Rich, bravo -- you have it all. In order to forbid Black to short-castle White has to long-castle. Now White has "re-written" the past of the position! By long-castling, White "proves" the promotion of one of its pawns into the d4 Rook. This problem is sometimes quoted as a quantum-chess one! à+ É. Catapulté de mon aPhone
Le 2 mai 2017 à 09:52, <rcs@xmission.com> <rcs@xmission.com> a écrit :
I expand Eric's position to
....k..r p.p.p.p. .p..P... ........ ...R.... ......P. .....P.P R...K...
If Black's castling move is unavailable, then R/a1-d1 followed by R/d4-d8 mates. I'm having trouble establishing that Black can't castle. Assuming that his k and r have never moved, his previous move to arrive at this position must have been p/b7-b6. To back up further, we must assume that White's previous move captured a Black piece (or pawn). Let's say it was R x n/d4. From this position, I think we can fairly mechanically work backwards, with Black's queenside pieces developing through d8 or d7, his knights developing normally, and White capturing Black's kb at f8 with a knight. The WN can get to f8 and back out without giving check to the bk/e8.
Incidentally, in the diagram position, White must have moved his K to allow the KR to escape -- again, assuming the bk and bkr have never moved. If White created a third rook by underpromotion, there's no escape path from the 8th rank that doesn't check bk.
Rich
------ Quoting Eric Angelini <eangelini@everlastingprod.be>:
Bill & alii, here is a really tricky mate in 2.
4k2r / p1p1p1p1 / 1p2P3 / 8 / 3R4 / 6P1 / 5P1P / R3K3 /
Best, Éric.
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