|
|
d6 (Old Steinitz Defense) .........................................................................................................Bb4(o)
4
d4
c3(p)
Bd7(a)
Ba5
5
Nc3
0-0
exd4................................................... Nf6 ...................................................Nge7
Nge7
6
Nxd4
0—0(g)
Bc4(m)
Bxc6(q)
g6
Be7
exd4
Nxc6
7
Be3..................... 0—0
Rel.................. Bxc6
Nxd4
b4
Bg7
Bg7
exd4(h) Bxc6
Nxd4
Bb6
8
Qd2
Bxc6
Nxd4
Qd3
Qxd4
b5
Nf6
bxc6
0—0
exd4
Nc6
Na5
9
f3(b)
Rel(e)
Bxc6
Nxd4
Qe3
Nxe5
0—0
Ne7
bxc6
Bd7
Ne5
0—0
10
Bxc6(c)
Bf4
Bg5(i)
b3(k)
Bb3
d4
bxc6
c5
h6
0—0
c6
Qe8
11
0-0-0
Nf3
Bh4
Bb2
Qg3
Nd2
Re8(d)
0—0(f)
Re8(j)
c6(l)
Ng6(n)
d6(r)
(a) 4 . . . exd4 5 Qxd4 Bd7 6 Bxc6 is a line from Philidor s Defense that is better for White.
(b) Archives consider 9 Bxcfi bxc6 10 Bh6 Bxh6 11 Qxh6 c5 12 Nde2 Bc6. White may be a bit better here.
(c) 10 0—0—0 Nxd4 11 Bxd4 Nxe4! 12 Nxe4! (12 fxe4 Bxd4 13 Qxd4 BxbS 14 NxbS Qg5t] 12 Bxb5 13 Bxg7 Kxg7 14 Qc3± f6 Comments.
(d) 12 Bh6 Hha 13 h4 Qb8 14 h5 Qb4! =, Tseitlin—Kimelfeld, USSR 1967.
(e) 9f4c510 Nde2 f5 11 e5 Bc6 oo, Comments.
(f) 12 e5 and Black has difficulties (keres).
(g) Note that this position may arise by the move order 3 . . . Nf6 4 0—0 d6 5 d4 Bd7 6 Nc3.
(h) A famous trap is 7. . . 0—0? 8 Hxc6 Bxc6 9 dxes dxe5 10 Qxd8 Raxd8 11 Nxe5 Hxe4 12 Nxe4 Nxe4 13 Nd3 f5 14 f3 Bc5± 15 Nxc5 Nxc5 16 Hg5 Rd5 17 Be7 and White wins the exchange since if the rook on f8 moves then 18 c4 wins a piece, Tarrasch—Marco, Dresden 1892.
(i) (A)
10 b3 d5! 11 e5 Bb4 is equal, Forgacs—Wolf, Nurnberg 1906. (B)
10 Bf4 Rba 11 Nb3 is given by ECO as leading to a slight advantage for
White, but this evaluation is
debatable.
(j) 12 Qd2 Nh7 13 Bxe7 Rxe7 =, Capablanca—Lasker, World Chp. 1921.
(k) 10 Bg5 0—0 11 Rael h6 12 Bh4~Nh7 13 Bxe7 Qxe7 14 Nd5 also gives White some advantage, Lasker—Capablanca, World Chp. 1921.
(l) 12 Radl Qc7 13 Rfel, Pillsbury—Steinitz, Vienna 1898. White has more space and a harmonious position. Black s position is solid but passive.
(m) Moving the bishop again is justified
here because f7 is weak. Also reasonable is 6 0—0
Ng6 7 b3 He7 8 Bb2 0—0 9 Nd5, Rogulj—Orlov,
Yugoslav Chp. 1980.
(n) The column is Lasker—Steinitz, World Chp. 1894. Now 12 Be3 gives White the advantage.
(o) This queer-looking move of Alapins is not that bad; it at least has surprise value.
(p) 4 0—0 Nge7 5 d4 exd4 6 Nxd4 leaves the Black bishop at b4 vulnerable and the kingside subject to attack, R. Burger—H. Steiner, Hollywood, 1953.
(q) On the natural 6 d4 then 6 ... exd4 7 cxd4 d5 equalizes.
(r) 12 Nd3 (C. Hansen—Dreyev, Kiljava
1984). According to Hansen, best is now 12... f5 13 e5 Qxb5 14 c4 Nxc4
15 a4 Qd5 16 Nxc4 Qxc4 17 a5 Bxd4 18 Ra4 Qd5 19 Nf4 Bxf2+ 20 Kxf 2 Qxdl
21 Rxdi dxes 22 Nd5 when White has some advantage.