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Very Hot Topic (More than 25 Replies) Two knights as black (Read 33172 times)
Stefan Buecker
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Re: Two knights as black
Reply #33 - 05/18/09 at 06:31:16
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MNb wrote on 05/17/09 at 20:26:24:
I suppose I should consult Kaissiber again for the zillionth time, but 15.a3 does not look like White's best to me - either 15.Qf3 at once or 15.Qh5.


Here is the full paragraph from issue 28 (a few posts ago I wrote #30, which was a mistake): "15.a3!? (15.Qf3 Kf7 16.Qh5+ costs time) 15...h6! (15...a5 16.Qf3 Kf7 17.Qh5+ g6 18.Qh6 Qf6 19.h3! Rhb8 20.Qxh7+ Qg7 21.Qh4 Rxb2 22.Nd5! =) 16.Qf3 Kg8 17.Qxc6 Kh7 =+ Bücker/Wind."
  
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Seth_Xoma
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Re: Two knights as black
Reply #32 - 05/18/09 at 04:11:39
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It seems like this thread would be a nice place to include the following game. Nakamura just smashed Friedel's Two Knight's Defense in the last round to win the US Championship. I am not familiar with the line, but it appeared that Josh was ill-prepared, and Nakamura was well-prepared. I remember hearing Emil Sutovsky on chess fm saying he was not impressed with 12...Rb4. Certainly the rook led a sad life.

[Event "US Championship"]
[Site "Saint Louis USA"]
[Date "2009.05.17"]
[EventDate "2009.05.08"]
[Round "9"]
[Result "1-0"]
[White "Hi Nakamura"]
[Black "J Friedel"]
[ECO "C58"]
[WhiteElo "2701"]
[BlackElo "2516"]
[PlyCount "44"]

1. e4 e5 2. Nf3 Nc6 3. Bc4 Nf6 4. Ng5 d5 5. exd5 Na5 6. Bb5+ c6 7. dxc6
bxc6 8. Bd3 Be7 9. Nc3 O-O 10. O-O Rb8 11. h3 c5 12. b3 Rb4 13. Re1 Bb7 14.
Ba3 Rf4 15. g3 Rd4 16. Nf3 Rxd3 17. cxd3 Qxd3 18. Nxe5 Qf5 19. g4 Qf4 20.
d4 Rd8 21. Qe2 Rxd4 22. Bc1 1-0

I winced in pain after the last move was played.
  
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MNb
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Re: Two knights as black
Reply #31 - 05/17/09 at 20:26:24
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I suppose I should consult Kaissiber again for the zillionth time, but 15.a3 does not look like White's best to me - either 15.Qf3 at once or 15.Qh5.
  
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Stefan Buecker
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Re: Two knights as black
Reply #30 - 05/17/09 at 15:06:01
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MNb wrote on 05/17/09 at 13:34:19:
Stefan Buecker wrote on 05/16/09 at 20:52:49:

By the way, 8...Qd8 was studied on 30 pages in Kaissiber #30 (2007) by Lev Gutman. After 9.Rxe4+ Be7 10.Nxd4 f5 Gutman doesn't recommend 11.Bh6, invented in 1920 by Carl Hartlaub, because of 11...Kf8! 12.Nxc6 Qxd1+ 13.Rxd1 fxe4 14.Nxe7 gxh6 which is slightly better for Black.


You don't think the exchange sac 12.Rxe7 Nxe7 13.Be3 correct? Black has problems finishing his/her development.

To quote from Kaissiber 28: "12.Rxe7 is playable, but doesn't equalize either." The main line goes: 12...Qxe7 13.Nxc6 bxc6 14.Be3 Be6 15.a3 h6! 16.Qf3 Kg8 17.Qxc6 Kh7 =+, analysis by Bücker and Maurits Wind. - That said, I admit that the whole variation, including the exchange sacrifice, might be a good surprise weapon. OTB the shock effect of such a move (Rxe7!??) should not be underestimated and should make up for a sacrifice of 0.2 in "Rybka Rating".
  
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MNb
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Re: Two knights as black
Reply #29 - 05/17/09 at 13:34:19
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Stefan Buecker wrote on 05/16/09 at 20:52:49:

By the way, 8...Qd8 was studied on 30 pages in Kaissiber #30 (2007) by Lev Gutman. After 9.Rxe4+ Be7 10.Nxd4 f5 Gutman doesn't recommend 11.Bh6, invented in 1920 by Carl Hartlaub, because of 11...Kf8! 12.Nxc6 Qxd1+ 13.Rxd1 fxe4 14.Nxe7 gxh6 which is slightly better for Black.


You don't think the exchange sac 12.Rxe7 Nxe7 13.Be3 correct? Black has problems finishing his/her development.

SWJediknight wrote on 05/16/09 at 22:35:08:
I agree with the above except that in the Anti-Lange line, the suggestion 8...Qh5 9.Nxe4 Be6 10.Bg5 Bb4 11.c3 dxc3 12.bxc3 Ba5 13.Ng3 is quite interesting (instead of 13.h4?! Qg4!- I've looked at that line with Fritz and it appears to refute 13.h4 outright).

After 13...Qg6 14.h4 play has transposed into the old 13.h4 line but with Black unable to answer it with ...Qg4.  Michael Goeller (Urusov) gives 14...h6 15.h5 Qh7 16.Qb3 with compensation, though I'm not sure White has enough after 16...0-0 17.Rxe6 fxe6 18.Qxe6+ Rf7!.  Fritz instead suggests chugging around with the knight on move 14 with Ne2-f4, and thinks White has enough compensation.  I don't entirely trust this for White but it's worth a look.


I certainly don't trust this for White. He/she should be satisfied with a minute edge after 11.Nxd4 Qxd1 12.Rxd1 Nxd4 13.Rxd4 Ba5 14.Nc5 Bb6 15.Rad1 0-0 16.Nxe6 fxe6 17.Rd2. Sure Black can hold this, but on the level this line is meant for White has still something to play for: the weakling on e6.
  
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SWJediknight
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Re: Two knights as black
Reply #28 - 05/16/09 at 22:35:08
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I agree with the above except that in the Anti-Lange line, the suggestion 8...Qh5 9.Nxe4 Be6 10.Bg5 Bb4 11.c3 dxc3 12.bxc3 Ba5 13.Ng3 is quite interesting (instead of 13.h4?! Qg4!- I've looked at that line with Fritz and it appears to refute 13.h4 outright).

After 13...Qg6 14.h4 play has transposed into the old 13.h4 line but with Black unable to answer it with ...Qg4.  Michael Goeller (Urusov) gives 14...h6 15.h5 Qh7 16.Qb3 with compensation, though I'm not sure White has enough after 16...0-0 17.Rxe6 fxe6 18.Qxe6+ Rf7!.  Fritz instead suggests chugging around with the knight on move 14 with Ne2-f4, and thinks White has enough compensation.  I don't entirely trust this for White but it's worth a look.

At GM level they have rightly given up on these 5.0-0 Nxe4 lines but I think at club level there's still scope for either side to play for a win- even after 8...Qh5, 10...Bd6.

Btw after 8...Qa5 9.Nxe4 Be6 I think 10.Neg5 is definitely the way for White to go.  10.Bd2 leads to combative play after 10...Bb4 or the risky 10...Qa4, but 10...Qh5 leaves White with nothing better than 11.Bg5 transposing to 8...Qh5 (Emms) while 10...Qd5 is more drawish than the ...Qh5 lines: 11.Bg5 Bd6 and after 12.Bf6 0-0 or 12.c3 0-0 a mass exchange on d4 ensues.
  
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Stefan Buecker
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Re: Two knights as black
Reply #27 - 05/16/09 at 20:52:49
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http://www.chesspub.com/cgi-bin/yabb2/YaBB.pl?num=1127771902/16#16

By the way, 8...Qd8 was studied on 30 pages in Kaissiber #30 (2007) by Lev Gutman. After 9.Rxe4+ Be7 10.Nxd4 f5 Gutman doesn't recommend 11.Bh6, invented in 1920 by Carl Hartlaub, because of 11...Kf8! 12.Nxc6 Qxd1+ 13.Rxd1 fxe4 14.Nxe7 gxh6 which is slightly better for Black. He prefers 11.Rf4 0-0 12.Nc6 Qxd1+ 13.Nxd1 bxc6 14.Ra4! Re8 15.Bd2. White's position is easier to play.
  
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MNb
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Re: Two knights as black
Reply #26 - 05/16/09 at 20:51:52
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If you insist, but you won't find more than those two game references. I can give you the two gamescores as well, but that will be tomorrow evening.
  
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BPaulsen
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Re: Two knights as black
Reply #25 - 05/16/09 at 20:41:10
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MNb wrote on 05/16/09 at 20:37:52:
BPaulsen wrote on 05/16/09 at 20:21:51:

I think 5.e5 offers better chances than 5.0-0 just because Nxe4 is completely sterile. That being said, it's like splitting hairs.


Aha! But in some other thread JediKnight and I have tried to find some ways to avoid sterility.
a) 8...Qa5 follow Engelbert-Kreutzkamp, Hamburg 1993.
b) 8...Qd8 8.Rxe4+ Be7 9.Nxd4 f5 10.Bh6 is a forced draw after some amazing complications.
c) 8...Qh5 9.Nxe4 Be6 10.Bg5 Bd6 (Bb4 might give White a minute endgame edge due to a weak pawn on e6)  follow Chan Yi Ren-Hasenohr, Vung Tau City 2008 and Yuksel-Yargici, Antala 2007.
Positions with opposite coloured bishops and all the heavy artillery still on the board are not sterile at all.


Could you link me to the thread? I'm definitely interested, since especially 8...Qh5 makes it boring. 8...Qa5/8...Qd8 never seemed half as sterile.
  
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MNb
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Re: Two knights as black
Reply #24 - 05/16/09 at 20:37:52
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BPaulsen wrote on 05/16/09 at 20:21:51:

I think 5.e5 offers better chances than 5.0-0 just because Nxe4 is completely sterile. That being said, it's like splitting hairs.


Aha! But in some other thread JediKnight and I have tried to find some ways to avoid sterility.
a) 8...Qa5 follow Engelbert-Kreutzkamp, Hamburg 1993.
b) 8...Qd8 8.Rxe4+ Be7 9.Nxd4 f5 10.Bh6 is a forced draw after some amazing complications.
c) 8...Qh5 9.Nxe4 Be6 10.Bg5 Bd6 (Bb4 might give White a minute endgame edge due to a weak pawn on e6)  follow Chan Yi Ren-Hasenohr, Vung Tau City 2008 and Yuksel-Yargici, Antala 2007.
Positions with opposite coloured bishops and all the heavy artillery still on the board are not sterile at all.
  
The book had the effect good books usually have: it made the stupids more stupid, the intelligent more intelligent and the other thousands of readers remained unchanged.
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BPaulsen
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Re: Two knights as black
Reply #23 - 05/16/09 at 20:21:51
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MNb wrote on 05/16/09 at 20:11:47:
BPaulsen wrote on 05/15/09 at 22:48:55:


I'm pretty sure 8. Be3 0-0 9. Nxc6 bxc6 10. Bxc5 Nxc5 11. Bxc6 Rb8 [if instead 11...Ba6 12. Nc3! (Estrin) d4! 13. Bxa8 +=] 12. Qxd5 Qe7 13. 0-0 Rxb2 13. Nc3 Ba6 14. Rfc1 is += (Jiangchuan - Svidler, Shanghai 2001)


And I think this is equal. In fact 12.0-0 might pose more problems.


Thanks for your response. 12. 0-0 is interesting as well.

MNb wrote on 05/16/09 at 20:11:47:
14...f6 15.f4 fxe5 16.fxe5 Rae8 Weeramantry-Brooks, 1994. White has nothing.
As White cannot really do without castling I am not sure if 12.Qc3 and 12.Nb3 have much point. But I admit immediately that I have never looked at it properly.
Of course it is still a whole game to play.  But from a purely theoretical point of view it is hard to maintain that 5.e5 offers better chances than 5.0-0 or 4.Ng5. And that's the only point I want to make.


I think 5.e5 offers better chances than 5.0-0 just because Nxe4 is completely sterile. That being said, it's like splitting hairs.

4. Ng5 is definitely better, but given that the 5.e5 variation can come up through different move orders, it's worth looking into a little.

I wonder if 12. Nb3 Bb6 13. 0-0-0 is worth anything, at least Sveshnikov ventured it, even though the game continuation was unimpressive.
  
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MNb
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Re: Two knights as black
Reply #22 - 05/16/09 at 20:11:47
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BPaulsen wrote on 05/15/09 at 22:48:55:


I'm pretty sure 8. Be3 0-0 9. Nxc6 bxc6 10. Bxc5 Nxc5 11. Bxc6 Rb8 [if instead 11...Ba6 12. Nc3! (Estrin) d4! 13. Bxa8 +=] 12. Qxd5 Qe7 13. 0-0 Rxb2 13. Nc3 Ba6 14. Rfc1 is += (Jiangchuan - Svidler, Shanghai 2001)


And I think this is equal. In fact 12.0-0 might pose more problems.

BPaulsen wrote on 05/15/09 at 22:48:55:

10...Nxd2 11. Qxd2 0-0 12. 0-0 (12. Qc3!?, 12. Nb3!? See below) Qe7 13. Qc3 Bb6 14. Nb3 unclear. I'd rather have white here though, as black I wouldn't want this, given that 12...Re8 seems to be just plain better.


14...f6 15.f4 fxe5 16.fxe5 Rae8 Weeramantry-Brooks, 1994. White has nothing.
As White cannot really do without castling I am not sure if 12.Qc3 and 12.Nb3 have much point. But I admit immediately that I have never looked at it properly.
Of course it is still a whole game to play.  But from a purely theoretical point of view it is hard to maintain that 5.e5 offers better chances than 5.0-0 or 4.Ng5. And that's the only point I want to make.
  
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Re: Two knights as black
Reply #21 - 05/15/09 at 22:48:55
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MNb wrote on 05/15/09 at 21:42:42:
Black, if prepared, may try 8.Be3 0-0.
Instead of 10...Qh4 I think 10...Nxd2 11.Qxd2 0-0 better, though I am not sure if I would prefer 12.0-0 Re8 or 12...Qe7.


I'm pretty sure 8. Be3 0-0 9. Nxc6 bxc6 10. Bxc5 Nxc5 11. Bxc6 Rb8 [if instead 11...Ba6 12. Nc3! (Estrin) d4! 13. Bxa8 +=] 12. Qxd5 Qe7 13. 0-0 Rxb2 13. Nc3 Ba6 14. Rfc1 is += (Jiangchuan - Svidler, Shanghai 2001) I'm not aware of any improvements for black, either. 

10...Nxd2 11. Qxd2 0-0 12. 0-0 (12. Qc3!?, 12. Nb3!? See below) Qe7 13. Qc3 Bb6 14. Nb3 unclear. I'd rather have white here though, as black I wouldn't want this, given that 12...Re8 seems to be just plain better.

12...Re8 (most accurate if black plays 10...Nxd2 with white responding 12. 0-0 from what I can tell) and white seems to have no better than 13. Qc3 Bb6 14. Nxc6 with a sterile equality (maybe the most microscopic endgame advantage). However, 13. Nb3!? Bb6 14. Qc3 is worth consideration because ...f6 now comes at a cost of weakning black's pawn structure.

12. Qc3!? Bb6 13. Nb3 (12. Nb3 can transpose) f6 is equal/unclear, but I'd rather have black. The ...f6 break does not compromise his pawn structure further.

12. Nb3!? Bb6 13. 0-0 f6! (13...Qe7 would transpose to the Qe7 line above after 14. Qc3, 13...Re8 14. f4 f6 transposes ) 14. f4 Re8 15. Qc3 fxe5 16. fxe5 Qg5! with an ending that only black could win. White could also try 14. Qc3 f6, but black should be at least equal.

I guess white could also consider 10. 0-0, but black should equalize more easily after 10...Qe7! 11. Re1 0-0 12. f3 Ng5 13. Qd2 f6 14. Nc3, and white has got nothing from the opening.

I guess if white's playing for an edge against black's best play (in my opinion) the 12...Re8 13. Nb3!? Bb6 14. Qc3 deserves attention.
« Last Edit: 05/16/09 at 00:04:44 by BPaulsen »  
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MNb
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Re: Two knights as black
Reply #20 - 05/15/09 at 21:42:42
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Black, if prepared, may try 8.Be3 0-0.
Instead of 10...Qh4 I think 10...Nxd2 11.Qxd2 0-0 better, though I am not sure if I would prefer 12.0-0 Re8 or 12...Qe7.
  
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Re: Two knights as black
Reply #19 - 05/15/09 at 21:16:16
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How does that line usually continue after 8. Be3 Bd7 9. Bxc6 bxc6 10. Nd2 Qh4?

11. 0-0 seems to score well (including a fairly recent win of Nakamura over Hebden from 2008). I'm guessing 11...0-0 is most accurate, or maybe 11...Bb6?

11. Nxe4 Qxe4 12. 0-0 Bb6 13. Re1 Qg6 14. Nb3 Bg4 15. f3 Bh3 16. Qd2 d4! seems equal.

There's also 14...0-0 15. Bc5 Rfe8 16. Re3 or 16. a4 with unclear chances, although I'd rather have white.
  
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