|(1) Posted by Jacques Rotenberg [Sunday, Oct 5, 2008 13:51]; edited by Jacques Rotenberg [08-10-05]|
An interesting helpmate to check
P. Moutecidis, H. Fouxiaxis, S. Pantazis
Tokaj Tourney, Budapest 1988, 2nd Prize
(= 12+4 )
h#2 a) b)Sf2<->f3
a) 1.Qa3 Ra7 2.K×f3 Qb7‡
b) 1.Qa8 Ba1 2.K×f2 Qb2‡
I like it very much.
The question is : Is it possible to get it without twins ?
I could manage only to obtain the following :
(= 9+6 )
1.Qh8 Ba8 2.K×b3 Qb7‡
1.Qc8 Rg8 2.K×c3 Qg7‡
Can anyone improve ?
|(2) Posted by Kostas Prentos [Sunday, Oct 5, 2008 15:05]|
Unfortunately, the wRa4 is not used in the Qg7 mate.
Apparently, the original problem had to sacrifice the two solutions form, so that wRe1 got a role.
|(3) Posted by Jacques Rotenberg [Sunday, Oct 5, 2008 15:24]|
Yes this is the point
In this special case, because of the Bristol, you have anyway mat nets with useless white pieces, so perhaps here it could be acceptable. The Rook a4 also prevents cooks with bK on a3.
|(4) Posted by Kostas Prentos [Sunday, Oct 5, 2008 17:36]|
OK, here is a version that takes care of everything, at the cost of some black plugs on the north-east corner:
(= 10+6 )
1.Qa3 Ra7 2.Kxf3 Qb7#
1.Qa8 Ba1 2.Kxf2 Qb2#
Note that a wPh3 would be a more economical way to control g4, but then the position would be illegal.
|(5) Posted by Jacques Rotenberg [Sunday, Oct 5, 2008 19:10]; edited by Jacques Rotenberg [08-10-06]|
Very well done Kostas ! Nice and tricky!
|(6) Posted by Jacques Rotenberg [Friday, Oct 10, 2008 15:49]|
What does Harry think of it ?
No more posts
MatPlus.Net Forum Helpmates An interesting helpmate to check