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| (1) Posted by Sarah Hornecker [Thursday, May 10, 2018 05:37] |
The other side to move first in composition
The Gady Costeff article in Variantim brought me to think: I have not seen, without trickery, self- or directmates with Black to move first, but commonly see White to move in helpmates, or Black to move first in studies.
So I would like to:
a) see such problems (without retro tricks etc.) in direct and selfmates
b) have your opinion about the other side moving first in direct checkmate play, i.e. directmates and selfmates
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| (2) Posted by seetharaman kalyan [Thursday, May 10, 2018 08:05] |
All direct mates with zugzwang play are black to play if you consider after keymove.
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| (3) Posted by Dmitri Turevski [Thursday, May 10, 2018 09:35] |
This kind of stipulation can be easily searched in the databases (all of them, I suppose).
Eg: http://www.yacpdb.org/#152206
There aren't many examples, as basically this means that composer could not find an acceptable key.
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| (4) Posted by Olaf Jenkner [Monday, May 14, 2018 21:50] |
All direct mates are black to play if you consider after keymove.
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| (5) Posted by Hauke Reddmann [Tuesday, May 15, 2018 20:46] |
Got an idea for a theme tourney :-) - #2, Black stands in a
triple (or higher) check. Clearly, retro tricks wouldn't
help, as having this illegal position as a starter is the
only way to set it up.
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| (6) Posted by Nikola Predrag [Tuesday, May 15, 2018 21:30] |
Such a position would not be chess but a sheer nonsense. For the nature of chess see 3.10.3, which applies to any position, including the 'starter' :-)
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| (7) Posted by Geoff Foster [Wednesday, May 16, 2018 00:28] |
Here is an a 12-check retro, using the Anticirce fairy condition.
Nikita Plaxin
dedicated to L.Loshinsky
Caissas Schlossbewohner I, 1983
 (= 1+13 )
Last move?
Anticirce
On the previous move Black played Qg8xa8(Qd8)++++++++++++. With g8 and a8 both occupied, a black Rook or Knight cannot capture on a light square.
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| (8) Posted by Hauke Reddmann [Wednesday, May 16, 2018 07:31]; edited by Hauke Reddmann [18-05-16] |
@Nikola - I love nonsense, in fact nonsense uncovers holes in the
FIDE rules. Remember 3.9? "The king is said to be 'in check' if it is
attacked by one or more of the opponent's pieces..." Earlier, this
read "one or two" since the FIDE had not the fantasy to find the
ingenious check defense problemists promptly came up with :-)
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| (9) Posted by seetharaman kalyan [Wednesday, May 16, 2018 12:40]; edited by seetharaman kalyan [18-05-16] |
Good position with 12 fold check. But as a Last move? problem the solution has to be unique including what was captured on a8.
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| (10) Posted by Miodrag Mladenović [Wednesday, May 16, 2018 13:45] |
To make unique solution it's possible to convert this problem to S#1 but then looking for solution is trivial. I do not mind that it's not unique what piece should be captured on a8.
 (= 2+15 )
S#1 Anticirce
1.a8R+! Qxa8(Qd8)#
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| (11) Posted by Frank Richter [Wednesday, May 16, 2018 15:42] |
Hm, do the black rooks give check in diagram position?
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| (12) Posted by Olaf Jenkner [Wednesday, May 16, 2018 18:26] |
Yes, but the last move could have been Sa8-b6+
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| (13) Posted by Miodrag Mladenović [Wednesday, May 16, 2018 19:20] |
@Olaf
Only Sa8-b6 (not a check because of the black Queen on g8). Or Sa8-c7 .
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| (14) Posted by Geir Sune Tallaksen Østmoe [Thursday, May 17, 2018 00:17] |
Isn't it Sa8-b6+ because Black's rooks now give check?
Another thing: Isn't the position illegal? Black has two pawns and seven promoted pieces.
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| (15) Posted by Miodrag Mladenović [Thursday, May 17, 2018 07:13] |
You are right. It's check by rook (not by knight but by rook). And there is no illegal position in fairy chess. For fairy problems there are different rules than for orthodox problems. And actually for fairy chess you do not have to resolve position at all. I could simply put wPc6 instead of bPb7 to eliminate bPb7 but I think that for fairy chess problems it's irrelevant.
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| (16) Posted by Geoff Foster [Thursday, May 17, 2018 08:46] |
Another possibility is to remove bPb7. A Queen promotion is then required, which will work if the board is reflected left/right.
 (= 2+14 )
S#1
Anticirce
1.h8=Q+! Qb8*h8[Qd8]#
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| (17) Posted by Geoff Foster [Thursday, May 17, 2018 08:48] |
Oops, bQ is reborn on d8!
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| (18) Posted by René J. Millour [Thursday, May 17, 2018 17:44] |
Here is another work showing 12 simultaneous checks, after a long K march and 8 promotions to Q. No capture at all!
René J. MILLOUR
Phénix, 1993
1st Prize
 (= 14+9 )
H++++++++++++ #12 MARS CIRCE
[ H++++++++++++ #12 means H#12, the mate being given by 12 simultaneous checks! ]
1.Sg6 h8Q 2.Kh7 g8Q 3.Rf4 Qfg7 4.Kh6 f8Q [Bf6 guards h6 and g5 via c1, hence 3.Rf4 allowing 4.Kh6 etc]
5.Bf7 e8Q 6.Be7 d8Q 7.Kg5 Bd7 8.Kh4 c8Q [Not 8.Kg4 as g4 is guarded by the wQs via d1]
9.Kg3 Kc7 10.Kf2 b8Q 11.Qb7 a8Q 12.Kf1 Re2++++++++++++ # [Not 9.Kh3 10.Kg2 as h3 and g2 are guarded by Bd7 via f1]
The final position!
 (= 14+9 )
12...Re2 puts the bK in check simultaneously by the 9 wQs via d1, by Kc7 via e1, by Ra1 via a1 and by Re2 via h1. This is mate because e1 and g1 are guarded by Qs and Rs, f2 by Kc7, g2 by Bd7 and, with e8 occupied, the bK cannot capture.
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