# No.355 (CF)

 No.355 Chris Feather (England) Original Problems, Julia’s Fairies – 2013 (II): May – August  →Previous ; →Next ; →List 2013(II) Please send your original fairy problems to: julia@juliasfairies.com

No.355 – Chris Feather – Interesting Rex-Solus Miniature with Symmetry Anti-Circe condition. The definitions below are taken from Fairings No.32 by Chris Feather. I have to mention that I’ve got this original by postal mail, as Chris doesn’t have internet now. But if you have some comments – please feel free to write them! After all, I can print them out and send to the author together with the publication! (JV)

Definitions (from Fairings No.32):

Symmetry Anti-Circe: As Anti-Circe except that the rebirth square for the capturing unit is that which is symmetrical in Symmetry Circe.

Anti-Circe: After a capture the capturing piece (Ks included) must immediately be removed to its game array square (necessarily vacant, else the capture is illegal). R, B & S go to the square of the same colour as the capture; Ps stay on the file of capture; fairy pieces go to the promotion square of the file of capture. [The default Calvet type (not usually specified) allows captures on the rebirth square, the rarer Cheylan variant (always to be specified) excludes them. In practice many problems can be of either type.]

Symmetry Circe: As Circe except that the rebirth square for the captured unit is that which lies at an equal distance (in a straight line) beyond the midpoint of the board. Thus a capture on c4 produces a rebirth on f5, a capture on g1 produces a rebirth on b8, and so on. Strictly speaking there are of course other types of symmetry: this one is rotational.

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

 No.355 Chris Feather England original-31.07.2013   h#4               2 solutions             (1+5)Symmetry Anti-Circe     Solutions: (click to show/hide)   I. 1.Se3-g2 Kh1×g2[wKg2->b7] 2.Kg1-h1 Kb7-b6 3.Ra8-a6 + Kb6×a6[wKa6->h3] 4.Bb8-a7 Kh3-h2 # II. 1.Ra8-a7 + Kh1-h2 2.Ra7-b7 + Kh2-g2 3.Rb7-c7 + Kg2-f2 4.Kg1-h1 Kf2-g1 # (С+ Popeye 4.63)

### One Response to No.355 (CF)

1. Geoff Foster says:

Initially neither king is in check because the rebirth squares (a8 and b8) are occupied. The bK will be mated on h1, so the bR must either be captured or else it must move away from a8 and be unable to return there.

A try is 1.Ra7+ (check to wK because a8 is vacant) Kh2 (wK not in check because a7 is occupied) 2.Rb7+ (wK in check because a7 is vacant) Kh3? 3.Ba7 ~ 4.Kh1 Kh2# (wK not in check because a7 is occupied, bK mated because bR can’t get back to a8). However White has no waiting move on move 3 (the bPh4 prevents 3…Kg3)! Therefore the wK gets to h3 in another way: 1.Sg2 Kxg2-b7 (like a Take&Make move!) 2.Kh1 Kb6 (a tempo move) 3.Ra6+ Kxa6-h3 4.Ba7 Kh2#.

In the second solution the bR helps the wK get to f2, by occupying the rebirth squares.

A very simple but very clever problem. It is worth looking at the other Symmetry Anti-Circe problems in Fairings No.32!