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1 problem(s) found in 6787 milliseconds (displaying 1 problem(s)). [COMMENTDATE>=20220810 AND G='Retro' AND NOT K='Wer ist am Zug?' AND A='Gruber, Hans'] [download as LaTeX]

1 - P1013376
Hans Gruber
3470 Die Schwalbe 68 04/1981
P1013376
(0+0)
Ergänze 1 Stein zu einem illegal cluster.
Wie viele Lösungen?
Kamikaze rex inklusive
Schlagschach
736.
play all play one stop play next play all
Henrik Juel: In Losing Chess the kings are not royal
In Kamikaze RI each capture removes one white and one black man
So the empty board is legal, and any position with one man (KDTLSP) is illegal
An Illegal Cluster is an illegal position that becomes legal when you remove any one non-king
My guess would be 2x6x64 = 768 or possibly 2x5x64 = 640
Where did I go wrong? (2020-07-12)
Henrik Juel: maybe 768 - 32 = 736
where 32 is the number of ways you can put a pawn on row 1 or 8 (2022-10-13)
A.Buchanan: In an “orthodox” illegal cluster, a pawn couldn’t be on the first or last rank unless it was the only non-king unit. There’s probably no principle to cover this rather dull corner-case KPvK. Ask HG! (2022-10-14)
A.Buchanan: And in orthodoxy, do two adjacent kings constitute an illegal cluster? There are no non-king units to remove, so the condition is vacuously satisfied (2022-10-14)
Henrik Juel: On your last comment, Andrew:
Two adjacent kings do not constitute an IC. The position is illegal, and when you remove any non-king you get the same illegal position (2022-10-14)
A.Buchanan: Define that a box is SAFE if it contains no radioactive objects. So is every empty box safe? Yes! :-) (2022-10-14)
comment
Keywords: Losing Chess, Kamikaze, Illegal cluster
Genre: Fairies, Retro
FEN: 8/8/8/8/8/8/8/8
Input: Hans Gruber, 2004-05-01
Last update: Kevin Begley, 2022-10-13 more...
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