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Over the Horizons Stefan Bücker
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Lower Life in the Latvian Gambit For the pawn deficit in the Latvian Gambit after 1 e4 e5 2 Nf3 f5 3 Nxe5 Black may not have compensation, but usually he can hope for complications. Suppose there is a variation which leads to an early exchange of queens and the prospect of further simplifications, while Black still is a pawn behind and has no immediate idea how to regain it – would a Latvian Gambiteer choose it? Probably not, in most cases he would rather give a second pawn to avoid frustrating simplifications, and to generate some vague chances.
This apparent handicap of the “Löwenthal Variation” 3…Nf6 must be the reason why the idea never became as popular as 3…Qf6 or 3…Nc6 and is usually attributed a ?! by theory. Introduced in B. Horwitz – Löwenthal, London 1851 [3], the move was later used by Blackburne. In 1907 even a well-known theoretician, Simon Alapin, recommended “3…Nf6!” in an article [2].
Kosten [6]: “This time there is no need for Black to worry about Qh5+, but on the other hand, he loses a whole pawn without embroiling White in complex calculations.” This doesn’t sound encouraging, and I am sure that my readers would now prefer a column on something more aggressive, for example 3…Qe7, an entertaining move from the 19th century which Lodewijk Prins once played against the readers of his column in the Amsterdam newspaper Het Parool. But after 4 Qh5+ g6 5 Nxg6 Qxe4+ 6 Be2 Nf6 7 Qh3 hxg6 8 Qxh8 Qxg2 9 Rf1 Kf7 the readers followed the old “Bilguer” line 10 Qh4 (“+/-“ [2]), running into problems. Instead, 10 d4! would have been a clear improvement.
The move was called “also sensible” by Kosten [6]. In fact it might refute 3…Qe7, for example 10…Nc6 11 c3 Bg7 12Qh4 d6 13 Nd2, threatening to catch Black’s queen (13 Bf3, resp. 12…Qd5 13 Bc4 +-). 13…g5 14 Qg3 +/- and Black has to exchange queens. Perhaps the text move (3…Nf6) isn’t such a bad idea, if one of the main alternatives results in such a bad ending… 4 exf5
“The simplest continuation”, writes Carl Schlechter 1916 in the “Bilguer”/Handbuch [2]. “Simple and good”, Tony Kosten in [6]. Which doesn’t mean that the text move is better (if we ignore that Kosten had attributed a “?!” symbol to the alternative 4 Bc4 in his first book on the Latvian Gambit [4]; in [6] this sign was changed to !?). After 4 Bc4, Black has to make a difficult choice between the relatively new Morgado Variation (which became popular in the early 1970s) 4…Qe7 5 d4 Nc6! and 4…fxe4 5 Nf7 Qe7 6 Nxh8 d5, an old line introduced by Blackburne. Both systems usually arise via the 3 Bc4 move-order, and if we look at their theoretical reputation, they are not a picnic for the Black player. But for the moment let’s concentrate on the text move. There are three reasons why 4 exf5 is much more popular (in 43 games of the database White continued 4 exf5, while 4 Bc4 happened only in 11 games in this situation): – 4 exf5 is simpler, as our experts already mentioned above. The 3 Bc4 system is really a jungle of variations. Here White has only to learn the Blackburne and Morgado Variations, but still… In comparison to said jungle trip, 4 exf5 looks like a walk on a sunny beach. – 4 exf5 is good (White scores 62%), while 4 Bc4, at least in this move-order, doesn’t impress (41%). – If White prefers the main line 3 Nxe5, why should his repertoire also include 3 Bxc4 with all those difficult variations which are often quite problematic for White? We might want to return to 4 Bc4 in another column, if the text move (4 exf5), which wins a whole pawn, does not guarantee a significant advantage for White. Which seems highly unlikely… However, by a lucky coincidence Joachim Hunstock from Göttingen contacted me in April, when I had just looked at early alternatives for Black in the Latvian Gambit. The Löwenthal Variation 3…Nf6 had been his specialty, which he had played in several correspondence games around 1990. He had come to the conclusion that the Löwenthal Variation was more than a desperate attempt to achieve a draw, he found it so promising that he started to play it even with White: 1 e3 e5 2 e4 Nf6 3 f4 Nxe4 4 Nf3
This fascinating and radical move order appeared in no less that five of his eight Löwenthal games. For a better overview, the games are here transformed to the “standard” version of the Latvian Gambit known to you and me from our sporadic use of it in our coffeehouse practice. If the German correspondence player is right to believe that the Löwenthal Variation gives sufficient compensation for the pawn, while the rest of the world thinks otherwise, the psychological pressure on White to prove an advantage must be immense. I am indebted to Joachim Hunstock for the permission to publish some of the most unusual Latvian Gambit games I have ever seen. Sometimes he added a hint, which move he regarded as better. The rest of the “theory” is mine, but of course it helps, if good games guide you into the right direction. 4…Qe7 5 Qe2
The piece sacrifice 5 d4!? d6 6 Qe2 dxe5 7 dxe5 deserves attention. 7…Nd5 8 Qh5+ Qf7 9 Qxf7+ Kxf7 10 Bc4 c6 (Kroonen – Diepstraten, Amsterdam 1989), and now Tony Kosten’s 11 e6+ [6] seems to be advantageous for White. Therefore Black should better decline the piece: 6…Bxf5 7 Nd3 Nc6 8 c3 0-0-0, when he has good play for the pawn. 5…d6 6 Nf3 Other possibilities: (a) 6 d4 transposes to the last note. (b) 6 Nc4 is Kosten’s main line (6…Bxf5 17 Qxe7+ Bxe7 8 Ne3 Be6 9 d4 Nc6 10 c3 “with advantage”, Tener – Denny, corr. 1972 [6]) and also mentioned by [3] and [5]. The most forcing reply is 6…d5!?, but it isn’t clear whether Black can equalize: 7 Ne5 (7 Ne3 d4 8 Nc4 Bxf5 9 d3 Nc6 and Black has no problems) 7…Bxf5 8 d4 (8 d3 Na6) 8…Nc6! 9 Nxc6 bxc6 10 c3 and now: (b1) 10…Kd7 11 Be3 Ng4 12 Nd2 Re8 [0-1, 51] Svaton – Cirulis, corr. WSTT/1/07/4, 2007, but 11 Nd2 comes into consideration (and seems stronger under these circumstances). (b2) 10…Kf7 11 Nd2 (else Black has active play for the pawn, e. g., 11 Qxe7+ Bxe7 12 Ba6 Bd6 13 0-0 Rab8 etc.) 11…Re8 12 Nf3 Qxe2+ 13 Bxe2 Bd6 14 h3 Rhf8 15 Be3 Ne4 16 Nh4 Be6 17 0-0-0 Kg8 18 Nf3 Bf7 19 Rde1 Ng3 20 fxg3 Rxe3 21 Kd2 Re7, and Black is only slightly worse. 6…Bxf5
In this important position White has to choose: 7 Nd4 (Variation A) or 7 d3 (Variation B). 7 Qxe7+ Bxe7 8 d3 Nc6 only transposes to Variation B. Variation A. 7 Nd4 1 e4 e5 2 Nf3 f5 3 Nxe5 Nf6 4 exf5 Qe7 5 Qe2 d6 6 Nf3 Bxf5 7 Nd4 Bd7
Because of this retreat Joachim Hunstock regards White’s seventh move as inaccurate. However it seems too early to draw a conclusion which of the two main lines offers more chances. 8 Nc3 c5! Kosten’s reference work on the Latvian Gambit only gives 8…Nc6 9 Nxc6 Bxc6 10 Qxe7+ Bxe7 11 Bb5 “and Black has no real compensation for the pawn, Müller – Diepstraten, corr. 1985” [6]. The text move gains space and forces the knight back – and hopes that it might be so optimistic to jump to b5. 9 Nf3 Relatively best. After 9 Ndb5 Kd8, Hunstock’s experiences in four games prove that Black has excellent compensation for the pawn:
(a) 10 Qxe7+ Bxe7 11 Be2 a6 12 Na3 b5 13 Bf3 Nc6 14 Ne2 d5 15 Nf4 b4 16 Nb1 Nd4 17 Bd1 Bd6
“A funny position, note the knight’s walking-tour from g1 to b1”, Hunstock. 18 d3 Re8+ 19 Kf1 Ba4 20 b3 Ra7 21 g3 Bd7 22 h4 Bg4 23 Bxg4 Nxg4 24 a3 Bxf4 25 gxf4 Re2 26 Be3 Rxc2 27 Bxd4 cxd4 28 axb4 Rxf2+ 0–1, Hamann – Hunstock, corr. 13. PZ-48, 1991 (b) 10 Qxe7+ Bxe7 11 d3 a6 12 Na3 b5 13 Ne4 d5 14 Ng5 Re8 15 Be2 b4 16 Nf7+ Kc8 17 Nb1 Bf8 18 Bf4 a5 19 Nd2 Bg4 20 f3 Bh5 21 Nd6+ Bxd6 22 Bxd6 Ra6 23 Bxb8 Rae6 24 0-0 Rxe2 25 Bf4. Here Klar believed he stood better.
25…g5!! 26 Bxg5 Rg8 27 Rae1 Rxe1 28 Bxf6 Re2 29 Rf2 Re6 30 Bh4 a4 31 Kf1 Rge8 32 Kg1 Re2 33 Bg5 Re1+ 34 Rf1 R8e2 35 Bf4 Kd7 36 b3 a3 37 Bg5 Kc6 38 Bf4 Kb5 0–1, Klar – Hunstock, Corr. Cup qualification, 1990. Played with reversed colors (1 e3 e5 2 e4 etc.). (c) 10 a4!? a6 11 Na3 d5 12 d3 b5? (12…Qf7!? Hunstock) 13 Nab1 b4 14 Nd1 Nc6 15 Qxe7+ Bxe7 16 Ne3 Re8 17 Be2 Bf8 18 h3 Ra7!?
“I loved the cooperation of Black’s rooks, introduced by this move”, Hunstock. 19 Kf1 g6 20 g3 Nd4 21 Bd1 Bc6 22 c3 Ne6 23 Ng2 d4 24 c4 Bd6 25 h4 Rf7! 26 f4 Nh5 27 Bxh5 gxh5 28 Rg1 Rg7 29 Kf2 Reg8 30 Nd2 Bxf4! 31 Nf1 Bxc1 32 Rxc1 Rf7+ 33 Ke2 Bxg2 34 Rxg2 Nf4+ 0–1, Štefan (CSSR) – Hunstock, corr. WT/M/482. Played with reversed colors. (d) 10 a4 a6 11 Na3 d5 12 b3!? Nc6 13 Bb2 Qf7! 14 0-0-0 Bd6 15 Qe1 Re8 16 Ne2 Nd4 17 Bxd4 cxd4 18 Nb1 d3
19 cxd3 Ng4 (19…b5!) 20 f3 Bc5 21 Qh4+ g5 22 Qxg5+ Be7 23 Qf4 Qxf4 24 Nxf4 Nf2 25 d4 Nxh1 26 Bxa6 Rxa6 27 Rxh1 Bg5 28 Nxd5 Be6 29 Nbc3 Bxd5 30 Nxd5 Rd6 31 Nc3 Rxd4 32 Ne4 Rexe4 33 fxe4 Rxe4 0–1, V. B. Grigorev (Leningrad) – Hunstock, World Cup /Gr 92. 9…Nc6 10 d4! More critical than the alternatives: (a) The stem game (for the aggressive 8…c5!?): 10 a3 0-0-0 11 Qxe7 Bxe7 12 Bb5 Bg4 13 Bxc6 bxc6 += (1-0, 44) De Jong – Stavast, NBC Beker 1981, semi-final [3]. A sensible alternative would be 12…a6!? 13 Bc4 Bg4, to keep more pieces on the board. (b) 10 Qxe7+ Bxe7 11 Bc4 Nb4 12 Bb3 c4
13 a3 cxb3 14 axb4 bxc2 15 Nd4 0-0 16 Nxc2 Bf5 17 Ne3 Bd3 18 f3 Rae8 19 Kf2 d5 20 Nexd5 Nxd5 21 Nxd5 Bh4+ 22 g3 Re2+ 23 Kg1 Rxf3 24 Nf4 Rxf4 25 gxf4 Be4 0–1, Nagel – Hunstock, 13. PZ-48, 1991. Played with reversed colors. The text move seems to be an improvement. White returns the pawn, but avoids a passive set-up with d3 and secures some advantage in the endgame: 10…Nxd4 11 Nxd4 cxd4 12 Nb5 Bxb5 13 Qxe7+ Kxe7! 14 Bxb5 Rc8
15 Ba4 Or 15 Bd3 Kf7 16 Bf4 Re8+ 17 Kf1 d5 18 g3 g6 19 Kg2 Bg7, e. g. 20 h4 a6 21 Rhe1 Ne4 22 Kf3 Nc5 +=, because of the doubled black d-pawn White has a slight advantage, but a draw seems the most probable result. 15…d5 16 0-0 Kf7 17 Bf4 Bc5 18 Rfe1 Rhd8 19 Rad1 Ne4 20 Re2 Kf6 21 h4 h6 22 Bb3 g5 21 hxg5+ hxg5 24 Bh2 g4!? =
White can win a pawn, but not the game: 25 Bxd5 Rxd5 26 Rxe4 Bb6 27 c3 (27 Re2 Rb5 28 b3 Rbc5) 27…Kf5 28 Re2 d3 29 Red2 Rcd8, and Black’s passed pawn fully compensates the pawn deficit. Variation B. 7 d3 1 e4 e5 2 Nf3 f5 3 Nxe5 Nf6 4 exf5 Qe7 5 Qe2 d6 6 Nf3 Bxf5 7 d3 Nc6 8 Qxe7+ Bxe7
The natural move 7 d3 is missing in [6]. According to Joachim Hunstock the diagram position should be considered as the main line, with the continuation 9 Be2 0-0-0 10 0-0 (so far already given by Diepstraten [3], but without any further analysis or an evaluation). 9 Be2 Alternatives: (a) 9 c3?! Ne5 10 Nxe5 dxe5 11 d4? exd4 12 cxd4 Bb4+ 13 Bd2 Bxd2+ 14 Nxd2 0-0-0
White has serious problems. 15 f3 Rxd4 16 Nb3 Rd6 17 Be2 Re8 18 Kf2 Bc2 19 Rhe1 Bxb3 20 axb3 Rd2 21 Kf1 Nd5 22 Bb5 Rxe1+ 23 Rxe1 c6 24 Bc4 Rxb2 25 Re2 Rb1+ 26 Kf2 Kd7 27 Rd2 Kd6 28 g4 h6 29 h4 a5 30 Rd3 b5 31 Bxd5 cxd5 32 h5 Rc1 33 Ke3 Rc2 34 Kd4 a4 0–1, Remus (USA) – Hunstock, corr. WT/M/482. Played with reversed colors. (b) 9 Nc3 0-0-0 (9…d5 10 Bf4 0-0-0 11 0-0-0 Rhf8 12 a3 d4 13 Ne2 Bg4 14 Neg1 Be6 15 Re1 Bd5 16 Bg3 Nh5, and Black has compensation for the pawn) 10 a3?! (“strange”, Hunstock; for 10 Be2 see 9 Be2 0-0-0 10 Nc3, below) 10…d5 11 Be2 d4 12 Nd1 Rhe8 13 Nd2 Rd5 14 0-0 Bd6 15 Bf3 Rc5 16 Bxc6 Rxc6 17 Nf3 Rxc2 18 Nxd4 Rxc1 19 Rxc1 Bxd3 20 Ne3 Bxf1 21 Rxf1 Be5
22 Nb5 Bxb2 23 Nxa7+ Kb8 24 Nc4 Bd4 25 Nb5 Bc5 26 Nc3 Ne4 27 Nxe4 Rxe4 28 Nd2 Re2 29 Nc4 Ra2 30 Ne5 Bxa3 31 Re1 Bb2 32 Nd7+ Ka7 33 Nf8 Bd4 34 Rf1 c5 0–1, Seack – Hunstock, Corr. Cup preliminaries, 1990. 9…0-0-0 10 0-0 10 Nc3 d5 (more logical than 10…Nb4 11 Nd4 [or 11 Bd1] 11...Bd7 12 a3 Nbd5 [0-1, 76] Jackson – Diepstraten, corr. Atars Memorial 1985/87, 13 Bf3 +/-) (a) 11 a3?! d4 12 Nb1 Rhe8 13 0-0
13…Bxa3 14 Nxa3 Rxe2 15 Bg5! Rd5 16 Bxf6 gxf6 17 Rfe1 Rxe1+ 18 Rxe1 Kd7 19 h3 b5 20 Re2 b4 21 Nc4 b3 22 Ne1 Nb4 23 cxb3 Bxd3 24 Rd2 Bb1 25 Rd1 Bg6 26 f4 Nc2 27 Nxc2 Bxc2 28 Ne3 Bxd1 29 Nxd5 Bxb3 30 Nxf6+ Ke6 31 Nxh7 Kf5 32 Kf2 Kxf4 33 g3+ Kf5 ½–½, Baranowsky – Hunstock, corr. World Cup/Gr 92. (b) After 11 Bd1 (or 11 0-0 d4 12 Nb1 followed by Nbd2) Black would still have to prove that he has sufficient compensation. For example: 11…h6 12 0-0 g5 13 Ne2 Bd7 14 Ned4 Nxd4 15 Nxd4 c5 16 Ne2 Rde8 17 f4 g4 18 a4 h5 and Black has almost equalized. 10…Rhe8 11 Nc3 d5 12 Bd1 12 Nh4 Bg4! 13 Bxg4+ Nxg4 14 Nf5 Bf6 15 Bd2 h5!? and Black exerts some pressure for the pawn. 12…h6 13 h3 g5 14 a3 a5 15 Re1 Bc5
Black controls more space and his pieces are placed on the “classical” squares, while White’s position is somewhat cramped. These factors may be less important in such a simplified position, where the queens have left the board. But still… it seems possible that there is enough compensation to hold the balance: 16 Rxe8 Rxe8 17 Bd2 g4!? 18 Nh4 Bd7 19 hxg4 Nxg4 20 Bxg4 Bxg4 21 Ng6 Nd4 22 Nxd5 Re2 +=
The rook on the second rank forces further simplifications. For example, 23 Bxh6 c6 24 Ngf4 Rxc2 25 Ne3 Ne2+ 26 Nxe2 Bxe3 27 Bxe3 Bxe2 28 b4 axb4 29 axb4 Bxd3, and Black should be able to achieve a draw. Many open questions remain, but Black’s practical chances after 3…Nf6 4 exf5 seem far better than the theoretical works suggest. Sources: Send your games or comments to redaktion@kaissiber.de. |
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