I was thinking about Nc6, but after a long variation it comes down to a bishop+ knight vs. rook endgame, which is easily winning for black. Then I started looking at
1. Bh6+ Kxh6
2. Qf7 Ng7?? Losing move leading to mate in 7
3. Ng4+ Kh5
4. Qxg7 Rd3
5. Qh6+ Kxg4
6. Rxd3 Kf5
7. Qh3+ Ke4
8. Re3+ Kd5
9. Qe6# 1-0
or
3. Ng4+ Kh5
4. Qxg7 Kxg4
5. h3+ Kf5
6. Rde1 e5
7. Rxe5+ fxe5
8. Qxe5# 1-0
2. ... Rh8 forced.
3. Ng4+ Kh5
4. Ne3 Rxd1
5. g4+ Kh4
6. Rxd1 Qc8
7. Rd5 Ng7
8. Qxg7 Qf8
9. Qxf8 Rxf8
and black is clearly better.
The position comes from the source below. My God, the analysis makes Kasparov's analysis look like Chernev's.
http://www.chesscafe.com/text/dvoretsky71.pdf
And I received the following email from Mark Dvoretsky proving that Thud did not Blunder:
Dear Thud,
Thank you for your letter. You are right: I failed to find any advantage for White after both your defences. I believe, chances are about equal in both cases.
I) 4...e6! 5.Rde1!? (5.Rb1!?) 5...Rd2 6.Rf3 B:f4!, and now either 7.R:f4
Qd7(c7) 8.R:f6 Q:f7 9.R:f7 Nd6 10.Re7 - with four rooks on the board White
perhaps is not worse, or 7.Q:e6!? Qd7 (7...Kh6 8.R:f4 Kg7) 8.Qe4 Kh6
(8...g5?! 9.Rh3+ Q:h3 10.gh R:h2+ 11.Kg1 R:h3 12.Nf5 with White's advantage)
9.Q:f4+ Kg7 - White keeps fair compensation for a pawn.
2) 4...Rd1! 5.Rd1 Qc8! 6.Rd5+ Kh6 7.h3 f5 (7...Nc7 8.Ng4+ Q:g4 9.hg N:d5
10.g5+ fg 11.fg+ K:g5 12.Q:d5+ Kh6 13.Qc4!? is worse for Black) 8.R:f5
(8.Q:e7 B:f4 9.Qh4+ Kg7 10.Qe7+ Kh6 - perpetual) 8...gf 9.N:f5+ Q:f5 10.Q:f5
Kg7 - equality.
Best regards,
Mark Dvoretsky