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Problem 759: Aleksandr Azhusin - Selfmate
aleksandr.azhusin(31.10.2016) A warm welcome to the Russian selfmate expert Aleksandr Azhusin! Here he presents a selfmate twomover with change of play and a Star.
 
 
 
 
 
 
 
 
759. Aleksandr Azhusin (Russia)
31.10.2016
759.1
 
Set: 1…Bxc5 2.Rd4+ Bxd4#
        1…Be5 2.Qd4+ Bxd4#
        1…d5 2.Qc4+ dxc4#
 
1.exd4+? A  Sxg3+ 2.Ke5?       
1.e4+? B, but 1…Be3!
1.Qd5? (2.e4+ B Sxg3#), but 1…Qb8!
 
   1.Qe5! (2.exd4+ A Sxg3#)
1…Bc5 2.Qd4+ Bxd4#, 1…Bxe3 2.Qd5+ Rxd5#
1…Bxe5 2.Rd4+ Bxd4#, 1…Bc3(xb2) 2.e4+ Sxg3#
1…d5 2.Qe4+ dxe4#                
1…Q(R)xa5 2.Q(x)b5+ Q(R)xb5#
 
 
Change of white continuations, bB Star (Author)
 

Comments  

 
+2 #1 Seetharaman Kalyan 2016-10-31 13:23
Nice changes, with correction play of the bishop!
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+4 #2 Seetharaman Kalyan 2016-10-31 13:32
The scheme has been used before, but the correction play and the additional change makes it original!

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+5 #3 Miodrag Mladenovic 2016-11-01 05:10
Nice problem. The triple defense on a5 (R,Q,Pxa5) can be avoided by slightly modifying position:

Now there is a defense 1...Bb7 2.Qxb5+ Rxb5#
The try 1.Qd5? is now defeated by 1...Qe6!
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+1 #4 Seetharaman Kalyan 2016-11-03 06:10
I like your version Misha... especially the added variation 1...Bb7! Definitely an improvement.
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+2 #5 Ivan Soroka 2016-11-03 17:19
See the nice Uri Avner's problem:

pdb.dieschwalbe.de/.../
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+1 #6 Seetharaman Kalyan 2016-11-04 19:41
Thanks Ivan Soroka for finding and quoting this classic. This is just amazing --- Four changes, Ideal Rukhlis and Dobrovskis effect! I am surprised no award is cited. An obvius winner in any tourney!

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0 #7 Diyan Kostadinov 2016-11-20 13:26
Here is the answer by Aleksandr Azhusin:
"Polagayu, chto moy otvet na kommentarii k moey 759 neobchodim.
Prekrasnaya zadacha U.Avner, po moemu mneniyu ne yavlyaersya predschestvennikom. Schemy zadach shozhie, a soderzhanie raznoe. V S#2 Avner idealny Ruchlis v dvuch fazach(2х4), v moey S#2 akzent napravlen ne na peremenu igry(2х3), a na vybor vstupleniya v trech popitkach i glavnoe - zvezdochka chernogo slona. Schitayu № 759 vpolne samostoyatelnoy zadachey.
Redakztiya M.Mladenovich interesna i ya videl eyo.Rechil otkasatsya, potomu cho ferz na a3 uchastvuet tolko a oproverzhenii popitki. Ya perestavil ego na a7 chtoby zagrusit b dopolnitelnom variante. Kakaya iz dvuch redakztiy luchschaya - eto delo vkusa.
Blagodaryu vsech, kto prislal kommentarii k moey S#2.
Aleksandr Azhusin"
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+1 #8 Diyan Kostadinov 2016-11-20 13:32
On english his comment is like:
"I think that I should answer to the comments on my problem 759. In my opinion the wonderful selfmate by Avner do not anticipate my, because even that the scheme is simmilar, the content is different. In Avner's problem there is Ideal Ruchlis (2x4), but in my 759 the main content is not the change of play (2x3), but the thematic tries (choice of the key) and the Bishop Star.
The version by Mladenovch is nice and I saw it too, but decide to not use it because the Qa3 is not active - needed only to refute a try. I put it on a7, because it play also in a variant. Which is better - it's a matter of taste.
I thank all who comment my S#2. Aleksandr Azhusin"
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+1 #9 Rodolfo Riva 2016-11-23 09:10
To #2
The pinned piece promenade in the Pankratiev's version is directly anticipated by

Z. Skarbnik, 1st HM, IRT 1960
3r2QK/3p1b1R/3PpkPR/4p1s1/4P2B/8/8/8 (8+7)
1.Qe8 zz with 3/4 Bishop star
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+1 #10 Rodolfo Riva 2016-11-23 09:24
T0 (#7) #8.
In the original the "active" 1...,Qxa5 is artificial: you can replace the wSa5 with a more economical wPb3.

In my opinion the Misha's version is preferable (the bQ stop a dual in the set play). but since does not eliminate the black dual 1....,Bxb2 it would be better to save the wSd2: -wSd2, -bBa6, +bPc4 (10+14), 1...,c3 2.Qxb5+, Rxb5#.
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