GM Sanan Sjugirov won the strong President of Uzbekistan Prizes Tournament by the proverbial skin of his teeth over American GM Han Moke Niemann after both finished the grueling 10-round Swiss classic event tied at 7½ points. The competition concluded November 18th in the City of Tashkent, Uzbekistan.
Sjugirov and Niemann were just two of the tournament’s 103 players that vied for the US$120,000 prize fund offered by the organizers. The event itself has the full support of Uzbekistan’s President who issued a decree in mid-January which mandated measures to further develop and popularize chess, as well as enhance the system of training chess players in Uzbekistan, and engage the youth, amateurs and qualified chess players to participate in chess competitions and tournaments, among other mandates. One can only dream of such a manfate in the Philippines.
Meanwhile, trailing Sjugirov and Niemann and sharing 3rd to 11th places is a logjam of 9 players with identical 7 points apiece. These are GM Maxim Matlakov, GM Vitaliy Bernadskiy, GM Javokhir Sindarov, GM Maksim Chigaev, FM Akmal Fayzullaev, GM Jakhongir Vakhidov, GM Timur Gareyev,
IM Mihail Nikitenko,and GM Mikhail Kobalia.
Here are some chessboard gems from the top finishers:
Matlakov,M (2691)-Muradli,M (2501) [B90]
Tashkent (3.1) 2021
1.e4 c5 2.Nf3 d6 3.d4 cxd4 4.Nxd4 Nf6 5.Nc3 a6 6.a4 Nc6 7.Nxc6 bxc6 8.a5 e6 9.Bd3 d5 10.0-0 Bd6 11.Re1 Qc7 12.h3 d4 13.Nb1 e5 14.Nd2 Bb4 15.Rf1 Be6 16.Nc4 Bxc4 17.Bxc4 Nxe4 18.Qg4 Nd2 19.Qxg7 0-0-0 20.Bxd2 Bxd2 21.Qg4+ Qd7 22.Qe2 Bf4 23.Ra3 Rhg8 24.g3 Qxh3 25.Bxf7 Rg7 26.Be6+ Qxe6 27.Qxa6+ Kd7 28.Qb7+ Kd6 29.Qxg7 Rg8 30.Qxh7 Rg6 31.a6 1-0
Bernadskiy,V (2592)-Makhnev,D (2513) [B12]
Tashkent (4.6) 2021
1.e4 c6 2.d4 d5 3.e5 Bf5 4.Nd2 e6 5.Nb3 a6 6.Nf3 c5 7.dxc5 Bxc5 8.Nxc5 Qa5+ 9.c3 Qxc5 10.Nd4 Ne7 11.Qa4+ Nd7 12.f4 Rc8 13.Be3 Qc7 14.Be2 0-0 15.0-0 Be4 16.Qd1 Nb6 17.Bf2 Nc4 18.Qc1 Rfe8 19.a4 Na5 20.b4 Nac6 21.Nb3 d4 22.Nc5 Bg6 23.Bxd4 a5 24.bxa5 Qxa5 25.Bf2 Qc7 26.Qb2 Nd5 27.Qxb7 Nxf4 28.Bb5 Nd5 29.Qxc7 Rxc7 30.Na6 Rcc8 31.Nb4 Nde7 32.a5 1-0
Sjugirov,S (2663)-Gareyev,T (2583) [C92]
Tashkent (6.2) 2021
1.e4 e5 2.Nf3 Nc6 3.Bb5 a6 4.Ba4 Nf6 5.0-0 Be7 6.Re1 b5 7.Bb3 d6 8.c3 0-0 9.h3 Re8 10.d4 Bb7 11.Nbd2 Bf8 12.d5 Nb8 13.Nf1 Nbd7 14.Ng3 h6 15.Nh2 Nc5 16.Bc2 c6 17.b4 Ncd7 18.dxc6 Bxc6 19.Bb3 a5 20.a3 Nb6 21.Ng4 Nc4 22.Ra2 axb4 23.axb4 Nxg4 24.Qxg4 Re6 25.Bxc4 bxc4 26.Rd2 Bb7 27.Qe2 Qc8 28.f3 Rg6 29.Qf2 Ra1 30.h4 h5 31.Rdd1 Qa8 32.Nxh5 d5 33.exd5 Bxd5 34.Ng3 Re6 35.h5 Bc6 36.Be3 Ra3 37.Rc1 Ra2 38.Re2 Ra1 39.Rxa1 Qxa1+ 40.Qe1 Qa3 41.Bf2 f6 42.Nf5 Be8 43.Ng3 Qb3 44.Bc5 Bxc5+ 45.bxc5 Qb5 46.Qf2 Ra6 47.Rd2 Qb1+ 48.Kh2 Kh7 49.Rd8 Bc6 50.Qf1 Qxf1 51.Nxf1 e4 52.fxe4 Bxe4 53.Ne3 Kh6 54.Rd4 Ba8 55.Rg4 Ra5 56.Nf5+ Kxh5 57.Rxg7 Ra2 58.Kh3 Bf3 59.g4+ Bxg4+ 60.Rxg4 1-0
Niemann,Hans Moke (2635) – Vakhidov,Jakhongir (2557) [D96]
International Tournament for the prizes Tashkent (8.4), 16.11.2021
1.d4 Nf6 2.c4 g6 3.Nc3 d5 4.Nf3 Bg7 5.Qb3 dxc4 6.Qxc4 0-0 7.Bf4 Bf5 8.e3 c6 9.Qb3 Qb6 10.Bc4 Nbd7 11.0-0 a5 12.Nh4 Be4 13.Nxe4 Nxe4 14.Nf3 Nd6 15.Be2 Qxb3 16.axb3 c5 17.Rfc1 cxd4 18.Rc7 Rfd8 19.exd4 Nf5 20.Rd1 Nf8 21.d5 Rd7 22.Rc2 Rad8 23.g4 Nh6 24.h3 Rxd5 25.Rxd5 Rxd5 26.Rc8 b5 27.Rb8 f5 28.Bxb5 Rd1+ 29.Kh2 fxg4 30.Bc4+ Kh8 31.Ng5 gxh3 32.Kxh3 e5 1-0
Chigaev,M (2626)-Yakubboev,N (2621) [B10]
Tashkent (10.4) 2021
1.e4 c6 2.Nf3 d5 3.d3 dxe4 4.dxe4 Qxd1+ 5.Kxd1 Bg4 6.Be2 Bxf3 7.Bxf3 e5 8.Nd2 Bc5 9.Ke2 Nd7 10.Nc4 Ngf6 11.Rd1 Ke7 12.a4 b5 13.Na5 Rhc8 14.g4 Bb6 15.Nb3 Nc5 16.Nxc5 Bxc5 17.g5 Ne8 18.Bg4 Rc7 19.axb5 cxb5 20.Rd5 f6 21.Bd2 Bd6 22.c3 Rc5 23.Rd3 Rc4 24.b3 Rxe4+ 25.Kf3 Rxg4 26.Kxg4 a5 27.c4 bxc4 28.bxc4 a4 29.f4 g6 30.fxe5 Bxe5 31.Re1 Kf7 32.Bb4 Ra7 33.h4 Ng7 34.c5 Ne6 35.Red1 f5+ 36.Kf3 Ra6 37.Rd7+ Kg8 38.Ba3 Rc6 39.R1d5 Bh2 40.Re7 Kf8 41.Rxh7 Ke8 42.Rh8+ Ke7 43.Ra8 Nc7 44.Rd6 Rxd6 45.cxd6+ Bxd6 46.Bxd6+ Kxd6 47.Rxa4 Ne6 48.Ra7 Ke5 49.Rd7 f4 50.Kg4 Nc5 51.Re7+ Ne6 52.Rf7 1-0
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PUZZLERS
WHITE TO MOVE, MATE IN TWO.
The key to our last problem is 1.a7!, and Black’s futile options are: 1…Nd7 2.Bd5#; 1…Na6 2.Qxa6#; 1…Nb7 2.axb8=N#; 1…Nf7/xe6 2.a8=Q#; 1…Kb7 2.a8=Q/B#; 1…Bb3/c2/xd1/b5 2.Q(x)b5#.
