Fairy chess composition

# No.1368 (BB)

 No.1368  Bojan Bašić  (Serbia) Original Fairy problems JF-2018/II: July – December’2018

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 No.1368 Bojan Bašić Serbia original – 30.12.2018 Solutions: (click to show/hide) white Kc2 Pa3b2e2e7 black Ka8 Rc6 Bc8 Pb5c4 h#2.5                                          (5+5) b) b5 → a7 c) = b) + c2 →h4 d) = c) + c4 →d4 Transmuted Kings Isardam a) 1...e7-e8=Q 2.Ka8-a7 Qe8-e7 + 3.Ka7-d4 e2-e3 # b) bPb5-->a7 1...e7-e8=B 2.Bc8-f5 + Kc2-a4 3.Bf5-c2 + Be8*c6 # c) bPb5-->a7 wKc2-->h4 1...e7-e8=R 2.Rc6-h6 + Kh4*c4 3.Rh6-h4 + Re8*c8 # d) bPb5-->a7 wKc2-->h4 bPc4-->d4 1...e7-e8=S 2.Ka8-b7 Se8-d6 + 3.Kb7-c5 b2-b4 # { (C+ by Popeye 4.79)} Two pairs of analogous solutions. In the solutions b) and c) the mating piece is unprotected but nevertheless the black king cannot capture it since the capture would (re)activate the check to the white king, which would then attack the black king and thus the kings would be in an Isardam-illegal position. (Author)

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Joost
December 31, 2018 08:07

The logic for (b) and (c) is wrong: the king wouldn’t be in an Isardam-illegal position, the king would place itself in check.

Bojan Bašić
January 1, 2019 15:17

You are right, the definition of Isardam actually does not forbid king-to-king checks; therefore, the reason why Black cannot capture in b) and c) is indeed because he would then put himself in (the reactivated) self-check. The point of the problem remains the same though.

Thank you for your comment and all the best in 2019!

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