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MatPlus.Net  Forum General #2 with a maximum of stalemate tries |
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| | | (1) Posted by Hauke Reddmann [Tuesday, Dec 17, 2024 21:11] | #2 with a maximum of stalemate tries
I already did that here (2021, 285.442 Albrecht),
but Black *was* already stalemated in the position.
It is much more difficult to do it if he is not.
Here is a first attempt (14+12+5+1+1=33 tries,
plus a lot of unthematic ones):
 (= 14+1 )
Beat me! (A less lame key is also thinkable,
and it's perfect if all tries are stalemate.)
| | | (2) Posted by Jacques Rotenberg [Wednesday, Dec 18, 2024 09:18] |
Less tries but with a changed mate (mutate (+ Dombrovskis paradox)) :
 (= 13+1 )
2#
1…K×g7 2.Rb6‡
1.Be6? [2.Rb8‡] but 1…K×g7!
1.Bg5!z K×g7 2.Rb8‡
1.Bf6??/1.Bf8??
1.Rb~??
1.Rg~??
1.Qg1??
1.f6??
1.h6??
| | | (3) Posted by Hauke Reddmann [Wednesday, Dec 18, 2024 10:12] |
26<<33 (mathematically speaking :-), but: nice!
Since it isn't widely known (even I had to google),
the corresponding onemover record(s) from the feenschach table
can be accessed online here: http://www.skcaissa.de/einzueger.htm
Relevant here is
https://pdb.dieschwalbe.de/search.jsp?expression=PROBID%3D%27P1178647%27
(Zagler, who else) or with promotion moves
https://pdb.dieschwalbe.de/search.jsp?expression=PROBID%3D%27P1178648%27
- Black is not stalemated, but 46 (48) White moves stalemate.
Since 33<<46, my attempt might be beaten.
Also note that Zaglers, Jacques and my position all have a set play.
This isn't necessary from the task aspect, the Cross/Cross doesn't.
Modifying it a bit, I get:
 (= 10+3 )
1.Qc7+, with B9+S6+S7+R11+Q5+P1=39 stalemate tries if I didn't miscount.
| | | (4) Posted by Frank Richter [Wednesday, Dec 18, 2024 10:13] |
For posts #1 and #2 - last black move?
| | | (5) Posted by Hauke Reddmann [Wednesday, Dec 18, 2024 18:08] |
Beats me :-) +bPa2? Loses 1 stalemate.
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MatPlus.Net Forum General #2 with a maximum of stalemate tries |
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