No.1688
Armin Geister & Daniel Papack
(Germany)

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


Definitions / Определения


No. 1688 Armin Geister &
Daniel Papack

Germany
original - 31.01.2022

black Ph4h3g2d2 Bh1a5 Sg1 Qf8 Re8c8 Kd8 white Ph2e2c7e7f7g5g6 Kh6 Qg7
ser-h#8                       9+11
Mars Circe

Solution: (click to show/hide)


Subscribe
Notify of
guest
2 Comments
Oldest
Newest Most Voted
Inline Feedbacks
View all comments
Seetharaman
Seetharaman
February 1, 2022 12:25

Four switchbacks to release WQ and a reciprocal Bristol as a bonus 😀!

Last edited 3 months ago by Seetharaman
Geoff Foster
Geoff Foster
February 12, 2022 00:25

Initially, 1…Qxd2?? is illegal because of check from bPg2 (via g7). The plan is to promote bPg2 to R and then move it back to g2, but 1.Sf3+?? moves the bS to a white square, from where it checks the wK via g8. Also, 1.Qg8? 2.Sf3 fails because 3.g1=R+?? checks the wK via h8. Therefore 1.Qh8! 2.Rg8! 3.Sf3 (allowed because g8 is occupied) 4.g1=R (allowed because h8 is occupied). After 5.Rg2, Black wants to restore the position, starting with 6.Sg1 (not 6.Se5?, 6.Sd4? or 6.Se1?, which would stop the intended mate). With the bS back on a dark square the bR can leave g8 without checking the wK, so 7.Re8 blocks e8, preventing the bK from moving there). The wQ is pinned by bBa5 (via f8), so finally 8.Qf8 unpins the wQ.
Very nice logic for all the switchbacks!