No.1683
Ľuboš Kekely
(Slovakia)

Original Fairy problems
JF-2022-I:
01.01.2022 - 30.06.2022


Definitions / Определения


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

Solution: (click to show/hide)



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Seetharaman
Seetharaman
January 20, 2022 19:16

Seems checklesss condition not needed for first solution

Joost de Heer
Joost de Heer
January 21, 2022 08:32
Reply to  Seetharaman

It is used for move order.

Seetharaman
Seetharaman
January 21, 2022 14:03
Reply to  Joost de Heer

True. Eliminated cooks by PAh1 prematurely

Nikola Predrag
Nikola Predrag
January 21, 2022 18:28
Reply to  Seetharaman

It doesn’t help the solution?! Fairy condition as a mere cookstopper, without a strong thematic meaning?

Janos Mikitovics
January 22, 2022 20:40

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

Janos Mikitovics
January 25, 2022 23:14
Reply to  Julia

Thank you, Julia!