SAM Shankland edged closer to winning the 3rd Prague International Masters Tournament by holding on to a half-point lead going to the final round of the Category 18 (average Elo: 2677) event.
The American grandmaster defeated erstwhile co-leader Nijat Abasov in Round 5, and then drew his penultimate round outing with tournament surprise Thai Dai Van Nguyen to maintain his slim lead over closest pursuers Radoslaw Wojtaszek and top seed Jan Krzysztof Duda, who both won their Round 6 outings. Wojtaszek whipped Abasov in 54 moves of a Queen’s Gambit, while Duda dumped Nils Grandelius in 36 moves of a French Defense. The other Round 6 match between Jorden Van Foreest and David Navara was drawn after 80 moves of an English Opening.
The important and crucial final round matches will pit Shankland against Van Foreest, while Wojtaszek and Duda square off. The 30-year old Shankland – the 2018 United States chess champion – only needs to draw to at least get a share of the championship, while both Wojtaszek and Duda will aim for a win and the chance to win the title themselves.
Standings after Round 6: 1. Shankland, 4½; 2-3. Wojtaszek, Duda, 4; 4-5. Nguyen, Abasov, 3; 6. Van Foreest, 2½; 7. Grandelius, 2; 8. Navara, 1.
Shankland, S (2691)-Abasov, N (2665) [E46]
Prague (5.2) 2021
1.d4 Nf6 2.c4 e6 3.Nc3 Bb4 4.e3 0-0 5.Bd2 d5 6.Nf3 b6 7.cxd5 exd5 8.Rc1 Bb7 9.Bd3 Be7 10.0-0 c5 11.dxc5 bxc5 12.Re1 Nbd7 13.e4 c4 14.Bb1 dxe4 15.Nxe4 Nxe4 16.Bxe4 Bxe4 17.Rxe4 Bf6 18.Bc3 Bxc3 19.Rxc3 Nb6 20.Qe2 Qd5 21.Re5 Qd6 22.g3 Rad8 23.h4 f6 24.Re4 Qd1+ 25.Kg2 Qxe2 26.Rxe2 Rd7 27.Rec2 Rfd8 28.Ra3 Kf7 29.Ra6 Rc7 30.Nd2 Rdd7 31.Ne4 Kg6 32.Ra5 Rd5 33.Ra3 Re5 34.Nd2 h5 35.Rac3 Ra5 36.a3 Ra4 37.Rc1 Kf5 38.b4 Kg6 39.Ne4 Nd5 40.Rf3 c3 41.Nxc3 Ra6 42.Ne2 Re7 43.Nd4 Re4 44.Nb5 Re2 45.Rd3 Ne7 46.Rc7 Rb6 47.Nxa7 1-0
Duda, J-K (2729)-Van Foreest, J (2701) [C80]
Prague (5.3) 2021
1.e4 e5 2.Nf3 Nc6 3.Bb5 a6 4.Ba4 Nf6 5.0-0 Nxe4 6.d4 Be7 7.Re1 b5 8.Rxe4 d5 9.Nxe5 Nxe5 10.Rxe5 bxa4 11.Qe2 Be6 12.f4 g6 13.Nc3 a3 14.b3 Bb4 15.Nxd5 0-0 16.Qe4 Bd6 17.Nc3 Bf5 18.Rxf5 gxf5 19.Qxf5 Re8 20.Ne4 Be7 21.Be3 Qc8 22.Ng3 Bh4 23.Qd3 Bxg3 24.hxg3 Qe6 25.Kf2 Qe4 26.Rd1 a5 27.g4 c6 28.Qxe4 Rxe4 29.g5 Rae8 30.Rd3 h6 31.gxh6 f5 32.c4 R4e6 33.d5 cxd5 34.cxd5 Rxh6 35.d6 Rd8 36.Bc5 Kf7 37.Rd5 Kf6 38.Bxa3 Rh4 39.Kf3 Rh1 40.Rxa5 Rf1+ 41.Ke3 Re8+ 42.Kd3 Rxf4 43.Bc5 Rg4 44.Bd4+ Ke6 45.Re5+ Kd7 46.Rxf5 Kxd6 47.a4 Rxg2 48.b4 Rg4 49.Bc5+ Ke6 50.Rh5 Reg8 51.Kc3 R8g5 52.Rh6+ Rg6 53.Rh8 Kd5 54.Rd8+ Kc6 55.a5 Rh4 56.Kb3 Rh1 57.Rc8+ Kb7 58.Rf8 Kc6 59.Bd4 Rd1 60.a6 Rb1+ 61.Ka4 Kd5 62.Bf6 Kc6 63.a7 1-0
Wojtaszek, R (2687)-Abasov, N (2665) [D38]
Prague (6.1) 2021
1.d4 Nf6 2.c4 e6 3.Nf3 d5 4.Nc3 Bb4 5.cxd5 exd5 6.Bg5 h6 7.Bh4 0-0 8.e3 Bf5 9.Qb3 Bxc3+ 10.bxc3 Nbd7 11.Be2 c5 12.dxc5 Nxc5 13.Qb4 Rc8 14.0-0 g5 15.Bg3 Qe7 16.Be5 Rfd8 17.Rad1 Nfe4 18.Bd4 Rd6 19.c4 Rb6 20.Qe1 dxc4 21.Bxc4 Be6 22.Be2 Rd6 23.Ne5 Bxa2 24.f3 Nf6 25.Qa5 Qe6 26.Qxa7 b6 27.Ba1 Rxd1 28.Rxd1 Qb3 29.Kf1 Re8 30.Bd4 Qe6 31.Ra1 Bd5 32.e4 Nb3 33.exd5 Qxd5 34.Bc3 Nxa1 35.Bc4 Qd1+ 36.Be1 Nd7 37.Bxf7+ Kf8 38.Bxe8 Kxe8 39.Nxd7 Qd3+ 40.Kf2 Qd4+ 41.Ke2 Qc4+ 42.Ke3 Nc2+ 43.Kd2 Nxe1 44.Nf6+ Kf8 45.Qb8+ Ke7 46.Qe5+ Kf7 47.Kxe1 Qc1+ 48.Kf2 Qd2+ 49.Kg3 Qf4+ 50.Qxf4 gxf4+ 51.Kxf4 Kxf6 52.Ke4 Ke6 53.g4 b5 54.f4 1-0
Grandelius, N (2670)-Duda, J-K (2729) [C02]
Prague (6.4) 2021
1.e4 e6 2.d4 d5 3.e5 c5 4.c3 Nc6 5.Nf3 Qb6 6.Be2 Nh6 7.Bxh6 gxh6 8.Qd2 Bd7 9.0-0 0-0-0 10.Na3 Rg8 11.dxc5 Bxc5 12.b4 Bf8 13.Nb5 Nxb4 14.Nxa7+ Qxa7 15.cxb4 Rg4 16.a3 Ba4 17.h3 Rg7 18.Rfc1+ Kb8 19.Qc3 Qb6 20.Nd4 Be7 21.Qe3 Rdg8 22.Bf1 Bg5 23.f4 Bd8 24.Kh1 Rg3 25.Qd2 Qa7 26.Rc3 Rxc3 27.Qxc3 Bb6 28.Nf3 Rc8 29.Qe1 Bf2 30.Qb1 Bc2 31.Qb2 Be3 32.a4 Be4 33.Qe2 Bxf4 34.a5 Rc2 35.Qd1 Qf2 36.a6 Be3 0-1
*******
PUZZLERS
WHITE TO MOVE,
MATE IN TWO.
The key to our last problem is 1.b8=N!, and Black’s futile options are: 1…Kd8, Rf8 2.Qd7#; 1…Kf8, Bd8 2.Qf7#; 1…Rg8/h7/h6/h5/h4/h3/h2/h1 2.Q(x)g8#; 1…Bd6/e5/f4/g3/h2/xb8 2.Qc8#./PN
