# No.1178 (KM)

 No.1178  Karol Mlynka (Slovakia) Original Problems, Julia’s Fairies – 2016 (II): July – December    →Previous ; →List 2016(II) Please send your original fairy problems to: julia@juliasfairies.com

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 No.1178 Karol Mlynka Slovakia original – 30.12.2016 Dedicated to Ján Golha 60* Solutions: (click to show/hide) white PPa2h2 black PPb2e4 hs=4            2 solutions             (2+2) Back-to-Back Chameleon Chess Reverse Pawn a2,h2,b2,e4 1.PPa2-a1=B PPb2-b3 2.Ba1-e5=R PPe4-e1 3.Re5-e2=Q PPe1-h1 4.Qe2-b5=S PPb3-b4 {=} 1.PPh2-h1=B PPe4-e5 2.Bh1-c6=R PPb2-b4 3.Rc6-e6=Q PPe5-a1 4.Qe6-b6=S PPb4-b5 {= (C+ by Popeye 4.75)} Chameleon echo model stalemates in Reverse-Pawn-Wenigsteiner without Kings by Back To Back. (Author)

### 2 Responses to No.1178 (KM)

1. Jan Golha says:

Karol, thank you for dedication

2. Juraj Lörinc says:

This problem brings into light an interesting question about equivalence of various fairy elements and their combinations. Totally same content can be shown using the following position:

White Pa7 Ph7
Black Pb7 Pe5
Face-to-Face + Chameleon Chess
hs=4
2.1.1…
C+ by Popeye 4.75

1.a7-a8=B b7-b6 2.Ba8-e4=R e5-e8 3.Re4-e7=Q e8-h8 4.Qe7-b4=S b6-b5 =
1.h7-h8=B e5-e4 2.Bh8-c3=R b7-b5 3.Rc3-e3=Q e4-a8 4.Qe3-b3=S b5-b4 =

Obviously, this version saves one fairy element (no fairy pieces on the board), thanks to reflecting pawn position on the board and choosing other suitable -t- condition. Is there any reason to prefer one of the choices of fairy elements, especially BtB in this case?