By: A.E. Boy Espejo Jr.
A virtually obscure grandmaster from India, whose only claim to big-time tournament fame was tying for first at the 2005 World Open, is likely to win the 2019-20 Hastings Masters in England.
GM Magesh Chandran Panchanathan, ranked just 10th in the 124-player, 9-round Swiss event, has 7 points and also a full point ahead of his nearest pursuers with only a round to go. The 37-year old Madurai native has beaten top seed GM David Howell, drew with 2nd seed GM Romain Edouard, and defeated 3rd seed GM Erik Van den Doel and 4th seed GM Deep Sengupta. He will face compatriot 6th seed GM G.A. Stany in the final round with worst that could happen to him is tying for first if he loses.
Sharing 2nd to 5th places are GM Gergely Kantor, Edouard, IM Mate Bagi, and Stany who tallied identical 6 points apiece. Eight players that included Van Den Doe and 9th seed GM Oleg Korneev are bunched in a tie for 8th to 13th places with zero chance of winning the tournament.
Panchanathan,MC (2479)-Howell,DWL (2676) [A10]
Hastings (4.1) 2019
1.c4 b6 2.b3 Bb7 3.Bb2 Nf6 4.e3 g6 5.Nf3 Bg7 6.Be2 0-0 7.0-0 c5 8.d4 d6 9.Nc3 e6 10.Qc2 Nc6 11.Rfd1 Qe7 12.a3 Rfd8 13.d5 exd5 14.cxd5 Ne5 15.Ne1 a6 16.a4 Re8 17.Rab1 h5 18.Ba1 Rac8 19.Bc4 h4 20.h3 Nfd7 21.Qe2 Ra8 22.Nc2 f5 23.f4 Nxc4 24.bxc4 a5 25.Re1 Ba6 26.Qd3 Nf6 27.Rxb6 Nh5 28.Nb5 Bxb5 29.cxb5 Bxa1 30.Rxa1 Ng3 31.Rc6 g5 32.fxg5 Rad8 33.b6 Qxg5 34.Rc7 Re7 35.Rb1 Rb8 36.Rxe7 Qxe7 37.Na3 Qe8 38.Nc4 Qxa4 39.Nxd6 Qd7 40.Nc4 a4 41.d6 Ne4 42.b7 Kh7 43.Rb6 a3 44.Nxa3 Qg7 45.Nc4 Rg8 46.Rb2 1-0
Panchanathan,MC (2479)-Van den Doel,E (2589) [C77]
Hastings (6.1) 2020
1.e4 e5 2.Nf3 Nc6 3.Bb5 a6 4.Ba4 Nf6 5.Qe2 b5 6.Bb3 Be7 7.c3 d6 8.0-0 0-0 9.Rd1 Na5 10.Bc2 c5 11.d4 Qc7 12.d5 Bd7 13.b3 g6 14.Nbd2 Nh5 15.Nf1 Bf6 16.g3 Bg7 17.Ne1 Qc8 18.Ne3 Qe8 19.Qf3 Bh6 20.N1g2 Ng7 21.g4 Qd8 22.Bd2 Bg5 23.Kf1 Nb7 24.Ke2 c4 25.b4 a5 26.a3 h5 27.gxh5 Nxh5 28.Rg1 Qf6 29.Nf5 axb4 30.Bxg5 Qxg5 31.h4 Qf6 32.axb4 Kh8 33.Ng3 Qxf3+ 34.Kxf3 Nf6 35.Ne3 Kh7 36.Ke2 Kh6 37.Kd2 Rfb8 38.Rh1 Nd8 39.Rag1 Ra2 40.h5 Nxh5 41.Nef5+ Bxf5 42.exf5 Nb7 43.fxg6 fxg6 44.Nxh5 Rxc2+ 45.Kxc2 gxh5 46.Rg2 Rf8 47.Rhg1 Rf6 48.Rg8 Rxf2+ 49.Kd1 Rf7 50.Rh8+ Rh7 51.Rgg8 h4 52.Ke2 h3 53.Kf2 e4 54.Rxh7+ Kxh7 55.Rb8 e3+ 56.Kg3 Kg6 57.Rxb7 Kf6 58.Rh7 1-0
Sengupta,D (2564)-Panchanathan,MC (2479) [C59]
Hastings (8.1) 2020
1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.Ng5 d5 5.exd5 Na5 6.Bb5+ c6 7.dxc6 bxc6 8.Be2 h6 9.Nh3 Bc5 10.d3 0-0 11.0-0 Nb7 12.Nc3 Bd4 13.Kh1 Nd6 14.Ng1 Re8 15.Na4 Nf5 16.Nf3 Ba6 17.Nxd4 Qxd4 18.c3 Qd6 19.Be3 Rad8 20.Bxa7 e4 21.d4 Bxe2 22.Qxe2 h5 23.Bc5 Qf4 24.Nb6 Qh4 25.Kg1 Re6 26.Nc4 Nd5 27.g3 Qh3 28.Kh1 h4 29.Rg1 Rh6 30.Qf1 Qxh2+ 0-1
Flear,GC (2440)-Kantor,G (2541) [E12]
Hastings (8.3) 2020
1.d4 Nf6 2.c4 e6 3.Nf3 b6 4.Bf4 Bb4+ 5.Nfd2 Nh5 6.Bg3 Bb7 7.a3 Bf8 8.Nc3 Nxg3 9.hxg3 g6 10.e3 Bg7 11.Qc2 d6 12.Rh2 c5 13.dxc5 dxc5 14.0-0-0 Qe7 15.Be2 Nc6 16.Nde4 0-0 17.g4 f5 18.gxf5 exf5 19.Ng3 Ne5 20.Nd5 Bxd5 21.Rxd5 Rad8 22.Rxd8 Rxd8 23.Rh1 Qb7 24.f3 b5 25.cxb5 c4 26.Bxc4+ Kh8 27.Rd1 Rc8 28.b3 Qe7 29.Kd2 Nxc4+ 30.bxc4 Qd6+ 31.Ke2 Qxg3 32.Kf1 h5 33.c5 Qe5 0-1
Wadsworth,MJ (2401)-Edouard,R (2653) [A04]
Hastings (7.2) 2020
1.Nf3 c5 2.b3 d5 3.Bb2 f6 4.c4 d4 5.b4 e5 6.bxc5 Bxc5 7.d3 Nc6 8.g3 f5 9.Bg2 e4 10.Nfd2 e3 11.fxe3 Nf6 12.Bxc6+ bxc6 13.exd4 Bxd4 14.Bxd4 Qxd4 15.Nb3 Qd6 16.Qd2 0-0 17.Nc3 a5 18.Na4 Ng4 19.c5 Qe7 20.0-0 Be6 21.Nd4 Bd5 22.Qf4 Rad8 23.Nb6 g5 24.Nxf5 Qxc5+ 25.Qd4 Qxd4+ 26.Nxd4 Rxf1+ 27.Rxf1 Ne3 28.Nxd5 Nxf1 29.Ne7+ Kh8 30.Ndf5 Nd2 31.Nxc6 Rd5 32.Nfd4 Rc5 33.a4 Nb1 34.e4 Nc3 35.e5 Kg8 36.Ne7+ Kf7 37.Ndc6 Nxa4 38.d4 Rc1+ 39.Kf2 Nb6 40.Ke3 a4 41.Kd2 Rc4 42.Kd3 a3 43.d5 a2 44.e6+ Kf8 45.Nf5 Rxc6 0-1
*******
PUZZLERS
WHITE TO MOVE,
MATE IN TWO.
The key to our last problem is 1.Ne6!, intending 2.Ng5#. Black’s futile options are: 1…Bxe3 2.Qd4#; 1…Qxc6 2.Qg2#; 1…Rxe3 2.Qd3#./PN
