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MatPlus.Net Forum Competitions Miroslav Henrych 60 JT C 27. 9. 2014
 
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(1) Posted by Michal Dragoun [Thursday, Jan 2, 2014 16:57]

Miroslav Henrych 60 JT C 27. 9. 2014


Helpmate of any length. All pieces will exchange their places (compared diagram and mating position). If all pieces exchange the places in one cycle, fairy condition or pieces are allowed (but not both). If there are more separated cycles or reciprocal exchanges, only orthodox helpmates are accepted. In the case of more solutions or twins is required exchange of places of all pieces in each of them. Special prize for the orthodox composition with exchange of places of all pieces in one cycle.
Entries to Jiří Jelínek, Španielova 1313, 163 00 Praha 6, Czechia, jjelinek@chello.cz. Judge: M. Henrych.
 
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(2) Posted by seetharaman kalyan [Friday, Jan 3, 2014 12:02]

Any examples?
 
 
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(3) Posted by Michal Dragoun [Friday, Jan 3, 2014 17:34]; edited by Michal Dragoun [14-01-03]

Miroslav Henrych
Šachová skladba 42/1994
(= 3+2 )

H#4, white begins
1.- Sb8 2.Kb4 Ba6 3.Ra4 Kb7 4.Ka6 Sc6#
white cyclic and black reciprocal exchange

Miroslav Henrych
109 Šachové umění 10/1999
(= 2+3 )

H#4, circe, white begins
1.- Kb3 2.Rxb4(Sg1)+ Kxa2(Ra8) 3.Ka4 Se2 4.Ra5 Sc3#
one cycle

Edit: sorry, I forgot to give author´s name and publication.
 
   
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(4) Posted by Nikola Predrag [Friday, Jan 3, 2014 17:58]

Well, after Michal was forced to provide these beautiful examples, the area of originality is pretty reduced. I wonder whether some technical static pieces or at least the Pawns are allowed?
 
   
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(5) Posted by Michal Dragoun [Friday, Jan 3, 2014 18:25]

> Nikola: No, static technical pieces are not allowed.
 
 
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(6) Posted by seetharaman kalyan [Friday, Jan 3, 2014 22:00]

Thanks for these two dishearteningly beautiful examples. My reason for asking is because I could not clearly understand when Fairy conditions are permitted.
 
 
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(7) Posted by Roland Ott [Monday, Jan 6, 2014 15:41]

There is a small type error in the solution of the first example:
4. Ka6 instead of 4. Ka5
The correct solution should read as follows:
1.- Sb8 2.Kb4 Ba6 3.Ra4 Kb7 4.Ka5 Sc6#
 
 
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MatPlus.Net Forum Competitions Miroslav Henrych 60 JT C 27. 9. 2014