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Problem 414: Sebastien Luce - Fairy (Hanovre)

sebastien(23.06.2014) A very nice Help-double-stalemate with super Durbar and the fairy condition Hanovre. Welcome to Sebastien Luce from France!

 

 

 

414

 

1…Kd5 2.Kg3 Ke4 3.Kg2 Ke3 4.Kh2 Kf2 5.Kxh3 Kf3==

1…Kd7 2.Ke5 Ke8 3.Kf6 Kf8 4.Kg6 Kg8 5.Kh6 Kh8==

1…Kc7 2.Ke4 Kd7 3.Kd5 Ke7 4.Kc4 Ke6 5.Kd4 Kd6==

 

Hanovre is a very uncommon condition. I found only 2 incorrect (!) problems in Winchloe.But it can become more popular because the definition is quite simple:A left square cannot be occupied anymore. (Author)

 

Comments  

 
+1 #1 Diyan Kostadinov 2014-06-25 01:08
Three equal lenght solutions with no repetitions in such a light construction is very nice. Hanovre condition looks interesting to me. Is it true that it is so rare used? Some examples by our readers?
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0 #2 Juraj Lorinc 2014-06-27 18:50
I think a significant number of problems with fairy condition Haan can be set as Hanovre. The conditions differ for linemovers, but e.g. for kings and pawns they tend to be the same. Even the published h==4,5 works as Haan too.
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0 #3 Bjorn Enemark 2020-04-30 17:56
The condition only has effect for linemovers. The name is HAANOVER. The definition is that squares which are PASSED by a move become holes. The departing square is not affected.
Long castling will create a hole on b1/b8.
So, for a kindergarten problem the condition is useless.
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