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Problem 825: Vitaly Medintsev - Helpmate
vitaly.medintsev(20.04.2017) An excellent Zilahi from the Russian master!
 
 
 
 
 
 
 
 
 
 
 
 
825. Vitaly Medintsev (Russia) 
826
H#2          2 Sol.          (9+9)
 
1.Bxc2+ Kxc2 (Kc4?) 2.Bb4 Rd7#
1.Bxd2 Kxd2 (Kd4?) 2.Ba4 Qh7#
 

Comments  

 
+3 #1 Dmitri Turevski 2017-04-20 20:03
Eccellent indeed.
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+2 #2 Rodolfo Riva 2017-04-20 21:43
Vitaly,
you forgot the principles of economy!

8/2p2k2/4SS2/prP4P/6s1/1bbK2s1/B1QR4/B7 (9+8)
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+2 #3 Seetharaman Kalyan 2017-04-21 06:34
Nice modification by Rodolfo Riva!



H#2 2 Sols.
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+2 #4 Yuri BILOKIN 2017-04-21 07:44
Можно экономичнее. Чёрный аристократ.
8/8/5k2/2P1NN2/1r5P/6n1/rbbK2n1/BBQR4
h#2 2.1... 9+7
1.Bxc1+ Kxc1 2.Ba4 Rd6#
1.Bxd1 Kxd1 2.Ba3 Qh6#
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+2 #5 Vitaly Medintsev 2017-04-21 07:51
Here is the version by Yuri BILOKIN

1.Bxc1+ Kxc1(Kc3?) 2.Ba4 Rd6#

1.Bxd1 Kxd1(Kd3?) 2.Ba3 Qh6#
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+2 #6 Yuri BILOKIN 2017-04-21 07:54
Оказывается, аристократ не обязателен
8/8/5k2/1pP1NN2/7P/6n1/rbbK2n1/BBQR4
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+2 #7 Vitaly Medintsev 2017-04-21 07:56
Here is the version by Nebojsa Joksimovic

1.Bxc2+ Kxc2 2.Bb4 Rd7#

1.Bxd2 Kxd2 2.Ba4 Qh7#


Unfortunatelly, dual avoidance (Kc4?) is lost in this the most economical version.
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+2 #8 Vitaly Medintsev 2017-04-21 08:00
Quote:
Rodolfo Riva 2017-04-20 21:43
Vitaly,
you forgot the principles of economy!

You are right Rodolfo! This is my sin :-)
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+2 #9 Vitaly Medintsev 2017-04-21 08:11
I thank all commentators for their versions. Let's the judge decides!
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+2 #10 Rodolfo Riva 2017-04-21 12:13
To #7

In my opinion the anti-logical tries (Kc4/Kd4) are not worthy of highlighting.
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+1 #11 Vitaly Medintsev 2017-04-21 14:09
Quote:
In my opinion the anti-logical tries (Kc4/Kd4) are not worthy of highlighting.

True, there is something anti-logical in those tries. Still, I think they worth to be indicated since they both fail by the same specific reason - line-closure of friendly linear officer.
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0 #12 Vitaly Medintsev 2017-04-25 16:21
Another version by Nebojsa Joksimovic:


1.Bxc2+ Kxc2 (Kc4?) 2.Bb4 Rd7#

1.Bxd2 Kxd2 (Kd4?) 2.Ba4 Qh7#


I like this version more than others
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+1 #13 Sre?ko Radovi? 2017-04-25 18:02
Blak king from f7 to e7 and pawn h5 is not need.9+5!
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+2 #14 Vitaly Medintsev 2017-04-25 19:08
Version suggested by Srecko Radovic in comment #13:


1.Bxc2+ Kxc2 (Kc4?) 2.Bb4 Rd7#

1.Bxd2 Kxd2 (Kd4?) 2.Ba4 Qh7#


Solution remains unchanged but here WR delivers mate with the support of WSf6

The best ecomomy on this moment.
Who can suggest a Meredith position?! :-)
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0 #15 Manikumar S 2017-04-28 07:27
In the version suggested by Srecko Radovic, a black pawn at f3 can replace BRe1 with the same solutions and better effects

8/4k3/4NN2/2P5/1P6/rbbK1p2/B1QR4/B7
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0 #16 Vitaly Medintsev 2017-04-28 08:43
Quote:
In the version suggested by Srecko Radovic, a black pawn at f3 can replace BRe1 with the same solutions and better effects 8/4k3/4NN2/2P5/1P6/rbbK1p2/B1QR4/B7

I know about this opportunity - Nebojsa Joksimovis has sent me this position on 26 April via email but I have reject it because the try Ke4? (false move along thematic line) hasn't its identical orthogonal try Kd4??
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0 #17 Manikumar S 2017-04-28 09:03
Quoting Vitaly Medintsev:
I know about this opportunity - Srecko has sent me this position on 26 April via email but I have reject it because the try Ke4? (false move along thematic line) hasn't its identical orthogonal try Kd4??


Okay sir. Thank you for the explanation. Maybe a black knight at g3 may preserve these effects instead of rook at e1. If my understanding is correct, hope this should be fine.
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0 #18 Viktoras paliulionis 2017-04-28 09:37
Versions with one white pawn only:
B7/B1QR4/rbbK2n1/8/p1p5/4NN2/2P1k3/8
B3r3/B1QR4/rbbK4/8/p1p5/4NN2/2P1k3/8
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0 #19 Vitaly Medintsev 2017-04-28 09:48
Quote:
Maybe a black knight at g3 may preserve these effects instead of rook at e1. If my understanding is correct, hope this should be fine.

Yes, it is pretty good.
I still prefer version by Nebojsa Joksimovich (#12) since it has no additional (accidental) effects/features such as supporting of mating piece
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0 #20 Vitaly Medintsev 2017-04-28 09:56
Version without technical WPs by Viktoras Paliulionis suggested in comment #18

1.Bxc7+ Kxc7(Kc5?) 2.Bb5 Rd2#

1.Bxd7 Kxd7(Kd5?) 2.Ba5 Qh2#


Wow, such a huge number of versions!
We are going to blow up the judge's mind :-)
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