|(1) Posted by Siegfried Hornecker [Saturday, Aug 27, 2016 10:32]; edited by Siegfried Hornecker [16-08-27]|
Build your own helpmate!
(= 3+1 )
Add the black king. How many possibilities are there to have a helpmate with a single solution?
Siegfried Hornecker, Weltenfern 2013
(Note that only the helpmates are counted where Black gets checkmated!)
|(2) Posted by Hauke Reddmann [Sunday, Aug 28, 2016 17:32]|
Being a mathematician, I give an upper bound of 55 :-)
|(3) Posted by Hannu Harkola [Monday, Aug 29, 2016 11:43]|
I found only these nine:
bKb8 1.Ka8 Rc8 #
bKh1 1...Se3 2.g1=R Rh7 #
bKa1 1.g1=R Rc5 2.Rb1 Ra5 #
bKa2 1.g1=R Rb7 2.Ra1 Rb2 #
bKf8 1.g1=R Sd6 2.Rg8 Rf7 #
bKg1 1.Kh1 Se3 2.g1=R Rh7 #
bKh3 1.g1=R Se3 2.Rg3 Rh7 #
bKh5 1.g1=R Se5 2.Rg5 Rh7 #
bKe8 1...Ka7 2.Kd8 Rd7 + 3.Kc8 Sb6 #
|(4) Posted by Dmitri Turevski [Monday, Aug 29, 2016 12:40]|
|(5) Posted by seetharaman kalyan [Monday, Aug 29, 2016 19:11]|
hm... or is it #1?
|(6) Posted by Siegfried Hornecker [Monday, Aug 29, 2016 20:11]|
Okay, Hannu Harkola found a h#2.5 that I hadn't seen. It also has bKa7/b7 with h#1 each, which however fail to an illegal position (and the legal Ka8 h#0.5). So it should have been a "9/11" (9 out of 11) symbol problem. Those were all intended solutions (except - as I said - the h#2.5 which I didn't see).
Now it is 10/12. Good find, Mr. Harkola!
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