|(1) Posted by Roddy McKay [Monday, Apr 26, 2021 21:25]|
An easy h#2
This problem was rejected by The Problemist & The Problemist Supplement, reasons included being too easily solved. I had hoped that the twinning might be slightly unusual....
(= 6+9 )
(= 9+6 )
|(2) Posted by seetharaman kalyan [Monday, Apr 26, 2021 22:04]|
I think the Solution is.1.Bc1 Bxc1 2.Qd2 Bb2#. White bishop is unnecessary for this mate and idle. The solution for the twin is definitely simple with queen mate on base rank and has no strategy.
|(3) Posted by Viktoras Paliulionis [Monday, Apr 26, 2021 22:12]|
This twinning type is known as Polish type. Definition from Encyclopedia of Chess Problems by Velimirovic&Valtonen (2012):
Polish type: twin that is created by changing the colors of all the pieces on the board. It is desirable to attain maximal economy of all pieces (even including Pawns) in mates which would be then - ideal mates, but this is not an explicit thematic requirement. Example:
I.pr Wola Gulowska 1988 / TT
(= 3+3 )
h#3 b) PolishType
1.Kd7 d6 2.Ba5 Kd5 3.Bd8 Bc6# (IM)
1.d4 Ke7 2.Ke5 Be6 3.Be4 Bd6# (IM)
|(4) Posted by Roddy McKay [Tuesday, Apr 27, 2021 11:33]|
The first diagram should be as posted plus :- remove Bb7, add Pa6
Accordingly in (b) remove bb7, add pa6
Apologies, this slightly more economical version is from a suggestion by Michael Mcdowell when I first composed this.
|(5) Posted by seetharaman kalyan [Tuesday, Apr 27, 2021 15:32]|
Still you persisted with the idle Bb7 version?
|(6) Posted by Roddy McKay [Wednesday, Apr 28, 2021 01:06]|
For some reason the diagram which I had on file hadn't been updated with the improved economy, but I remembered when the comment about redundant Bb7 was made. The problem is from some years ago.
|(7) Posted by Siegfried Hornecker [Wednesday, Apr 28, 2021 16:00]|
EDIT: Ok, b is correct. I see now.
So the solutions are (mark to see):
a) 1.Bc1 Bxc1 2.Qd2 Bb2 mate
b) 1.Bxf4 Qh3 2.Kc1 Qf1 mate
Yes, indeed fast and easy to solve.
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MatPlus.Net Forum Helpmates An easy h#2