Henrik Juel: After a couple of hours Natch 3.1 had not even found the symmetrical solution (2018-12-7)
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Henrik Juel: C+ by Natch 3.1 (16 min.)
The colored position has black Ta1,h1 and Pb5,g6, and white Ta7,h8, otherwise 'natural' (2018-12-7)
François Labelle: Henrik, did you really test all 2^30 possible colorings with Natch, or did you use human deductions to reduce the possibilities (making this HC+), or did you only test the intended coloring (making this a partial C+)? (2018-12-9)
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Henrik Juel: C+ by Natch 3.1
The colored position is 'natural' except for wTa8, wDd8, and sDd1 (2018-12-7)
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s. P0000133
NN: It seems we fall short of one retro-move and fail to keep either the bK or bR still.
Let's analyse, the distribution of bPs indicate 5 captures (bPh7×g6×f5×e4×d3×c2) plus 3 captures (bPc7×d6, bPb7×c6, bg7×f6) equalling a total of 8 captures, thus accounting for all of white's missing pieces. Next, wPs captured at least 3 (fxexdxc) plus wPc2×b3. Now, in order to keep the bK & bR still, we require sufficient supply of retro-moves for black. Thus, we must strive to release the bNa1 (as pawns are not enough!) at the earliest. But this can only be done through uncapturing the bishop on c1 followed by retracting bPd3×c2, wPd2-d3. An optimal way of doing this may be: -1.bPb7×wNc6 wNc6-d4, -2.bPa4-a3 wPc6-c7, -3.bPc7×wBd6 wBd6-b4, -4.bPa5-a4 wBb4-d2 -5.bPe6-e5 wBd2-c1 and here we can't retract anymore moves without violating bK's right to castle. Retracting bPa6-a5 for instance would mean that bRa8 got out of the pawn to be captured by a wP (as bBc8 never left its homesquare) forcing the bKe8 to move.
Thus, Black can't castle and the solution is 1.Kxf6 Rh6+ (1...0-0 is illegal) 2.Nxh6 ... 3.c8R/Q# (2018-11-30)
Mario Richter: If we assume that Black can still Castle, White needs a fifth capture to provide wPh2 as a capture object.
Furthermore, before d2-d3 can be retracted, both wRa1 and wBc1 have to return home (to b1 and c1) (2018-12-1)
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Henrik Juel: Here is one of the three similar cooks
1.Sg1-f3 Sb8-c6 2.Sf3-e5 Sc6-d4 3.Se5xd7 Be7-e5
4.Sd7-b6 Ba7xb6 5.Sb1-c3 Ta8-a3 6.Sc3-b1 Ta3-g3
7.Bh2xg3 Sd4-f3 8.Bg2xf3 Lc8-f5 9.Th1xh7 Lf5xc2
10.Lf1-h3 Lc2xh7 11.Lh3-c8 Bf7-f5 12.Bd2-d3 Ke8-f7
13.Lc1-e3 Kf7-g6 14.Sb1-d2 Kg6-h5 15.Ta1-c1 Bg7-g6
16.Tc1-c4 Lf8-h6 17.Tc4-e4 Dd8-f8 18.Le3-d4 Lh6-e3
19.Lc8-e6 (2018-12-7)
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Henrik Juel: The long promenade is made by k, not K (2018-12-7)
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Henrik Juel: C+ by Natch 3.1 (5 min.) (2018-12-7)
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Henrik Juel: Unfortunately cooked
1.Sg1-h3 Sb8-c6 2.Bf2-f4 Sc6-e5 3.Bf4-f5 Bg7-g5
4.Sh3-f4 Bg5-g4 5.Bf5-f6 Bg4-g3 6.Bf6xe7 Bf7-f5
7.Sf4-h5 Bf5-f4 8.Bh2-h3 Bf4-f3 9.Th1-h2 Bg3xh2
10.Ke1-f2 Bh2-h1=T 11.Kf2-g3 Th1xf1 12.Sb1-c3 Tf1-h1
13.Dd1-g1 Se5-g4 14.Dg1xa7 Sg4-h2 15.Da7-b8 Ta8-a3
16.Sc3-a4 Ta3-c3 17.Bb2xc3 Bf3-f2 18.Lc1-a3 Bf2-f1=D
19.La3-b4 Df1-f5 20.Ta1-f1 Df5-e6 21.Tf1-f3 Sg8-f6
22.Tf3-e3 Sh2-f1 23.Kg3-f3 Bc7-c5 24.exd8=T+ (2018-12-7)
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Henrik Juel: C+ by Natch 3.1 (4 min.) (2018-12-8, hidden) more ...
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Renny Bosch: I believe the minimum number of moves since the last pawn move is 68, based on the following PG (or SPG?): 1. g4 Nc6 2. Bg2 Na5 3. Bc6 Nb3 4. axb3 Nf6 5. Nc3 Ne4 6. Nd5 Nc3 7. dxc3 Rg8 8. Qd4 Rh8 9. Qb6 axb6 10. Be3 Rg8 11. Bc5 Rh8 12. Bb4 Rg8 13. Ra5 Rh8 14. Rc5 bxc5 15. Ba5 Rg8 16. Bb6 Rh8 17. Ba7 Rg8 18. Bb8 Rh8 19. Ba4 Ra6 20. Kd2 Re6 21. Kd3 Re5 22. Kc4 Re4+ 23. Kb5 Rc4 24. bxc4 Rg8 25. Bb3 Rh8 26. Nh3 Rg8 27. Ra1 Rh8 28. Ra7 Rg8 29. Nb6 Rh8 30. Na8 Rg8 31. Nf4 Rh8 32. Nd5 Rg8 33. Ndb6 h6 34. Na4 b6 35. Ba2 Bb7 36. Bb1 Be4 37. g5 Bh7 38. g6 Rh8 39. gxh7 { the last pawn move } g6 40. Rb7 Bg7 { move bB, bK, bQ out of row 8 } 41. Ba7 Be5 42. Rb8 Kf8 43. Ka6 Kg7 44. Kb7 Qg8 45. Rf8 Kf6 46. Kc8 { allow bR to pass wK } Qg7 47. Rg8 Ke6 48. Kd8 Qf6 49. Ke8 Qf5 50. Kf8 Bf4 51. Kg7 Qg5 { reverse order of the two rooks }52. Rb8 Rc8 53. Rb7 Rd8 54. Bb8 Rc8 55. Ra7 Rd8 56. Ra5 Rc8 57. Ba7 Rb8 58. Rb5 Rb7 59. Bb8 Ra7 60. Rb3 Ra5 61. Ra3 Rb5 62. Ra1 Rb3 63. Ba2 Ra3 64. Bb3 Ra2 65. Rd1 Ra1 66. Rd2 Re1 67. Rd1 Rf1 68. Ra1 Rb1 69. Ra3 Rc1 70. Ba2 Ra1 71. Bb1 Ra2 72. Rb3 Ra3 73. Rb5 Rb3 74. Ra5 Ra3 75. Ra7 Rb3 76. Rb7 Ra3 77. Ba7 Rb3 78. Rb8 Ra3 79. Rh8 Qf6+ 80. Kf8 { move bK past wR } Qg7+ 81. Ke8 Kf6 82. Kd8 Be5 83. Kc8 Qg8+ 84. Kb7 Kg7 85. Ka6 Kf8 86. Kb5 Ke8 87. f3 Kd8 88. Ba2 Kc8 89. Bb1 Kb7 90. Ba2 Qg7 91. Rg8 Qf6 92. Bb1 Qe6 93. Ba2 Bh8 94. Rg7 { put wR into pocket, and final cleanup } Kc8 95. Ka6 Kd8 96. Kb7 Ke8 97. Kc8 Kf8 98. Kd8 Rb3 99. Bb1 Rb5 100. Ba2 Ra5 101. Bb8 Ra7 102. Bb1 Rb7 103. Ba7 Qc6 104. Ba2 Qb5 105. Bb1 Qb4 106. Ba2 Qa3 107. Rg8# (2007-4-24)
Michel Caillaud: the exact stipulation ends "mindestens geschehen".
(in english : "minimum number of moves since the last Pawn move?").
the last Pawn move is f2-f3,that must be interposed before g7-g6 as otherwise the 50-moves rule would have been violated.
the answer is a minimum of 15,0 moves with a retro-play running as :
-1.Tg7-g8# ...-16.f2-f3! ... -65...g7-g6
with 49,5 moves between g7-g6 and f2-f3 the 50-moves rule has been respected (2007-9-12)
Michel Caillaud: more detail on intended retro-play :
-1.Tg7-g8# … -4…De6-c6 … -8…Tb3-b5 -9.Kb5-a6 Ta3-b3+ -10.La7-b8 … -13…Kb7-c8 -14.Tg8-g7 Ld4-h8
-15.Tf8-g8 Df6-e6 -16.f2-f3! Dg7-f6 -17.Th8-f8 Dg8-g7 -18.Lb1-a2 Ta1-a3 -19.La2-b1 … -24… Kf6-g7 -25.Kb7-a6 Dg7-g8 -26.Tb8-h8 Ke5-f6 -27.Kc8-b7 … -28… Dh4-f6 -29.Ke8-d8 … -35… Td2-d1 -36.Ta3-b3 … -37.Lb3-a2 … -39.Th1-a1 Td1-d2 -40… Ta1-d1 -41.Kg7-f8 Ta3-a1 -42.La2-b3 …-48… Th8-b8 -52.Tb7-a7 … -53.La7-b8 … -55.Tg8-b8 … -59.Kç8-d8 Kf6-e5 -60.Kb7-ç8 Kg7-f8 -61.Tb8-g8+ Kf8-g7 -62.Kc8-b7+ Ke8-f8 -63.Tb7-b8 Lg7-d4 -64.Lb8-a7 Lf8-g7 -65.Ta7-b7 g7-g6 (2007-9-18)
Henrik Juel: After -1.Tg7-g8# the main parts of the resolution are
(s/w mean black/white)
sDa3 out, sTb7 to a3, wKe8 to b5, sKf8 to b7, wTg7 to h8 and sLh8 to d4, sD to g8, sKb7 via g7 to e5, sDg8 out, wKb5 to g7, sTa3 to d2, wTh8 to e1, sTd2 to h8, wTe1 to g8,
(the rooks have switched place)
wKg7 to b7, wTg8 to b8, wKb7 to c8, sKe5 to e8, sLd4 to f8,
-65... g7-g6 -66.g6xLh7
(so -16.f2 was needed to avoid draw by the 50 moves rule)
wTb8 to h1, wKc8 to b5, sD to d8, sLh7 to c8,
-88... b7-b6
wSa8 out, wLa7 to a5,
-99... b6xTc5
wTc5 to a1, wLa5 to c1,
-111.d2xTc3
sTc3 to a8 and wSa4 to b1, wLa2 to a4
-222.b3xSc4 a7xDb6 -223.a2xSb3 (2018-12-5)
Henrik Juel: It is difficult to avoid typos in such a mess
in line 3, wKd8 to b5, not e8
Michel also seems to have made some typos:
49.0 moves between -16.f2 and -65... g7, not 49.5
-25... Kf6, not 24 (may have implications for the solution) (2018-12-5)
Renny Bosch: Correction: The last pawn move is of course not 39. gxh7, but 39 ... g6. Therefore the answer is 67.5 moves since the last pawn move. (2007-9-4, hidden) more ...
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NN: The above is basically two parts of a single solution depending on the history of the game and not two different solutions.

Proof of why castling rights are mutually exclusive: Let's assume black CAN castle. Now, the bPs piled up on h-file implies six captures of white units and these can only be wRBBNN & wPe2. Because all other possibilities involve promotion of wPs on d8/e8/f8 followed by their capture by bPs, interfering with black's castling right. Next, we observe that wRh1 getting captured by bPs implies one of the following two things: (i) wKe1 moved to let wRh1 out of the pawn-chain or (ii) wPg2,h2 cross-captured on h3,g3 respectively to facilitate the escape of wRh1. Furthermore, option (ii) necessitates bPd7 to be promoted to d1 so that it can be captured by wPs, implying wKe1 had to move anyway to let bPd7 promote.
Thus, Black can castle implies wKe1 moved and white can't castle. Contrapositively, white can castle will imply that black can't. (2018-11-29)
Henrik Juel: Nice analysis, Satanick (2018-11-29)
NN: Thanks Henrik Juel. Hope there are no mistakes. (2018-11-29)
Yoav Ben-Zvi: The claim that retaining Black castling requires an uncapture of a wR by Black's h-file pawns is not accurate. Consider the case that all 6 of these bP uncaptures are of wBs and wNs. In this case 2 of the uncaptured White pieces unpromote. Both could be unpromoted (on f8 or d8) by an uncapture from e7. This scenario requires a third wP uncapture (on the e file) so one of White's uncaptures is accounted for by [bPd7]. The direct uncapture of [bPd7] (on the e file) or its unpromotion without stepping on d2 requires this pawn to uncapture the wR, confirming the conclusion of mutually exclusive castling. A similar argument works for the case that the Black h file pawns play bPe7xwPf6 along with 5 uncaptures of WNs and wBs. (2018-12-3)
Henrik Juel: Now we can test
C+ by Popeye 4.61 (and analysis)
The main content is the retro analysis, of course, but as a helpmate problem the repeated 1.Td7 detracts (2018-12-3)
NN: Solutions: If white can castle: 1.Rd7 0-0-0 2.Kd8 Qxd7# & if black can castle: 1.Rd7 Rxa7 2.0-0-0 Ra8# (2018-11-29, hidden) more ...
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NN: Solution: If Black can castle long: 1.Qxc7! (Threat: Qxe7#) 1...N moves (1...Kf8 2.Qxe7+ Kg8 3.Qxg7#) 2.Qd7+ Kf8 3.Qf7# & if Black can castle short: 1.Qxg7! (Threat: Qxe7#) 1...N moves (1...Kd8 2.Qxh8+ Ng8 3.Qxg8#) 2.Qf7+ Kd8 3.Qd7#

The bBe1 is obviously a promotee. This implies bPd7,e7 made one capture each namely, dxe & fxe. Consequently, either the missing wPg2 or wPb2 (but not both!) had to be captured by bP on the e-file. But neither wPg2 nor wPb2 could have gone to the e-file capturing white units because that, together with wPe2×d3, make three captures in all and black is missing only two units. Hence, either one of them promoted to h8/a8 first prior to be captured on the e-file. This shows exactly one of bRh8 and bRa8 has moved before facilitating the promotion and therefore, black has right to castle only on one side. Accordingly,1.Qxc7 is the key if black has the right to castle long and 1.Qxg7 is the key if otherwise. (2018-11-29)
NN: Castling for black is legal only on one side. The above solution cover the two mutually exclusive cases. (2018-11-29, hidden) more ...
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Henrik Juel: C+ by Natch 3.1 (5 min.) (2018-12-6, hidden) more ...
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Henrik Juel: C+ by Natch 3.1 (2 min.)
Tempo loss at 13.f2-f3-f4 and 18.Lf1-h3-g2-h1 (2018-12-6)
Joost de Heer: IMO f2-f3-f4 isn't a tempo loss, as the pawn is needed for shielding. (2018-12-7)
Henrik Juel: I see no shielding by [Pf2], Joost
White could play 13.f4?, but he would run out of moves
White plays 17.gxf3, and this wP will later shield, so wLg2 does not check sKb7, but that is a different story (2018-12-7)
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Henrik Juel: C+ by Natch 3.1 (28 min.) (2018-12-8)
Henrik Juel: No solution in 15.0 or 14.5, so C++... (2018-12-8)
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Henrik Juel: C+ by Natch 3.1
No solution in 18.5 or 17.5, so C++ (2018-12-8)
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Henrik Juel: Natch 3.1 tests the first 20.0 moves OK in 13 seconds, but the first 20.5 moves would take more than an hour, and the entire problem more than a couple of hours, because of the possibility of [Pd7] promoting
So we may never know for sure whether this proof game with 16 oscillations by Ta8 is correct (2018-12-8)
paul: 18.Tca3 & 19.T3a4 (2010-4-30, hidden) more ...
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Henrik Juel: The content is concealed pawn captures
Despite no visible captures White captured a2xPb3xTa4 to open the a-line for [Ta1,Pa7], and h2xTg3xLh4 with black h7xLg6xDh5 to open the h-line for [Th1,Th8]
This makes life difficult for the human solver and for Natch 3.1, who tested the first 15.0 moves correct in 40 sec.
Testing the entire proof game would need more than a couple of hours by Natch (2018-12-8)
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paul: C+ (Euclide 1.11) for the first 13 moves (in about 51 hours). I do not dare to check for 13.5 moves. (2018-3-8)
Henrik Juel: Tiny typo in Mario's cook: 10.Txh7 (2018-3-8)
Mario Richter: The typo comes from a "by hand conversion" of the output of several tools I'm using (here: rawbats and chessbase) - unfortunately each tool has it's own output format, incl. the PDB solution field format ... (2018-3-9)
A.Buchanan: I have a janitor's tool which allows me to change the representation of positions & solutions.
Try to copy the file here: https://www.dropbox.com/sh/ld64s00zm0f9tx8/AAA7r8BsA0vuLcZOOKNvKkmua?dl=0
Click on it, and then hit the download button on the right.
The most useful is the one which converts from FEN to PDB "pieces" format.
Stick the input string in the yellow cell, and pick up the result from the green cell.
If you want to see how it's done, unhide the hidden rows and columns: no VBA just simple Excel.
Only 1% as clever as rawbats, but it still saves me a lot of time.
Feel free to use and to improve. (2018-3-9)
Mario Richter: @Andrew: Thanks for your offer - but I have such tools myself. My fault was not to use them (I thought that I'll be able to transform such a short PG by hand without errors ...)
Would be nice if the PDB-System would be more helpful with respect to language support.
I've decided to use 'english' in the language settings, but that doesn't help here, since solutions still have to be given in german piece notation.
Might be difficult to change this, especially when thinking of fairy piece names, so I can live with that, but there are other aspect which sometimes drive me crazy, e.g. the handling of source names in different languages ... (2018-3-9)
A.Buchanan: I thought you might have such a tool, but I thought it was a nice opportunity to announce it for others who might find it useful. (2018-3-9)
Henrik Juel: this problem should be labelled cooked (2018-12-8)
Mario Richter: The sequence of moves is not unique, e.g.:
1. f3 e5 2. Kf2 e4 3. De1 e3+ 4. Kxe3 b5 5. Dg3 b4 6. Kf2 Lc5+ 7. Ke1 Lxg1 8. Kd1 Lxh2 9. Txh2 d6 10. Lxh7 Le6 11. Txg7 Th1 12. Th7 Lxa2 13. Th8 Lxb1 14. Txa7 Ke7 15. Tb7 Ta1 16. Ta7 De8 17. Ta8 Kd8 18. De1 (2018-3-8, hidden) more ...
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Version A. Frolkin
Henrik Juel: Cooked by
1.Ba2-a4 Bd7-d5 2.Ta1-a3 Bd5-d4 3.Bh2-h4 Lc8-h3
4.Ta3-f3 Be7-e6 5.Tf3-f6 Lf8-d6 6.Sb1-a3 Ld6-h2
7.Bf2-f4 Bd4-d3 8.Ke1-f2 Dd8-d4 9.Kf2-f3 Dd4xg1
10.Dd1-e1 Dg1xf1 11.De1-f2 Df1xc1 12.Df2xa7 Dc1xb2
13.Kf3-f2 Db2-e5 14.Da7-e3 Sb8-d7 15.Th1-b1 O-O-O
16.Kf2-e1 Kc8-b8 17.Tb1xb7 Kb8-a8 18.Sa3-b1 De5xe3
with lots of variations (2018-12-6)
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Henrik Juel: C+ by Natch 3.1 (59 min.)
The content is given by the diagram position: Get wKe1 to h8 and interchange the black royals
Lf8 makes the roundtrip f8-g7-f6!-e5-f4-h6-f8
[Dd8] goes d8-e8-f7-d5-e6!-f7-e8
Resentful of the dissipation by some of his officers [Ke8] wastes no time going e8-f7-e6-d6-e6-f7-e8-d8 (2018-12-8)
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Henrik Juel: The capture was fxDe
If White made it, Dd1 is white and so is Ke1, but he would stand in an illegal check from sPf2
So Dd1 and Ke1 are black (and Ke8 is white), and the rest of the men have a 'natural' color because of the shortish proof game, which is
C+ by Natch 3.1 (2 sec.) (2018-12-8)
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Henrik Juel: The fastest captures are d2xLe3 and d7xLe6, so Lb5/Ld6 are black/white respectively
If Ka4 were white, so that Ka1,Tb1 would be black, more than 19 black moves would be needed
So Ka1, Tb1 (and Tb6) are white, needing white castling, which entails that Dd8 is white, because this saves a white move; then Dd1 and Ta6 are black, and the rest are 'naturally' colored, with Sb4/Se8 white/black
The colored position is C+ by Natch 3.1 (0.3 sec.)
The position after switching the colors of Sb4/Se8 cannot be reached in 19.5 moves (2018-12-8)
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Henrik Juel: The decoding is easy, with letters ABCDEF representing the men DKPLST; the coloring is natural, with Ld7 being white
The shortest proof game lasts 9.5 moves and is
C+ by Natch 3.1 (2018-12-8)
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Henrik Juel: C+ by Natch 3.1 (1 min.) (2018-12-8)
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Henrik Juel: C+ by Natch 3.1 (3 min.) (2018-12-8)
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Henrik Juel: C+ by Natch 3.1 (9 hours)
The two Belfort officers are wSg8 and sLc1
Here is the added feature that [Lf8] goes to c1 (to capture wLc1), then back to f8 (to make room for [Dd1] and [Ke1] to interchange, while clearing the way for [Ke8]'s march to a1), and then to c1 a second time (2018-12-9)
Mario Richter: Warum Genre h#? (2011-5-1, hidden) more ...
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Henrik Juel: C+ by Natch 3.1 (57 min.)
The author made 64 proof games, all ending in Black being mated, on the 64 squares of the board
This is a neat model mate (2018-12-9)
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corrected version by V. Onitiu
Henrik Juel: Natch 3.1 did find a strategy, but otherwise nothing happened in more than an hour (2018-12-6)
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NN: White can't have rights to castle on both sides as his only legal last moves are with wK or wRs. (2018-11-30)
VL: Reprints: (3) StrateGems SG19, p.156, 07/2002 (2017-12-1, hidden) more ...
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Henrik Juel: Jacobi got nowhere in more than an hour (2018-12-6)
Mario Richter: Cook found by rawbats:
1. a4 h5 2. a5 Th6 3. Ta4 Tb6 4. axb6 Sf6 5. bxc7 h4 6. cxd8=L h3 7. Lxe7 hxg2 8. Lxf6 gxh1=L 9. Lxg7 Lg2 10. Lxf8 Lxf1 11. Ld6 Lxe2 12. Lxb8 Lxd1 13. Lg3 Lxc2 14. f4 Lxb1 15. Kf2 Lg6 16. Te4+ Kf8 (2018-12-7)
Henrik Juel: Thanks for bringing your versatile program out, Mario
I was pretty sure that this proof game must be cooked, but could not show it by my own tools (2018-12-7)
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Henrik Juel: The given solution implies that wSg1 is missing in diagram
With this added, there are lots of cooks, e.g.
1.Sb1-c3 Sb8-c6 2.Ba2-a4 Ta8-b8 3.Ba4-a5 Sg8-f6
4.Ta1-a4 Sf6-d5 5.Sc3-a2 Sd5-b6 6.Ba5xb6 Bd7-d5
7.Bb6xa7 Bd5-d4 8.Ba7-a8=S Bd4-d3 9.Sa8-b6 Bd3xe2
10.Sb6-c4 Be2xf1=S 11.Ke1-e2 Tb8-a8 12.Ke2-f3 Sc6-e5
13.Kf3-f4 Sf1-e3 14.Kf4-g5 Bh7-h5 15.Dd1xh5 Se3xc2
16.Dh5xh8 Sc2-a3 17.Dh8xf8 Ke8xf8 18.Sc4xa3 Lc8-h3
19.Ta4-g4 Bc7-c6
With the given diagram position, the problem is also badly cooked (2018-12-6)
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paul: Dual: a) 6.- Sh6! 7.Tf6 gxf6 8.Txh6 Lxh6 9.d3 Lxc1. (2010-7-30)
Henrik Juel: Part b) is also ruined, by the same dual (2018-12-9)
Henrik Juel: ... almost the same, namely
6... Sh6 7.Tf6 gxf6 8.Txh6 Lxh6 9.Dc1 Lxc1 10.e4 (2018-12-9)
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Henrik Juel: Mario, can your rawbats test this one? (2018-12-9)
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Gerald Ettl: Wichtig Typ Calvet. Nicht schlecht. (2009-2-10)
Alfred Pfeiffer: Beim Typ Calvet sind Schläge auf das eigene Wiedergeburtsfeld erlaubt. Wäre deshalb hier nicht der Typ Cheylan zu fordern, um 2. Kxe8 auszuschließen ? (2018-12-6)
Henrik Juel: You are right, Alfred (2018-12-7)
Gerald Ettl: ja das stimmt. (2018-12-7)
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Renny Bosch: The answer given as "a) 1. Sc3 0. 0?" was probably intended to be "a) 1. Sc6 0-0 2. ?", implying that in a) B may still have castling rights. But regardless of B's castling right, 1. Qd6 ... 2. Bb5# works in either case. Clearly there must be a reason why that is impossible in one of the two cases.

Most likely there is a misprint in the given position: wBc4 should be at b3. Then 1. Qd6 ... 2. Bc4# works in b) but not in a). And in a), the retro play shows that B can no longer castle, because he has to make a move after wKa4.

Also, the retro play as given leads to an initial position which is illegal, because with bPs at c7, d6, f6 and g7, the wB created on d8 would not be able to get out. And if the bPd6 were still at d7, then the bBc8 would not have been able to get captured by the wPh2. So bPb7 must already be at b6.

Here is a PG for a) that preserves B's castling right as long as possible, but in the end must lose it, leading to the #2.

1. Nc3 Nf6 2. Nb5 Ng4 3. h3 a6 4. hxg4 b6 5. Rh6 Bb7 6. Rf6 Be4 7. Nf3 Bf5 8. gxf5 exf6 9. Nh4 Qe7 10. Ng6 Qe6 11. fxe6 Nc6 12. e7 Rd8 13. exd8=B Bd6 14. Be7 Bh2 15. Bd6 Bg1 16. Bh2 Nd4 17. Na7 Nb3 18. Nb5 Nxa1 19. Na7 Nb3 20. Nb5 Nxc1 21. Na7 Nb3 22. Nb5 Nd4 23. Na7 Nf5 24. Nb5 Ng3 25. Na7 Nh1 26. g3 d6 27. Bh3 d5 28. Be6 d4 29. Kf1 d3 30. Kg2 c6 31. Kf3 c5 32. Ke3 c4 33. Qf1 c3 34. Qg2 b5 35. Qe4 b4 36. Qd4 b3 37. Kxd3 a5 38. Kxc3 a4 39. Kb4 a3 40. Kxa3 hxg6 41. Bxb3 g5 42. Kb4 Rg8 43. Ka4 Rh8 {the given position} 44. Nc6 Rg8 45. Qd8# (2008-5-16)
Michel Caillaud: in a), 1.Qd6? ~ 2.Lb5# doesn't work because of 1...Rh4! (2008-5-19)
Henrik Juel: Typos in the solution for part a)
Here are the correct solutions and tries:
a) 1.Sc6 thr. 2.Dd8# (1... 0-0?? is illegal), not 1.Dd6? Th4!
b) 1.Dd6 thr. 2.Lb5#, not 1.Sc6? 0-0!

Very elegant twinning (2018-12-4)
Henrik Juel: Thanks for your update, Alfred
There still are errors in the given retroplay
The black b-pawn should retract to b6 only, and the d-pawn all the way to d7
Then e7xTd8=L can be retracted, and [Lc8] is not locked out (2018-12-4)
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Henrik Juel: C+ by Natch 3.1 (1 hour 40 min.) (2018-12-9)
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Henrik Juel: The fastest pawn captures are a2xb3, b2xc3, g2xf3, h2xg3, and a7xb6, h7xg6
Surprisingly, in the intended coloring, the four men on row 8 are white; the rest is 'normal' (2018-12-9)
Henrik Juel: The colored position is C+ by Natch 3.1 (1 sec.)
No solution in 17.5 or 17.0 moves (2018-12-9)
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Henrik Juel: C+ by Natch 3.1 (1 min.) (2018-12-9)
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Henrik Juel: C+ by Natch 3.1 (21 min.)
A fantastic idea: By move 10 Black has promoted on a1,h1, so he has rooks in all four corners; during the next 18 moves the rooks interchange cyclically clockwise, motivated by the need for white to play a8=D-g8-g6
This leaves an ugly promoted wD on the board, whereas the promoted wLh3 is more discrete; maybe the author had a8=T in the first version, which was then cooked (2018-12-9)
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Henrik Juel: Diagram error: sKg5, not g6
Then C+ by Natch 3.1 (3 min.) (2018-12-9)
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Autorangabe: V. Z. Balanovskij SSSR,
Forderung: "Shortest game?" (kein Zusatz "2 Lösungen", und die mit 1. c4 beginnende BP ist explizit als NL gekennzeichnet).
Henrik Juel: Natch 3.1 found thousands of solutions, so badly cooked, regardless of the number of solutions stipulated (unless this number is 8605, but that would be silly) (2018-12-6)
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Henrik Juel: C+ by Natch 3.1 (7 sec.) (2018-12-9)
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Henrik Juel: The moves edified by an exclamation mark are not particularly good or surprising
The marks just highlight the four white promotions to L and the black pawn captures getting rid of them (2018-12-6)
Henrik Juel: Many cooks, e.g.
1.Bg2-g4 Sb8-c6 2.Bh2-h4 Ba7-a5 3.Bh4-h5 Ba5-a4
4.Bh5-h6 Ta8-a5 5.Bh6xg7 Bh7-h5 6.Bg4-g5 Sg8-h6
7.Bg5-g6 Ta5-g5 8.Bg7-g8=L Lf8-g7 9.Th1-h2 Lg7-e5
10.Bg6-g7 Bf7-f6 11.Lg8-b3 Sc6-b4 12.Bg7-g8=L Ba4xb3
13.Ba2-a4 Sb4-a2 14.Lg8-c4 Bd7-d5 15.Ba4-a5 Bd5xc4
16.Ba5-a6 Dd8-d3 17.Ba6-a7 Dd3-g3 18.Be2-e3 Sa2xc1
19.Lf1-e2 Sc1-a2 20.Ke1-f1 Bb7-b5 21.Bf2-f4 Lc8-b7
22.Bf4xg5 Lb7-h1 23.Bg5-g6 Bc4-c3 24.Bg6-g7 Bh5-h4
25.Bg7-g8=L Bh4-h3 26.Lg8-c4 Sh6-g8 27.Ba7-a8=L Bb5xc4
28.La8-g2 Bh3xg2 (2018-12-6)
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Henrik Juel: C+ by Natch 3.1 (2018-12-9)
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Henrik Juel: Paradoxical twins
a) needs 31 single moves, but the position obtained after 27 single moves can be reached from the initial array in just 26 single moves, as shown in b)
There is no real paradox of course, in a) you must use an extra white move to preserve the castling right
Gerd may know the reason for not publishing, maybe it is because the first 20 single moves are identical in the twins (2018-12-6)
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Henrik Juel: C+ by Natch 3.1 (2 min.) (2018-12-9)
paul: It took 1st Prize in the tourney. (2014-3-26, hidden) more ...
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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 (2005-6-20)
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 (2018-12-9)
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Henrik Juel: Part 1) is C+ by Natch 3.1 (2018-12-9)
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Henrik Juel: Seems to be cooked by
1. e4 f5 2. Ne2 f4 3. h4 f3 4. h5 fxe2 5. f4 exf1=B 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. g8=B Bg7 15. Bxh7 Kf7 16. Qf3 Raf8 17. Bg8+ Kg6 18. Qe2 Rf7 19. Qd1 a6 20. e8R Qd8 (2018-12-9)
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Henrik Juel: C+ by Jacobi using the constraints
Ke8-c2-d1-e1xf1-e1-d1-c2 Ke1-d1-c2-d3-g3-h2-g1-f1-e1
which are valid, because [Ke8] is the only black man who could capture [Lf1] in time, so [Ke1] would have to get out via c2-d3 and back home via g3-h2
It is recommended that HC+ is used, but this is not available on the PDB (yet)
Ke1 made a round trip (which is stronger than a promenade) (2018-12-9)
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Henrik Juel: C+ by Natch 3.1 (12 sec.) (2018-12-10)
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Alfred Pfeiffer: Aufgabe nicht gefunden. Eine Quelle "11393 feenschach 06/1978" gibt es nicht. (2018-12-3)
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Popeye findet 277 Lösungen: klären
Anton Baumann: Alybadix und Gustav 4.0e finden ebenfalls 277 Lösungen:
1.Sh3: 20 1.Se4: 94 1.Sf7: 76 1.Sh7: 87 total = 277 (2018-11-24)
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Alexej Oganesjan: Additional try: 1...Kxf6? 2.Sf3 Le7 3.Sg5 Lf8 4.Sh7 Kf7 5.?? Lg7# - Black has no a tempo-move! (2018-11-19)
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Sally: Thema :
Verzicht auf Fesselung einer schwarzen Figur.
"Diese ambitiöse Darstellung bedurfte die die Mitwirkung von fünf Personen und einer Flasche Longrow; eines der seltensten Single Malt Scotch Whiskys".
Zu bewundern ist nicht zuletzt die "Nachtwächter - vermeidende Zwillingsbildung". (2018-12-5)
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Sally: Unter Beibehaltung der Mustermatts wird hier schwarzweiße Bahnung einheitlich in beiden Lösungen gezeigt. Das Ganze ökonomisch mit nur elf Steinen und schlagfrei. (2018-12-6)
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Mario Richter: vgl. mit P0533984 (2018-11-15)
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klären Autor
: Autor: A. Solowej (2006-11-22)
Alfred Pfeiffer: No.1649 in SK 32, 11/1999, ist eine andere Aufgabe (#2) von anderen Autoren. (2018-11-28)
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Henrik Juel: wPb7 is orig. Pg2. The NE corner can release only by -1.Kh5 g7, and the SW corner can release only by -9.d2xc3 (after wL uncaptured on g5 has been brought home), so sLa3 is orig. Pc7 promoted on g1 after f3xg2 (not f4xg3, because the lightsquared wL was captured by the sP). Wonderful lightweight. (2004-4-26)
Henrik Juel: Theoretically, the NE corner could also be opened by retracting f6xg5, later Kg5-h6 and the uncheck g7xf6, so sLa3 is promoted on g1 no matter what; but g7xf6xg5 would require too many black captures (thus the somewhat unsightly Sd8 makes the NE opening unique) (2018-11-30)
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Published in issue 206-210. Solution never published !
Keyword=Solution Unknown !
Gerald Ettl: steht der wSa4 wirklich auf dem Brett?
1.Sc3 Kg6 2.Df5+ Kf7 3.Dh5+ Kg8 4.Dd5+ Kh7 5.Sb1 9.Sd2 13.Sf3 17.Sg1 Kg6 18.Lg2 Sf2# (2018-12-6)
Gerald Ettl: 1.Sc5 5.Kg2 9.Kf3 13.Ke3 17.Kd2 21.Kc3 25.Kb4 29.K5 33.Ka6 37.Kb7 41.Kc8 45.b7 49.b8L 53.Sb6 57.a8T 61.Ta1 65.Kb7 69.Ka6 73.Ka5 77.Kb4 81.Kc3 85.Kd2 89.Ke3 93.Kf3 97.Kg2 101.Kh1 105.Lg2 106.Tg1 Sf2#
aber warum nicht 53.Sc7? muss ich mir nochmal ansehen. (2018-12-6)
Gerald Ettl: auch 49.b8S geht in der Loesung. (2018-12-7)
Gerald Ettl: Die Autorabsicht koennte 45.Sc7 49.Sa6 53.Sb8 57,a8T usw. gewesen sein. Waere eindeutig.
1.Sc5 Kg6 2.Df5+ Kf7 3.Dh5+ Kg8 4.Dd5+ Kh7 5.Kg2 9.Kf3 13.Ke3 17.Kd2 21.Kc3 25.Kb4 29.K5 33.Ka6 37.Kb7 41.Kc8 45.Sc7 49.Sa6 53.Sb8 57.a8T 61.Ta1 65.Kb7 69.Ka6 73.Ka5 77.Kb4 81.Kc3 85.Kd2 89.Ke3 93.Kf3 97.Kg2 101.Kh1 105.Lg2 106.Tg1 Sf2#
(siehe aber Cook mit 45.b7 49.b8S,L 53.Sc7,Sb6 usw.) (2018-12-7)
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TT: proof game with capture of Anti-Pronkin piece
paul: Cooked by: 1.Sc3 h5 2.Sd5 Th6 3.Sxe7 Te6 4.Sd5 Lb4 5.Sc3 Dh4 6.Sb1 Lc3 7.dxc3 g6 8.Lf4 Kf8 9.e3 Kg7 10.La6 bxa6 11.h3 Lb7 12.Lh2 Kh8 13.g3 Lxh1 14.Dg4 d5 15.Dg5 Sd7 16.Dd8 Tb8 17.De7 Tb3 18.axb3 Dxh3 19.Ta4 Dg2 20.Th4 c6 21.Tg4 hxg4 (M.Caillaud) (2018-11-17)
paul: See P1067990 as a correction. (2018-11-17)
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No. 656 HN
Henrik Juel: Note that -1... e6xYd5?? fails, because the black capture exd must have been e7xLd6, as the dark-squared [Lc1] could have been captured on g6,f5
In fact, after the position has been opened, the further retroplay must include
Retract Lf3 to c8, d7-d6, c5xd6xe7, Kf4 to e8, Dg3 to d8, Lh2 to f8, e7xLd6-d5 (2018-11-26)
Henrik Juel: The white pawn-play could also have been d5xe6-e7, so the black play e7xLd6 and d7-d5 is also possible... (2018-11-26)
Henrik Juel: ... and of course, in my first comment,
[Lc1] could NOT have been captured on g6,f5 (2018-11-26)
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Originalforderung war "Kürzeste Beweispartie?"
Henrik Juel: C+ by Natch 3.1 (2018-12-7, hidden) more ...
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Henrik Juel: C+ by Natch 3.1
4... Ke8-d8-c7-b6-b5-c5-d6-c7-d8-e8 to correct the tempo
Ez. is short for single moves, nowadays 12.5 (2018-12-7)
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Henrik Juel: C+ by Natch 3.1
Fake castling
now 8.0, not 16 single moves (2018-12-7)
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1230
SCHRECKE: C+, Gustav 4.0f, Brute Force (2018-12-8, hidden) more ...
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Felber, Volker: Original hatte Nebenlösung 1. Lg3! Wann und wo wurde der sBa2 hinzugefügt. Im Originaldiagramm und in der Notation der Lösungsbesprechung ist er nicht vorhanden. (2018-11-16)
Rosalie Fay: 1 Sd3+ Kxe4 2 Sc5+ Ke5 3 Ld4+ Kxd4 4 Te6 ~ 5 Te4#
C+ Popeye Windows-64Bit v4.69
Sacrifice (LS) (2015-10-6, hidden) more ...
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Alfred Pfeiffer: Fehlenden wBe7 ergänzt. (2018-11-28)
Henrik Juel: Something must be wrong (besides the promoted Lb3)
Popeye 4.61 reports 1.Lc2,Ld3,Le4,Lxe6 (2018-11-29)
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Felber, Volker: Die AL soll lt. NBSZ mit 1. De8! beginnen. Die 2. NL mit 1. Tcxa3! wurde von den Preisrichtern ebenfalls nicht erkannt. (2018-12-2)
Martin Ramsauer: 159, The English Mechanic and World of Science, 19/06/1874, p. 367
1. Tcxa3! droht 2. Sd3 ... 3.Tb2#
2. ... Tc2 3. Txa1#
2. ... Tb4 3. Sc3#
1. ... Txc1 2. Da4 droht 3. Txa1+ Kb2 4. Db3#, 4. Ta2#
usw.(sehr viele Duale)
NL 1. De8! droht 2. Tb3+ Sxb3 3. Dxg6+ Tc2 4. Dxc2#
1. Dxa7! droht 2. Dxa3 ... 3. Db2#, 3. Txa1#, 3. Tb2# (2018-8-28, hidden)
Henrik Juel: should be labelled cooked (2018-8-28, hidden) more ...
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Sally: Satz:
1.... - Lxc6, 2. Sbd7+ (A) - Lxd7/Kd6, 3. Sfxd7 (B)/Txc6#.
1.... - Lxe6, 2. Sfd7+ (B) - Lxd7/Kd6, 3. Sbxd7 (A)/Txe6#
Lösung : 1. d6! [2. Lf4+ und 3. Dd5#)
1.... - Lxc6, 2. Sfd7+ (B) - Lxd7/Kxd6,3. Sbxd7 (A)/Lf4#
1.... - Lxe6, 2. Sbd7+ (A) - Lxd7/Kxd6,3. Sfxd7 (B)/Lf4#.
(1... - Kxd6/Sxd1/Kxe6, 2. Lf4+/Sfxd7+/Se8#)
(1... - Ta1, 2. Lf4 - Kxe6, 3. Dd5#.
(1... - Lf4, 2. Lxf4+- Kxe6, 3. Dd5#.
Wohl die erste , einzige und vielleicht letzte Kombination von Umnow I und dem Tura-Thema! Eine konstruktive und kompositorische Sonderleistung. (2018-12-5)
Henrik Juel: C+ by Popeye 4.61 (2018-12-5, hidden) more ...
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paul: Version of P1011752. (2018-11-17)
paul: C+ (Euclide 0.96) (2010-5-4, hidden) more ...
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Henrik Juel: C+ by Natch 3.1
Cyclical interchanges
La8-Td8-Db8-Se8-Kc8(-La8), Sf8-Lh8-Tb8(-Sf8) (2018-12-7)
hans: 1. h4 Nf6 2. h5 Rg8 3. h6 gxh6 4. Nf3 Bg7 5. Ne5 Bh8 6. Nxd7 Nbxd7 7. a4 Nf8 8. a5 Qd4 9. a6 Qxb2 10. axb7 Bxb7 11. f4 O-O-O 12. f5 Ne8 13. f6 Ba8 14. fxe7 Qb8 (2011-3-1, hidden) more ...
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Sally: Hier sieht man eine reziproke L-L-Bahnung, die allerdings zurücktritt gegenüber dem dominierenden Rehm-Thema und dem Rundlauf des wL. Es ist dem Autorenduo gelungen, eine sehr elegante, ausgefeilte Darstellung gelungen; durch die für den sT freizuhaltenden Linien wird eine überraschende ökonomische Stellung ermöglicht. (2018-12-9)
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Sally: Eine sowohl visuell wie auch logisch attraktive Bahnungsidee. (2018-12-9)
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Sally: Ein schwieriges, aber sehr strategisch, wunderschönes Problem. Die wD verlässt das kritische Feld, um den sT durchzulassen (und dem wK das Feld c2 zu geben) und kehrt zurück, um den dort verstellten wB zu schlagen. (2018-12-9)
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Sally: Einheitlich beginnt die sD in jeder Lösung.
Zwei mal wird sie geopfert.
Dreimal muss der wT fesseln. Eine klare Darstellung. (2018-12-9)
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Sally: Der sB muss den wL ins Spiel bringen und wandelt sich dreimal um. (2018-12-9)
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Sally: Alte Idee in neuen Gewändern. Gefällige gemischte Allumwandlung mit zwei s Phönix Figuren, und dies in Meredith Fassung. (2018-12-9)
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