Fritz gives 23... Qf3 as equal. I guess there's perpetual, but I don't know if a human could see it in every line. I doubt it's important, though. Unless it's touted as a refutation, I doubt this position will turn up often, so the important thing is to prove it's NOT a refutation. At least, that's my opinion.
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I agree
Apparently 9...Qd6 10.d3 Bg4 11.Nf7 Qb6 12.Qd2 Be2+ 13.Kxf2 Ng4+ 14.Ke1 Qf6 15.Qxe2 Nxe2 16.Kxe2! Qf2+ 17.Kd1 Qxg2 18.Re1 Nf2+ 19.Kd2 Ne4+ 20.Ke3 Qf2+ 21.Kxe4 Qxe1+ 22.Be3 Qh4+ is a perpetual.
1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.Ng5 Bc5 5.Nxf7 Bxf2+ 6.Kf1 Qe7 7.Nxh8 d5 8.exd5 Nd4 9.d6 Qxd6 10.c3 Bg4 11.Qa4+ Nd7 12.Kxf2 Qf6+ 13.Ke1 0-0-0 14.Rf1 Qh4+ 15.Rf2 Nb6 16.g3 Qh3 17.Nf7 Nxa4 18.cxd4 Rxd4 19.d3 Qh5 20.Be3 Rd7 21.Ng5 Rd8 22.Nc3 Nxc3 23.bxc3 and white is much better.
I believe the book move is 11... Qh4+ with perpetual.
12. g3 Qf6+ 13. Ke1 (13. Kg1 Ne2+ 14. Nxe2 Bxe2 and Qf1# looms, since 15. h4 Qf1+ 16. Kh2 Qf2+ 17. Kh3 Bf3 wins) 13...Nf3+ 14. Kd1 (actually, all other fifteenth moves lose) Nd4+ etc. with perpetual
12. Ke3 0-0-0 13. cxd4 exd4+ 14. Kxd4 Bd1+ wins
12. Kf1 0-0-0 with a winning attack
I will update the critical variations
I have eliminated all lines with 13.Nc3 since we have analysed them to be equal.
I have two 7.Ke3 variations that are the future of the entire variation.
An additional 6.Kf1 line has been added.
7.Kg1 has been proven drawn, so I removed white's trys.
7.Ke3 Lines
E: 1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.Ng5 Bc5 5.Nxf7 Bxf2+ 6.Kxf2 Nxe4+ 7.Ke3 Qh4 8.g3 Nxg3 9.hxg3 Qd4+ 10.Kf3 d5 11.Rh4 e4+ 12.Kg2 0-0 13.Qh5 Be6 14.Be2 Rxf7 15.Qxh7+ Kf8 16.Rf4 Rxf4 17.gxf4 Qf6 18.d3 Nd4 19.Bd1 exd3 20.Qxd3 Bf5 21.Qa3+ Ke8 22.Nd2
F: 1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.Ng5 Bc5 5.Nxf7 Bxf2+ 6.Kxf2 Nxe4+ 7.Ke3 Qh4 8.g3 Nxg3 9.hxg3 Qd4+ 10.Kf3 d5 11.Rh4 e4+ 12.Kg2 0-0 13.Bb3 Rxf7 14.Rf4 Rxf4 15.gxf4 Be6 16.Nc3
6.Kf1 Lines
A: 1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.Ng5 Bc5 5.Nxf7 Bxf2+ 6.Kf1 Qe7 7.Nxh8 d5 8.exd5 Nd4 9.d6 cxd6 10.Kxf2! Bg4 11.Qf1 0-0-0 12.Nf7 d5 13.Kg1! Rf8 14.c3 Nc2 15.Na3 Nxa1 16.Bxd5 Nxd5 17.Qc4+
B: 1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.Ng5 Bc5 5.Nxf7 Bxf2+ 6.Kf1 Qe7 7.Nxh8 d5 8.exd5 Nd4 9.d6 Qxd6 10.c3 Bg4 11.Qa4+
SirDavid wrote:
I believe the book move is 11... Qh4+ with perpetual.
12. g3 Qf6+ 13. Ke1 (13. Kg1 Ne2+ 14. Nxe2 Bxe2 and Qf1# looms, since 15. h4 Qf1+ 16. Kh2 Qf2+ 17. Kh3 Bf3 wins) 13...Nf3+ 14. Kd1 (actually, all other fifteenth moves lose) Nd4+ etc. with perpetual
12. Ke3 0-0-0 13. cxd4 exd4+ 14. Kxd4 Bd1+ wins
12. Kf1 0-0-0 with a winning attack
Which line are you referring to?
7. Ke3 E: 13... Rxf7, threatening Qf2+ with perpetual. White can play Qxh7+, but after Qh8+ the checks run dry. White is forced to play Rf4, and then Black trades and takes the Bishop with an equal endgame. Rf4 right after Rxf7 allows Be6. White will save the Bishop, Black trades Rooks and plays Rf8. Basically, to avoid perpetual White must either enter an even endgame or play 14. Rf4 Be6 15. Bb3 Rxf4 16. gxf4 Rf8, where yet again Black has two pawns for a piece with compensation.
F: Same situation- it basically transposes. Fritz gives slight advantage to White.
6. Kf1 A: OK, I give up on 9... cxd6 for now.
B: 11... Nd7, and then what? 12. Kxf2 Qf6+ 13. Ke1 0-0-0
By the way, I'm going on vacation for a week tomorrow, so I won't be here to discuss the lines. Also, I think there are some games with B. I'll look for them. EDIT Hmm, the database I usually use seems to be down.
Oh, excuse me, I somehow skipped over d6 Qxd6 and ignored those moves. Ignore that post.
I think I am finally going to concede on the 7.Ke3 lines. Black equalizes with best play.
In your second line, 11...Nd7 12.Kxf2 Qf6+ 13.Ke1 0-0-0 14.Rf1 Qh4+ 15.Rf2 Nb6 16.g3 Qh3 17.Nf7! Nxa4 18.cxd4 Rxd4 19.d3 Qh5 20.Be3 Rd7 21.Ng5 Rd8 22.Nc3 Nxc3 23.bxc3
11... Nd7 12. Kxf2 Qf6+ 13. Ke1 0-0-0 14. Rf1 Qh4+ 15. Rf2 Qxh2
Fritz says this gives perpetual, and my brain is too tired to check- sorry :P
Looks like it, another variation equalized. I think we have gone through all the critical variations. Do you think it is time to move on to the Bxf7+ variations?
That would be fine with me.
We can start with 1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.Ng5 Bc5 5.Bxf7+ Ke7 6.Bd5 d6 7.d3
I believe 6... Rf8 is the main line. At least, that's what I have and Fritz's fairly small database has 5 games with it opposed to 1 with d6. It also looks a bit more logical.
5.Bxf7+ Ke7 6.Bd5 Rf8 7.0-0 d6 8.Bxc6 bxc6 9.Nf3
Wait a minute, I'm not sure we're finished with Nxf7 Kf1 yet.
I'm still looking at (or planning to):
1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.Ng5 Bc5 5.Nxf7 Bxf2+ 6.Kf1 Qe7 7.Nxh8 d5 8.exd5 Nd4 9.d6 Qxd6 and here:
A 10.c3 Bg4 11.Qa4+ Nd7 12.Kxf2+ Qf6+ 13.Ke1 0-0-0 14.Nf7
B 10.c3 Bg4 11.Qa4+ Nd7 12.Nf7
C 10.Nf7
I assume we're satisfied with the line that Bücker mentions in response to 10.Be2 and 9.h3?
I still wonder about 9.Be2 Bh4 10.c3 Nxe2 11.Qxe2 Bg4 12.Qb5+ Nd7 13.g3. Apparently Heisman has written about it, but I don't have his DVD.
About line A, I think black has sufficient compensation, e.g.
(14.Nf7) Nb6 15.Nxd8 Nxa5 (apparently sort of a standard trade off in these lines) 16.cxd4 Qh4+ 17.Kf1 Qxd8 18.d3 Nxb2!? 19.Bxb2 Qf6+ 20.Ke1 Qh6 21.Nc3 Qe3+ perpetual.
However, I'm having some trouble with a line I did not mention yet:
14.cxd4 (instead of Nf7 above). So far I looked at this, but I think white is better in the final position here:
14.cxd4 exd4 15.Nc3 dxc3 16.Be2 cxb2 17.Rb1 bxc1Q 18.Rxc1 Bxe2 19.Kxe2 Kb8 20.Qc4 Qe5+ 21.Kf2 Rf8+ 22.Nf7 Nb6 23.Kg1 Nxc4 24.Nxe5 Nxe5. Can black improve, or is he ok here?
9...Qd6 10.d3 Bg4 11.Nf7 Qb6 12.Qd2 Be2+ 13.Kxf2 Ng4+ 14.Ke1 Qf6 15.Qxe2 Nxe2 16.Kxe2! Qf2+ 17.Kd1 Qxg2 18.Re1 Nf2+ 19.Kd2 Ne4+ 20.Ke3 Qf2+ 21.Kxe4 Qxe1+ 22.Be3 Qh1+ 23.Kxe5 Qxh2+ 24.Kd4 is one line.