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Problem 472: Sebastien Luce - Fairy (Circle SneK, Alphabetical chess)
sebastien(09.12.2014) Here is the first original in the website with the new fairy condition Circle SneK! Very nice longmover Help-stalemate with AUW by Sebastien Luce.
1.a1Q+ Kb3 2.Qh1 Ka2 3.b1S Kxb1(Sh1) 4.c2+ Kb2 5.c1S Kxc1(Sh3)
6.d2+ Kc2 7.d1B+ Kxd1(Bh3) 8.e2+ Kd2 9.e1R Kxe1(Rh3) 10.f2+ Kf1=
The author said that probably the problem can be sound also without the Alphabetical condition, but without it looks impossible the problem to be computer checked.
   ABC (Alphabetical Chess):The squares are considered in the order a1, a2...a8, b1...b8, c1 and so on to h8. The player whose turn it is may move only his unit standing on the square which comes earliest in this order. However check and mate are normal.
   Circle SneK:
When a Queen is captured - a Rook (or Royal Rook) of the same color (if exists on the board) becoming Queen;
When a Rook is captured – a Bishop (or Royal Bishop) of the same color (if exists on the board) becoming Rook;
When a Bishop is captured - a Knight (or Royal Knight) of the same color (if exists on the board) becoming Bishop;
When a Knight is captured – a Queen (or Royal Queen) of the same color (if exist on the board) becoming Knight.
Only one piece may change its type after a capture. In case of option – the capturing side choose which piece will be transformed.
The capture and the change of type is a single move. If this full move result a selfcheck - the capture is forbidden. 
The capture of a pawn is normal. The capture is normal also in the case when there is no piece on the board which should be transformed. Castling with Royal piece is not allowed.


+4 #1 ThemisArgirakopoulos 2014-12-09 18:12
Without ABC chess, the problem is not sound :
1.a1=S Ka2 2.b1=B+ Kxa1(Sh3) 3.Sg1 Kxb1(Bg1) 4.c2+ Kb2 5.d2 Kxc2 6.e2 Kxd2 7.f2 Kd3 8.Kh1 Ke3 9.h3 Kxe2 10.h2 Kf1=

Unfortunately, this solution have duals...
Without fairy conditions at all, we have the following line (full of duals too):
1.c2 Kxa2 2.d2 Kxb2 3.e2 Kxc2 4.f2 Kxd2 5.Qd7+ Kxe2 6.Qd1+ Kxd1 7.Kh1 Kd2 8.h3 Ke2 9.h2 Kf1=
0 #2 Diyan Kostadinov 2014-12-10 19:48
Thanks Themis. So, all looks fine with the problem.

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