Valladao Task
Genre: Retro, Fairies, h#
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, 2022-03-22 more...
13 - P0001136
Edgar Fielder
The Fairy Chess Review 1941
P0001136
(13+10)
Darf Schwarz rochieren?
Schwarz hat bereits rochiert!
play all play one stop play next play all
See P1398939
Henrik Juel: Black captured b7xLc6 and axbxc; White captured [Lf8], e2xd3, and fxexdxc6xb7-b8=D.
Ke8 to b7, Th8 to b8, Kb7 to g8, Tb8 to f8, Da4 to b8 for unpromotion, -22.c6xDb7 Da8 -23.d5xLc6 Dd8 -24.d4 Lb7 -25.e3xTd4 Lc8 -26.f2xSe3 c6 -27.h5 b7xLc6, Lc6 to f1, f2xSe3 etc.
Black may not castle, because he already did.
It is not possible to avoid the early castling:
-11... Ka7 -12.Db4 Ka6 -13.Db8 Ka7 -14.D=b7 Kb8 -15.c6xDb7 Kc8 -16.d5xLc6 Kd8 -17.d4 Ke8 -18.e3xTd4 Da8 -19.f2xSe3 Dd8 -20.h5 Lb7 -21.h4 Lc8 and two white pawn retractions are missing (2012-07-23)
Yoav Ben-Zvi: The solution does not require a full analysis since if Black King and Rook never moved then Black is almost immediately in retro-stalemate. An alternative stipulation is "First move of Black King?". (2014-06-03)
A.Buchanan: Maybe there is intentional irony? (2014-06-03)
Henrik Juel: Nowadays composers are not afraid to use the real stipulation in this type of resolution retro, like 'Release the position'.
Formerly a formal stipulation was popular, like 'Mate in 1', with the real stipulation understood but not mentioned.
Here, the stipulation question may well be ironic, but you will get no solution points, if you just answer 'No'; you still need to explain how the position arose (2014-06-03)
more ...
comment
Keywords: Castling (sk), Castling Paradox (sk hidden)
Genre: Retro
FEN: 4k2r/2pppppp/7P/2p5/Q1P5/PPRP4/RBpP2P1/N1K5
Reprints: 71 32 personaggi e 1 autore 1955
12 Europe Echecs 12 08/1959
222 FIDE Album 1914-1944/III 1975
341 Eigenartige Schachprobleme , p. 110, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-23 more...
14 - P0001191
Leon Loewenton
64 Europe Echecs 38 10/1961
P0001191
(15+14) C+
h#2
b) wSd3 nach e1
a) 1. Kd8 Se5 2. Te8 Sxf7#
b) 1. 0-0 Sd5 2. Sh8 Sxe7#
play all play one stop play next play all
Originalquelle?

nicht sicher, ob Zwilling b) auch schon im Original.
Artikel von Petrovici zum "Thema Than" 03/2016 gibt als Originalquelle an: "Europe Echecs 10/1951"
hans: Counting problem
1. Kd8 Se5 2. Te8 Sxf7# !!
1. 0-0 Sd5 2. Sh8 Sxe7# ??
Black needs a tempo to get on move, and the only piece to do that is Th8, so 0-0 is illegal. (2010-06-20)
Mario Richter: The term usually used to describe this kind of problems is "Parity problem". The bRh8 might have been on h8 all the time, since the tempo move might also have been made by the black Queen, but in this case, the bK must have already moved, thereby losing the right to castle too.. (2010-06-21)
Ladislav Packa: You both are right. Black would be in this position made an uneven number moves.
This can be achieved by using Rook moves (eg Rg8-f8-h8), or by the Queen moves while she was still alive (eg Kf8-g8, then Qf8-e8-d8 and then the King back). But both possibilities preventing black castling. (2013-10-04)
A.Buchanan: Parity change could also be achieved without triangulation by e.g. SxDe8 after the black queen has moved once (2022-03-15)
comment
Keywords: Castling (sk), Parity Argument, Cant Castler, Than theme
Genre: h#, Retro
Computer test: rawbats
FEN: r1b1k2r/1ppppppp/p5n1/8/8/P1NN3P/1PPPPPPR/nRBK1B2
Reprints: 787 Themes-64 10-12/1961
(7) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2018-03-22 more...
15 - P0001263
Luigi Ceriani
135 Europe Echecs 78 06/1965
P0001263
(15+11)
h#2
1. bxa5 Sb6 2. axb4 Ta8#
play all play one stop play next play all
Henrik Juel: analysis
Black captured a3xb2 (not a7xb6, as the SW corner then cannot be released)
The missing white man is [Pe2], which must have promoted on b8
Now the white captures can be seen: axb, exdxcxb, and [Lf8] on f8
The only free white officer is Dd8, so the retroplay must include unpromotion of it on b8
It is impossible to shield Ke8 from check by Db8, so Ke8 has moved and Black may not castle (2022-04-27)
Henrik Juel: solution
1.bxa5 Sb6 2.axb4 Ta8#
not 1.0-0? Lxb6 2.Txa8 Txa8
HC+ Popeye 4.61 (2022-04-27)
Henrik Juel: Andrew, I should have been more precise in recommending your suggestion to introduce the new keyword Organ Pipes
I was talking about the standard meaning of Organ Pipes, which dates back to Sam Loyd, 1859: black LTTL in a problem where each L interferes with each T and vice versa, forming four Grimshaws
You have probably used a pattern search to find all occurrences of black or white LTTL, and you have neglected the interferences
This problem has LTTL in the wrong color, and there are no interferences (2022-04-27)
James Malcom: I agree with Henrik, although White organ pipes still exist, few and far. (2022-04-28)
James Malcom: Until Andrew bumbles back, I've updated the English definition to "A problem *utilizing* the pattern bishop, rook, rook, bishop in a straight line to create multiple Grimshaws." This specifies the purpose and formation of the Organ Pipes. (2022-04-28)
Henrik Juel: Thanks, James (2022-04-28)
A.Buchanan: Hi Guys. I wondered about other uses of organ pipes. In a first pass population of the 500-odd records with the current PDB interface, there’s not much time for thinking. I have multiple tabs open to eliminate the wait time associated with PDB refresh. One could go back and eliminate those which are not grimshaws. Alternatively (and this would be my preference) accept that this is a visual pattern which may occur in non-Grimshaw context. Then use the keyword Grimshaw(4x) to indicate when it’s really Grimshaw. I think it can also occur with white pattern, as a problem by Pal Benko shows. So I blitzed through the d# records, if someone wants to complete the rest that’s good. I think the existence of a few false transient positives is an acceptable price even if we take the narrow definition of Organ Pipes. (2022-04-28)
A.Buchanan: By the way, these days Deepl is good enough to give us decent translation to Feench & German and as a matter of policy whenever I make a change to a definition I try to align the other two. Other users who do not maintain the glossary are encouraged to propose definitions where there is a gap (E.g. Grimshaw). There are a lot of undefined terms, many very recent. (2022-04-28)
A.Buchanan: Hi Henrik feel free to respond to my response to your message, say thanks for the tags I have added, or add tags yourself to complete the work. The ones I’ve added were mostly the d# and it was a deliberate decision to blast through as a first pass and just add them for now anyway, not “neglected”. Now, I really don’t feel like continuing (2022-04-28)
Henrik Juel: Andrew, I believe that most PDB users appreciate your contributions to the site
I certainly do, so please continue your good work (2022-04-28)
comment
Keywords: Castling (sk), Cant Castler, Promotion
Genre: h#, Retro
FEN: N3k2r/2pppppp/1p6/B7/RP5P/RP3P2/BpPP2P1/NK1Q3r
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-04-28 more...
16 - P0001420
Branko Koludrovic
294 Europe Echecs 197 04/1975
1. ehrende Erwähnung
P0001420
(13+12) C+
h#3*
1. ... cxd6ep 2. 0-0-0 0-0-0 3. Kd7 Sa7#
1. De7 c6 2. Th8 c7 3. Tf8 Sd6#
play all play one stop play next play all
White pawn caps: axb,dxe,gxf,hxg.
Black: fxg,bxc,cxb.
wPb4 came from b3 to release wBa3, so bPb3 captured to reach that square.
All pcs accounted for means bPd never captured.
In the set play, there are 13 retro tries in which one or both players do not castle. The intention is that both castling rights are needed in order to imply the pawn double hop.
A.Buchanan: White pawn caps: axb,dxe,gxf,hxg definite.
Black: fxg and two to resolve c-file. But that may be c&d cross-capture, so in set play last move might have been c6xd5. So I think this problem is cooked. What am I missing? (2022-03-21)
Mario Richter: If Black's last move was c6xd5, how do you get white Bishop a3 out of his cage? (In that case, black pawn b7 never left the b-file). (2022-03-21)
A.Buchanan: I agree Mario thanks (2022-03-21)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wgsg)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro-logic.
FEN: r3k3/6p1/4p3/1NPp4/BPp4q/Bp2PPPr/pP2PPpp/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
17 - P0001940
Nikita M. Plaksin
Shakhmaty v SSSR 1980
Spezialpreis
P0001940
(13+16)
Remis
hans: 1. b4 a5 2. Bb2 Na6 3. Bd4 Nc5 4. Nc3 Na4 5. Bc5 Nb6 6. b5 Nh6 7. a4 Nf5 8.
Na2 Nh4 9. Nb4 axb4 10. a5 Rb8 11. a6 Na8 12. a7 Nb6 13. a8=Q Rg8 14. Qa2
Ra8 15. Qe6 Ra3 16. Qg6 hxg6 17. Qb1 Na8 18. Ba7 b6 19. Qb3 Bb7 20. Qd3 Bf3
21. Qf5 gxf5 22. Nh3 Bh5 23. Nf4 Bg6 24. e3 Rh8 25. Bc4 Rh6 26. f3 Bh7 27.
Ng6 Bg8 28. Nh8 Rd6 29. h3 g6 30. Rf1 Bh6 31. Rf2 Bf4 32. Rf1 Bh2 33. f4
Rdd3 34. Rf3 Rac3 35. Rg3 Kf8 36. Rg5 Kg7 37. Rh5 Kf6 38. Rh7 Qc8 39. Rg7
Bh7 40. Rg8 Qa6 41. Rb8 Qa3 42. Rb7 Qb2 43. Bb8 Rb3 44. Rba7 Qd4 45. R7a2
Ra3 46. Rb2 Ra7 47. Rba2 Rb7 48. Ba7 Rb8 49. Rb2 Rg8 50. Rba2 Rg7 51. Bb8
Bg8 52. Rb2 Rh7 53. Rba2 Rh5 54. Rb2 Rg5 55. Rba2 Rg3 56. Rb2 Rf3 57. Rba2
Rf2 58. Rb2 Ra3 59. Rba2 Ra7 60. Rb2 Rb7 61. Ba7 Rb8 62. Rba2 Bh7 63. Ra6
Rg8 64. Bb8 Rg7 65. Ra7 Bg8 66. Rb7 Rh7 67. Ba7 Rh5 68. Rb8 Bh7 69. Rg8 Rg5
70. Rg7 Bg8 71. Rh7 Rg3 72. Rh6 Rgf3 73. Bb3 Kg7 74. Ba4 Kf8 75. Rh7 Ke8
76. Rg7 Bh7 77. Bb8 Kd8 78. Bb3 Kc8 79. Rg8+ Kb7 80. Ba4 Qg7 81. Re8 Bg8
82. Rf8 Bg3 83. O-O-O {50 moves rule} 1/2-1/2 (2012-11-14)
Olaf Jenkner: Warum nicht z.B. 83. Te8 remis?
Was bedeutet das Schlüsselwort unused? (2012-11-14)
Henrik Juel: The castling shows that White never moved his king before; otherwise the position could be reached faster, and the 50 moves rule could not be applied.
The keyword unused seems non-sensical here and should be deleted (2012-11-14)
A.Buchanan: This problem is very interesting. It's orthodox 50M, so that castling does not reset the counter, and the timing works out nicely. But it makes me wonder... suppose we have a problem where it is B32 which was the last reset. Then W83 0-0-0 would prove that the position had already been at 50.0. If 50M convention (which needs to be rewritten because it's a mess) operated like 3Rep, then the game end would have been mandatory after B82. So castling W83 would be illegal. Is this how one would want 50M convention to work? Or should there be a carve-out to say: if you can definitely prove that no-one claimed, then no-one claimed. And should there be a similar carve-out for 3Rep rule? See https://www.thehoppermagazine.com/AA010 (2022-01-20)
A.Buchanan: Personally, I think no carve-outs. The 50M convention should be rewritten as: "A position is considered as a draw if it can be proved that the last 50.0 moves in the proof game combined with the solution did not contain a capture or pawn move. Unless expressly stipulated, this applies only to retro-problems." (2022-01-20)
Thomas Volet: What if the composer intends the retroplay to go beyond 50 non-P and non-capturing moves? (2022-01-20)
A.Buchanan: Hi Thomas - thanks for your question. I think it’s best if we take this offline. I will email you, if that’s ok (2022-01-21)
James Malcom: This is the Volet problem in question: P0008399 In hindsight, it is quite humorous that the 75 move rule was later introduced in the 21st century. (2022-01-21)
Thomas Volet: The question was not directed specifically to P0008399, which is just one of several compositions with the property at issue. (2022-01-22)
A.Buchanan: I've emailed Thomas. My earlier comment should have include checkmate as a third mechanism to zero the count. (2022-01-22)
James Malcom: You can always carry it on over to MatPlus. (2022-01-24)
more ...
comment
Keywords: 50 move rule, Castling (wl)
Genre: Retro
FEN: nB3RbN/1kppppq1/1p4p1/1P3p2/Bp3P1n/4PrbP/2PP1rP1/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-25 more...
18 - P0001941
Nenad Petrovic
2nd International Team Match 1967-1970
3. (2.?) Platz
P0001941
(10+12)
Gewinn
AP
Henrik Juel: If White may castle, last move was g7-g5 to avoid retrostalemate. 1.fxg6ep followed by 0-0 to legitimize the en passant capture. (2004-09-23)
A.Buchanan: In V&V Encyclopedia, which I admire more for its scope than for its precision, this problem is given to illustrate "Petrovic Theme". The definition given is: "PETROVIC THEME Also called 'retroproblem of Petrovic Type'. Mutual dependence of en passant capture and castling. By playing an en passant capture other retro elements of position are legalized (usually castling)."
The definition makes no reference to A Posteriori. I am trying to get my head around the text here, because the e.p. does not "legalize" castling, rather it mandates it. I also encountered Öffner for this castling/e.p. AP (but not in V&V), but that might be someone's confusion in that Types Öffner vs Keym exist in PRA. (2022-02-15)
comment
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic), En passant
Genre: Retro, Studies
FEN: b7/p4P2/2kbPp1p/3ppPp1/n3pp2/8/P1PPP2P/4K2R
Reprints: (2) Problem 141-143 08/1971
(77) Problem 144-147 12/1971
(C) Die Schwalbe 16 10/1972
Encyclopedia of Chess Problems 2012
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
19 - P0002056
Samuel Loyd
Musical World 1859
P0002056
(2+4) C+
#2
1. Da1! droht (unparierbar) 2. Dh8#
1. ... 0-0-0? ist illegal, da sK oder sTa8 schon gezogen haben muss
play all play one stop play next play all
more ...
comment
Keywords: Cant Castler, Castling (sg), Minimal, Miniature, Homebase
Genre: Retro, 2#
Computer test: C+ Popeye 4.61
FEN: r3k3/p1p5/Q3K3/8/8/8/8/8
Reprints: 73 150 Schachkuriositäten 1910
43 64 Schach-Scherze 1915
32 Retrograde Analysis 1915
168 Allgemeine Zeitung Chemnitz 27/11/1927
Arbeiter-Zeitung (Wien) 27/11/1932
8 Comoedia 09/07/1933
(II) Problem 37-40 09/1956
(D10) feenschach 27 04/1975
84 100 Classics of the Chessboard 1983
(7a) Die Schwalbe 145 08/1995
Thema Danicum 95 1999
52 Opfer-Opfer-Matt Gaudium 21 10/2000
Outrageous Chess Problems 2005
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-27 more...
20 - P0002075
Thomas R. Dawson
Pittsburgh Gazette Times 21/12/1913
P0002075
(16+10)
#2
1. 0-0-0?? illegal
1. Td1! ... 2. Td3#
play all play one stop play next play all
Henrik Juel: 1.Td1, not 1.0-0-0? -1... c7 -2.Kd2 Tg3 etc. (2004-03-09)
more ...
comment
Keywords: Cant Castler (wl), Castling (wl)
Genre: Retro
FEN: 8/1p1ppppp/2p1B3/P6R/pP4PP/2N4k/PP2PPrN/R3KQB1
Reprints: 43 Retrograde Analysis 1915
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-08 more...
21 - P0002353
Theodor Tauber
Michel Caillaud

The Problemist 1991
P0002353
(14+11) C+
ser-h#3 (AP)
1. dxe3ep 2. fxg1=L 3. 0-0 Tg4#
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
play all play one stop play next play all
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
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...
22 - P0002471
Gerd Rinder
(G) Die Schwalbe 48 12/1977
Lob
P0002471
(4+3)
#2 (AP)
BTM: 1. ... Lxb7+ 2. Ke3 0-0-0 3. Sb6#
WTM: 1. Sf6+! Kd8,Kf8 2. Dc7#,Ld6#
play all play one stop play next play all
If WTM, b000 rights are already lost. Under Keym AP, Black attempts to steal the move. White disruption of castling now counts as win for White, so the only chance is 1. … BxQb7+ 2. Ke3! thr 3. Sf6+ disrupting castling but 2. … 0-0-0 3. Sb6#. 2. Kf5? Be4+ 3. ~ 0-0-0! or 2. K~? 0-0-0! directly. As usual when flip of player to move in d#n, Black gets an extra move rather than White losing one (c.f. Codex Article 15). Sublime miniature!
VL: AP after Keym. Solution:
I: 1.Sf6+.
II: Bl's try to be on move. 0... Bxb7+! 1.Ke3! O-O-O (legalizing!) 2.Sb6#. (2007-01-26)
A.Buchanan: Very nice problem. But why didn't VL say "Omigod! Sorry chaps: this problem doesn't work without PRA! You need to add it!" Answer: because PRA is irrelevant. So can't we get rid of it from most of the stipulations? (2022-02-15)
comment
Keywords: Castling (sg), a posteriori (AP) (Type Keym), Homebase (s), Aristocrat, Miniature
Genre: Retro, 2#
FEN: r1b1k3/1Q1N4/8/8/4K3/8/7B/8
Reprints: Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-24 more...
23 - P0002476
Werner Kuntsche
1777v Problem 73-78 06/1961
P0002476
(5+7) cooked
h#3 AP
Einheitslösung!
1) 1. cxd3ep Ta5 2. Sf5 0-0 3. Sd4 Txe5#
2) 1. gxf3ep Th5 2. Sd5 0-0-0 3. Sf4 Txe5#
play all play one stop play next play all
Cook: 707 cooks which do not begin with e.p., e.g. 1.Sg3-e2 Ra1-a2 2.Ke4-d3 Rh1-h3 3.e5-e4 Ra2-d2 #
What is the obscure intent behind "Einheitslösung!"? (= "One size fits all solution"?)
Mario Richter: Im Nachdruck in problem 144-147 steht nur auf e5 ein sB (also keiner auf e3). Allerdings gibt es in allen drei Varianten (sBe5+e3, nur sBe5, nur sBe3) jede Menge NL, die ohne Rochade und e.p.-Key auskommen - am wenigsten dann, wenn nur auf e3 ein sB steht. (2010-10-10)
VL: Cf. P0002475. (2012-08-26)
A.Buchanan: Possibly "Einheitlösung!" is a request to unify retro implications derived across separate parts of the PRA solution. But cooked chessically irrespective of these philosophical issues, which still wait to be resolved 60 years on. (2022-02-16)
comment
Keywords: a posteriori (AP) (Type Petrovic - ccee), En passant as key (2), Castling (wb), Symmetrical position, Symmetrical solution, Superseded by (P1399112)
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/8/8/4p3/2pPkPp1/2n1p1n1/8/R3K2R
Reprints: (66) Problem 144-147 12/1971
Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-17 more...
24 - P0003269
Tivadar Kardos
149 British Chess Magazine 10/1956
P0003269
(5+17) cooked
h#3
1. exf3ep e3 2. 0-0-0 Tb4 3. Tg4 Tb8#
I think that this cooked problem is an early A osteriori. I can see ways that it might be made sound, but I would like to see the published version. San anyone help please? Thanks.
play all play one stop play next play all
Cook: 17 Black pawns, retro cook and numerous forward cooks
klären wK im Schach, vielleicht wBb2?
Alfred Pfeiffer: 9 schwarze Bauern! (2012-02-07)
Ladislav Packa: Auch mit wBb2 NL, z.B. 1.exf3 e.p. Txg5 2.0-0-0 Tb5 3.Tg~ Tb8# (2012-02-07)
A.Buchanan: And adding to earlier comments, if we do swap sBb2 for wB, the retraction is still not unique with R: 1. f3-f4 Lh6xg5+ (2021-11-26)
A.Buchanan: I think that this cooked problem is an early A Posteriori. I can see ways that it might be made sound, but I would like to see the published version. Can anyone help please? Thanks. (2022-04-25)
Gerald Ettl: Verbesserungsvorschlag: -sBa2, -sBb2, -sSg7, +sBg7
r3k3/3b2p1/5p2/6b1/4pPRp/2pq2rp/2p1P1pB/2K3n1/ (2022-04-25)
Gerald Ettl: und +sSg8 (wegen exf4 Vermeidung) (2022-04-25)
Gerald Ettl: +sSa4 nicht g8 (2022-04-25)
A.Buchanan: Hi Gerald. Thanks for this. Your final proposed version is r3k3/3b2p1/5p2/6b1/n3pPRp/2pq2rp/2p1P1pB/2K3n1, yes? There are 16 candidate solutions, all using castling. But wPa cannot leave the a-file, and is required for capture balance, so Black cannot castle. Five of the candidates begin with e.p.: play might have just been R: 1. Kb1-c1 b3xPc2+, so e.p. is not permitted in any case. What am I missing? (2022-04-25)
Gerald Ettl: Hi Andrew,
Those were 2 mistakes I made. I improved it again.
r3k3/3b4/4pp2/6b1/4pPRp/n1pq2rp/PPp1P1pB/2K3n1/
(den wBa2 habe ich auf das Brett gestellt, da sonst wieder Lh6xXg5 geht.) (2022-04-26)
A.Buchanan: Hi Gerald: Deine Retro-Logik ist gut. Aber es gibt 18 Lösungsvorschläge für die Zukunft. Ich habe heute im Discord meinen eigenen Vorschlag zur AP-Korrektur von Kardos veröffentlicht. Ich werde ihn hier im PDB hinzufügen. (2022-04-26)
comment
Keywords: En passant as key, Castling
Genre: h#, Retro
FEN: r3k3/3b2n1/5p2/6b1/4pPRp/2pq2rp/ppp1P1pB/2K3n1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-26 more...
25 - P0003297
Wilhelm Hagemann
Deutsche Nachrichten (Sao Paolo) 1960
P0003297
(5+9)
h#2
1. cxd3ep f4 2. Lf3 Kxc5#
play all play one stop play next play all
Henrik Juel: C+ Popeye 4.61 (2022-04-18)
Henrik Juel: Obviously, last move was d2-d4 (2022-04-18)
comment
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/8/p7/P1r2p2/RKpPk3/p1p1p3/5P2/3b4
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2022-04-18 more...
26 - P0003411
Norman Alasdair Macleod
3970 Themes-64 04-06/1982
P0003411
(4+6) C+
h#2
2.1...
1. Kc3 0-0-0 2. Txc4 Txd3#
1. bxc3ep e4 2. Kc4 Ta4#
play all play one stop play next play all
Two solutions, with no retro tries or set play. The question is how info about game state should spread between one solution and another.
Mario Richter: How is the (AP) to be interpreted here? Is the intention as follows: Since in solution 1) White castles, Black is allowed to capture e.p. in solution 2)? (2011-05-28)
Henrik Juel: It seems to be an unusual situation, not covered by keywords like AP or PRA.
If last move was c2-c4, both solutions work; if not, there is no solution, because White may not castle.
(In the second solution, Kxc4 should be Kc4). (2011-05-28)
VL: This a generalized type of AP, which I call "consolidate AP": both solutions are considered as parts of one complete solution. However the order of both parts is significant. I know two similar problems: P0003437 and P0003186. (2011-06-01)
A.Buchanan: @Valery: I've started to classify all the AP problems in PDB. This and its kin are "Type Petrovic - cons" (standing for consolidated, but there's only a limited number of characters for a PDB parameter). I truly hope that the theory in the end will not have this as a special case, but rather is part of a more general pattern.
It's not just that PRA & RS each scales up within its own paradigm to handle seamlessly very complex collections of conditional move dependencies. They also scale *down* to handle situation when there is just one conditional move, or even zero conditional moves. And when n=0 or n=1, PRA & RS are in complete agreement. This is the kind of robust seaworthy behaviour we need if are going to set sail in the stormy seas of fairydom.
But small steps. The first thing is to classify all the problems in PDB into buckets, to find out what we have. I doubt there are 37 categories, so will probably need the results of your scholarship too. (2022-02-17)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic - cons), En passant as key, Castling (wg)
Genre: h#, Retro
Computer test: HC+ Forward play proved by Popeye v4.87 AP logic requires some framework but I think we know we want this cool idea to end up being sound.
FEN: 8/8/8/1nr5/1pPk4/1p1p4/4P3/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-17 more...
27 - P0003417
John Frederick Keeble
2206 The Problemist FCS 16, p. 173, 02/1936
P0003417
(4+5) C+
h#2.5 (AP)
1. ... exd6ep 2. 0-0-0 dxe7+ 3. Tf8 exf8=T,D#
play all play one stop play next play all
H.Juel: If Black may castle, last move was not made by Ke8 or Ta8, but by Pd5, so it was d7-d5, and hence White may capture en passant. 1... exd6ep 2.0-0-0 dxe7 3.Tf8 exf8=D,T# where the castling (belatedly) justifies the e.p. capture.
Not 1... exd6ep 2.Kd8? dxe7 3.Kc8 e8=D,T
AB: Another 120 retro tries (=60x2 due to tolerated dual promotion) of general form 1. exd6ep 2. T~ dxc7 3. T~ c8=D,T.
No set play
Nachdruck in "Die Schwalbe" 33 mit Diagrammfehler (wBe6 fehlt).
VL: The first AP-problem! Thematic illegal try:
1. ... exd6ep 2.Kd8? dxe7+ 3.Kc8 e8=T(D)#?? (2007-02-11)
Mario Richter: 46 years between this Problem and P1366663 - perhaps this was not Keeble's first AP-Problem?! (2019-08-29)
Ladislav Packa: Of course I know the idea a posteriori. But let my heresy be forgiven - I do not consider it logical. In this particular example, I consider e.p. for proof that black can castling - not vice versa. (2019-08-29)
A.Buchanan: Hi Ladislav. If ep is ok, then Black just moved a pawn with double hop. That doesn't stop Black having e.g. moved bK right at the beginning of the game. So you can't prove that Black definitely can castle this way. What am I missing here? (2019-08-29)
Ladislav Packa: Your statement is perfectly fine. However, the AP condition is usually interpreted as castling is evidence of e.p. However, the second party does not have to do castling - the possibility has already been demonstrated by the existence of e.p. (2019-08-29)
A.Buchanan: Hi again Ladislav. Please try again, I am obviously being very stupid. I don't understand how 1. exd6ep is any more indicative that Black can castle than 1. K~. (2019-08-30)
Ladislav Packa: Hi, Andre!
e.p. is just a result of the d7-d5 move. Black can castle only if it was the last move of black (even without wPe5). Therefore, it does not have the logic of joining the e.p. with castling. For example, ask yourself if black is allowed to castle in the position without wPe5. (2019-08-30)
A.Buchanan: @Ladislav: ep is not "proof that black can castle", rather it is a *necessary* condition for Black to be able to castle. The logical implication is the other way round. We can castle because of the optimistic convention, and because sPd might just have moved. Without wPe, there is no solution, but whatever White plays, Black can castle (2020-12-11)
Hans-Jürgen Manthey: was heißt hier Thematisch Illegal ??
Mir völlig wurscht ob ep möglich oder nicht, nach dem weißen Zug ist natürlich Kd8 möglich !!
also ist auch 1. ... exd6ep 2. Kd8 dxe7+ 3. Kc8 e8DT# eine Lösung, sonst müßte auf die erzwungende Rochade in der Aufgabenstellung hingewiesen werden ! (2020-12-11)
A.Buchanan: Can one tolerate “tolerated dual promotion” in Valladao? Well Valladao himself did, so it must be ok (2022-03-26)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sg), Promotion (D/T), Tolerated dual promotion (D/T), Valladao Task
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: r3k3/2p1p3/2P1P3/2KpP3/8/8/8/8
Reprints: V. Die Schwalbe 33, p. 323, 06/1975
2001 CHM Themes 12/2000
3 Die Schwalbe 215, p. 240, 10/2005
A1 Problemkiste 169 02/2007
14 Die Schwalbe 241, p. 374, 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-26 more...
28 - P0003428
Imre Sztankovszky
130 Die Schwalbe 09-10/1953
P0003428
(10+11) C+
h#1.5 AP
1. ... fxg6ep 2. 0-0 gxh7#
play all play one stop play next play all
Henrik Juel: Black captured c7xd6x..x.h2 and once more with an officer
White captured a2xb3, b2xa3, f2xg3, and e.g. exf and once more
If Black may castle, last move was g7-g5 (not b7-b6, because of Lh3) (2022-04-26)
Henrik Juel: HC+ Popeye 4.61
The castling serves two purposes:
enabling the mate and legitimizing the ep capture
so the a posteriori legitimizing is impure (2022-04-26)
A.Buchanan: Yes Henrik. The absence of other candidate solutions (beginning e.p. but excluding 0-0) declines an opportunity to embed additional content, and is arguably an artistic defect. However this very early AP problem is sound, and purity of motive is just one school. IMHO, there is space for such problems, if other content compensates: e.g. 4k2r/p2ppp1p/p7/5PpK/8/1PBB2Pb/2PP2Pp/8 h#2* AP. (2022-04-27)
more ...
comment
Keywords: En passant as key, Castling (sk), a posteriori (AP) (Type Petrovic), Volet Pawn
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 4k2r/3ppp1p/1p6/4BPpK/P7/pPPB2Pb/3P2Pp/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-27 more...
29 - P0003434
Jozsef Bajtay
2432 Problem 101-102 09/1966
P0003434
(10+11)
h#2
1. fxg3ep 0-0 2. Lg4 hxg3#
play all play one stop play next play all
A.Buchanan: Surely diagram typo. Change to sBh3, then everything works (2022-03-29)
Mario Richter: Yes, Pawn h3 is black (2022-03-29)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk)
Genre: h#, Retro
FEN: 8/b3p3/4p3/6pp/2P2pPk/1pPP3p/2PP1P1P/r2bK2R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-03-29 more...
30 - P0003442
Janko Furman
3. Makuc-Moder-Gedenkturnier 1971-1973
1. Preis
P0003442
(12+13) C+
h#2
b) wBd4 nach d5
a) 1. cxd3ep Sd5 2. 0-0 Se7#
NOT 1. ... Lf6? 2. Kf8 Tx8# because no AP justification
b) 1. Kd7 Lf6 2. Te8 Sxb6#
NOT 1. 0-0? Tf6 2. Kh8 Txf8# because rights lost
play all play one stop play next play all
Assume that bPb6 is really on b7:
Bl captures: dxe, exf, fxg, a|
Wh captures: bxc, g|, cxd=
So all pawn captures are accounted for.
a) If Bl 00 rights remain, then only way to give Black a prior move is by d2-d4.
b) No way to give Black a prior move, so Bl 00 rights must be lost.
Cook: NL
b) 1. 0-0 Tf6 2. Kh8 Txf8#
A.Buchanan: There is definitely something wrong here, with both (a) & (b). I think it's a simple diagram error: bPb7 has been misplaced on b6. Then the AP logic for (a) works great, and the castling "NL" for (b) is seen to be a thematic retro try. Can anyone confirm? (2022-03-21)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sk)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/7p/1pR5/2P5/NNpP4/KB2PPPP/p3pppq/B4bnr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
31 - P0003444
Janko Furman
Miroslav Stosic

7273 Schach-Echo 11/1972
P0003444
(8+14) cooked
h#2
b) Gespiegelt (a1<->h1) & wKd1->e1
a) 1. dxe3ep 0-0 2. Te4 Txf3#
b) 1. exd3ep 0-0-0 2. dxe2 Ld5#
play all play one stop play next play all
Cook: a) 1. Kg3 Kf1 2. Df4 Th3#
1. Kg3 0-0 2. Dh4 Txf3#
A.Buchanan: This is a very heavy position to prevent the possibility of R: 1. c2xb3. I'm not sure why this was done. A much lighter position 8/8/8/5np1/1r1pPkr1/2Bp1p2/1p1P2P1/4K2R with 7 less units achieves the mates soundly. Am I missing something? (2022-02-16)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk,wg), Superseded by (P1399967), Twinning by board reflection
Genre: h#, Retro
FEN: 8/1p1pp1p1/8/5nq1/1r1pPkr1/1PBp1p2/Pp1P2P1/2n1K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
32 - P0003451
Knud H. Hannemann
3823 Stella Polaris 03/1971
P0003451
(6+8) cooked
h#2 AP
1) 1. gxf3ep 0-0-0 2. Te7 Th4#
2) 1. cxd3ep 0-0 2. Da4 Txa4#
play all play one stop play next play all
AB: I think this is cooked. I don't see how AP castling can justify earlier ep here.
1. Kf3 Bf5 2. Kg2 Be4#
1. Kf3 Bxd7 2. Kg2 Bc6#
1. cxd3 Rf1 2. Qa4 Rxa4#
1. gxf3ep Ra5 2. Kf4 Rh4#
1. gxf3ep Rd1 2. Re7 Rh4#
Also, second solution given has typos 1. *c*xd3ep & 2. Q*a*4. (2002-03-21)
V.Liskovets: Indeed this problem is cooked, and I
failed to correct it preserving symmetry.
Here is a possible correction:
W: Ke1 Ra1 Rh1 Be6 Pd4 Pf4
B: Ke4 Qd7 Rf7 Pc3 Pc4 Pe3 Pg3 Pg4 Bh7 Rb2 Pb3

Another story, justifying its contents (e.p.).
In my opinion, all published treatments are
insatisfactory (the same concerns P0004295
(corrected) by Werner Kuntsche as well).
There is a way to make it sound under the
sophisticated HYBRID stipulation 'AP, pRA':
2 partial solutions legalized JOINTLY by
both castlings (details to be published). (2002-04-02)
A.Buchanan: Valery's suggestion doesn't seem to work: 8/3q1r1b/4B3/8/2pPkPp1/1pp1p1p1/1r6/R3K2R. Have I got something wrong? (2022-02-14)
VL: Sorry, Andrew, I see no issues with my version (excepting the mentioned special retro-convention/genre for justifying its soundness, of course!). h#2 (pRA&AP). The full solution consists of 2 partial AP-based ones: I 1.gxf3 e.p.(!?) 0-0-0! 2.Te7 Th4#; II 1.cxd3 e.p.(!?) 0-0! 2.Da4 Txa4#.
C+(popeye): h#2 & two h#1.5 after the keys (added Tb2 & Lh7 are cookstoppers).
The main features of the corresponding suitable AP-genre ("consolidated")are rather clear. However, not all subtle aspects have been analyzed thoroughly yet. I have got only a draft manuscript with few examples, still. (2022-02-15)
A.Buchanan: Hi Valery: I sent you an email couple of days ago - please check your spam folder if you haven’t received. It will shortly be the 20th anniversary of your 2002-04-02 post above where you talk about “details to be published”. It’s hard to get low urgency stuff done I know, but I’ve been patient all these decades. I want to help you! Please can you send me the 37 patterns that you have (another post of yours somewhere here). Happy to have a zoom call too if that works for you :-) (2022-02-17)
more ...
comment
Keywords: Castling (wb), En passant as key (2), a posteriori (AP) (Type Petrovic - ccee), Superseded by (P1399178), Quasi-symmetrical position
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/3q1r2/4B3/8/2pPkPp1/2p1p1p1/8/R3K2R
Reprints: (65) Problem 144-147 12/1971
(8) StrateGems SG19, p. 156, 07/2002
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-17 more...
33 - P0003736
Giuseppe Brogi
1249 Sinfonie Scacchistiche 07-09/1972
P0003736
(12+12) cooked
h#2
b) wSa5
a) 1. f5 Le5 2. 0-0 Th8#
b) 1. Kd8 0-0-0 2. Te8 Lf8#
play all play one stop play next play all
Try to deduce the diagram error, as we do have the intended solutions!
a) wRh7 is original, but cannot have escaped from cage by white cross-capture as there are too many captures needed. On the other hand, Black only has to make 4 captures so the cross-capture is ok.
b) changing the colour of Sa5 shifts the capture balance: now White can cross-capture but not Black, so White but not Black can castle.
Thus the retro-logic works with the diagram as it is, but the forward logic does not. How can we repair the diagram?
See P0000899 a companion problem.
A.Buchanan: Something odd going on here. There are numerous h#2 in both a&b. PRA not normally found in a problem with single solution for each twin. White has 8 pawns so trivially wRh7 must be original. (2018-10-13)
A.Buchanan: There are three similar problems by Brogi which are all twinned RS rather than PRA, but don't have diagram error. P0003743, P0003746 & P0003747. (2018-10-13)
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
Ladislav Packa: Cooked a) and b):
1...b8S and 2...R:h8# (2021-02-10)
A.Buchanan: I think the composer simply forgot that wPb7 can promote. Most of the cooks come from S promotions, but it's also possible to have QR. I've fixed it, rejigging the captures to make them still add up. I've posted in Discord, as I did with the partner composition, and will create entries for them here. (2022-03-15)
comment
Keywords: Castling (wgsk), Cant Castler (wgsk), Cross-capture (s,w), Superseded by (P1399806)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: 4k2r/pPp2p1R/n1pB1ppp/npP5/1P6/5PPP/2P2P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-15 more...
34 - P0003826
Radu Dragoescu
RA29 diagrammes 20 03-04/1976
P0003826
(9+9)
Wieviel Bauern muß man entfernen, damit die Stellung legal ist?

Zero - the position is legal.
A.Buchanan: Isn't this a legal position? In each of the 4 pairs of adjacent files, one cross-capture suffices. So a total of 8 captures is necessary, with up to 14 available. (2022-03-17)
Henrik Juel: Yes, any master of the 20 bishops problem can see that the position is legal
The diagram must be wrong (2022-03-17)
A.Buchanan: Maybe it's a correct composition, with a humorous solution: "zero"? All that's wrong is that it shouldn't be tagged with keyword "illegal position". I will fix that (2022-03-17)
comment
Keywords: Kindergarten Problem, Remove pieces
Genre: Retro
FEN: 4k3/8/PPPPPPPP/8/8/pppppppp/8/4K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-17 more...
35 - P0003833
Gaston Balbo
L'Echiquier de Turenne 1958
P0003833
(10+5) C+
#3
1. d3 ... 2. Tc2 ... 3. Tc8#
play all play one stop play next play all
Henrik Juel: Black just moved K or T, so he cannot defend with 1... 0-0
White moved his king to allow Te2 to enter, so he cannot mate with 1.0-0 thr. 2.Tc1 thr. 3.Tc8# (2022-05-10)
more ...
comment
Keywords: Castling (wksk), Cant Castler (wksk), Homebase
Genre: Retro, 3#
Computer test: C+ Popeye 4.61 and analysis
FEN: 4k2r/N3p1pp/4P3/8/8/4P3/3PRP1P/B3K2R
Reprints: RA38 diagrammes 21 05-06/1976
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-11 more...
36 - P0003839
Radu Dragoescu
RA44 diagrammes 23 09-10/1976
Jean Oudot gewidmet
P0003839
(16+16)
Wieviele Figuren muß man entfernen, damit die Stellung legal wird?
A.Buchanan: In each of the 4 pairs of adjacent files, one cross-capture suffices. So removing 8 officers is enough. However, they need to be an even number from each side.
However it's possible to do better by removing pawns: wPadgh bPcf = 6 units. Now wPaxb bPcxb wPdxe bPfxe.
Is it possible with 5 or less? (2022-03-17)
more ...
comment
Keywords: Remove pieces, Illegal position
Genre: Retro
FEN: rnbqkbnr/8/PPPPPPPP/8/8/pppppppp/8/RNBQKBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-17 more...
37 - P0003960
Arthur Ford Mackenzie
467 Jamaica Gleaner 1891
P0003960
(3+1) C+
#3
1. La7 Kxa7 2. c8=T Ka6 3. Ta8#
play all play one stop play next play all
Max J. Meyer in 'Brighton and Hove Society', 1904: "It will be seen that the ideas of these two problems [dieses und P1143870] are the same. The addition of the White B. enables Mr. Mackenzie to get a much better key for this position, a point in which Steinweg's problem does not excel; but, of course, the use of an extra piece renders the miniature less remarkable from the point of view of smallness in size."
Paulo Roque: .
Solution:
1) 1. La7 KxLa7 2. c8=T! Ka6 3. Ta8#

2) Proof da legality of the answer:
Retro: 0... Ka7-a8 -1. b7-b8=L etc (2009-09-06)
Sally: Der letzte weiße Zug von Weiß kann nur Ba7xb8L gewesen sein, demgemäß enthält die Aufgabe 2 Bauernumwandlungen
in verschiedenen Figuren: in der lösung in einem turm und retrospektiv in einen Läufer.
406 Mattaufgaben mit 3 und 4 Steinen (Teil 1) Speckmann 1976 (2016-08-30)
Henrik Juel: There is even a third promotion in the threat 2.c8=D+
C+ by Popeye 4.61 (2016-08-30)
A.Buchanan: Here the threat play (1) is sound & (2) can never occur as Black has no neutral move. But if threat play is *unsound* yet can never occur, can it be safely ignored or will some directmate purists regard that as a defect? (2022-04-02)
Henrik Juel: I think not, Andrew
When I test problems, I ignore plays that do not occur (2022-04-02)
comment
Keywords: under-promotion (T), Stalemate avoidance, Promotion in the retro play (L)
Genre: 3#, Retro
Computer test: Popeye 4.61
FEN: kB6/2P5/2K5/8/8/8/8/8
Reprints: 64 Chess Lyrics , p. 100, 1905
1359 Wiener Hausfrauen-Zeitung 5, p. 77, 29/01/1905
5 Vliegend Blaadje 22/07/1911
Szachy 01/1958
The Problemist (20) 03/1969
406 Mattaufgaben mit drei und vier Steinen 1976
diagrammes 64 01-02/1984
(1) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2020-02-26 more...
38 - P0004199
Werner Frangen
Problem 41-44, p. 52, 03/1957
P0004199
(14+10) cooked
#5
1. axb6ep+ Kb5 2. bxc7+ Tb6 3. 0-0-0 ...
play all play one stop play next play all
K. Fabel: "Von Interesse ist, dass diese strittige Idee [Beweis des e.p.-Schlagrechts durch spätere Ausführung der Rochade] auch im direkten Mattproblem dargestellt werden kann, vergl. das Diagramm [diese Aufgabe P0004199]. (Vielleicht ist diese Aufgabe noch nicht korrekt doch es wird nicht schwer sein, sie zu verbessern). Falls Schwarz zuletzt Bb6-b5, Tb6-c6 oder Lb6-c7 gezogen hat, ergibt sich, dass Weiss im letzten oder vorletzten Zug den K oder T bewegt haben muss. Falls jedoch K und T noch nicht gezogen haben, muss b7-b5 der letzte Zug gewesen sein. Weiss spielt daher 1. ab6e.p.+ Kb5 2. bc7+ Tb6. Jetzt könnte Weiss mit Dxb6 sofort mattsetzen, aber er muss ja den e.p.-Schlag noch legalisieren. Daher 3. 0-0-0! und Matt erst im 5. Zuge."
Cook: 1. Sxc7+!
1. Lxc7!
Innerhalb des 2. Teils einer Artikelserie "Die Konventionen im Problemschach" von Karl Fabel.
A.Buchanan: The retro logic is fine, but forward play is savagely cooked. I wonder about changing wSc8 to sS. The retro stuff still works OK, indeed sSc8 is a fourth thematic unit to retract in the try. Ignoring proof of ep legality, there is a unique #4 beginning with ep, and no #5. Promising: but when can we interpolate w000? Any solution must begin 1. axb6ep+ Kb5. There are now numerous #4, but none include w000, so I think we must stick with the still unique line beginning 2. Dxc8 thr 3. 0-0-0. There are 3 black defenses to refute this. 2. ... Txb6,axb3,d5. So maybe #6 needed for White to prevail? (2022-03-28)
A.Buchanan: But if we shift to #6 then 1. Sxc7!,Qxc8+! mate without ep. Can anyone retrieve the original diagram for this one, please? (2022-03-28)
Mario Richter: The position here is identical to the original diagram, but it should be regarded more as a schematic example than a "real problem". Fabel in his quote above: "... perhaps the problem is still not correct, but it should be easy to improve it ..." (2022-03-28)
Henrik Juel: unfortunately K. Fabel forgot to give the easy correction... (2022-03-28)
A.Buchanan: Thanks for this. Valery Liskovetz, an AP expert, was kind enough to send a pdf of the relevant Problem issue, so I can confirm. It's easy enough to remove the two cooks (e.g. sTc8) but the difficulty is in validation of the intended solution. Is there any "exact" option in Popeye that forces the solution to include a waiting move? The actual solution might be expected to be included in that set. (2022-03-29)
Henrik Juel: Yes, Andrew
'stip exact-#5' would disregard #4 as a solution, I believe (2022-03-29)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Non-standard material (TLL)
Genre: Retro, n#
FEN: NQNB4/B1brpp2/k1rp4/Ppp5/Rp6/BP6/RPP5/R3K3
Reprints: (8) Problem 101-102 09/1966
(52) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-29 more...
39 - P0004296
Luis Alberto Garaza
1809 Problem 73-78 06/1961
P0004296
(16+13) cooked
h#1? h#2?
1. fxe3ep? Dxa8# No justification for the ep, so just retro try.
play all play one stop play next play all
White pcs: dxexf5,hxBg. Black none. Assume that Black can still castle. White can't have just played h4xg5 as sBh2 would be blocked. White might apparently just have moved D,Se5,Td4,Tc3,g4-g5,a5-a6,e3-e4 or e2-e4 but what might Black have played before? Only the last allows a move sDf1-e1. So set up for AP Type Petrovic is OK.
Cook: 446 candidate h#2
However 27 have no ep, while 59 have both ep & castling.
So a total of 86 viable solutions, 85 of which must be cooks.
s.a. Version P0004341
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004341)
Genre: h#, Retro
Computer test: Popeye v4.87 indicates cook
FEN: r3k3/p1ppp3/Pp6/4NPP1/2PRPp2/2RK2PN/1PBn1PQp/2Brq1b1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
40 - P0004341
Luis Alberto Garaza
(9) Problem 101-102 09/1966
P0004341
(15+12) cooked
h#1 oder h#2
1. fxe3ep Dxa8#? because no AP justification for ep

1. 0-0-0 g7 2. Tf8 gxf8=D#
play all play one stop play next play all
62 apparent h#2, so seems cooked.
Cook: 415 candidate solutions for h#2. 401 begin with ep, of which 48 contain 0-0-0 to justify. There are also 14 solutions without ep, of which 13 begin with 0-0-0. The odd one is 1. Kd8 Sc6+ 2. Kc8 Sge7#
Version zu P0004296

Autor: "In Nr.9, I intended to do something similiar to No. 8 [P0004199], but with a different key: the solution 1. PxP "e.p.", Qxa8 checkmate will not go, for if blackside does not castle, the "en-passant" capture cannot be justified. Though, well considering, there is the demolition 1. PxP "e.p." -any 2. 0-0-0! Qb7 or a8 checkmate.
A.Buchanan: Thanks Mario for retrieving the author's intent with this. So he knew it was cooked! Assume s000 rights remain. R: 1. b5xa6? as sBb promoted on b1. R: 1. h4xg5? as sBh retro-blocked. R: 1. Ke3-d3? impossible check from sBf4. R. 1. K~-d3 d3-d2? illegal check. I like these! So by elimination, R: 1. e2-e4 e3xTd2 2. T~d2 L~e1. So must have sLe1. (2022-03-22)
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004342)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k1N1/p1pp4/P5P1/4PPp1/2PNPprB/3K1QRp/P2p1PB1/4b2n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
41 - P0004342
Luis Alberto Garaza
(10) Problem 101-102 09/1966
P0004342
(15+11) cooked
h#2
1. bxc3ep Dg8#? (thematic try)
1. bxc3ep Dd5 2. 0-0-0 Db7#
play all play one stop play next play all
Cook: 1. bxc3ep Ld6 2. 0-0-0 Dc4#
Author: "From the former [d.h. P0004341], the No. 10 was born, something similar to No. 8 [P0004199] but with Black-castling for it is a help-mate. The mating move 1. ... Qg8? will not go."
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P1399966)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k3/p2p4/8/NP6/BpPPN1PP/B2K1Ppp/QP1p1pP1/Rrb5
Reprints: (55) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
42 - P0004454
Immo Fuß
Die Schwalbe , p. 463, 03/1939
Allen Teilnehmern herzlich gewidmet
Internationaler Lösungswettkampf 1938
P0004454
(13+11)
#3
1. Tf1? 0-0! Thematic retro try
1. Dxd7+,Sf6+,exd7+,exf7+? Kf8!
1. Dxa8+? Sd8!
1. Sh7? Txh7!

1. 0-0! (droht 2. Txf7,Dxa8+) 0-0? illegal
1. ... Lxb2,Lxb4,b5,c6 2. Dxa8+ Sd8 3. Sxc7#
1. ... Kf8 2. Txf7+
1. ... Tg8 2. Dxd7+,Dxa8+
play all play one stop play next play all
Suppose both players can castle, and derive a contradiction.
White captures: a3xb4,fxe,g2xh3,Bc8 & QR in cage a8--d8.
Black captures: a7xb6,fxe/g,Bc1. bPg promoted. bPf promoted or was captured by wPfxe.
So all captures accounted for. Pieces captured by pawns were wRB & bQXX
b6 is dark, so light wBf1 was not captured there. By elimination, a7xRb6, which released bQ.
Was an original officer captured on b4 to release wR?
bQ not yet released
bRh never moved, bRa captured in cage
bB wrong shade, bBc8 captured in cage
bS couldn't escape g1, and two others on board.
So it must have been a promoted officer captured on b4 earlier.
What was captured on h3, to open the line for promotion on g1? Must be original as wPa & bPa have not yet captured. For the same reasons as axb4, we can eliminate all 4 possible officer types.
Contradiction! So at least one player cannot castle.

We apply the Retro Strategy (RS) convention, and White is permitted to castle, while the range of possible histories shrinks to exclude all cases where Black might have castled. This common special case of RS is also known as Mutual Exclusion.
Kees: Only one castling is legal With black castling there's no #3
1. 0-0!
1. ... Kf8 2.Txf7+ Kg8 3 Dxa8#
1. ... c6 2. Dxa8+ Pd8 3. Pc7#

(axNb3, and for f1=N or g1=N wK must move. so bD must pass bK)
Somebody can better explain than me. (2022-02-14)
A.Buchanan: Hi Kees thanks for the solution which grabs the essence - I have used more words, please point out any slips I might have made! :) (2022-02-15)
comment
Keywords: Retro Strategy (RS), Castling, mutual exclusive (wksk)
Genre: Retro, 3#
Computer test: Popeye v4.87 for forward play Non-trivial thinking for retro logic
FEN: n3k2r/1ppppn2/1p2P3/3N2Np/QP6/b6P/1PPPP2P/4K2R
Reprints: (14) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
43 - P0004481
Matjas Zigman
Länderkampf Mazedonien-Slowenien 1969
7. Platz
P0004481
(13+10) cooked
ser-h#6**
* 1. ... Sb5#
* 1. ... dxc3#
1. fxe3ep 2. e2 3. exf1=S 4. Sxd2 5. Sxb1 6. Sxa3 0-0-0!#
play all play one stop play next play all
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: 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)
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)
comment
Keywords: Castling (wg), Seriesmover, a posteriori (AP) (Type Petrovic), Illegal position, En passant as key, Promotion (s), Valladao Task
Genre: Retro, Fairies
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, 2022-03-20 more...
44 - P0004657
Mato Prikril
(27) Problem 206-210 07/1981
1. ehrende Erwähnung
P0004657
(5+7) cooked
h#2
1. fxg3ep Ta4#? (no proof of ep right)
1. fxg3ep 0-0-0! 2. g2 Td4#
play all play one stop play next play all
A short retro try h#1 and 16 h#2 tries. B2 is unique tempo move
Cook: 1. Kxg4 Ta5 2. h4 Tg5#
1. Kxg4 Ta4 2. Kh4 Txf4#
61. TT (Pavlovic Memorial), Gruppe A
Ladislav Packa: NL 1.Kxg4 Th4 2.Kh4 Txf4# (2012-02-07)
A.Buchanan: Easy to patch the cook with wPa2, but harder to do while retaining Meredith and motivation for 000. (2022-03-09)
A.Buchanan: Maybe the easiest resolution is to *remove* wPf2 instead of adding anything, and declare 2 solutions with now just 4+7 units. Flipping left-right, and nudging wK back to e1 generates more retro tries. (2022-03-09)
comment
Keywords: Castling, En passant as key, a posteriori (AP) (Type Petrovic), Tempo Move
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 8/6p1/6pB/7p/5pPk/5p1p/5P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
45 - P0004778
Hans Klüver
30 Schachkongress Teplitz-Schönau im Oktober 1922 , p. 484, 1923
P0004778
(13+14)
Weiß nimmt 1 Zug zurück, dann #2
R: 1. Lb6-d8, dann 1. bxc6ep
play all play one stop play next play all
Henrik Juel: White captured g6xDg7 and hxLg
Black captured b5xa4, fxe, and once more with an officer
Following the retraction, it is clear that last move was c7-c5
Possible retroplay R: 1.Lb6-b8 c7-c5 2.Kd4-d5 e6-e5+ 3.Sa2-b4 Kc6-b7 4.b4-b5+ etc. (2022-04-18)
comment
Keywords: Help retractor, En passant
Genre: Retro
FEN: nrbBB1n1/rk1pp2P/p4Ppp/PPpKpN2/pNP1P1P1/3P4/8/8
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2022-04-18 more...
46 - P0004917
Andrej N. Kornilow
(C) Die Schwalbe 80 04/1983
P0004917
(5+8)
#2 (AP)
1. ... Tfxg8 2. Lg6! h5,~ 3. Lxf7+
2. ... fxg6=?,hxg6=?,Txg6#?
2. ... Tf8,Tg7 3. Lxf7+ Txf7#

2. Lxe6? Tg6+! 3. Kf5 0-0!
2. ... dxe6? 3. d7+!
2. ... fxe6=?

2. Lxh7? Tg6+! 3. Lxg6 0-0!

Therefore it's WTM
1. Dg2 h5,~ 2. Da8#
play all play one stop play next play all
Black cannot steal the move, as White can prevent the castling justification.
A.Buchanan: Ingenious play, but Black cannot execute the castling. Lines include a good try, checkmate by Black and pat by Black. (2022-04-16)
A.Buchanan: There are two kinds of directmate Type Keym. In one White successfully pushes the move to Black, in the other (as here) Black unsuccessfully pulls it. (2022-04-17)
Ladislav Packa: The logic of this problem is foreign to me, but it is incorrect: 2.Bxe6 Rg6+ 3.Kf5 0-0 (2022-04-18)
A.Buchanan: Hi Ladislav thanks for this. You're right. So 2. Lxe6? Tg6! 3. Kf5 0-0! is another try. The solution must be 2. Lg6! I'll post the solution above (2022-04-18)
more ...
comment
Keywords: Castling (sk), a posteriori (AP) (Type Keym)
Genre: Retro, 2#
FEN: 4krQr/3p1p1p/3PpK1p/4PB2/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-18 more...
47 - P0004920
Valery Liskovets
(F) Die Schwalbe 80 04/1983
P0004920
(5+4)
#2 (AP, pRA)
BTM 1. ... Txh6? 2. Sd7! Txc6+ 3. Kb5! (Kb7?) Td6 4. Sf8 Tf6 5. Sh7! Th6 6. Sf8 no castling
5. a6? Tfxf8 6. a7 Tf5+ 7. Kc6,K~ 0-0!
1. ... 0-0? 2. Se7+! Kh8 3. S5g6#
WTM 1. Td6 droht 2. Td8#
play all play one stop play next play all
A.Buchanan: A key feature of adversarial A Posteriori is that any castling must be forced in a finite number of moves (but not necessarily limited by the number of moves in the stipulation goal). If the other side can prevaricate indefinitely, then that is sufficient to defeat the A Posteriori "steal" (2022-02-16)
comment
Keywords: a posteriori (AP) (Type Keym), Cant Castler, Castling
Genre: Retro, 2#
FEN: 4k2r/6pr/K1N4R/P3N3/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-16 more...
48 - P0005637
Osmo Ilmari Kaila
Hannu Harkola

(D) Die Schwalbe 48 12/1977
P0005637
(6+1)
#3
AP
WTM: ???
BTM: 1. ... Kxf3 2. 0-0+! Ke3 3. Td1 Kf3 Td3#
not 2. Kd2? Kg2 3. Sf4+ Kf2,Kf3 4. Tf1#
play all play one stop play next play all
A.Buchanan: I don't get the regular part of this. White can mate in 3 in various ways: 1.Tf1,Se2+,Sf2. Black just played R: 1.Kh2g2 Td~h1+, so White can't castle but doesn't need to. What's going on? (2022-02-15)
comment
Keywords: a posteriori (AP) (Type Keym), Castling, Rex solus, Miniature
Genre: Retro, 3#
FEN: 8/8/6R1/8/8/5PNN/6k1/4K2R
Input: Gerd Wilts, 1995-06-06
Last update: A.Buchanan, 2022-02-15 more...
49 - P0006046
Niels Høeg
8474 The Fairy Chess Review 28/11/1949
P0006046
(11+11)
Schwarz nimmt 1 Zug zurück, dann h#1
R: 1. ... Ka3-b2, dann 1. Ka4 Lb5#
play all play one stop play next play all
The pawns have captured 8 times to get around each other, plus 2 times to create the square color imbalance (6 light-squared white bishops and only 4 black ones). Thus all captures were done by pawns, so retractions like Ka3xb2, Kb3-b2, or Kc2-b2 are illegal.
Henrik Juel: The problems in this issue were dedicated to T.R. Dawson to celebrate his 60th birthday on 28/11/1949
The solution (FCR 06/1950 p.109) offered no real explanation
Dawson just wrote: Cooks claimed to this masterpiece involve illegal retractions (2022-04-28)
more ...
comment
Keywords: Help retractor, Non-standard material, Symmetrical position, Asymmetrical solution
Genre: Retro
FEN: bb3BB1/bBb3KB/1B5B/8/2B5/3B2b1/bk3BBb/1b4bb
Reprints: 1494 FEENSCHACH 10/1952
7 Rund um das Schachbrett 1955
(267) feenschach 52 10-12/1980
Thema Danicum 94 1999
H34 ASymmetrie 2013
Input: Gerd Wilts, 1995-07-16
Last update: Alfred Pfeiffer, 2017-09-27 more...
50 - P0006067
Nikita M. Plaksin
Andrej N. Kornilow

(2) diagrammes 15 07-09/1994
P0006067
(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
play all play one stop play next play all
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
Keywords: Monochrome, 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...
51 - P0006423
Andrey Frolkin
9128 Die Schwalbe 157, p. 283, 02/1996
Leonid Borodatow gewidmet
P0006423
(9+4)
#3 (AP)
White pushes the move (Keym AP)
1. ... f5! (f6?,fxe6?,fxg6? 2.0-0! ~ 3.Tf3#) 2. gxf6ep! exf6 3. 0-0! f5 4. Tf3#
(2. 0-0=?)
Valladao Task via the try: 1. exf7? e5! 1. ... e6? 2. f8=D e5 3. Da3#
Another try shows fully differentiated black Albino:
1. Kf1? fxg6!
1. ... f5? 2. Lxf5! h3 3. Txh3#
1. ... fxe6? 2. Lxe6! h3 3. Txh3#
1. ... f6? 2. Lf5! h3,fxg5 3. Txh3#,Th3#
And retro tries:
1. 0-0?? f5,~ 2. Tf3# (short solution)
1. f5? f6,~ 2. 0-0?? fxg5,~ 3. Tf3#
play all play one stop play next play all
Wenn Schwarz zuletzt gezogen hat, dann kann Weiß nicht mehr rochieren: R: 1. Kh2-g3 Tf1(g1)-h1+. Da aber Weiß in einem direkten Matt beginnt, ist die Rochade nicht mehr zulässig. Der Zusatz "AP" in der Forderung ist überflüssig: es könnte höchstens der Anzug a posteriori dem Schwarzen übertragen werden. Das geht aber nicht, da Weiß in einem direkten Matt beginnt.
Guus Rol: This is apparently AP after Keym. The move goes to black:
0. ... f5! 1.gxf6ep exf6 2.0-0(justifies the handover) f5 3.Tf3 (2007-02-13)
A.Buchanan: There are two kinds of directmate Type Keym. In one Black unsuccessfully pulls the move, in the other (as here) White successfully pushes it to Black. (2022-04-16)
A.Buchanan: I disagree with the German comment in the solution text. I think including "AP" is advisable in the stipulation. Keym AP riffs off Codex Article 15, but it's not the default. (2022-04-17)
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comment
Keywords: a posteriori (AP) (Type Keym), Castling, En passant, Promotion (D), Valladao Task
Genre: Retro, 3#
FEN: 8/4pp2/4P1N1/6PP/5P1p/6kB/6P1/4K2R
Input: Gerd Wilts, 1996-06-12
Last update: A.Buchanan, 2022-04-18 more...
52 - P0006428
Andrey Frolkin
9133v Die Schwalbe 157, p. 283, 02/1996
P0006428
(13+12) cooked
BP in 20.0
May White castle?
Weiß darf noch rochieren: 1. h4 Sa6 2. h5 Sc5 3. h6 a6 4. hxg7 h5 5. f4 h4 6. f5 h3 7. f6 h2 8. fxe7 Sf6 9. g8=L Lh6 10. Lxf7 Kxf7 11. e8=T Kg6 12. Te6 Dg8 13. Tc6 Db3 14. axb3 dxc6 15. Ta4 Lh3 16. Te4 Tag8 17. Te8 hxg1=D 18. e4 Dd4 19. Lc4 Tg7 20. Lg8 Dd8
Ohne die Bedingung, daß Weiß rochieren darf, gibt es andere BPs: 1. f4 Sa6 2. f5 Sc5 3. f6 a6 4. fxe7 f5 5. e4 f4 6. Lb5 f3 7. Lc6 f2+ 8. Ke2 fxg1=D 9. h4 dxc6 10. h5 Lf5 11. h6 Sf6 12. hxg7 Dd5 13. g8=L Db3 14. Lxh7 Kf7 15. axb3 Lh6 16. Ta4 Tag8 17. Td4 Dxd4 18. e8=T Tg7 19. Lg8+ Kg6 20. Ke1 Dd8
play all play one stop play next play all
SH: Cooked?:
1. e4 f5 2. Ne2 f4 3. h4 f3 4. h5 fxe2 5. f4 exf1B 6. f5 Bc4 7. f6 Bb3 8. axb3 Na6 9. fxe7 Nc5 10. Ra6 Nf6 11. Rc6 dxc6 12. h6 Bh3 13. hxg7 Qd7 14. g8B Bh6 15. Bxh7 Kf7 16. Qf3 Rag8 17. Qe2 Rg7 18. Bg8+ Kg6 19. Qd1 a6 20. e8R Qd8 (2005-06-20)
Henrik Juel: The stipulation question was probably intended as an implicit condition; when answered affirmatively, there should be a unique solution
But SH's solution shows that the problem is cooked (2018-12-09)
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comment
Keywords: Anti-Pronkin, Constrained problem, Unique Proof Game, Castling
Genre: Retro
FEN: 3qR1Br/1pp3r1/p1p2nkb/2n5/4P3/1P5b/1PPP2P1/1NBQK2R
Input: Gerd Wilts, 1996-06-12
Last update: James Malcom, 2022-02-08 more...
53 - P0006447
Gerd Wilts
Norbert Geissler

Retros mailing list 05/1996
P0006447
(15+16) C+
BP in 7.0
1. f4 e5 2. Kf2 Dh4+ 3. Kf3 Df2+ 4. Kg4 h5+ 5. Kh3 h4 6. e4 d5+ 7. g4 hxg3ep#
play all play one stop play next play all
A.Buchanan: Hi the keyword "En passant as mating move" occurred three times, so I've removed two of them. Also the prefix condition means that a search for "En passant" will always catch an "En passant as mating move", so I've removed the vaguer term. (2022-05-07)
A.Buchanan: Does anyone know a trick to find duplicate occurrences of a keyword? My only technique is to stumble across them from time to time (2022-05-07)
Henrik Juel: I cannot help you, of course, Andrew
But it is curious that the error occurred in a problem by the father of the PDB (2022-05-07)
A.Buchanan: Recent changes can be seen by clicking on the "more...", but it doesn't matter (2022-05-07)
comment
Keywords: Unique Proof Game, En passant as mating move
Genre: Retro
Computer test: Natch 2.1 Copyright (C) 1997,98,99,2001,2002 Pascal Wassong
FEN: rnb1kbnr/ppp2pp1/8/3pp3/4PP2/6pK/PPPP1q1P/RNBQ1BNR
Input: Gerd Wilts, 1996-06-13
Last update: A.Buchanan, 2022-05-07 more...
54 - P0008246
Alexander Kislyak
4 Shakhmaty v SSSR 04/1974
P0008246
(12+12)
#2
1. bxc6ep
play all play one stop play next play all
Lösetip beim Originalabdruck: beim Abzählen der geschlagenen Figuren beachte man die Farbe der Felder, auf denen sie fielen.

AL
Le4 ist ein UW-L, der nur so entstanden sein kann: h2-h7xg8=L.
Schwarz zog h7xg6x5, um ihn vorbeizulassen.
Somit ergibt sich folgende Schlagbilanz:
Weiß: axb, dxc, fxg (als Schlagobjekt für den sBh7) [Anmerkung: statt fxg auch fxe mit UW auf e8 möglich, ändert aber nichts]
Schwarz: bxc, h7xg6xh5 sowie den wLf1, der nicht von den sBB geschlagen wurde

Unter den geschlagen weißen Figuren befindet sich der schwarzfeldrige Lc1.
Die Schläge h7xg6xh5 fanden auf weißen Feldern statt, also wurde der Lc1 mittels b4xc3 geschlagen. ("Effekt Zvetnosti")
Als letzte schwarze Züge kommen also nicht infrage: b3xc2, e7-e5 (sLf8 wird als Schlagobjekt gebraucht), g6xh5 (dann käme der wUW-L nicht von g8 nach e4).
Einziger legaler letzter schwarzer Zug war also c7-c5, und es löst: 1. bxc6ep
Des Autors (geb.27.12.1938)"`Erstling"', also mit 36 Jahren.

abgedruckt in der Rubrik: "Redkije Shanry"
Henrik Juel: Solution: 1.bxc6 ep (2.cxd7#). White captured axb, dxc, fxe, h7xg8=B and promoted on e8; Black captured orig. Bf1, bxc, h7xg6xh5 and promoted on a1. So last move must be c7-c5. (2003-05-27)
Mario Richter: @all PDB activists: Would the introduction of a new keyword: "Farbbalance / color balance"
or something similiar make any sense? (describing the fact, that a certain capture must have happened on a square of a determined color) (2022-04-26)
A.Buchanan: The pieces have colour: black & white. But the squares and the bishops have *shade*. Informally of course one can say anything, but if we are glossarizing then maybe be a bit more formal. And what is “balance” here? I think of balance in terms of pawn captures required vs available. Typically the issue you are talking about is where almost all pawn captures are in one shade square, allowing us say something interesting about a bishop capture. Maybe: “shade logic”? (2022-04-26)
Henrik Juel: I use Andrew's terminologi: men are white or black, and squares are light or dark
But my non-retro friends (both of them) use white/black for squares also
So 'Shade logic' may be too esoteric
What about 'Square color argument'?
It is used far more frequently in retro analysis than, say, 'Parity argument', so a keyword is needed
Thanks for the suggestion, Mario (2022-04-26)
A.Buchanan: I am convinced by Henrik, but I will continue to use “shade” in my own writing. (2022-04-27)
comment
Keywords: En passant as key, Obvious promotion (wLe4)
Genre: Retro
FEN: brkn2R1/Rn1p1pp1/N7/1PpKp2p/QPP1B3/2P5/2p1P1P1/8
Reprints: Caissas Schloßbewohner 3 1987
Input: Gerd Wilts, 1996-09-16
Last update: Mario Richter, 2022-04-26 more...
55 - P0008399
Thomas Volet
Rex Multiplex 1983
1. Preis
P0008399
(14+12)
=
R: 1. Th1-g1 Lb8-a7 2. Th5-h1 La7-b8 3. Td5-h5 Lb8-a7 4. Td2-d5 La7-b8 5. Tc2-d2 Lb8-a7 6. Tc1-c2 Kc2-b3 7. Ta1-c1 Kd2-c2 8. Lb4-a3 Ke1-d2 9. Ta3-a1 Kd2-e1 10. Tb3-a3 Ke1-d2 11. La3-b4 Kd2-e1 12. Tb5-b3 Ke1-d2 13. Ta5-b5 Kd2-e1 14. Ta7-a5 Ke1-d2 15. Tb7-a7 La7-b8 16. Tb8-b7 Kd2-e1 17. Kg8-f8 Ke1-d2 18. Tf8-b8 Te8-e7 19. Kh8-g8 Tb8-e8 20. Tc8-f8 Tb7-b8 21. Kg8-h8 Lb8-a7 22. Kf8-g8 Ta7-b7 23. Ke7-f8 Ta5-a7 24. Td8-c8 La7-b8 25. Tb8-d8 Tb5-a5 26. Tb7-b8 Lb8-a7 27. Ta7-b7 Tb3-b5 28. Lb4-a3 Ta3-b3 29. Ta5-a7 Ta1-a3 30. Tb5-a5 Tc1-a1 31. La5-b4 Tc2-c1 32. Tb3-b5 Td2-c2 33. Ta3-b3 Td6-d2 34. Ta1-a3 Td5-d6 35. Tc1-a1 Tf5-d5 36. Tc2-c1 Tf4-f5 37. Td2-c2 Te4-f4 38. Td6-d2 Td4-e4 39. Td5-d6 Td2-d4 40. Td4-d5 Tc2-d2 41. Td2-d4 Tc1-c2 42. Tc2-d2 Ta1-c1 43. Tc1-c2 Ta3-a1 44. Ta1-c1 Tb3-a3 45. Ta3-a1 Tb5-b3 46. Lb4-a5 Ta5-b5 47. Tb3-a3 Ta7-a5 48. La3-b4 Tb7-a7 49. Tb5-b3 La7-b8 50. Ta5-b5 Tb8-b7 51. Ta6-a5 Tg8-b8 52. Ta5-a6 Lb8-a7 53. Ta7-a5 Th8-g8 54. Tb7-a7 La7-b8 55. Tb8-b7 Tg8-h8 56. Tf8-b8 Lb8-a7 57. Kd8-e7 La7-b8 58. Kc8-d8 Th8-g8 59. Kb7-c8 Tg8-h8 60. Tb8-f8 Te8-g8 61. Kc8-b7 Te7-e8 62. Tb7-b8 Lb8-a7 63. Ta7-b7 Kd2-e1 64. Ta5-a7 Ke1-d2 65. Tb5-a5 Kd2-e1 66. Tb3-b5 Ke1-d2 67. Lb4-a3 Kd2-e1 68. Ta3-b3 Kc2-d2 69. Ta1-a3 Kb3-c2 70. Tc1-a1 Ka2-b3 71. Tc2-c1 Kb1-a2 72. Td2-c2 Ka2-b1 73. Td5-d2 Kb1-a2 74. Th5-d5 Ka2-b1 75. Th1-h5 Ka1-a2 76. h3xLg4
play all play one stop play next play all
Henrik Juel: Following R: 76.h3xLg4, the resolution could continue with
Retract wKc8 to b5, sLg4 to c8, b7-b6, sSa8 to g8, wSa4 to g1, wKb5 to e1, sLb8 to a5, b6xTc5, wTc5 to d4, sKa1 to b5, a2xTb3xDc4+, etc. (2020-08-13)
Henrik Juel: A little analysis may be in order
Pawns captured all missing men
Kb3 is now confined to the SW corner, but if we retract b3xc4 he is locked in for good, so [Lc8] was captured on g4
Before we retract h3xLg4, sLg4 to c8, and b7-b6, wKf8 must be liberated
The plan for this is: Retract wKg8-f8, sTg1 to f8, sTe7 out, wTf8 out past the sT, wKg8 to e7, sT to h8, wT to e8, and further as per my old comment (2022-01-22)
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comment
Keywords: 50 move rule, Move Length Record
Genre: Retro
FEN: n4K2/b1pprppp/1p2p3/2p5/N1P3P1/BkP1P3/1P2PPP1/3N1BR1
Reprints: Redkiye zhanry-plus 1996
Input: Gerd Wilts, 1996-09-17
Last update: James Malcom, 2022-01-21 more...
56 - P0008465
Thomas Volet
9375 Die Schwalbe 161 10/1996
Nikita Plaksin gewidmet
P0008465
(14+9) cooked
Die Partie ist beendet. Weshalb?
Thomas Volet: Problem P0008465 is faulty. (2000-11-27)
James Malcom: What is the author's solution and the cook? (2022-01-21)
Mario Richter: I do not the "official cook", but the following should work:
R: 1. Th1-g1 Kf5-e4 2. Th4-h1 Kg6-f5 3. Td4-h4 Kh6-g6 4. Td1-d4 Kg6-h6 5. Ta1-d1 Kh6-g6 6. Ta3-a1 Kg6-h6 7. Tb3-a3 Kh6-g6 8. Tb5-b3 Kg6-h6 9. Ta5-b5 Kh6-g6 10. Lb8-a7 Kg6-h6 11. Ta7-a5 Kh6-g6 12. Tb7-a7 Kg6-h6 13. La7-b8 Kh6-g6 14. Kg8-f8 Kg6-h6 15. Tb8-b7 Kh6-g6 16. Tf8-b8 Te8-e7 17. Lb8-a7 Tc8-e8 18. La7-b8 Tb8-c8 19. Te8-f8 Tb7-b8 20. Lb8-a7 Ta7-b7 21. Kf8-g8 Ta5-a7 22. Ke7-f8 Tb5-a5 23. Kd8-e7 Tb4-b5 24. Kc8-d8 Th4-b4 25. Kb7-c8 Th5-h4 26. Ka6-b7 Tg5-h5 27. Kb5-a6 Tf5-g5 28. Kc4-b5 Te5-f5 29. Kb3-c4 Te4-e5 30. Ka2-b3 Tb4-e4 31. Kb1-a2 Tb5-b4 32. Te7-e8 Ta5-b5 33. Kc1-b1 Ta7-a5 34. Kd1-c1 Tb7-a7 35. La7-b8 Tb8-b7 36. Ke1-d1 Th8-b8 37. Te8-e7 Kg5-h6 38. Tb8-e8 Tg8-h8 39. Tb7-b8 Th8-g8 40. Lb8-a7 Tg8-h8 41. Ta7-b7 Th8-g8 42. Ta5-a7 Tg8-h8 43. Tb5-a5 Th8-g8 44. Tb4-b5 Tg8-h8 45. Th4-b4 Tf8-g8 46. Th1-h4 Th8-f8 47. h2xLg3 (2022-01-23)
Thomas Volet: Mario, I do appreciate your tactfully lengthy unwind, but this effort at economy is hugely flawed. (2022-01-23)
comment
Keywords: 50 move rule
Genre: Retro
FEN: N4K2/B1pprPpp/1p2p3/2p5/N3k3/2P3P1/1PP1PPP1/5BR1
Input: Gerd Wilts, 1996-10-26
Last update: A.Buchanan, 2017-03-22 more...
57 - P0008592
Leonid M. Borodatow
9504 Die Schwalbe 163 02/1997
P0008592
(15+7)
#3. Two solutions (AP)
An earlier stipulation in PDB was "a) #3 b) Weiß setzt in 3 Zügen # (AP)"
hans: a) white pawns capture all black pieces, including h-pawn. If last move was fxLe4, the h-pawn promotes without capture, so 0-0 is illegal.
1. Da3!
1. … Dxb2 2. Ke2+ Kc2/Dc1 3. Sb4/Txc1#
1. … Dxa2 2. Ke2+ Kc2 3. Tc1#
1. … Kc2+ 2. Sxa1+ Kd3/Kb1 3. Da6/Ke2#

b) If last move was a white pawn capture, and black capture hxLg, there must be 0-0 in the solution to prove this.
0. … Dxb2 1. 0-0+ Dc1 2. Txc1+ Kb2 3. Le1#
0. … Dxa2 1. 0-0+ Kc2 2. Tc1+ Kxb2/Kd3 3. Le1/Db5#
0. … Kc2+ 1. Sxa1+ Kxb2/Kd3 2. Db3+/0-0 Ka1/Lxd7 3. 0-0/Sc1#
0. … Lxd7 1. 0-0+ Kc2 2. Lc1+ Kb1/Kd3 3. Sd2/Rd1#

0. … Kxb2+ 1. Lc1+ Kb1 2. 0-0 Dxc3/Dxa2 3. Sxc3/Lb2#
(1. Sxa1? Lxd7 2. Dc2+ Ka3 3. Lc1# but no castling, so illegal) (2015-09-08)
A.Buchanan: This is a good problem with varied and mostly accurate play in both a & b. How would one translate the stipulation of b into English, please? (2015-09-09)
Henrik Juel: What about the stipulation: #3
and the twinning: b) Black to move (AP)
(the German stipulation text is not clear, either) (2015-09-09)
VL: Stipulations like "Black to move (AP)" make no sense because using the AP-logic is a right rather than a duty: if Black to move were stipulated explicitly then nothing would need to be proven and therefore this twin would be cooked. In my opinion, the most appropriate stipulation
is the following paradoxical one: "#3. Two solutions (AP)". "AP" indicates that the AP-logic is permitted, and at least one solution (or maybe only a thematic try) does use it. An excellent problem, anyway. In Die Schwalbe, H.167, its detailed solution is published.

Here a kind of AP-logic, called sometimes "typ Keym" or "ad libitum", is employed: the justification of the improper side's turn to move (rather than of an e.p.-key) a posteriori. (2015-09-20)
A.Buchanan: In the V&V Encyclopedia, "Type Keym"/"ad libitum" is described in the context of PRA rather than AP. It is contrasted there with "Type Offner"/"a priori". I still feel it's all rather cloudy. How are these accurately defined, and exactly how does the distinction carry across to AP, please? (2022-02-15)
A.Buchanan: I've looked at V&V encyclopedia carefully, and in the absence of definitive information, I am going to make the assumption that for AP we distinguish between Types Petrovic & Keym, and this has nothing to do with the terms "Type Keym"="ad libitum"/"Type Offner"="a priori" in PRA. Werner Keym has two types, is all. If someone has an authoritative specification, then I would be grateful. This is sufficient to clear up the island which is A Posteriori to some extent. (2022-02-15)
A.Buchanan: I have adopted VL's suggestion for the stipulation. Apart from anything else, in set play, Black would *lose* a move, while in a retropat situation (like Codex Article 15, and here) Black *gains* a move. (2022-02-15)
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comment
Keywords: a posteriori (AP) (Type Keym), Castling
Genre: Retro, 3#
FEN: 4b3/p1pP1Pp1/3P4/4P3/Q2Pp3/1NP1P3/NP1B1R2/qk2K2R
Input: Gerd Wilts, 1997-03-19
Last update: A.Buchanan, 2022-02-16 more...
58 - P0008793
Andrej N. Kornilow
Shakhmaty v SSSR 1978
P0008793
(15+8)
#1.5
a) 1. "Ta1-d1+?" ... Td4 2. Txd4# (Completing the illegal castling!)
b) 1. "xc5ep+?" ... Tb5 2. Txb5# (Completing the illegal e.p.!)
c) 1. "xe8=S+?" ... e6 2. Sf6# (Completing the illegal promotion!)

0) 1. Sxe7+! Lxe7 2. f8=B# (Truncating the promotion move!)
play all play one stop play next play all
n in directmate stipulation #n means that White has n moves to do the job. With n=1.5, therefore, one of the White moves is fractional, so we know we are in the realm of jokes, ho ho!

If White's first move is the fractional one, there are three retro tries which attempt to complete: castling, ep & promotion. However all are illegal:
castling: 8 white pawns, so wTa5 came from h1, dislodging wK.
ep: retracting sBc5 to c7 means wLb8 is promoted, but 8 white pawns.
promotion: white made 7 pawn captures, while sBgh were waylaid.

So we consider that White's second move was the fractional one, by omitting the replacement of wBf8 by an officer.
Henrik Juel: Completing the key isn't legal, 1.'Rd1'/'-bPc5'/'e8S'+, but omitting the promotion in the mating move is, 1.Sxe7+ Bxe7 2.'f8'#. Excellent joke. (2003-09-19)
James Malcom: A very witty joke Valladao! (2020-09-24)
A.Buchanan: Really like this joke (2020-09-25)
Henrik Juel: To see the illegality of completing 1.fxe8=S+?, note the captures:
Black captured fxDe, so [Pg7,h7] were captured on their files, while the other six missing black men were captured by white pawns
The illegality of completing 1.0-0-0,bxc6ep? is rather obvious (2020-09-26)
James Malcom: And for those you don't find it obvious: If the White king hasn't moved, then where did wRa5, and wBb8 if bPb5 has just done a double-step, come from? Neither can be promoted pieces, as White still has all eight pawns. Trying to finish castling and en passant therefore both produce illegal positions and thus cannot be the solution. (2020-09-26)
Henrik Juel: Continuing beating the dead horse...
How does the white player actually perform an entire move?
1.0-0-0+: Ke1-c1 and Ta1-d1
1.bxc6ep+: Pb5-c6 and remove sPc5
1.fxe8=S+: remove sYe8 and Pf7-e8 and replace wPe8 with wSe8 (three fractional actions)
2.f8=Y#: Pf7-f8 and replace wPf8 with wYf8
So a marginally better stipulation might be: 'White to move mates in less than 2 moves' (2020-09-26)
A.Buchanan: Hurray I've got the animation working! I agree with Henrik's stipulation. There is a dummy pawn on f8, but not by the Dummy Pawn rule. Instead it's the joke that does it. If Dummy Rule applied, the move would be full length! :) (2022-02-09)
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comment
Keywords: En passant, Start a move but do not finish it, Castling, Promotion (S,B), Valladao Task (half!), Joke (End move, Start move), Dummy Pawn (Start move), Complete an unfinished move, waylaid (sBgh)
Genre: Retro, 2#
FEN: 1B1N1bB1/p2ppP2/2P5/R1pk1N2/1r2p3/1P2P1P1/4PP1P/R1K5
Reprints: (III) Quartz 4 1997
Input: Gerd Wilts, 1997-06-21
Last update: A.Buchanan, 2022-02-09 more...
59 - P0008872
Denis M. Saunders
C9206 The Problemist 03/1998
P0008872
(12+14) C+
#2?
b) -sBe7
a) 1. 0-0-0? illegal
1. Td1! droht 2. Dc4#
1. ... Se3 2. Dxe3#
1. ... Se5 2. Dxe5#
1. ... cxb5 2. Txb5#
1. ... Tf4 2. Dxe7#
1. ... Dxc2 2. Txc2#
1. ... Dd3 2. Sxd3#
1. ... De4+ 2. Sxe4#
b) 1. Td1? Te8! pinning
1. ... Te7? 2. Dxe7#
1. 0-0-0! then all variations as in a)
play all play one stop play next play all
(a) sLa7 is promoted. If White castling rights remain, then bPf captured fxgxh2xg1=L, which with dxc6 makes 4 captures. White has lost Bd, Be, Bf/h & Lc. However, wBe can't have captured or promoted (h/fxg & Lf at home are the only Black casualties), so can't have contributed to Black captures. Therefore impossible, and wK did lose castling rights before. Now trivial for sBf to have promoted.
(b) sLa7 can be original, so castling is still ok. Even if sLa7 is promoted, wBe can have promoted, and again White can still castle.
The forward play is more varied than is usual in these kinds of problems.
Kees: a) 1.Td1! [2.Dc4‡]


b) 1.0-0-0! [2.Dc4‡]

La7 is promoted.
In a) fxg/e=L King must move (e2 cannot promote) so no 0-0-0.
In b) fxgxhxg=L and 1. 0-0-0 for 1. Td1? Te8! (2022-02-02)
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comment
Keywords: Castling, Castling as key, Twin
Genre: Retro
Computer test: HC+ Popeye 4.61 and analysis
FEN: b6r/b1p1prpq/1pp1Q2p/1Pk5/p5n1/P4NP1/1RP2NP1/R3KB2
Input: Gerd Wilts, 1998-06-26
Last update: A.Buchanan, 2022-02-03 more...
60 - P0009099
Klaus Wenda
4123 diagrammes 127 10/1998
P0009099
(11+5)
s#12
Längstzüger
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#
play all play one stop play next play all
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)
more ...
comment
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...
61 - P0500165
Gino Mentasti
1721v Die Schwalbe 33-34, p. 561, 09-10/1964
Lob
P0500165
(3+8) cooked
h#4
1. Lf2 e3 2. Kf3 Txf7 3. Kg3+ Tb7 4. Tf3 Tg7#
play all play one stop play next play all
Cook: NL:
1. Kg4 Kg2 2. f5 Kf1 3. Kg3 Tb5 4. f4 Tg5#
1. Kf2 e4 2. Ke1 Kg2 3. Tf3 Kxf3 4. Kf1 Tb1#
Korrektur Timo Koistinen: +sDa2, +sBh6h7, ST 2001, p.179
Yuri Bilokin: Correction: bBd4-g1, bPh3 b4r2/1R3p2/8/8/7p/6kp/4P2p/6bK (3+8) (2022-04-21)
comment

Genre: h#
FEN: b4r2/1R3p2/8/8/3b3p/6kr/4P2p/7K
Reprints: 584 FIDE Album 1965-1967
Input: Gerd Wilts, 1996-06-06
Last update: Marcin Banaszek, 2016-01-22 more...
62 - P0501022
Zdravko Maslar
Problem 1958
Sonderpreis
P0501022
(1+13)
h=34
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=
play all play one stop play next play all
James Malcom: This is outdone by P0501023. (2022-04-27)
more ...
comment
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...
63 - P0501043
Theodor Steudel
AHK-Memorial 1983-1985
2. Preis
P0501043
(1+4)
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#
play all play one stop play next play all
in feenschach Verbesserung?!?!?!?
SCHRECKE: a)+b): C+, popeye 4.87 (2022-04-12)
comment
Keywords: Seriesmover
Genre: Fairies
FEN: 1K6/8/8/8/8/8/6br/6nk
Input: Gerd Wilts, 1996-06-06
Last update: hpr, 1999-05-21 more...
64 - P0501256
Karl Fabel
Theodor Steudel

274 The Problemist 07/1971
P0501256
(3+14) C+
h#4
1. Dxf4 0-0-0 2. Sf2 Kb1 3. Sd2+ Ka1 4. Td4 Te1#
play all play one stop play next play all
Yuri Bilokin: Version: -bPe4 8/8/2r5/3b4/r1n2P2/pp1pkp2/6pq/R3K1bn (3+13) (2022-04-23)
comment
Keywords: Interchange (KT (16)), Castling (wg), Tempo Move (K)
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/2r5/3b4/r1n1pP2/pp1pkp2/6pq/R3K1bn
Input: hpr, 1996-06-14
Last update: Alfred Pfeiffer, 2018-12-03 more...
65 - P0501419
Stanko Milenkovic
101 Problemas 04-06/1964
P0501419
(3+8) cooked
h#4
1. f2 Tb1 2. Sf3 Tb2 3. Ke1 Txc2 4. Sd2 Tc1#
play all play one stop play next play all
Cook: NL
1. Sg2 Th1 2. f2 Th8 3. Ke1 Th2 4. Kf1 Th1#
Adrian Storisteanu: Possible fix:
Kh1 Rb1 / Kc2 Sd1 ppb2 d2 e3 g3 h2 h3 (2+8) h#4
1.e2 Ra1 2.Se3 Ra2 3.Kd1 Rxb2 4.Sc2 Rb1# (2015-07-24)
milan: milan frelih: i've found key solution,your move? (2015-07-24)
milan: Milan Frelih:[Ka1 h1,+bBb1,wPe3=bPe4] (2015-08-31)
Yuri Bilokin: Correction: wKa1-h1, wRc1-b1, bKd2-c2, bPc2-b2, bNe1-d1, bPe2-d2, bPf3-e3, -bPb3, -bPd3, -wPe3, -bPe4, +bQg3 8/8/8/8/8/4p1q1/1pkp4/1R1n3K (2+6) h#4 1.e2 Ra1 2.Se3 Ra2 3.Kd1 Rxb2 4.Sc2 Rb1# (MM) (2022-04-21)
comment
Keywords: Interchange (ks (2)), Pure Round Trip (T)
Genre: h#
FEN: 8/8/8/8/4p3/1pp1Pp2/2pkp3/K1R1n3
Input: hpr, 1996-07-01
Last update: Alfred Pfeiffer, 2015-07-24 more...
66 - P0501467
Fritz Hoffmann
7030v Schach 09/1972
P0501467
(3+7) cooked
h#4
1. Db6 b5 2. Ld8 bxc6 3. Ke7 c7 4. Df6 c8=S#

NL:
1. Da5 b5 2. Ld8 bxc6 3. Ke7 c7 4. f6 c8=S# ua
play all play one stop play next play all
Yuri Bilokin: Correction: -bPf7, +bRe1 3qb1K1/3n4/2pk1b2/5P2/1P6/8/8/4r3 (3+7) (2022-05-02)
comment
Keywords: Interchange (dl (8))
Genre: h#
FEN: 3qb1K1/3n1p2/2pk1b2/5P2/1P6/8/8/8
Input: hpr, 1996-07-01
Last update: hpr, 1999-04-06 more...
67 - P0503612
Gligor Denkovski
9884v FEENSCHACH 12/1970
P0503612
(3+4)
h#4
1. Se3 Tf7 2. Sg4 Txd7 3. Kf4 Td5 4. e4 Tf5#

NL
1. Se3 Tf3 2. Lg4 Th3 3. Kf4 Kf6 4. e4 fxe3#
play all play one stop play next play all
Yuri Bilokin: Correction: sNd5-g8 6n1/3b4/6K1/4pR2/4k3/8/5P2/8 (3+4) h#4
1.Sh6 Rf7 2.Sg4 Rxd7 3.Kf4 Rd5 4.e4 Rf5# (2022-04-20)
comment
Keywords: Pure Round Trip (T)
Genre: h#
FEN: 8/3b4/6K1/3npR2/4k3/8/5P2/8
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-13 more...
68 - P0503613
Hermann Lücke
1925v Schach-Echo 05/10/1959
P0503613
(3+7) C+
h#4
1. Sb3 Tg2 2. Tf1+ Tg1 3. Ta1 Tc1 4. b1=S Tc2#
play all play one stop play next play all
Yuri Bilokin: Redaction: Yuri Bilokin & Hermann Lücke -wPh2 8/8/8/8/5b2/p4r1p/kpR5/2n4K (2+7) h#4 2.1…
1.Sb3 Rg2 2.Rf1+ Rg1 3.Ra1 Rc1 4.b1=S Rc2# (MM)
1.Ka1 Rh2 2.Se2 Rxh3 3.Rf1+ Kg2 4.Rb1 Rxa3# (MM) (2022-04-25)
comment
Keywords: Pure Round Trip (T)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/8/8/5b2/p4r1p/kpR4P/2n4K
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-13 more...
69 - P0503938
Janko Furman
7058v Probleemblad 09-10/1971
P0503938
(7+13)
h#4
b) sBa6 nach a5
a) 1. Dxe4 La7 2. Dxe3+ Kb4 3. Dxf2 Lc6+ 4. Df3 Lxf3#
b) 1. Dxf2 Lc6 2. Dxe3+ Kb5 3. Dxe4 La7 4. Df3 Lxf3#
play all play one stop play next play all
SCHRECKE: a)+b): C+, Gustav 4.2a, Brute Force (2022-04-05)
comment
Keywords: Pure Round Trip
Genre: h#
FEN: 1Br2b2/1pp5/p2r4/2K5/B3P3/1b2Pqnp/4PPkp/5n2
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-25 more...
70 - P0505147
Philip Leonard Rothenberg
41 To Alain White , p. 30, 1945
Alain C. White
zum 65. Geburtstag
P0505147
(2+8) C+
h#3
1. Dh3 Kg8 2. Tg3 Kf7 3. f3 Lc1#
play all play one stop play next play all
Frank Müller: Der korrekte Quellenname ist "To Alain White". Das ist ein sehr gesuchtes Buch aus der Overbrook-Serie. (2010-12-01)
Dieter Berlin: The Overbrook-Press: Stamfort, Connecticut
Three hundred copies of this volume have been printed
to commemorate the sixty-fifth birthday of
Alain White, March 3, 1945 (2022-04-08)
comment
Keywords: Clearance, Minimal
Genre: h#
Computer test: (Popeye C-Version 3.47 (1024 KB))
FEN: 7K/7p/p6k/7p/5pp1/r2q4/1B6/8
Input: hpr, 1996-12-26
Last update: Dieter Berlin, 2022-04-08 more...
71 - P0505381
Lajos Riczu
Magyar Sakkelet 11/1970
1. Preis
Turnier des Ungarischen Schachbundes
P0505381
(3+6) cooked
h#4
1. De3 Te1 2. De8 e4 3. Le7 e5 4. Tf8 e6#

NL:
1. Dg5 e4 2. Df5+ exf5 3. g5 Te1 4. Tg6 fxg6#
play all play one stop play next play all
Yuri Bilokin: Correction: bPg7-g5, +bNb7, +bPg3, +bPg4, +bPh2, +bRh3, +bNh4 5br1/1n3k1K/5p2/6p1/6pn/6pr/4P2p/6qR (3+12) h#4
1.Qe3 Re1 2.Qe8 e4 3.Be7 e5 4.Rf8 e6# (MM) (2022-04-20)
comment
Keywords: Clearance
Genre: h#
FEN: 5br1/5kpK/5p2/8/8/8/4P3/6qR
Input: hpr, 1996-12-27
Last update: hpr, 1999-07-18 more...
72 - P0505422
Paul Leibovici
292 Revista Romana de Sah 03/1934
2. -3. ehrende Erwähnung
P0505422
(3+6) C+
h#4
1. Te3 c5 2. Tf1 Tf2 3. Sd3 Tc2 4. Tf3 Tc4#
play all play one stop play next play all
Yuri Bilokin: Redaction: Yuri Bilokin & Paul Leibovici bPh5-h4 8/4p3/4KR2/5r2/1nP1k2p/5r2/8/8 (3+6) h#4 2.1…
1.Re3 c5 2.Rf1 Rf2 3.Sd3 Rc2 4.Rff3 Rc4# (MM)
1.Kf4 Rxf5+ 2.Kg4 Rxf3 3.Kh5 Kf5 4.h3 Rxh3# (MM)
Active sacrifice (black). Chumakov theme (rr, 2). Model mate × 2 (2022-04-26)
comment
Keywords: Clearance
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/4p3/4KR2/5r1p/1nP1k3/5r2/8/8
Input: hpr, 1996-12-27
Last update: hpr, 1999-07-18 more...
73 - P0509786
Beni Snaider
9020 The Fairy Chess Review (17), p. 148, 04/1951
P0509786
(3+6) C+
h#4
2.1...
1) 1. Lb2 Kg3 2. Ke2 Txb2+ 3. Kf1 Th2 4. Le2 Th1#
2) 1. Lf4 Txa6 2. Ke2 exf4 3. Kf1 Kg3 4. Kg1 Ta1#
play all play one stop play next play all
Yuri Bilokin: Version: bQf5, bPg4-e4, -bPh5 8/8/b7/4bq2/4p2K/3kP3/R7/8 (3+5) (2022-04-26)
more ...
comment
Keywords: Check Protection
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/b7/4br1p/6pK/3kP3/R7/8
Input: Markus Manhart, 1997-06-28
Last update: Erich Bartel, 2010-12-17 more...
74 - P0510599
Ricardo C. Galli
885 Ajedrez Magico 24 01/1971
P0510599
(3+5) cooked
h#4
1. Sf4+ Te6 2. Lg4 gxf4 3. Kf5 Te7 4. De4 Tf7#

NL:
1. Sf4 gxf4 2. Lg4 Tg6 3. Kf5 Ke7 4. De4 Tf6#
play all play one stop play next play all
Yuri Bilokin: Correction: b1=a1, +bPa7, +bPf7, +bPg3 8/p1K2p2/3nRn2/3qb3/3k4/5Pp1/8/8 (3+8) h#4
1.Sde4+ Rd6 2.Bf4 fxe4 3.Ke5 Rd7 4.Qd4 Re7# (MM) (2022-04-20)
comment
Keywords: Check Protection
Genre: h#
FEN: 8/3K4/4nRn1/4qb2/4k3/6P1/8/8
Input: Markus Manhart, 1997-06-27
Last update: hpr, 1999-10-20 more...
75 - P0510719
Albert Heinrich Kniest
1216 Allgemeine Zeitung Chemnitz 25/10/1931
P0510719
(2+3) C+
h#5
1. f1=L c4 2. e1=T c5 3. Te7 c6 4. Lc4 c7 5. Lf7 c8=D#
play all play one stop play next play all

Duplicate Diagram: P0509698

Erich Bartel: steingetreu nachempfunden: Kahl,Trautmann 7596 FS (Bl.629) XII 1965.--

Nachdruck 1) 485 Ajedrez Magico (13) XII 1969 (2007-01-05)
Klaus Funk: weiterer Nachläufer: P1979776 (gespiegelt);
siehe auch P1095134 (2022-05-05)
YM: P1318489, P1318490, P1320464 (2022-05-06)
comment
Keywords: Check Protection, Promotion, Excelsior, white (auto key)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 4k3/8/8/8/8/8/2P1pp2/7K
Input: Markus Manhart, 1997-06-27
Last update: hpr, 1999-05-09 more...
76 - P0516143
Jan Ingvar Hannelius
Kauko Virtanen

Suomen Tehtäväniekat 01/1976
P0516143
(8+12)
h#4
1. Da1 Lxf3 2. Da8 Lxa8 3. Lb7 Txb7 4. Lxh2 Txb1#
play all play one stop play next play all
Verstellung TL
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-04-05)
comment
Keywords: Line closure
Genre: h#
FEN: Bbn4q/7R/7K/5PrP/6p1/4Pp1p/4pPbP/1r4nk
Input: Andreas Mokosch, 1997-07-23
Last update: hpr, 1999-10-25 more...
77 - P0516432
Tatomir S. Matovic
236 Novi Temi 1964
P0516432
(4+2)
h#3
1. Sc7 Lh8 2. Sd5 Sg7 3. Kd4 Se6#
play all play one stop play next play all
Verstellung ScL
SCHRECKE: C+, popeye 4.87 (2022-04-22)
more ...
comment
Keywords: Line closure
Genre: h#
FEN: 8/8/4n3/2k2N2/8/3P4/4K3/B7
Input: Andreas Mokosch, 1997-07-25
Last update: hpr, 2009-07-10 more...
78 - P0519078
Edgar D. Holladay
137v Ideal-Mate Review 07-08/1983
P0519078
(5+2) C+
h#4
1. Ke4 Kb2 2. Db5+ Kc2 3. Kd5 e4+ 4. Kc4 d3#
play all play one stop play next play all
YM: The symmetric final position and wK tempo move. (2022-04-10)
comment
Keywords: Symmetrical position
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/8/8/3P4/1KPqPk2/3P4/8
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: hpr, 1999-10-28 more...
79 - P0519843
Alfred Gschwend
7788 Schach-Echo 1, p. 14, 01/1974
P0519843
(10+11)
h#4
2.1...
1. Lb8 Lxe7 2. Td6 Txf2 3. gxf2 Lf6 4. Kg3 Le5#
1. Txf8+ gxf8=D 2. Lc7 Db8 3. Ld8 Dxg3+ 4. Kxg3 Le5#
play all play one stop play next play all
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-04-05)
comment
Keywords: Sacrifice of white pieces
Genre: h#
FEN: r2r1RK1/p3p1P1/4pB2/4b1P1/3P3p/4p1pP/4PpPk/5B2
Input: Hans-Jürgen Schäfer, 1997-10-11
Last update: Felber, Volker, 2015-09-15 more...
80 - P0520611
Fadil Abdurahmanovic
8695v Schach-Echo 24, p. 383, 12/1975
P0520611
(9+13)
h#4*
* 1. ... c8=D 2. Txh7 Dxc6 3. Th3 Dxg2+ 4. Kxg2 Db7#
1. cxb1=L c8=D 2. Lxf5 Dxc6 3. Lh3 Dxg2+ 4. Kxg2 Le4#
play all play one stop play next play all
siehe P0520614
Felber, Volker: Korrektur 06/1976, Heft 12/1976, Seite 192: +wBe2 (2015-10-25)
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-04-05)
comment
Keywords: Sacrifice of white pieces, Promotion
Genre: h#
FEN: 8/prP4B/P1p5/2p2P2/2P5/6p1/2pPPprb/nQK2bqk
Input: Hans-Jürgen Schäfer, 1997-10-11
Last update: Felber, Volker, 2015-10-25 more...
81 - P0524167
Carl Erik Israel Lind
1161 Scacco! 01/1978
P0524167
(2+4) C+
h#2
2.1...
1) 1. Sc5 Tb6 2. Kd4 Tb4#
2) 1. Sg7 Ke2 2. Sf5 Tg4#
play all play one stop play next play all
HBae: Vorgänger P0524171 (2022-04-04)
comment
Keywords: Echo
Genre: h#
Computer test: (Popeye C-Version 3.47 (1024 KB))
FEN: 8/8/4n1R1/3pp3/4k3/8/3K4/8
Input: Markus Manhart, 1998-02-16
Last update: hpr, 1999-11-03 more...
82 - P0527647
Michel Caillaud
4827 Mat (Belgrade) 11-12/1983
P0527647
(5+13)
h#4*
*) 1. ... b8=D 2. Sc4 Dh8 3. Sa3 Dxb2+ 4. Kxb2 Tb4#
1) 1. Sb3 Tc4 2. Tf8 Txc2 3. Tc8 bxc8=L 4. Kxc2 Lf5#
play all play one stop play next play all
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-04-06)
Ladislav Packa: Zilahi theme (2022-04-07)
comment
Keywords: Sacrifice of white pieces
Genre: h#
FEN: 8/1P6/3p4/nP1p4/5R2/8/ppp1prp1/qkrbB1K1
Input: Hans-Jürgen Schäfer, 1998-03-11
Last update: hpr, 1999-11-06 more...
83 - P0530886
Albert Heinrich Kniest
John Niemann

922 Schachmatt 69 22/02/1948
P0530886
(3+4)
s#9
Längstzüger
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#
play all play one stop play next play all
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
Keywords: Maximummer
Pieces: du = 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...
84 - P0531884
Jan Hartong
197v Problem 03/1952
1. Preis
P0531884
(3+11) C+
h#4
b) sBa5 nach g5
a) 1. Dd8 Tc5+ 2. dxc5 Ke3 3. Kc4 Ke4 4. Dd3+ cxd3#
b) 1. Df2 Ta6 2. Kc5 Ke4 3. Kc4 Ta3 4. Dc5 cxb3#
play all play one stop play next play all
Yuri Bilokin: Version: bPa5-f2, bPb3, -bNg2, -bNh2, +bPg3, +bPh5 8/8/2Rp4/1p1k3p/1b5q/1ppK1pp1/2P2p2/8 (3+11) h#4 2.1…
1.Qg5 Ra6 2.Kc5 Ke4 3.Kc4 Ra3 4.Qc5 cxb3# (MM)
1.Qd8 Rc5+ 2.dxc5 Ke3 3.Kc4 Ke4 4.Qd3+ cxd3# (MM) (2022-04-24)
comment
Keywords: Sacrifice of white pieces
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/2Rp4/pp1k4/1b5q/1bpK1p2/2P4n/6n1
Reprints: 1340 FIDE Album 1945-1955 1964
Input: Ralf Krätschmer, 1998-03-30
Last update: Alfred Pfeiffer, 2016-02-15 more...
85 - P0536129
Rudolf Svoboda
198 Mat-Pat 7 12/1985
P0536129
(8+9)
h#2
b) in Mattstellung a) wird die Farbe des mattsetzenden Steins gewechselt
a) 1. Tc6 Kf3 2. Sa6 a4#
b) 1. Sb4 Kf4 2. Ta6 Sa3#
play all play one stop play next play all
Henrik Juel: Part a) is C+ Popeye 4.61 (2022-04-04)
comment

Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 3B1b2/8/r5p1/1kq4b/1n3KR1/P2pP1P1/2N1r1B1/8
Input: Michal Dragoun, 1998-04-06
Last update: Marcin Banaszek, 2022-04-04 more...
86 - P0545310
Dieter Linden
?
P0545310
(3+1) cooked
h#4
0.1...
1. ... c6 2. Kd6 c7 3. Kd7 Te8 4. Kxe8 c8=D#

NL:
1. ... Ta8 2. Kc6 Ta1 3. Kd7 c6+ 4. Ke8 Ta8#
uvm
play all play one stop play next play all
klären: Veröffentlichung??
Yuri Bilokin: Correction: +bPb5, +bPe6, +bRf3, +bBf4, +bRg2, +bPg6, +bBg8 2R3b1/8/4pKp1/1pPk4/5b2/5r2/6r1/8 (3+8) (2022-04-20)
comment

Genre: h#
FEN: 2R5/8/5K2/2Pk4/8/8/8/8
Input: hpr, 1998-04-26
Last update: hpr, 1999-11-10 more...
87 - P0546226
György Bakcsi
4811 L'Italia Scacchistica 1969
1. Preis
Informalturnier 1969
P0546226
(8+15) C+
h#3
1. Te5 b3 2. Sf6 c4 3. Df7 d5#
play all play one stop play next play all
F147
Verst-4
Sally: 1. Preis 1969.
Schwarze und weiße Selbstentfesselungen und Selbstfesselungen. (2017-09-15)
Mario Richter: Welche Stellung ist die richtige (diese mit sTh4 oder die mit sTh4 [P0574415])? (2022-05-14)
Yuri Bilokin: Version: More economical -wPf2, -bPf3, +bNh5 8/3pb2p/p2pk1qR/p2r1ppn/K2Pn2r/P7/1PP5/3bR3 (7+15) (2022-05-15)
comment

Genre: h#
Computer test: Popeye C-Version 3.41 (1024 KB)
FEN: 8/3pb2p/p2pk1qR/p2r1pp1/K2Pn2r/P4p2/1PP1RP2/3b4
Input: HBae, 1998-02-01
Last update: Alfred Pfeiffer, 2017-09-17 more...
88 - P0546565
Jozsef Balazs
5154 Die Schwalbe 03/1966
P0546565
(6+10)
h#5
1. a1=T Lxf7 2. c1=L Ke6 3. Ka2 Kd5 4. b1=L Kc4 5. Lxa3 Kc3#
play all play one stop play next play all
Verst-4
Felber, Volker: Quelle: Schach, 03/1966, Nr. 5154 (2008-01-23)
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-04-05)
comment

Genre: h#
FEN: n7/b1pK1p2/1pP2pB1/1P6/1P6/P7/ppp5/1k6
Input: HBae, 1998-03-10
Last update: hpr, 1999-11-10 more...
89 - P0546719
Joao Baptista Santiago
Charles Edward Kemp

4309 Probleemblad 09-10/1961
P0546719
(4+13)
h#4
1. Tf3 dxe4 2. Dg2 e5 3. Lf2 Lf5 4. Se3 e6#
play all play one stop play next play all
F111
Verst-4
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-04-06)
comment

Genre: h#
FEN: K1B5/4p1B1/3p4/2pn2p1/1pbbp3/3P4/p2q1r2/k7
Input: HBae, 1998-04-12
Last update: hpr, 1999-11-10 more...
90 - P0548143
Jan A. Rusek
14 Zadaniowiec 10-12/1955
1. ehrende Erwähnung
Sektion h#2
P0548143
(4+5)
h#2
b) sBc6 nach b4
a) 1. Sd6 La2+ 2. Sbc4 Te5#
b) 1. Kd4 Td1+ 2. Kc3 Td3#

NL
a) 1. Sd3+ Lxd3 2. Sd6 Te5#
play all play one stop play next play all
Marcin Banaszek: Mit sTa2 statt sBa4 ist die Aufgabe korrekt ohne NL (C+) (2022-04-20)
comment
Keywords: Fesselungsspiel (222 202 000)
Genre: h#
FEN: 8/8/2p5/3k4/p1n5/8/1n3K2/BB2R3
Input: Markus Manhart, 1998-05-11
Last update: Marcin Banaszek, 2022-05-15 more...
91 - P0550888
Niharendu Sidkar
(III) British Chess Magazine 05/1968
P0550888
(3+3) C+
h#3
Duplex
s) 1. Lf5 Kf2 2. Kf4 Lf3 3. Le5 Le3#
w) 1. ... Lf3 Kf6 2. Kf4 Lf5 3. Le3 Le5#
play all play one stop play next play all
Totalsymmetrie, Idealmatts
A.Buchanan: What's a polar echo? (2022-03-08)
Henrik Juel: Good question
It must refer to the two mate positions:
The bishops stand on e3,f3 and e5,f5; with the kings on f2 and f4, Black is mate, and with kings on f4 and f6, White is mate
In non-duplex problems echo/chameleon echo mates are pretty well defined (black king on same/different square color)
So polar echo is just a third type, maybe only occurring in duplex problems (2022-03-08)
A.Buchanan: Thank you, Henrik. A non-duplex mechanism for Polar Echo is shown by P1305339 & P1305340. (2022-03-08)
comment
Keywords: Polar Echo, Aristocrat, Miniature
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/B6b/8/4k3/8/4K3/8/b6B
Reprints: 1823 Ideal-Mate Encyclopedia Vol.1 1999
Input: Ronald Schäfer, 1998-07-01
Last update: A.Buchanan, 2022-03-08 more...
92 - P0552996
Tamas Lefkovits
723 Die Schwalbe 26 08/1961
P0552996
(7+15) C+
h#4
1. bxa1=L e7 2. Lxe5 e8=S 3. Lxf6 Sxf6 4. exd1=S Sxg4#
play all play one stop play next play all
Erich Bartel: die richtige Problem Nr. ist 723.----
Nachdruck:
1) G13 Problemkiste (89-90) X 1993.--- (2007-04-29)
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-04-06)
comment
Keywords: Sacrifice of white pieces, Promotion (l,S,s)
Genre: h#
Computer test: SCHRECKE (2022-04-06): C+, Gustav 4.2a, Brute Force
FEN: 8/8/4PQ2/4B3/5Pb1/p2p2pp/1p1ppprk/RK1Rbrqn
Reprints: 723 Die Schwalbe 08/1961
G13 Problemkiste 89-90 10/1993
Input: Hans-Jürgen Schäfer, 1998-07-01
Last update: Mario Richter, 2022-04-07 more...
93 - P0554697
Sandor Boros
1019 Il Problema 10/1932
P0554697
(7+12)
h#4
1. f1=T+ Ke3 2. e1=T+ Kxd3 3. d1=T+ Kxc3 4. c1=L Lxb3#
play all play one stop play next play all
SCHRECKE: ZU möglich: 3.c1L+ K:c3 4.d1T L:b3# (2022-04-06)
comment
Keywords: Fesselungsspiel (111 101 000)
Genre: h#
FEN: 8/8/8/8/B2PP2P/nppp1KP1/krpppp1R/qb6
Input: Markus Manhart, 1998-06-10
Last update: hpr, 1999-04-11 more...
94 - P0554705
Ziva Tomic
1784 Problem 06/1961
P0554705
(6+16)
h#4
1. Lb6 Ta2 2. Tf7 Txa6 3. Se7 d5 4. Kf6 Ld4#
play all play one stop play next play all
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-04-06)
comment
Keywords: Fesselungsspiel (111 101 000)
Genre: h#
FEN: 8/qrb3k1/p5pn/1p1n1br1/3Pp3/1Pp2pp1/1RP1pB2/6K1
Input: Markus Manhart, 1998-06-10
Last update: hpr, 1999-04-11 more...
95 - P0554803
David Robert Wertheim
1831 Die Schwalbe 02/1976
P0554803
(10+9)
h#4
1. Sxc4 Sxc6 2. Sd2 Sxd4 3. Sxf3 Se2 4. Sd2 Tc1#
play all play one stop play next play all
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-04-06)
comment
Keywords: Fesselungsspiel (112 101 101)
Genre: h#
FEN: 2RN1K2/1p4p1/pPp3P1/4p2B/2PpPp2/3P1P2/3n4/3k4
Input: Markus Manhart, 1998-06-12
Last update: hpr, 1999-04-11 more...
96 - P0554859
Walter Supp
6193v Die Schwalbe 110 04/1988
P0554859
(4+13)
h#4
1. Sb5 Lg1 2. Da1 g4 3. Da8 g5 4. Sa7 Lh2#
play all play one stop play next play all
SCHRECKE: Es gibt eine zweite eindeutige Lösung:
1.Tf5 Lf6 2.Kc8 L:d8 3.De5 L:b6 4.Db8 L:f5# (2022-04-06)
Mario Richter: Im Original stand der sSd8 auf f8 mit der dann möglichen 2. Lösung 1. Ka7 Lxc5 2. Ka6 Lxb6 3. Ta5 Lc5 4. b5 Lc8#.
Durch die Versetzung nach d8 ('Schwalbe' Heft 114, Dezember 1988, S. 550) wollte der Autor die
Eindeutigkeit wieder herstellen - das hat offensichtlich nicht geklappt ... (2022-04-07)
comment
Keywords: Fesselungsspiel (112 110 110)
Genre: h#
FEN: 1k1n3q/1p3p2/1p5b/2p4r/1prBp3/2n4B/6P1/7K
Input: Markus Manhart, 1998-06-13
Last update: hpr, 1999-04-11 more...
97 - P0554909
Rudolf Leopold
Erwin Masanek

5195 Deutsche Schachzeitung 08/1987
P0554909
(8+15)
h#4
1. Tff1 b5 2. S1f2+ Lxe1 3. Sxe5 Kxb2 4. Lf5 Ld2#
play all play one stop play next play all
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-04-06)
comment
Keywords: Fesselungsspiel (112 110 110)
Genre: h#
FEN: 8/1p4pb/1Pp1p3/B1P1P3/1P1p1k2/1p1n1Pb1/1p3rP1/1K1nr1q1
Input: Markus Manhart, 1998-06-16
Last update: hpr, 1999-04-11 more...
98 - P0567560
Gyula Bebesi
Jozsef Pogats

Magyar Sakkelet 1952
P0567560
(11+12)
h#25
1. Kxc6 f5 2. Lc2 f4 3. Lxd1 Kxd1 4. Kd7 12. Kxf4 13. Kxf5 14. Kxf6 15. Kg5 Kd1 16. f5 20. f1=S Ke1 21. Sxd2 Kxd2 22. Kh4 Ke3 23. d2 Kf4 24. d1=L Kf5 25. Lh5 Lg5#
play all play one stop play next play all
Cook: Dual 12. Kxf5 Ke1 13. Kxf6 f5 14. Kxf5 Kd1 15. Ke4 Ke1 16. Kxd4 Kd1 17. f5 21. f1=S Ke1 22. Sxd2 Kxd2 23. Kc5 Ke3 24. Sf3 Kf4 25. Kd4 Le3#
Dieses Problem ist eine Version/Verlängerung eines H#24 von Bebesi: P0572173. Quelle und Datum der Version sind nicht bekannt und damit auch nicht, wie Pogats zur Co-Autorschaft kam.
Ion Murarasu: May be ,is this problem :
Gyula Bebesi
619 Magyar Sakkelet 4.1952, 2° dicseret
H#24 (9+12)
White: Pf6 Pf5 Pd4 Pf4 Pc3 Pb2 Pd2 Bc1 Kd1 (9)
Black: Pf7 Pd5 Ka4 Pc4 Pb3 Pd3 Pg3 Ph3 Pg2 Rh2 Sg1 Rh1 (12)
SOLUTION:
1.Ka4-b5 Kd1-e1 2.Kb5-c6 Ke1-d1
3.Kc6-d7 Kd1-e1 4.Kd7-e8 Ke1-d1
5.Ke8-f8 Kd1-e1 6.Kf8-g8 Ke1-d1
7.Kg8-h7 Kd1-e1 8.Kh7-h6 Ke1-d1
9.Kh6-h5 Kd1-e1 10.Kh5-g4 Ke1-d1
11.Kg4*f4 Kd1-e1 12.Kf4*f5 Ke1-d1
13.Kf5*f6 Kd1-e1 14.Kf6-g5 Ke1-d1
15.f7-f5 Kd1-e1 16.f5-f4 Ke1-d1 17.f4-f3
Kd1-e1 18.f3-f2 + Ke1-d1 19.f2-f1=S
Kd1-e1 20.Sf1*d2 Ke1*d2 21.Kg5-h4
Kd2-e3 22.d3-d2 Ke3-f4 23.d2-d1=B Kf4-f5
24.Bd1-h5 Bc1-g5 #

C+ Popeye WINDOWS-32Bit-Version 3.77 (4
MB) (2003-12-12)
Gerd Wilts: Stellung korrigiert (+sSg1), Anmerkung zu Quelle und Autorschaft ergänzt; Dual ergänzt (Hinweise von HG). (2022-04-09)
more ...
comment

Genre: h#
FEN: 8/5p2/2Qk1P2/3p4/1p1P1P2/1P1p1Ppp/1P1P2pr/1bBRK1nr
Input: hpr, 1999-03-21
Last update: Gerd Wilts, 2022-04-09 more...
99 - P0567876
Romeo Bedoni
1978 Europe Echecs 02/1973
P0567876
(3+7) C+
h#4
2.1...
1) 1. Dd5 Tc1 2. Td4 Txc2 3. Ke4+ Kg4 4. Sd3 Te2#
2) 1. De5 Txe1 2. Te4 Te2 3. Kf4+ Kxh4 4. Se3 Tf2#
play all play one stop play next play all
Yuri Bilokin: Version: -wPa2, -bPh4, bPg7-b2, +bBc5 8/8/8/2b2k1K/4qr2/8/1pn5/R3n3 (2+7) (2022-04-26)
comment
Keywords: Reihen-Echo
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/6p1/8/5k1K/4qr1p/8/P1n5/R3n3
Input: Ralf Binnewirtz, 1999-01-05
Last update: hpr, 1999-03-25 more...
100 - P0569828
Evgeny Sorokin
564 Problemas 01-03/1975
A.F.Argüelles gewidmet
P0569828
(4+1) cooked
h#2*
2.1...
*) 1. ... Te8 2. Kf7 Lh5#
1) 1. Kg5 Tg1 2. Kh4 Lf6#
2) 1. Kf7 Le6+ 2. Kf8 Tf1#

NL:
1. Kf7 Lh5+ 2. Kf8 Te8#
play all play one stop play next play all
Yuri Bilokin: Correction: rotate 180, then wKe2-e1, +bPe2 3R4/8/5B2/1B6/8/1k6/4p3/4K3 (4+2) h#2* 2.1…
1...Rd1 2.Kc2 Ba4#
1.Kc2 Bd3+ 2.Kc1 Rc8#
1.Kb4 Rb8 2.Ka5 Bc3# (MM)
Orthogonal-diagonal transformation. (2022-05-11)
comment

Genre: h#
FEN: 8/3K4/6k1/8/6B1/2B5/8/4R3
Input: hpr, 1999-04-01
Last update: hpr, 2012-05-07 more...
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The problems of this query have been registered by the following contributors:

Gerd Wilts (63)
hpr (9)
Hans-Jürgen Schäfer (8)
Markus Manhart (10)
Andreas Mokosch (2)
Ralf Krätschmer (2)
Michal Dragoun (1)
HBae (3)
Ronald Schäfer (1)
Ralf Binnewirtz (1)