Markovich wrote on 07/29/11 at 01:49:55:
Well then, after 5...d5 and 7...Qd5 8.c8 Qd6 9.Nxa7+ Bxd7 10.Qh5+ Kd8 11.Bxd7 Qxd7, what's the problem with Sokolov's 12.Nb5?
From my reply #103:
Stefan Buecker wrote on 07/27/11 at 10:53:37:
In my opinion Black should play
11...Kd8 (instead of g6), e.g. 12.Nb5 Nf6 13.Qe2 Bc5 14.0.0 Re8 15.Nc3 Ra6 +=.
There is some compensation for the two (!) pawns. Maybe not enough, but the chances are better than after 11...g6.
Edit: Black scores 50% (in 32 games) with 11...Kd8 and even 75% (6 games) with 13...Bc5. Surprisingly, Sokolov (who enters this position in the different move order 10.Qh5+ Kd8 11.Bxd7+ Qxd7 ...) considers only 13...c6 (with insufficient comp.), which looks illogical.
In the 12.Qa5 Ke8 13.Qe5+ Kf7 14.Nb5 c6 15.Qd4 Qg4 alternative suggested by MNb, after
16.Qb6 Be7 17.Qxb7 Nf6 18.Nd6+ Ke6 19.Qxc6 Bxd6 20.c5 Rhd8 21.cxd6 Black also has alternatives like
21...Rac8!? (instead of 21...Qxg2 22.d7+ +0.30) 22.Qa6 Qxg2 23.Qf1 Qf3 24.Rg1 Nh5 (or 24...Qf5, or 24...Nd5?!) 25.d7 Rc7 26.Qe2 (26.Qg2 Qxg2 27.Rxg2 Nf4 = [0.08]) 26...Qxe2+ 27.Kxe2 Rdxd7 28.Kf1 (28.Kd1 Rf7 29.Rf1 Nf4 = [+0.08]) 28...g6 29.Rg5 Rc2 30.b3 Rf7 31.Ke2 Nf4+ when Black should have enough play to hold the ending.
So in my opinion we have the choice between the diagram above and
16.Nc7 Ra5! 17.Qb6 Re5 18.Qxb7 Qd7 19.c5 (19.0-0 Re7 20.d3 with four pawns for the knight, but Black has active play; or 20.d4 resp. 20.Re1; or 19.a4, with unclear endings) 19...Bxc5 (19...Kg6? 20.b4 Nf6 21.Qb8) 20.Qb3+ Kg6 21.Na6 Ba7 (probably best)