Summation of Threads---4.Ng5 Two Knights' Defense

The main line of the Fritz can be thought of as 4.Ng5 d5 5.exd5 Nd4 6.c3 b5 7.Bf1 Nxd5 8.Ne4 & the main tries are the Berliner Variation 8...Qh4 9.Ng3 Bg4 10.f3 e4 (The Berliner Gambit) 11.cxd4 Bd6 12.Qe2! (The Muir Variation). This is the winning move. 12...Bxb5ch is weaker.

First of all what doesn't work is 12...Bxg3ch? 13.hxg3 Qxh1 14.Qxb5ch +-

Black could also try 12...O-O, but is down too much material after 13.fxg4 Bxg3ch 14.Kd1 Nb4 15.Nc3 Rfe8 16.a3 Nd3 17.Kc2 c5 18.Nxe4 c4 19.Kb1 Rad8 20.Qf3 Nxc1 21.Kxc1 Rxd4 22.Nxg3 Re1ch 23.Kc2 Rxa1 24.Bxc4! Rxc4ch 25.Kb3+-

The second try by Black is even weaker: 12.Qe2 Be6? (Book---Now the blunder 13.fxe4?? which merely drives the Knight right where it wants to go is book!

If White had nothing better, then 13.Qf2 holding the piece is far complicated but winning for White.

Correct is 13.Nc3!! Game over! This is already +- 13...Nxc3 14.dxc3 Bxg3ch 15.hxg3 Qxh1 16.Qxb5ch Kf8 17.fxe4 Rc8 18.Bf4 c6 19.Qc5ch Kg8 20.O-O-O Qh2 21.Ba6 Rf8 22.d5 Qh5 23.Qxc6 +- Black has Bishop and three pawns for the Rook. From the starting position, it took Fritz 8 only 17 moves to Queen a pawn and win a Rook here.

In a post coming soon, I hope to address the critical variation, 8.Ne4 Ne6!?

Here is the potential fatal flaw in the variation 8.Ne4 Ne6!? 9.Bxb5ch Bd7---10.Ba6!! (It took me months to find this concept---White maintains control over d3 & the retreat Bf1 allows White to play Ng3 while still defending g2). Here are three tries for Black

10...Ndf4 11.O-O Bc6 12.Re1 Nd3 13.Re3 Nef4 14.Qf3! Nxc1 (What else?) 15.Qxf4! & White is much better i.e. exf4?? 16.Nf6 doublecheck and mate!

10...f5 11.Ng3 Nc5 12.Bc4 Nb6 13.Bf1 e4?! (recommended by Fritz 8) 14.d4 exd3 e.p. 15.Bxd3 Nxd3 16.Qxd3 Qe7ch 17.Ne2 +/-

10...f5 11.Ng3 Nc5 12.Bc4 Nb6 13.Bf1 Qe7!? (A suggestion of GM Lev Alburt) 14.b4 Nb7 (Nca4 looks awful) 15.Ba6! Nd6 16.O-O f4 17.Nh1 Qg5 18.d4 Bc6 (Bh3? 19.Ng3+-) 19.f3 +/-

Thus, it would appear that 8...Ne6 is no better as an equalizing try than 8...Qh4.

To prepare the Two Knight's player for all contingencies, here is an obscure variation of the main line:

4.Ng5 d5 5.exd5 Na5 6.Bb5ch Bd7?! 7.Qe2 Be7

(7...Bd6 8.Bxd7ch Qxd7 9.Nc3 O-O 10.O-O Bb4 11.Rd1 Rad8 12.Nf3 Nxd5 13.Nxd5 Qxd5 14.Qxe5 Qxe5 15.Nxe5 Rfe8 16.Nf3 Re2 17.c3 Bd6 18.d3 +/-)

8.Nc3 O-O 9.Bxd7 Qxd7 10.O-O Rfe8 11.d3 Nxd5 12.Qxe5 Nxc3 13.Qxc3 Nc6 14.Nf3 Bd6 15.Be3 Ne5 16.Nxe5 Bxe5 17.d4 +/-

@ sloughterchess

You've made eight posts in a row, containing quite lengthy variations, not a diagram in sight. Is this thread intended for general discussion or just a repository for your thoughts?

I'm not looking to quarrel, just curious as to what's going on here.

EDIT: Let me explain, I'm very interested by the discussion but haven't had a board set-up in my home for years. I feel that a diagram in each post would add value.

The summation of threads was designed so that players who have been following the general discussion will not have to wade through pages and pages of non essential information, to get to the critical theory. It is hoped that most of the diagrams were covered in the other posts. If you want diagrams, please go to the original posts for each variation e.g. the Wilkes-Barre has two posts, one for the theory, and one for the game I played against Fritz 8. It is the intent of this thread to be pretty much provide you with what you would get in ECO, MCO or BCO.

A correction is necessary to one thread and an additional variations are provided here. My attempt to salvage the 4.Ng5 d5 5.exd5 Nd4 6.c3 b5 7.Bf1 Nxd5 8.Ne4?! Ne6! was unsuccessful.

After 9.Bxb5ch Bd7 10.Ba6?! doesn't work as pointed out by a post member. The cook is that after 10...f5 11.Ng3 Nc5 12.Bc4 Nb6 13.Bf1 f4!! (I didn't analyze beyond Qh5ch thinking that picking up the e pawn was good for a plus. This is a superficial assessment.) 14.Qh5ch g6 15.Qxe5ch Qe7 16.Qxe7ch Bxe7 17.d4 fxg3 18.dxc5 gxf2ch 19.Kxf2 Bxc5ch 20.Ke1 O-O -/+

In the Fritz, Black can try to equalize by another means, but it doesn't appear to work. First:

4.Ng5 d5 5.exd5 Nd4 6.c3 b5 7.Bf1 Nxd5 8.cxd4 Qxg5 9.Bxb5ch Kd8 10.Qf3 exd4?! This requires precise calculation:

11.Bc6 Nb4 12.Bxa8 Nc2ch 13.Kd1 Bg4 14.Kxc2 Bxf3 15.Bxf3 d3ch 16.Kd1 Qc5 17.Nc3 Qxf2 18.Ne4 Qd4 19.Re1 Bd6 20.Nxd6 Qxd6 21.Re3 Re8 22.b3 Rxe3 23.dxe3 Qxh2 24.Kd2 Qe5 25.Rb1 Qa5ch 26.Kxd3 Qxa2 27.Rb2 Qa6ch 28.Kc2 +/= to +/- The Bishops and Rook should overpower the Queen despite the extra pawns.

11.Bc6 Nb4 12.Bxa8 Nc2ch 13.Kd1 Nxa1 14.d3 Qc5 15.Qd5ch Qxd5 16.Bxd5 Ba6 17.Ke2 Nc2 18.a3 Bd6 19.Bc6 Na1 20.Nd2 Bc8 21.b4 Bg4ch 22.f3 Bd7 23.Bxd7 Kxd7 24.Kd1 a5 25.Bb2 axb4 26.Bxa1 bxa3 27.Bxd4 c5 28.Bc3 +-

So, unless there is a clear path to equality that has been missed, it would appear that 5.Nd4 favors White.

When faced with a new variation OTB it is tough to always get the right idea. Here, for example, 4.Ng5 d5 5.exd5 Nd4 6.c3 Nf5?! 7.Qe2! (My trademark move, but I missed it OTB!)  Nxd5 8.d4 Be7 9.Qxe5 c6 10.Qe2 O-O 11.Nf3 Re8 12.O-O +/=