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1 - P0000498

3097 Die Schwalbe 62 04/1980

3. ehrende Erwähnung

(10+13)

Wie weit muß der sK maximal nach links gelaufen sein?

**Thomas Volet**3097 Die Schwalbe 62 04/1980

3. ehrende Erwähnung

(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

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

**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...

**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...

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#

Alle Versuche, das Satzspiel 1. ... 0-0! 2. cxd4 Tc1# durch einen schwarzen Vorschaltzug aufrecht zu erhalten, scheitern.

*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#

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)

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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

3876 Die Schwalbe 74 04/1982

(4+3)

h=2 (AP)

Circe, Monochromes Schach

**Nikita M. Plaksin**

Andrej N. KornilowAndrej N. Kornilow

3876 Die Schwalbe 74 04/1982

(4+3)

h=2 (AP)

Circe, Monochromes Schach

Intended solution:

1. dxc3ep[+wBc2] 0-0 2. axb1=S Kh2=

2. ... Kf2? 3. Kf3!

1. ... Tf1? 2. axb1=S Kf2? 3. Kf3!

1. dxc3ep[+wBc2] 0-0 2. axb1=S Kh2=

2. ... Kf2? 3. Kf3!

1. ... Tf1? 2. axb1=S Kf2? 3. Kf3!

**Erich Bartel**: vom Dual 2.-- Kf2/Kh2 abgesehen C+ PY V4.41.--- (2008-11-07)

**A.Buchanan**: This is cooked in Popeye because White can't castle as long as bK controls f1! This is quite reasonable behaviour in monochrome: compare to en passant! So no solution (2022-02-18)

comment

**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...

1. ... Kg5 2. 0-0?? Se7#

1. ... Txh7 2. Kf8 Txh8#

1. cxb3ep Txh7 2. Kf8?? Txh8#

1. cxb3ep Kg5 2. 0-0! Se7#

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.

1. ... Txh7 2. Kf8 Txh8#

1. cxb3ep Txh7 2. Kf8?? Txh8#

1. cxb3ep Kg5 2. 0-0! Se7#

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)

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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...

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

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.

b) also 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#

is clearly the intention

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...

1. bxc3ep Sa6 2. 0-0-0 Tc4#

**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...

a) 1. T2xh3 Txh3 2. 0-0 Th8#

b) 1. La4 0-0 2. Tf8 Te1#

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)

b) 1. La4 0-0 2. Tf8 Te1#

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)

**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...

R: 1. Kg8xDh8

**A.Buchanan**: Solved in YouTube: https://www.youtube.com/watch?v=Do_rkLQnrpk (2022-02-17)

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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...

R: 1. a7xTb8=L

**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

Problem 5-6, p. 92, 12/1951

6. ehrende Erwähnung

(4+1)

Welches war der letzte Zug?

**Peter Kahl**

Jan R. Mortensen

Sveto StambukJan R. Mortensen

Sveto Stambuk

Problem 5-6, p. 92, 12/1951

6. ehrende Erwähnung

(4+1)

Welches war der letzte Zug?

R: 1. Dh3xSh2+

im Original ist als Dritter Autor D. Suboticanec angegeben, habe entsprechend geändert.

Zu P. Kahl heißt es ergänzend: [aus] Osterholz-Scharmbeck

I know that 'Jan Robert Mortensen' is identical to 'Jan Mortensen'; he never used his middle name (2017-03-08)

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)

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)

comment

Zu P. Kahl heißt es ergänzend: [aus] Osterholz-Scharmbeck

**HBae**: Ist "Klaus Peter Kahl" identisch zu "Peter Kahl" ? (2010-09-22)**Henrik Juel**: Yes, probablyI 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, AndrewI 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 PlaksinBut 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...

1. ... hxg6ep[+sBg7] 2. a1=T+ Txa1[+sTh8] 3. 0-0 Ta3 4. bxa3[+wTa1] 0-0-0 5. Td8 Txd8[+sTh8]#

Wenn die beiden Könige noch nicht gezogen haben (was a.p. durch Ausführung beider Rochaden bewiesen wird!), kann der letzte Zug nur g7-g5 gewesen sein, Sonst ginge z.B. auch R: 1. Kd8-e8 c2xDb3

Wenn die beiden Könige noch nicht gezogen haben (was a.p. durch Ausführung beider Rochaden bewiesen wird!), kann der letzte Zug nur g7-g5 gewesen sein, Sonst ginge z.B. auch R: 1. Kd8-e8 c2xDb3

**Henrik Juel**: A possible retroplay is -1... g7 -2.f3xP(-Pg7) b5 -3.f2 c6xP(-Pb2) -4.b2 a6 -5.Rc3 a7 -6.Rc1 b6xB -7.Bf8 b7 -8.Bg7xB etc. (2003-04-22)

**Michel Caillaud**: The problem is not cooked as indicated, as under A Posteriori convention both castles must be played in the solution to justify the en passant key. (2022-02-04)

**A.Buchanan**: If wK never moved, then Black has not just moved another non-capturing pawn, as White has no prior move. The only possible exception is R: 1. g7-g5 f/h3xPg4[+Pg7]. (2022-02-04)

comment

**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...

Schwarz hat bereits rochiert!

See P1398939

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)

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)

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comment

**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)

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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...

a) 1. Kd8 Se5 2. Te8 Sxf7#

b) 1. 0-0 Sd5 2. Sh8 Sxe7#

b) 1. 0-0 Sd5 2. Sh8 Sxe7#

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"

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)

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)

comment

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 problem1. 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...

1. bxa5 Sb6 2. axb4 Ta8#

**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

1. ... cxd6ep 2. 0-0-0 0-0-0 3. Kd7 Sa7#

1. De7 c6 2. Th8 c7 3. Tf8 Sd6#

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.

1. De7 c6 2. Th8 c7 3. Tf8 Sd6#

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...

**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)

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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...

**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...

1. Da1! droht (unparierbar) 2. Dh8#

1. ... 0-0-0? ist illegal, da sK oder sTa8 schon gezogen haben muss

1. ... 0-0-0? ist illegal, da sK oder sTa8 schon gezogen haben muss

**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...

1. 0-0-0?? illegal

1. Td1! ... 2. Td3#

1. Td1! ... 2. Td3#

1. dxe3ep 2. fxg1=L 3. 0-0 Tg4#

1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt

1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt

**Henrik Juel**: If Black may castle, his latest move must have been e3xf2, so last move was e2-e4. (Orig. wPg2 promoted on g8, so g7xh6 happened early). 1.dxe3 e.p. 2.fxg1=B 3.0-0 Tg4# (2003-03-21)

**James Malcom**: Is this the only know one-sided Valladao in a regular series-mover helpmate? Also, can wPh2 be shifted to h4, and the White king moved as well to h2, so that the solution must have 2. fxg1=S? (2020-12-08)

**James Malcom**: Well, the only known one that is in minimum form, i.e. 3 moves, I mean. (2020-12-08)

**A.Buchanan**: This is nice. I don't know of any other AP ser-h#3 with Valladao, but there could be. Just been cleansing PDB data today. Yes I think wKh3 and wBh2 can be transposed, and then promote to S instead of L. (2020-12-08)

**A.Buchanan**: The legality of intermediate positions in seriesmovers is of course irrelevant. However, each proof game that leads to the diagram position with BTM can be appended with a black move and then a "flip" that changes the player to move. We can repeat this process, and then the fact that Black retains castling rights implies that they had castling rights in the diagram position. I think this is the right formal way to handle AP with seriesmovers. (2022-03-21)

comment

**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...

BTM: 1. ... Lxb7+ 2. Ke3 0-0-0 3. Sb6#

WTM: 1. Sf6+! Kd8,Kf8 2. Dc7#,Ld6#

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!

WTM: 1. Sf6+! Kd8,Kf8 2. Dc7#,Ld6#

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...

1) 1. cxd3ep Ta5 2. Sf5 0-0 3. Sd4 Txe5#

2) 1. gxf3ep Th5 2. Sd5 0-0-0 3. Sf4 Txe5#

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"?)

2) 1. gxf3ep Th5 2. Sd5 0-0-0 3. Sf4 Txe5#

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...

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.

Cook: 17 Black pawns, retro cook and numerous forward cooks

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.

Cook: 17 Black pawns, retro cook and numerous forward cooks

klären wK im Schach, vielleicht wBb2?

r3k3/3b2p1/5p2/6b1/4pPRp/2pq2rp/2p1P1pB/2K3n1/ (2022-04-25)

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)

comment

**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, +sBg7r3k3/3b2p1/5p2/6b1/4pPRp/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

1. cxd3ep f4 2. Lf3 Kxc5#

**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...

1. Kc3 0-0-0 2. Txc4 Txd3#

1. bxc3ep e4 2. Kc4 Ta4#

Two solutions, with no retro tries or set play. The question is how info about game state should spread between one solution and another.

1. bxc3ep e4 2. Kc4 Ta4#

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)

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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...

1. ... exd6ep 2. 0-0-0 dxe7+ 3. Tf8 exf8=T,D#

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

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).

1. ... exd6ep 2.Kd8? dxe7+ 3.Kc8 e8=T(D)#?? (2007-02-11)

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)

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)

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**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...

1. ... fxg6ep 2. 0-0 gxh7#

**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)

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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...

1. fxg3ep 0-0 2. Lg4 hxg3#

**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

3. Makuc-Moder-Gedenkturnier 1971-1973

1. Preis

(12+13) C+

h#2

b) wBd4 nach d5

**Janko Furman**3. Makuc-Moder-Gedenkturnier 1971-1973

1. Preis

(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

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#

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

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

7273 Schach-Echo 11/1972

(8+14) cooked

h#2

b) Gespiegelt (a1<->h1) & wKd1->e1

**Janko Furman**

Miroslav StosicMiroslav Stosic

7273 Schach-Echo 11/1972

(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#

Cook: a) 1. Kg3 Kf1 2. Df4 Th3#

1. Kg3 0-0 2. Dh4 Txf3#

b) 1. exd3ep 0-0-0 2. dxe2 Ld5#

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...

1) 1. gxf3ep 0-0-0 2. Te7 Th4#

2) 1. cxd3ep 0-0 2. Da4 Txa4#

2) 1. cxd3ep 0-0 2. Da4 Txa4#

**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)

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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...

a) 1. f5 Le5 2. 0-0 Th8#

b) 1. Kd8 0-0-0 2. Te8 Lf8#

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?

b) 1. Kd8 0-0-0 2. Te8 Lf8#

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.

1...b8S and 2...R:h8# (2021-02-10)

comment

**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

RA29 diagrammes 20 03-04/1976

(9+9)

Wieviel Bauern muß man entfernen, damit die Stellung legal ist?

**Radu Dragoescu**RA29 diagrammes 20 03-04/1976

(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...

1. d3 ... 2. Tc2 ... 3. Tc8#

36 - P0003839

RA44 diagrammes 23 09-10/1976

Jean Oudot gewidmet

(16+16)

Wieviele Figuren muß man entfernen, damit die Stellung legal wird?

**Radu Dragoescu**RA44 diagrammes 23 09-10/1976

Jean Oudot gewidmet

(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)

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**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...

1. La7 Kxa7 2. c8=T Ka6 3. Ta8#

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."

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)

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)

C+ by Popeye 4.61 (2016-08-30)

When I test problems, I ignore plays that do not occur (2022-04-02)

comment

**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 Bauernumwandlungenin 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, AndrewWhen 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...

1. axb6ep+ Kb5 2. bxc7+ Tb6 3. 0-0-0 ...

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!

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.

'stip exact-#5' would disregard #4 as a solution, I believe (2022-03-29)

comment

**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...

1. fxe3ep? Dxa8# No justification for the ep, so just retro try.

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.

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.

**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...

1. fxe3ep Dxa8#? because no AP justification for ep

1. 0-0-0 g7 2. Tf8 gxf8=D#

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#

1. 0-0-0 g7 2. Tf8 gxf8=D#

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.

more ...

comment

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...

1. bxc3ep Dg8#? (thematic try)

1. bxc3ep Dd5 2. 0-0-0 Db7#

Cook: 1. bxc3ep Ld6 2. 0-0-0 Dc4#

1. bxc3ep Dd5 2. 0-0-0 Db7#

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 ...

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**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

Die Schwalbe , p. 463, 03/1939

Allen Teilnehmern herzlich gewidmet

Internationaler Lösungswettkampf 1938

(13+11)

#3

**Immo Fuß**Die Schwalbe , p. 463, 03/1939

Allen Teilnehmern herzlich gewidmet

Internationaler Lösungswettkampf 1938

(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+

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.

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+

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...

* 1. ... Sb5#

* 1. ... dxc3#

1. fxe3ep 2. e2 3. exf1=S 4. Sxd2 5. Sxb1 6. Sxa3 0-0-0!#

* 1. ... dxc3#

1. fxe3ep 2. e2 3. exf1=S 4. Sxd2 5. Sxb1 6. Sxa3 0-0-0!#

**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...

1. fxg3ep Ta4#? (no proof of ep right)

1. fxg3ep 0-0-0! 2. g2 Td4#

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#

1. fxg3ep 0-0-0! 2. g2 Td4#

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

comment

**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

30 Schachkongress Teplitz-Schönau im Oktober 1922 , p. 484, 1923

(13+14)

Weiß nimmt 1 Zug zurück, dann #2

**Hans Klüver**30 Schachkongress Teplitz-Schönau im Oktober 1922 , p. 484, 1923

(13+14)

Weiß nimmt 1 Zug zurück, dann #2

R: 1. Lb6-d8, dann 1. bxc6ep

**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...

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#

Black cannot steal the move, as White can prevent the castling justification.

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#

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...

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#

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#

**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...

WTM: ???

BTM: 1. ... Kxf3 2. 0-0+! Ke3 3. Td1 Kf3 Td3#

not 2. Kd2? Kg2 3. Sf4+ Kf2,Kf3 4. Tf1#

BTM: 1. ... Kxf3 2. 0-0+! Ke3 3. Td1 Kf3 Td3#

not 2. Kd2? Kg2 3. Sf4+ Kf2,Kf3 4. Tf1#

**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

8474 The Fairy Chess Review 28/11/1949

(11+11)

Schwarz nimmt 1 Zug zurück, dann h#1

**Niels Høeg**8474 The Fairy Chess Review 28/11/1949

(11+11)

Schwarz nimmt 1 Zug zurück, dann h#1

R: 1. ... Ka3-b2, dann 1. Ka4 Lb5#

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.

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)

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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

(2) diagrammes 15 07-09/1994

(2+7)

Ist die Stellung legal?

Monochromes Schach

**Nikita M. Plaksin**

Andrej N. KornilowAndrej N. Kornilow

(2) diagrammes 15 07-09/1994

(2+7)

Ist die Stellung legal?

Monochromes Schach

R: 1. ... 0-0 2. Kf8-e7 Lg8-h7 3. Kg7-f8 Lh7-g8 4. Le7-d8 Lg8-h7 5. La3-e7 Lh7-g8 6. Kh6-g7 e7xTf6 7. Kg5-h6

**Kees**: R: -1. … 0-0 -2. Kf8-e7 Lg8-h7 -3. Kg7-f8 Lh7-g8 -4. Le7-d8 Lg8-h7 -5.Le7-a3 Lh7-g8 -6. Kh6-g7 e7xTf6 Kg5-h6 (2022-02-16)

**A.Buchanan**: The final position in Kees' solution has wTf6 which must be promoted. This could have been wBb, which captured e.g. bBa5, bBb6 e.p., bLa7, bSb8.

wTf7 is also promoted, and might be sBh having capturing wBg5, wBh6 e.p., wLg2, wTh1. sBfxLg6 completes the picture. I don't see any difficult captures e.g. of S remaining, so the position looks to be legal. (2022-02-16)

comment

**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

9128 Die Schwalbe 157, p. 283, 02/1996

Leonid Borodatow gewidmet

(9+4)

#3 (AP)

**Andrey Frolkin**9128 Die Schwalbe 157, p. 283, 02/1996

Leonid Borodatow gewidmet

(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#

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.

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#

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

9133v Die Schwalbe 157, p. 283, 02/1996

(13+12) cooked

BP in 20.0

May White castle?

**Andrey Frolkin**9133v Die Schwalbe 157, p. 283, 02/1996

(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

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

**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...

1. f4 e5 2. Kf2 Dh4+ 3. Kf3 Df2+ 4. Kg4 h5+ 5. Kh3 h4 6. e4 d5+ 7. g4 hxg3ep#

**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...

1. bxc6ep

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

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"

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)

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)

comment

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 darkBut 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...

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

**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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**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

9375 Die Schwalbe 161 10/1996

Nikita Plaksin gewidmet

(14+9) cooked

Die Partie ist beendet. Weshalb?

**Thomas Volet**9375 Die Schwalbe 161 10/1996

Nikita Plaksin gewidmet

(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...

An earlier stipulation in PDB was "a) #3 b) Weiß setzt in 3 Zügen # (AP)"

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)

and the twinning: b) Black to move (AP)

(the German stipulation text is not clear, either) (2015-09-09)

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)

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comment

**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: #3and 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 stipulationis 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)more ...

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...

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!)

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.

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!)

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...

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)

(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.

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)

(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.

1. a8=T+ Kd7 2. e8=L+ Ke6 3. f8=S+ Kf5 4. c8=D+ Se6 5. 0-0-0 Ke4 6. Dxc6+ Kf5 7. g4+ fxg3 e.p. 8. Df3+ Sf4 9. Lb5 axb5 10. Ta1 bxc4 11. Tab1 cxb3 12. Dd3+ Sxd3#

**paul**: Cooked by 1.f8=Q Kd7 2.c8=B Kd6 3.e8=S Ke5 4.Ra5+ Kd4 5.Qc5+ Ke4 6.Re3+ fxe3 7.Qxc6+ Kd3 8.Bf5+ Se4 9.Kd1 exd2 10.Qf6 Kxc4 11.Be6+ Kd3 12.Qc3+ Sxc3#

2... Kc7 3.a8=S+ Kd6 4.e8=S++ Ke5 5.Re3+ Se4 6.d4+ Kxd4 7.Rd1+ Sd2 8.Rf3 Ke5 9.Qf5+ Kd4 10.g3 fxg3 11.Re3 Kxe3 12.Qf2+ gxf2# (Jacobi)

See P1287941 as correction. (2022-04-06)

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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...

1. Lf2 e3 2. Kf3 Txf7 3. Kg3+ Tb7 4. Tf3 Tg7#

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#

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

comment

**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...

1. f3 Kb8 2. f2 Ka8 3. f1=L Kb8 4. Ld3 Ka8 5. Lb1 Kb8 6. La2 Ka8 7. Db1 Kb8 8. f5 Ka8 9. f4 Kb8 10. f3 Ka8 11. f2 Kb8 12. f1=L Ka8 13. Tf2 Kb8 14. Kf7 Ka8 15. Ke6 Kb8 16. Kd5 Ka8 17. Kc4 Kb8 18. Kc3 Ka8 19. Kb2 Kb8 20. Ka1 Ka8 21. Tb2 Kb8 22. T8f2 Ka8 23. Lf5 Kb8 24. Lc2 Ka8 25. d3 Kb8 26. Le3 Kc7 27. e5 Kxd6 28. Lc1 Ke6 29. Td2 Kf5 30. e4 Kg4 31. Le2+ Kxh3 32. e3 Kg2 33. Led1+ Kf1 34. e2+ Ke1=

**Keywords:**Rex solus (w), Superseded by (P0501023), Move Length Record

**Genre:**Fairies

**FEN:**K1b2rk1/4pr2/1b1n1p2/5q2/3p1p2/1p5n/8/8

**Reprints:**The Problemist 07/2008

**Input:**Gerd Wilts, 1996-06-06

**Last update:**James Malcom, 2022-04-27 more...

63 - P0501043

AHK-Memorial 1983-1985

2. Preis

(1+4)

ser-h#5

magischer Kb8

b) wKb8 nach g8

**Theodor Steudel**AHK-Memorial 1983-1985

2. Preis

(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#

**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...

1. Dxf4 0-0-0 2. Sf2 Kb1 3. Sd2+ Ka1 4. Td4 Te1#

**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...

1. f2 Tb1 2. Sf3 Tb2 3. Ke1 Txc2 4. Sd2 Tc1#

Cook: NL

1. Sg2 Th1 2. f2 Th8 3. Ke1 Th2 4. Kf1 Th1#

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...

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

NL:

1. Da5 b5 2. Ld8 bxc6 3. Ke7 c7 4. f6 c8=S# ua

**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...

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#

NL

1. Se3 Tf3 2. Lg4 Th3 3. Kf4 Kf6 4. e4 fxe3#

**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...

1. Sb3 Tg2 2. Tf1+ Tg1 3. Ta1 Tc1 4. b1=S Tc2#

**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...

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#

b) 1. Dxf2 Lc6 2. Dxe3+ Kb5 3. Dxe4 La7 4. Df3 Lxf3#

**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

41 To Alain White , p. 30, 1945

Alain C. White

zum 65. Geburtstag

(2+8) C+

h#3

**Philip Leonard Rothenberg**41 To Alain White , p. 30, 1945

Alain C. White

zum 65. Geburtstag

(2+8) C+

h#3

1. Dh3 Kg8 2. Tg3 Kf7 3. f3 Lc1#

**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

71 - P0505381

Magyar Sakkelet 11/1970

1. Preis

Turnier des Ungarischen Schachbundes

(3+6) cooked

h#4

**Lajos Riczu**Magyar Sakkelet 11/1970

1. Preis

Turnier des Ungarischen Schachbundes

(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#

NL:

1. Dg5 e4 2. Df5+ exf5 3. g5 Te1 4. Tg6 fxg6#

**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

72 - P0505422

292 Revista Romana de Sah 03/1934

2. -3. ehrende Erwähnung

(3+6) C+

h#4

**Paul Leibovici**292 Revista Romana de Sah 03/1934

2. -3. ehrende Erwähnung

(3+6) C+

h#4

1. Te3 c5 2. Tf1 Tf2 3. Sd3 Tc2 4. Tf3 Tc4#

**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

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#

2) 1. Lf4 Txa6 2. Ke2 exf4 3. Kf1 Kg3 4. Kg1 Ta1#

**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...

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#

NL:

1. Sf4 gxf4 2. Lg4 Tg6 3. Kf5 Ke7 4. De4 Tf6#

**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...

**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...

1. Da1 Lxf3 2. Da8 Lxa8 3. Lb7 Txb7 4. Lxh2 Txb1#

**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...

1. Sc7 Lh8 2. Sd5 Sg7 3. Kd4 Se6#

**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...

1. Ke4 Kb2 2. Db5+ Kc2 3. Kd5 e4+ 4. Kc4 d3#

**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...

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#

1. Txf8+ gxf8=D 2. Lc7 Db8 3. Ld8 Dxg3+ 4. Kxg3 Le5#

**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...

* 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#

1. cxb1=L c8=D 2. Lxf5 Dxc6 3. Lh3 Dxg2+ 4. Kxg2 Le4#

1) 1. Sc5 Tb6 2. Kd4 Tb4#

2) 1. Sg7 Ke2 2. Sf5 Tg4#

2) 1. Sg7 Ke2 2. Sf5 Tg4#

**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...

*) 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#

1) 1. Sb3 Tc4 2. Tf8 Txc2 3. Tc8 bxc8=L 4. Kxc2 Lf5#

**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...

1. Kb2 Ga8 2. Td7 Gc6 3. Kc3 Gf3 4. Kd4 Ga8 5. Ke5 Gc6 6. Kf6 Gg2 7. Kg7 Ga8 8. Kh8 Gc6 9. Tg7 Gh1#

**Anton Baumann**: NL: 1.Ta4+ Kb8! 2.h8=D+ Kc7 3.De8 Ga8 4.Dh5 Ga3 5.Tb4 Ga8 6.Tb2 Lh1 7.Dh7+ Kd6,Kd8 8.Dd3+ Ld5 9.Db1 Ga2# (2022-02-02)

**HBae**: Korrekturvorschlag (Neufassung)

weiß: Ka1 Gd7f4h7 (4) schwarz: Kb7 Gd5e4 (3)

1.Kb2 Ga8 2.Kc3 Gc6 3.Kd4 Gf3 4.Ke5 Ga8 5.Kf6 Gc6 6.Gf7 Gg2 7.Kg7 Ga8 8.Kh8 Gc6 9.Gg7 Gh1 #

C+ Popeye v4.37 (2022-02-03)

comment

**Keywords:**Maximummer

**Pieces:**= Grasshopper (G)

**Genre:**Fairies

**FEN:**8/kb5P/8/3*2q4/3R*2q3/8/8/K7

**Input:**Ralf Krätschmer, 1998-03-30

**Last update:**hpr, 1999-11-07 more...

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#

b) 1. Df2 Ta6 2. Kc5 Ke4 3. Kc4 Ta3 4. Dc5 cxb3#

**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

198 Mat-Pat 7 12/1985

(8+9)

h#2

b) in Mattstellung a) wird die Farbe des mattsetzenden Steins gewechselt

**Rudolf Svoboda**198 Mat-Pat 7 12/1985

(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#

b) 1. Sb4 Kf4 2. Ta6 Sa3#

**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...

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

NL:

1. ... Ta8 2. Kc6 Ta1 3. Kd7 c6+ 4. Ke8 Ta8#

uvm

klären: Veröffentlichung??

comment

**Yuri Bilokin**: Correction: +bPb5, +bPe6, +bRf3, +bBf4, +bRg2, +bPg6, +bBg8 2R3b1/8/4pKp1/1pPk4/5b2/5r2/6r1/8 (3+8) (2022-04-20)comment

87 - P0546226

4811 L'Italia Scacchistica 1969

1. Preis

Informalturnier 1969

(8+15) C+

h#3

**György Bakcsi**4811 L'Italia Scacchistica 1969

1. Preis

Informalturnier 1969

(8+15) C+

h#3

1. Te5 b3 2. Sf6 c4 3. Df7 d5#

F147

Verst-4

Schwarze und weiße Selbstentfesselungen und Selbstfesselungen. (2017-09-15)

comment

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...

1. a1=T Lxf7 2. c1=L Ke6 3. Ka2 Kd5 4. b1=L Kc4 5. Lxa3 Kc3#

Verst-4

comment

**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...

1. Tf3 dxe4 2. Dg2 e5 3. Lf2 Lf5 4. Se3 e6#

**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

14 Zadaniowiec 10-12/1955

1. ehrende Erwähnung

Sektion h#2

(4+5)

h#2

b) sBc6 nach b4

**Jan A. Rusek**14 Zadaniowiec 10-12/1955

1. ehrende Erwähnung

Sektion h#2

(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#

b) 1. Kd4 Td1+ 2. Kc3 Td3#

NL

a) 1. Sd3+ Lxd3 2. Sd6 Te5#

**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...

s) 1. Lf5 Kf2 2. Kf4 Lf3 3. Le5 Le3#

w) 1. ... Lf3 Kf6 2. Kf4 Lf5 3. Le3 Le5#

w) 1. ... Lf3 Kf6 2. Kf4 Lf5 3. Le3 Le5#

Totalsymmetrie, Idealmatts

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)

comment

**A.Buchanan**: What's a polar echo? (2022-03-08)**Henrik Juel**: Good questionIt 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...

1. bxa1=L e7 2. Lxe5 e8=S 3. Lxf6 Sxf6 4. exd1=S Sxg4#

**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...

1. f1=T+ Ke3 2. e1=T+ Kxd3 3. d1=T+ Kxc3 4. c1=L Lxb3#

**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...

1. Lb6 Ta2 2. Tf7 Txa6 3. Se7 d5 4. Kf6 Ld4#

**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...

1. Sxc4 Sxc6 2. Sd2 Sxd4 3. Sxf3 Se2 4. Sd2 Tc1#

**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...

1. Sb5 Lg1 2. Da1 g4 3. Da8 g5 4. Sa7 Lh2#

**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...

1. Tff1 b5 2. S1f2+ Lxe1 3. Sxe5 Kxb2 4. Lf5 Ld2#

**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...

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#

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#

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.

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)

more ...

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**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 ...

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**Genre:**h#

**FEN:**8/5p2/2Qk1P2/3p4/1p1P1P2/1P1p1Ppp/1P1P2pr/1bBRK1nr

**Input:**hpr, 1999-03-21

**Last update:**Gerd Wilts, 2022-04-09 more...

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#

2) 1. De5 Txe1 2. Te4 Te2 3. Kf4+ Kxh4 4. Se3 Tf2#

**Yuri Bilokin**: Version: -wPa2, -bPh4, bPg7-b2, +bBc5 8/8/8/2b2k1K/4qr2/8/1pn5/R3n3 (2+7) (2022-04-26)

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**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

564 Problemas 01-03/1975

A.F.Argüelles gewidmet

(4+1) cooked

h#2*

2.1...

**Evgeny Sorokin**564 Problemas 01-03/1975

A.F.Argüelles gewidmet

(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#

1) 1. Kg5 Tg1 2. Kh4 Lf6#

2) 1. Kf7 Le6+ 2. Kf8 Tf1#

NL:

1. Kf7 Lh5+ 2. Kf8 Te8#

**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)

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