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Peter Svidler vs Peter Leko
Linares (1999), Linares ESP, rd 4, Feb-25
Spanish Game: Classical Variation. Zukertort Gambit (C64)  ·  1-0

ANALYSIS [x]

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Given 9 times; par: 67 [what's this?]

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Kibitzer's Corner
< Earlier Kibitzing  · PAGE 2 OF 2 ·  Later Kibitzing>
Dec-30-04
Premium Chessgames Member
  johnwgoes: What's the continuation if 38. ...♗d4?
Dec-30-04  who: 39.Bd5+ I think
Dec-30-04  newold: It seems that 36. Qe4 (threat Qh7 and Qh7 mate, and the rook in c6) is simpler : 36. ... Rc7 37. Qh7+ Kf8 38. Rf6 and it's over for black (a piece down and f7 under quadruple attack) 36. ... Qd7 37. Qh7+ Kf8 38. Rhf5 and the pressure on f6 / f7 wins for white
Dec-30-04  cade: I saw the final move as 40 Rd1, winning the Queen.
Dec-30-04  orior: I think that 40.Rd1 is met by .. Rc1 pinning the rook.
Dec-30-04
Premium Chessgames Member
  Marius: hey crafty
what happens after 38. Qe4+ Kd7
Dec-30-04  Silvanel: I am not crafty but i think that after

38. Qe4+ Kd7
39. Rxd8
39. ... Kc7 40. Rxd2

39. ... Kxd8 40. Qxc6

39. ...Bxd8 40. Rxf8+

Dec-30-04  Nickisimo: Boy, I really swung and missed on this one. Well, Thursday is usually the cutoff point for me.
Dec-30-04  Assassinater: Why not just take the rook with the bishop on Rh8+?
Dec-30-04  percyblakeney: <Assassinater> Then Rxf7+ and Qg8# would follow.
Dec-30-04  dac1990: An okay puzzle. I got the first coupla moves.
Dec-30-04  noone2: <newold> after 36 Qe4 how about... Qd3! and black avoids any immediate danger.
Dec-30-04  noone2: Isn't 37 Rh7 at least as good as the game continuation? i.e. 36 Qg4+ Kf8 37 Rh7

37 ...R8d7 Rd1 and black will lose at least R for Q.

37 ...R6c7 R:f6

With no way to guard f7 black should resign (he could try running with Ke8 lol)

Dec-30-04  GreenDayGuy: That would work also, but 37. Rh8+ makes things more forcing.
Dec-30-04  newold: noone2 : fine, you're right !
Dec-30-04  newold: <noone2> on 37 ... R8d7 38. Rd1 black has 38. ... Rc1 ! and it's not so clear ...
Dec-30-04
Premium Chessgames Member
  kevin86: A toughy today! Black was a hard nut to crack in this one.
Dec-30-04  Castle In The Sky: got it for once <Dick Brain> thanks for the Harvard math and chess reference in one of your other postings, the online material about a "pure mate" is pretty interesting
Dec-30-04  shortsight: why not black respond 36. ... Qd4?
Dec-30-04  aw1988: Because he is in check.
Jan-05-05  shortsight: oh, i mean 35. ... Qd4?
Jan-05-05
Premium Chessgames Member
  penarol: shortsight: 35...Qd4 36 Qh6+ Kg8 37. Qh7+ kf8 38. Qxf7 mate.
Jan-06-05  shortsight: oh, yeah, the mate attack. didn't see that, thx.
Apr-03-08
Premium Chessgames Member
  tpstar: The defensive try 40 ... Be7 preventing any Discovered Check creates another good puzzle:


click for larger view

There are at least two good continuations:

1) 41. Rh5 threatening 41 ... Rd5+; not 41 ... Rd6? 42. Qb7#, or 41 ... Rxf7?! 42. Rxf7 Qd1+ 43. Kh2 Qxh5 44. Qxe7+ mates. Yet since the Rh7 vacated the 7th rank, Black has 41 ... Bd6 holding (42. Qe6+ Kd8).

2) 41. Qg4+ looks better; 41 ... Kc7 (41 ... Kd6/Kd8 42. Rd1) 42. Rd1 Qb4 (defending the Be7). Now the prosaic 43. Qxb4 Bxb4 44. Bd5+ wins the exchange, or White has 43. Qd7+ Kb6 44. Bd5 playing for mate (44 ... Rc7? 45. Rh6+ Bf6 46. Qe6+).

Oct-16-12
Premium Chessgames Member
  plang: Leko prepared the hybrid Berlin system with 3..Nf6 and 4..Bc5 for this tournament and obtained short draws against Topalov and Ivanchuk in the two other games he used it. Svidler's 7 Qd3!? is very rarely played possibly because of the gambit response 7..d5!? as played by Short against Ehlvest in Belgrade 1989. For this reason 7 Bg5 is the most popular continuation. Five months later at Dortmund Topalov tried 13 Nxg5!? against Leko with interesting complications which, in that game, ended in Black's favor. Another alternative 13 e5 would have been good for Black after 13..d5 14 Bb3..gxh 15 exf..Qxf6 16 Bxd5..Nxd4. 13..g4 14 e5..gxf 15 Bh4 would have backfired badly for Black. Leko erred with 22..Rad8? and after 23 Bb3! he was clearly worse. Svidler gave the best defense as 22..Red8 23 Ne4..Qg6 24 Nf6..Qxf5 25 Rxf5..Rd8 26 Be2 with equality. Note that the straightforward 25..Bxb2 would have led to a clear advantage for White after 23 Bb3..Qxd2 24 Rxe5..Bxe5 25 Bc2!..Qxc2 26 Qxc2. Svidler did not fall for the trap 28 Bxf7..Bc3! with equality. A good example of an attack in an opposite-colored bishop position.
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