# No.277,278 (CF)

 No.277, 278 Chris Feather (England) Original Problems, Julia’s Fairies – 2013 (I): January – April ?Previous ; ?Next ; ?List 2013(I) Please send your original fairy problems to: julia@juliasfairies.com

No.277, 278 – Two very exotic problems by Chris Feather! Enjoy a combination of fairy conditions Anti-Kings+Anti-Circe and author’s explanations to the problems! I’m also grateful to Chris for the definitions he’s sent with his problems! (JV)

Definitions:

Anti-Kings: A king is in check if he is not attacked. Mate occurs when a  king is not attacked and his side has no move which exposes him to attack. (Of course kings may not be captured.)

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.

 No.277 Chris Feather England original-05.03.2013   h#2              3 solutions        (2+1+2n)Anti-Kings+Anti-Circe     Solutions: (click to show/hide)   I. 1.nPc1=nR Pa3 (Pa4? … 3.nRxa4[nRa8]!) 2.nRa1 nPe8=nQ# II. 1.nPc1=nS+ Ke2 2.nSxa2[nSg8]+ nSxe7[nSg1]# III. 1.nPc1=nB+ Kd2 2.nBa3+ nBxe7[nBc1]# Explanation: Neither king is in check initially because both are attacked. Solution I: The nR must hide so that it cannot attack the BK. The nQ does not attack the BK because the rebirth square d1 is occupied. Solution II: The nS must keep the WK under attack while removing the nPe7 which attacks the BK. The mate answers the check when the WK is briefly left unattacked. Solution III: Now the nB keeps the WK under attack while removing the nPe7. This nB incidentally performs a round trip. Again the mate answers the check when the WK is briefly left unattacked. The unifying theme is of course the neutral Allumwandlung. (Author) No.278 Chris Feather England original-05.03.2013   h#2              2 solutions        (3+1+1n)Anti-Kings+Anti-Circe     Solutions: (click to show/hide)   I. 1.nPg1=nQ Kf2 2.nQf1 Pg8=B# {3.Kf7??} II. 1.nPg1=nR Pg8=S+ 2.Kg7 Kg2# {3.Kf6/h6??} Explanation: Neither king is in check initially because both are attacked. The WPd5 is only a cookstopper. There are no antiCirce captures in this problem but antiCirce is exploited in each of the mates. As indicated in { … } in the solutions, the promoted B & S cannot attack the BK because their rebirth squares (f1 & g1) are occupied. In each solution the WK shields the BK from attack by standing in the way. This also keeps the WK safely under attack himself. This time the unifying theme is a mixed Allumwandlung (neutral/white). The antiKings+antiCirce combination may possibly be new in the helpmate. (Author)

### 2 Responses to No.277,278 (CF)

1. Geoff Foster says:

What a terrific idea! Another interesting possibility is:

h#2 (b)bKg8>e4
white kg1
black kg8
neutral pf7 ph2

(a) 1.nPh1=nR nPf8=nR+ 2.nRh8+ Kh1#

Now Black cannot play 3.Kh7??, because the nRh8 doesn’t attack the bKh7, because the nR’s rebirth square (h1) is occupied. For the same reason the bK cannot play 3.Kxf8?? or 3.Kxh8??, because it would be reborn on e8, which is a white-coloured square.

(b) 1.nPh1=nQ nPf8=nB 2.nBb4 Kxh1[wKe1]#

2. seetharaman says:

Anti-king is a very interesting condition. Chris has shown two excellent ideas in combination with Anti-Circe ! Thanks Chris.