No.1375 
Sébastien Luce
(France)

Original Fairy problems
JF-2019/I:
January – June’2019


Definition: (click to show/hide)


No.1375 Sébastien Luce
France

original – 20.02.2019

Solution: (click to show/hide)

White Kg3 Black Kh1 Pa6c6d5e4b3b2e2h2 Neutral Pg2

h=9                                  (1+9+1) C-
Neutral pawn g2


No.1375.1 Sébastien Luce
France

version of No.1375 – 13.03.2019

Solution: (click to show/hide)

White Kg3 Black Kh1 Sa8 Pc6d5e4b3b2e2h2 Neutral Pg2

h=8                                  (1+9+1) C-
Neutral pawn g2


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ganapathi
ganapathi
1 year ago

Stalemate scene well planned! The double jump of the neutral P comes in handy and AUW is bonus!
Very enjoyable!

Georgy Evseev
Georgy Evseev
1 year ago

1.g1NS Kf2 2.a5 NSxe2 3.d4 NSxd4 4.b1Q NSxb3 5.Qa1 NSxa5 6.Qb1 NSxc6 7.Qf1+ Kxf1 8.NSd4 NSf3 9.ef Kf2 =

Luce
Luce
1 year ago

Aie aie aie (in french) Georgy !
Thank you for your help.
Please Julia, replace it by another problem waiting i re work it if possible.

Luce
Luce
1 year ago

Maybe it is possible to put a black bishop in e4 instead of a pawn but it is not so nice.
I think the best solution seems to inverse black g2 pawn and neutral e2 pawn with the sol. 1.g1=S Kf2 2.Sf3…
What do you think about it Georgy and Julia ?

Georgy Evseev
Georgy Evseev
1 year ago
Reply to  Luce

Inverse e2&g2 definitely will not work – black have free moves.
Be4 need more testing.

Luce
Luce
1 year ago

Yes you are right Georgy. To inverse e2 and g2 is cook with 1.e1=nS ?

Luce
Luce
1 year ago

Grrr… it is also cook with a black bishop in e4 with 1.g1=nS Kf2 2.b1=S Snxe2 3.d4 Snxd4 4.a5 Snxc6 5.Sd2 Snxa5 6.Snac4 Snxd2 7.Bg2 Snxb3 8.Snd2 Snf1 9.Bxf1 Kxf1=
Dear Georgy Julia I think the idea of the problem is definively cook. ?

Aleksandr Bulavka
Aleksandr Bulavka
1 year ago

bPe2nPg2 and wKg3->e1?

Aleksandr Bulavka
Aleksandr Bulavka
1 year ago

Exchange bPe2 and nPg2. And wKg3->e1?

Luce
Luce
1 year ago

Dear Alexandre unfortunately your proposition is also cook. The last attempt is to put a black bishop in d5 in the original position. I have to check it…

Luce
Luce
1 year ago

To save the idea of the problem 1375,
I propose the following version with a solution one move shorter. I invite all the composers of JF to see if there is a cook. Best to all

h=8
White : Kg3
Black : Kh1 Sa8 Pc6d5e4b3b2e2h2
Neutral : Pg2
Sol. :
1.e1=S Kf2 2.Sf3 gxf3 3.b1=Q fxe4 4.b2 exd5 5.Qh7 dxc6 6.Qb7 cxb7 7.b1=R bxa8=Bn+ 8.Rb7 Kf1 =