THE 4th Sharjah Masters ended Sept. 25 with four grandmasters – Pavel Eljanov, M. Amin Tabatabaei, A.R. Saleh Salem, and Parham Magsoodloo – tying for first place, each finishing with identical 6½ points apiece.
However, by virtue of a better tiebreak, Eljanov – the current number four player of Ukraine – clinched the 4th Sharjah Cup, a gold medal, and the $10,000 champion’s purse. Iran’s Tabatabaei has the best tiebreak quotient of the four, but Eljanov beat him in Round 7, and the first tie-break criteria will be the result of their direct encounter gave the nod to the 38-year-old Eljanov. Tabatabaei settled for the silver and $8,000 prize money, Salem pocketed the bronze and $7,000, while Maghsoodloo went home with $5,000.
Meanwhile, five GMs shared 5th to 9th places with identical six points apiece: Mikheil Mchedlishvili, Rinat Jumabayev, S.L. Narayanan, Nodirbek Yakubboev, and Abhijeet Gupta, who took home $4,500, $4,000, $3,500, $3,000, and $2,500, respectively, after tiebreaks. GM Baskaran Adhiban led an eight-man group of grandmasters who finished with 5½ points apiece and shared 10th to 17th places.
Filipino IM Oliver Dimakiling wound up in a share of 26th to 38th places with 4½ points and out of the money circle, as only the top 15 finishers are given prize money.
Eljanov, P (2680)-
Tabatabaei, M.A (2624) [D71]
Sharjah (7.1) 2021
1.d4 Nf6 2.c4 g6 3.g3 c6 4.Bg2 d5 5.Nf3 Bg7 6.b3 dxc4 7.bxc4 c5 8.Bb2 cxd4 9.Nxd4 0-0 10.Nd2 e5 11.Nb5 a6 12.Nc3 Nc6 13.0-0 Be6 14.Rb1 Bh6 15.e3 Bg4 16.Bf3 Bxf3 17.Nxf3 e4 18.Nd2 Re8 19.Ba1 Bg7 20.Kg2 Qa5 21.Rxb7 Rad8 22.Qe2 h5 23.Nd5 Nxd5 24.Nb3 Nf4+ 25.gxf4 Qf5 26.Bxg7 Kxg7 27.h3 Re6 28.Rd1 Rd3 29.Qb2+ Kg8 30.Rd7 g5 31.Rd5 Qh7 32.f5 Re8 33.R1xd3 exd3 34.Qf6 Ne7 35.Qxg5+ Kh8 36.Qf6+ Kg8 37.Rxd3 Nxf5 38.Rd5 Qg6+ 39.Qxg6+ fxg6 40.Nd4 Ne7 41.Rd7 a5 42.c5 a4 43.c6 Kf7 44.c7 Kf8 45.Ne6+ 1-0
Yakubboev, N (2604) –
Nisipeanu, L-D (2657) [A13]
Sharjah (7.7) 2021
1.c4 e6 2.d4 d5 3.Nf3 dxc4 4.e4 b5 5.a4 Bb7 6.axb5 Bxe4 7.Bxc4 Nf6 8.Nc3 Bb7 9.0-0 Be7 10.Ne5 0-0 11.Qe2 c6 12.Nxf7 Rxf7 13.Qxe6 Qf8 14.Re1 cxb5 15.Nxb5 Nc6 16.Bg5 Bb4 17.Bxf6 gxf6 18.Re3 Ne7 19.Rb3 Bd5 20.Bxd5 Nxd5 21.Qxd5 Rd8 22.Qc4 a5 23.Na7 Kh8 24.Nc6 Rc7 25.d5 Qd6 26.Rd3 Re8 27.g3 Re2 28.b3 Bc5 29.Rxa5 Bxf2+ 30.Kf1 1-0
Tabatabaei, M.A (2624)-
Naiditsch, A (2663) [C65]
Sharjah (8.3) 2021
1.e4 e5 2.Nf3 Nc6 3.Bb5 Nf6 4.d3 Bc5 5.Bxc6 dxc6 6.Nbd2 Be6 7.0-0 Nd7 8.Nb3 Bxb3 9.axb3 0-0 10.Qe2 Re8 11.Bg5 Be7 12.Be3 Bf6 13.g3 g6 14.b4 Nf8 15.Nd2 Bg7 16.c3 Qd7 17.Nc4 f5 18.Ra5 a6 19.Re1 Qf7 20.f3 Nd7 21.Kg2 h5 22.h4 Re6 23.Raa1 Bf8 24.g4 Kh7 25.exf5 gxf5 26.Bg5 b5 27.Ne3 fxg4 28.fxg4 e4 29.Rf1 Qg8 30.d4 hxg4 31.Qxg4 Rae8 32.Qh5+ Bh6 33.Rf7+ 1-0
Dimakiling, O (2409)-
Fawzy, A (2488) [A05]
Sharjah (8.21) 2021
1.Nf3 Nf6 2.g3 b5 3.Bg2 Bb7 4.Na3 b4 5.Nc4 e6 6.0-0 d5 7.Na5 Ba6 8.a3 bxa3 9.d3 Bb4 10.Nb3 0-0 11.bxa3 Bc3 12.Rb1 Qd6 13.Bf4 Qxa3 14.Bxc7 Nc6 15.Bf4 Nb4 16.Bd6 Rfd8 17.Bc5 Qa4 18.Ra1 Qc6 19.Nfd4 Qc8 20.Be7 Rd7 21.Bxf6 gxf6 22.Ra4 Rb8 23.e3 Rc7 24.Qg4+ Kh8 25.Qh4 Qd8 26.Ne2 Nxc2 27.Rxa6 Rxb3 28.Nxc3 Rbxc3 29.Ra4 Qe7 30.Rb1 Na3 31.Rbb4 d4 32.Rb8+ Rc8 33.Be4 1-0
Maghsoodloo, P (2695)-
Narayanan, S.L (2632) [A36]
Sharjah (9.2) 2021
1.c4 c5 2.g3 g6 3.Bg2 Bg7 4.Nc3 Nc6 5.a3 Nf6 6.Rb1 d6 7.b4 Bf5 8.d3 Qc8 9.h3 0-0 10.b5 Nd8 11.e3 e5 12.Nf3 Bd7 13.Nd2 Ne6 14.a4 Rb8 15.Nd5 Qd8 16.0-0 Ne8 17.Bb2 f5 18.f4 exf4 19.exf4 Bxb2 20.Rxb2 Nf6 21.Nb3 Nxd5 22.Bxd5 Qf6 23.Re2 Kg7 24.Rfe1 Rfe8 25.Na5 b6 26.Nc6 Ra8 27.a5 Kf8 28.Qa1 Qxa1 29.Rxa1 Nc7 30.Rxe8+ Bxe8 31.Bf3 d5 32.Bxd5 Nxd5 33.cxd5 Rc8 34.axb6 axb6 35.Ra7 1-0
Salem, A.R. S (2679)-
Abdusattorov, N (2648) [D32]
Sharjah (9.4) 2021
1.d4 d5 2.c4 e6 3.Nc3 c5 4.cxd5 cxd4 5.Qa4+ Bd7 6.Qxd4 exd5 7.Nf3 Nf6 8.e3 Nc6 9.Qd3 Bg4 10.Be2 Bd6 11.0-0 0-0 12.Rd1 Qe7 13.h3 Bxf3 14.Bxf3 Rad8 15.Nxd5 Nxd5 16.Bxd5 Nb4 17.Qc4 b5 18.Qxb5 Qe5 19.f4 Qh5 20.Rd2 Rb8 21.Qc4 Rfc8 22.Qe4 Bc5 23.Kh2 Bxe3 24.g4 Qg6 25.Qxe3 Nc2 26.Qe5 Nxa1 27.b3 Nc2 28.Be4 Ne1 29.Bb2 f6 30.Qe6+ 1-0
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PUZZLERS
WHITE TO MOVE, MATE IN TWO.
The key to our last problem is 1.Qd6!, and Black’s futile options are: 1…Rd5/xe5/b4 2.Q(x)d5#; 1…Qxe5/xh4 2.Qg6#; 1…Bc3/b6/a7/xe5 2.N(x)c3#; 1…Nxe5 2.Bc2#./PN
