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(21) Posted by Hauke Reddmann [Wednesday, Oct 21, 2009 11:18] |
I am always right. Except when I'm wrong :-)
Hauke
P.S.
(= 4+4 )
Another idea for #1. *Any* black or white move mates.
Surely you can top that. (Stalemated Black is cheating,
at least one mate of both colors is mandatory :-) |
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(22) Posted by Frank Richter [Wednesday, Oct 21, 2009 12:28] |
9 white + 8 black moves.
More should be possible, I think.
(= 9+13 )
#1, Duplex |
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(23) Posted by Sarah Hornecker [Wednesday, Oct 21, 2009 12:48] |
1.bxa3 is also possible without checkmate... |
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(24) Posted by Frank Richter [Wednesday, Oct 21, 2009 13:19] |
Sorry for blindness.
A new position (hope without cooks):
(= 8+12 )
Any move mates (10w + 13b) |
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(25) Posted by Kostas Prentos [Wednesday, Oct 21, 2009 22:37] |
Interesting challenge.
Here is an improved version:
(= 12+13 )
Any move mates (14w+17b) |
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(26) Posted by Hauke Reddmann [Thursday, Oct 22, 2009 12:24]; edited by Hauke Reddmann [09-10-22] |
Nice job, but I want to see a retro legality proof of that one
before I declare it current record holder :P
Hauke
(EDIT: If retro fails - I'm not good at proof games - Ba5 instead
of Pa5 might help.) |
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(27) Posted by Georgy Evseev [Thursday, Oct 22, 2009 12:40] |
Definitely legal: white captured black knight exd and black pawn dxc; black captured white bishop bxa3 and white knight dxe. Remaining pawns were captured on source files. |
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(28) Posted by Kostas Prentos [Thursday, Oct 22, 2009 19:44] |
One of my concerns was to make the position legal.
Otherwise, the following symmetrical position:
(= 14+14 ) would raise the number of forced mates to 17 for each side.
Note that the position is illegal even with promoted rooks instead of the pawns on a6/h3 (unless I am missing something). |
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(29) Posted by Hauke Reddmann [Sunday, Nov 8, 2009 15:16] |
(= 3+3 )
Double forced stalemate, 7+1. Improve :-)
Hauke |
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(30) Posted by Sarah Hornecker [Sunday, Nov 8, 2009 19:32] |
(= 7+7 )
SH, original
Win
Try: 1.Bf8?
Solution: 1...Bxg7! |
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(31) Posted by Hauke Reddmann [Monday, Nov 9, 2009 14:06] |
Was it so hard to improve? :-)
(= 10+4 )
11+1 stalemates
(I don't doubt that the question has been asked before but I
have to contact a fairy specialist.)
Hauke |
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(32) Posted by Joost de Heer [Monday, Nov 9, 2009 15:39] |
What's the exact definition of the task? 'Any move stalemates'? If so, you can do things like
(= 8+2 ) where any white move stalemates (and yes, this can easily be improved)
Someone like Bernd Schwarzkopf has probably already looked into this task. |
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(33) Posted by Frank Richter [Monday, Nov 9, 2009 15:52] |
The task should show in my opinion "active" stalemate moves from both sides. Means, that neither White nor Black already is stalemate in initial position.
And the next question is of course: Any move selfmates.
Possible? |
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(34) Posted by Michael McDowell [Monday, Nov 9, 2009 17:02] |
Going off at a slight tangent, I'm reminded of a problem where White cannot avoid mating in two. It features in a story entitled "Mr.Brown done Brown" by Jacob Elson, published in the first issue of "Lasker's Chess Magazine" in November 1904.
(= 12+8 )
Mate in 2
I wonder if four variations are possible. |
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(35) Posted by Joaquim Crusats [Monday, Nov 9, 2009 20:22]; edited by Joaquim Crusats [09-11-09] |
You can give a choice to the wS
(= 12+10 )
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(36) Posted by Hauke Reddmann [Tuesday, Nov 10, 2009 12:29] |
Elementary, my dear Frank (OK, not so elementary,
I needed a whopping ten minutes, as always s# is
much more complicated :-)
(= 11+9 )
Hauke |
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(37) Posted by Jacques Rotenberg [Friday, Nov 13, 2009 07:27]; edited by Jacques Rotenberg [09-11-13] |
like that ?
(= 4+8 )
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(38) Posted by Hauke Reddmann [Friday, Nov 13, 2009 14:54] |
Crafty. I wonder if anyone can come up with
more than 1+1 first moves. (NOT variants after
1st move - although that is already hard enough.)
Hauke |
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(39) Posted by Frank Richter [Friday, Nov 13, 2009 18:10] |
You may move the bPh3 to h2 and add a black queen h3. That position gives 1.Qh4,h5...h8 B:g2# and 1.B:g2+ Q:g2# (1+5 moves). |
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(40) Posted by Hauke Reddmann [Sunday, Nov 15, 2009 17:50] |
Courtesy of Bernd Schwarzkopf:
Heinz Hünerkopf
(CXLII) feenschach 11, S. 397
X 1972
(= 8+7 )
Subtle, eh?
I might also find the current (@1970 :-) record for the # duplex
on the bottom of a giant heap of SCHWALBE issues. Stay tuned.
Hauke |
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