|(1) Posted by Miodrag Mladenović [Saturday, Feb 22, 2020 16:58]|
29th Branko Atanacković MT - 2020
I am directing this tournament on March 14th. Here is announcement:
Up to now, there are 21 solvers planning to participate:
This will be very strong tournament (so far 4 GMs, 1 IM and 1 FM). There is still time to enroll. There are three locations for this tournament: Belgrade, Riga and Oberglatt.
I am directing this tournament since 2007 but this year it's the first time multi location tournament. Here are problems from last year: https://www.wfcc.ch/wp-content/uploads/2019-03-24-SRB-Zemun-BAM-Problems.pdf
I hope to see more solvers participating. This year there will be much more of my originals. So far I composed seven originals but I am planning to make few more.
|(2) Posted by Alain Villeneuve [Saturday, Feb 22, 2020 22:44]|
Unfortunately, I am too far from these 3 towns but I can advise to every solver to go there. I solve every year (thank to WFCC site) the Atanacković competition and enjoy it, especially Miodrag's problems, all interesting and sometimes masterpieces (for instance the S#3 of 2016 & 2019).
Even studies are a good choice, they are human (!) unless many in the usual solving contests. You know what I mean !
|(3) Posted by Miodrag Mladenović [Sunday, Feb 23, 2020 22:51]|
Thanks for a nice words. I am glad you like my originals. You may want to try to solve this S#5 that I composed for last WCSC. Luckily, Ryszard did not select my problem. Based on the feedback by some other strong solvers (I shown them this problem after I learned that problem was not selected for this WCSC tournament) it turned out that problem would be too difficult to solve with a time constraint (3 selfmates for 50 minutes). Actually only Piotr solved this problem but it took him 35 minutes to do this. Here is a problem without solution:
3rd Pr. Gennady Kozyura JT 64 2019
(= 10+7 )
To the participants of 29th Branko Atanackovic MT:
Do not worry, there will be only S#2 and S#3 for solving. Not so difficult problems to solve.
|(4) Posted by Andrey Selivanov [Monday, Feb 24, 2020 08:53]|
I know this problem! This is a beautiful masterpiece! Congratulations!
|(5) Posted by Alain Villeneuve [Monday, Feb 24, 2020 16:37]|
I agree ! It took me about 80 minutes (the 1st encouraging variation after 40 mn) : there's only one Piotr ! The diversity of model mates is amazing.
To Mr Selivanov about the Championship CFO in Tula : the Troitzky's study has a main line much more interesting than the line given in the official solution : 2...a1Q! 3 Rxa1 Kb2 4 Rf1!! (black correction !) a2 5 Kc4! a1Q 6 Nd3+ Ka2 7 Nb4+ Kb2 8 Rf2+ Kb1 9 Kb3 : no 9...Qa8 (WRh2), no 9...Qa7 (WRg2) and no 8...Kc1 (WRe2).
It seems this one, although beautiful, is inappropriate (like 75% of the existing studies) for a solving contest, where we need only one clear main line.
And Kralin's study is simply wrong : in 9...Qb7, the Knight is lost : 12...Kc5! (instead of Re4) 13 Ka2 (13 Ne3 Re4! 14 Nc2 Kc4! 15 Kb2 Re2! 16 Kb1 Kc3) Rf4 14 Ne3 Kd4! 15 Nc2+ Kc3 wins.
Congratulations for your S#6 of the same contest.
|(6) Posted by Siegfried Hornecker [Monday, Feb 24, 2020 19:30]; edited by Siegfried Hornecker [20-02-24]|
I am not sure what was told you to be the solution to the Troitzky study (probably the line ending in the mate wKc1 Nc2 bKa1 Pa2?), but you just gave (with the exception of the final move which was intended as 8.-Kc1 9.Sa2+ Kb1 10.Kb3 wins) the solution intended by Troitzky.
(= 3+3 )
Shakhmatnoe obozrenie 1910, 4th/5th prize (version)
1.Rc2+ Kb3 2.Rc1 a1Q! (2.-Kb2 3.Kd2 a1Q 4.Sd3+ Ka2 5.Sb4+ Kb2 6.R:a1, etc.) 3.R:a1 Kb2 4.Rf1!, etc.
|(7) Posted by Alain Villeneuve [Monday, Feb 24, 2020 20:23]|
Yes, I considered 8...Kb1 9 Kb3 as main line because it explains clearly the difference with the "thematic tries" 4 Rg1 & 4 Rh1 : the a6 square is controlled.
|(8) Posted by Dragan Stojnić [Monday, Feb 24, 2020 21:44]|
SOME RECORDS OF BRANKO ATANACKOVIĆ MEMORIAL 1992-2019 AND HIS PREDECESSOR
Branko Atanacković memorial is succesor the Serbian Open solving tourney, held at yearly Serbian chess problemists meeting in Belgrade since 1972. The winners were – XII 1972.Stošić,
I 1974.Kovačević 1975-76.not held 1977.Velimirović 1978.Kovačević 1979.Velimirović 1980.Kovačević 1981.Velimirović 1982.Velimirović 1983-not held 1984.Velimirović 1985,1986,1987.Kovačević 1988,1989,1990.Mladenović 1991-? (unknown)
Top solvers at previous 28 issues MBA, Belgrade 1992-2019:
-Independent Solving contest, since 1992
28.3.1992 1.Mladenović 2.Velimirović 3.Kovačević
27.3.1993 1.Kovačević 2.Mladenović 3.Đurašević
23.4.1994 1.Mladenović 2.Kovačević 3.Đurašević
18.3.1995 1.Kovačević 2.Đurašević 3.Gađanski
-Since 1999 MBA is involved in Yugoslav Solvers League cycle(now Serbian SL)
27.6.1999 1.Đurašević 2.Peruničić 3.Velimirović
6.5.2000 1.Kovačević 2.Čarnić S. 3.Velimirović
21.4.2001 1.Kovačević 2.Velimirović 3.Đurašević
6.4.2002 1.Peruničić 2.Paunović 3.Kovačević
20.4.2003 1.Vučković 2.Gađanski 3.Šoškić
IV 2004 1.Vučković 2.Šoškić 3.Sibinović
2.4.2006 1.Velimirović 2.Mladenović 3.Radomirović
18.5.2008 1.Kovačević 2.Vučković 3.Podinić
28.3.2009 1-2.Spirić i Sibinović 3.Vučković
27.3.2010 1.Kovačević 2.Vučković 3.Sibinović
19.3.2011 1.Kovačević 2.Vučković 3.Velimirović
17.3.2012 1.Velimirović 2.Kovačević 3.Mladenović
23.3.2013 1.Vučković 2.Kovačević 3.Podinić
22.3.2014 1.Vučković 2.Kovačević 3.Đurašević
28.3.2015 1.Kovačević 2.Đurašević 3.Radomirović
2.4.2016 1.Vučković 2.Kovačević 3.Podinić
1.4.2017 1.Vučković 2.Spirić 3.Roland
24.3.2018 1.Vučković 2.Kovačević 3.Roland
24.3.2019 1.Vučković 2.Serafimović 3.Podinić
According to incomplete list Bojan Vučković and Marjan Kovačević both won MBA at least 8 times.
However, I believe to Marjan have more.
Please if some of problemists have save the results for MBA 1996-98, 2005 and 2007 to help.
|(9) Posted by Alain Villeneuve [Sunday, Mar 22, 2020 13:26]|
Very interesting problems (bravo Miodrag) but the 2nd study is wrong : Black draws by 5...Kg4! (instead of g1Q) 6 g7 Kh3 7 g8R Kh2=.
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