Problem 324: Alberto Armeni

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Problem 324: Alberto Armeni - Fairy (Circe)

armeni (25.01.2014) Fairy problem with Retro-analysis elements by Alberto Armeni.

324

Try: 1…0-0-0? with for example 2.Kb3 Rd3+ 3.Ka2 Ra3#

Retro-analysis: the last black move was NOT a capture of a white piece, as due to Circe condition said piece would be reborn on its initial position, white square (white has only two pieces, both on black squares).

As a consequence, the previous white move was of the Ke1 or the Ra1, and after that, castling is illegal.

There are many apparent cooks in 3.5 moves, but they begin with an illegal castling, therefore there is only the following solution:

1…Ra7 2.Sxa7(Ra1) [now castling is legal] 0-0-0 3.Kb3 Rd3+ 4.Ka2 Ra3#

Circe: Captured units (not Ks) reappear on their game-array squares, of the same colour (pieces), on the file of capture (pawns), or on the capture file’s promotion square (fairy pieces). If the rebirth square is occupied the capture is normal.

Comments

+1 #1 Adrian Storisteanu 2014-01-25 06:32

The same retro 'dual avoidance' trick could be done perhaps with one piece less (and a cleaner mate picture):
Ke1 Ra1 / Kc4 Ra6 h#3 circe assassin
There is a promising set-play, 1...0-0-0 2.Kb3 Rd3+ 3.Ka2 Ra3#. But random attempts to keep it in the actual solution fail, eg 1.Ra6-~ 0-0-0?? etc. - it is B to play, and W's last move must have been with K/R, disallowing castling (see *Note). In the set-play (W to move), on the other hand, B's last move legitimizes the castling (regardless of any prior play) via a suicide such as bBf6xwRa1(+wRa1), giving birth to a brand-new ready to castle wRa1. Same strategy must be used in the solution:
1.Rxa1(+wRa1)! 0-0-0 2.Kb3 Rd3+ 3.Ka2 Ra3#
*Note: there is no B piece on its own rebirth square in the diagram. If the bR stood on a8 (rather than a6), white would have a non-K/R last move available - again a suicide (but white now), e.g., wBh1xbRa8(+bRa8) -, which would allow the dualistic 0-0-0 solutions...

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0 #2 Adrian Storisteanu 2014-01-25 14:19

That guy. Again.
In the proposed h#3 scheme above, if we ask "How many solutions?", then in
a) orthodox
the answer would be 0 (not 1: 1.Kb3 0-0-0?? 2.Ka2 Rd6 3.Ka1 Rxa6# fails because of a similar last-move reason), while in

b) circe assassin
the answer is 1 (not many -- cf all that jazz from before...).

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0 #3 Diyan Kostadinov 2014-01-27 18:18

The version proposed by Adrian in comment #1 looks interesting.

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+1 #4 Adrian Storisteanu 2014-01-27 18:30

Diyan,
Thanks for your interest. The twins (comment #2) also 'grew on me' in the meantime - the orthodox solution is otherwise fully correct, has a brand-new mate picture, and the 0-0-0 illegality is similar but, of course, different than the one under circe-assassin conditions.
Also, in both solutions the bR manages to disappear (economically) from the final mate tableau, one way or another...

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0 #5 Adrian Storisteanu 2014-02-07 17:05

"If a tree falls in a forest and no one is around to hear it, does it make a sound?" Too deep a philosophical question for a cold Friday afternoon...