No.1683
Ľuboš Kekely
(Slovakia)
Original Fairy problems
JF-2022-I:
01.01.2022 - 30.06.2022
No. 1683 Ľuboš Kekely
Slovakia
original - 19.01.2022
white Rd4 PAh5 Kh3
black Bb8 Qc2 Ka1 Pg7g6g5
h=3 2 solutions 3+6
ChecklessChess
PAO h5
ChecklessChess
PAO h5
Solution: (click to show/hide)
Seems checklesss condition not needed for first solution
It is used for move order.
True. Eliminated cooks by PAh1 prematurely
It doesn’t help the solution?! Fairy condition as a mere cookstopper, without a strong thematic meaning?
Same theme with another fairy conditions.
White PAa8 Rb2 Kh1
Black Rc2 Bh2 Ka1
H=5* ( 3 + 3 )
Circe Maximummer
PA<pao>a8
Set play 1…PAc8 2.Bb8 Rxb8[+bBf8] 3.Ba3 PAc5 4.Rh2 + Kg1 5.Ra2 PAa5 =
1.Bb8 PAa2 2.Bh2 PAxc2[+bRa8] 3.Bb8 PAc7 4.Ra2 Rxb8[+bBf8] 5.Ba3 PAa7 =
C+ by Popeye 4.87
The diagram for the example by János
Thank you, Julia!