g- and h-pawns versus b-pawn and a-pawn. Does it make a difference? Part 1 D. 1
Kostic-Janosevic 1951,
in Chéron, Handbuch, 2. edition, n. 392, p. 259-260.
The situation with g- and h-pawns versus b-pawn is similar (but not identical) to the structure g- and h-pawns against a-pawn. Chéron began the study of chapter “King, Rook and 2 connected pawns versus King, Rook and 1 pawn” in the 1. German edition of his Handbuch (1955) with the famous game Tarrasch-Chigorin (1893), n. 392 (see D. 2, with Black pawn on a3 and White rook on a5). Chéron considered the moves 1…Ra2 and 1…a2 and believed that in both cases Black loses. Initially Tarrasch showed in his “Dreihundert Schachpartien” (Leipzig 1895, p. 453-454) that 1…a2 was drawing. But 1896 Karstedt declared the analysis of Tarrasch as incorrect and his opinion has prevailed in the theory until mid of 1950s. Tarrasch himself accepted the analysis of Karstedt in the late editions of his 300 games as well as in his manual “Das Schachspiel” of 1931 (p. 81). The 2. Edition of Berger in 1922 endorsed the verdict of Karstedt. It was only in the second half of 1950s that Maizelis proved that the position after 1…a2 was drawn (unfortunately I was not able to find the original analysis of Maizelis. In the 1. Edition of Averbakh work (1958) Kopayev wrote that the draw was proved by Maizelis. I am familiar only with the analysis of Maizelis from 1960 in his manual “Shakhmaty”). In the 2. edition of his Handbuch 1960 Chéron corrected the evaluation of the position (1…Ra2 loses and 1…a2 draws) but replaced his n. 392 Tarrasch-Chigorin by the ending Kostic-Janosevic (where Black had a b-pawn) and proved that with correct play Black could have saved the game. Chéron put Tarrasch-Chigorin in the 2. edition as n. 392a and did not provide for any lines after 1…a2. He just noted that the draw was similar to the game Kostic-Janosevic. As I will show, first, the analysis of Chéron of 1960, n. 392, contains some serious slips, and, second, the positions n. 392 (Kostic-Janosevic) and n. 392a (Tarrasch-Chigorin), despite similarities, are not fully analogous. Chéron pointed rightly out that with the a-pawn the defence is easier but did not provide for any details.
1.h5+ D. 1.1
I. 1...Kh6? This move was played in the game.
2.Kh4 Kostic played 2.g4? which throws away the win because of 2…Rc5! 3.Rxb2 Kg5!=. But Janosevic returned the compliment and went on to lose.
D. 1.2
1) 2...Kg7 2…Rh2+ see in 2). Other moves are weak.
a) 2...Rc5 3.Rb6+ Kg7 4.Rxb2+-.
b) 2...Rc4+ 3.g4 Rc2 4.Rb6+ Kg7 (4...Kh7 5.Rb7+ Kg8 (5...Kh6 6.g5#) 6.h6) 5.h6+ Kf7 6.Kh5+-.
c) 2...Rd2 3.Rb6+ Kg7 4.g4 Kf7 +- (see D. 1.4).
A) 3.Rb6 Kf7 3...Kg8 4.Rb7+- or 3...Rc6 4.Rxb2+-.
4.g4 D. 1.3
This is a position of mutual zugzwang discovered by Chéron. Black to move loses, White to move is draw.
a) 4...Kg7 5.h6+ Kh7 6.Kh5+-.
b) 4...Ke7 5.h6 Rc6 6.Rb7+ Kf6 7.h7 Rc8 8.Kh5+-.
c) 4...Kg8 5.Rb7 5.h6? Rc6=.
5...Kh8 6.h6 Kg8 a) 6...Rc6 7.Kh5+-.
b) 6...Rc7 7.Rxb2 Kh7 8.Kh5 Ra7 9.g5 Ra5 10.Rb7+ Kg8 11.Re7 Rb5 12.Kg6 Rb6+ 13.Kf5 Rb5+ 14.Kf6 Rb6+ 15.Re6 Rb8 16.g6+-.
7.Kh5 Kh8 8.g5 Kg8 9.g6+-.
d) 4...Kf8 5.Rb7+-.
e) 4...Rd2 The game Kostic-Janosevic transposed to this position after 57.g4? Rd2? White wins here due to the fact that the distance between the rook and the g4-pawn is only 2 files.
D. 1.4
5.Kg5 Not 5.h6? Rd6=. Not 5.g5? Rc2=.
5...Rd5+ 6.Kf4 Rd4+ 7.Kf5 Rd5+ 8.Ke4 Rd2 9.Ke5 Kg7 9...Re2+ 10.Kf5 Rf2+ 11.Kg5+-.
9...Rc2 10.Rb7+ Kg8 11.Kf6 Rc6+ 12.Kg5 Rc5+ 13.Kh4 Rc2 14.h6+-.
10.g5 Re2+ 11.Kf5 Kostic played here 11.Kf4. The game went on 66...Kf7 67.Rb7+ Ke6 68.h6 Rf2+ 69.Kg3 Rc2 70.h7 (70.g6 was stronger but 70.h7 also wins) 70...Rc8 71.Kg4 b1Q 72.Rxb1 Kf7 73.Rb6 Kg7 74.g6 Ra8 75.Re6 Rb8 76.Kf5 Ra8 77.Re7+ Kh8 78.Re5 Kg7 79.Kg5 Ra7 80.Rf5 Rb7 81.Rf7+ Kh8 82.Rf8+ 1–0.
11…Rf2+ 12.Kg4 Rg2+ 12...Kf7 13.h6 Rg2+ 14.Kf4 Rf2+ 15.Kg3 Rc2 16.g6++-.
13.Kf4 Rf2+ 14.Kg3 Rc2 15.Rb7+ Kg8 16.h6+-.
Back to
D. 1.2 after 2…Kg7
B) 3.g4? Chéron proves that this move recommended by Vukovic throws away the win.
D. 1.5
3…Kf6 a) 4.Rb6 Kf7 Mutual zugzwang position of Chéron. Black draws.
D. 1.6
5.g5 5.Kg5 Rc5+ 6.Kf4 (6.Kh6 Rc6+ 7.Rxc6 b1Q) 6...Rc4+ 7.Kf5 Rc5+ 8.Ke4 Rc4+ 9.Kf3 Rc3+ 10.Kf4 Rc4+ 11.Kg3 Rc3+ 12.Kh4 Rc2=.
5...Rc4+ 6.Kg3 Rc5 7.g6+ 7.Rb7+ Kg8=.
7...Kg7 8.Rb7+ Kh6 9.Rh7+ Kg5 10.Rb7 10.g7? Rc8 11.Rh8 b1Q 12.Rxc8 Qb3+ 13.Kh2 Qb2+ 14.Kg3 Qxg7–+.
10...Kxh5= (Chéron).
Back to
D. 1.5 after 3…Kf6
b) 4.h6 Rh2+ 5.Kg3 Rc2 6.Rb6+ Kf7 7.g5 Rc6 8.Rxb2 Kg6. Back to
D. 1.5 after 3…Kf6
c) 4.g5+ Kf7 5.g6+ Kg7 6.Rb7+ Kf6 7.g7 b1Q 7...Rc8 8.h6 Kg6 9.Rb6+ Kh7=.
8.g8N+ Kf5 9.Rxb1 9.Ne7+ Ke6 10.Rxb1 Kxe7=.
9...Rh2+ 10.Kg3 Rxh5=.
Back to
D. 1.5 after 3…Kf6
d) 4.g5+ Kf7 5.h6 5.Rb7+ Ke6 6.h6 Kf5 7.h7 Rh2+ 8.Kg3 Rxh7=.
5...Kg6 6.Rb6+ Kf5 7.Rb8 Kg6=
Back to
D. 1.5 after 3…Kf6
e) 4.Rb7 Rh2+ 5.Kg3e1) 5...Rc2?This suggestion of Chéron loses.
6.Rb6+ Kg7 7.Kh3! Chéron considers here 7.Kh4? and 7.h6+? 7.Kh4? leads after 7...Kf7 to the mutual zugzwang position D. 1.6.
7.h6+? Kh7 8.g5 Rc6 9.Rxb2 Kg6 10.Rb8 Rc7 11.Rg8+ Kh7 12.Rf8 Kg6 13.Kg4 Rc4+ 14.Rf4 Rc6 15.Rf6+ Rxf6 16.gxf6 Kxf6.
7…Rd2 7...Kf7 8.Kh4+- mutual zugzwang D. 1.3.
8.Kh4! Kf7+- (D. 1.4).
e2) 5...Rd2 Black can draw also by 5…Re2.
6.Rb6+ Kf7 7.g5 Rd3+ 8.Kf4 Rd2 9.h6 Rg2 10.Kf5 Rf2+ 11.Kg4 Rg2+ 12.Kf4 Rg1! 13.Rxb2 Rf1+!=.
Back to
D. 1.22) 2...Rh2+ 3.Kg4 Rc2 4.Rb6+ Kg7 4...Kh7 5.Kg5 Rc5+ 6.Kh4 Rc2 7.g4 Kg7 8.h6+ Kf7 9.Kh5+-.
5.Kh4 Kf7 6.g4 +-. (D. 1.3, mutual zugzwang).
II. 1…Kf6D. 1.7
2.Kh4 2.g4 Rc5! 3.Rxb2 Kg5!=.
A) 2...Kf7? Chéron considers only this move which is a result-changing error.
D. 1.8
1) 3.g4? Another slip in Chéron’s analysis who does not consider the obvious 3.Rb7+ (2).
D. 1.9
3...Kg7 a) 4.Rb6 Kf7= (D. 1.6, mutual zugzwang).
Not 4...Rd2? 5.h6+ Kh7 6.Kh5! Rh2+ 7.Kg5 Rc2 8.Rb7+ Kg8 9.Kh5 Kh8 10.g5+-.
Not 4...Kg8? 5.Rb7+-.
Back to
D. 1.9 after 3…Kg7.
b) 4.g5 D. 1.10
4…Rc4+ 5.Kg3 Rc5 6.Rb7+ Kg8 7.Kf4 Rc4+ 8.Kf3 Rc3+ 9.Ke4 Rc5 9...Rc4+? 10.Kd5 Rg4 (10...Rh4 11.Ke5 Rxh5 12.Kf6 Rh2 13.Rb8+ Kh7 14.g6+) 11.g6 Rg5+ 12.Ke6 Rxh5 13.Kf6+-.
10.Rb8+ 10.g6 Rxh5=. 10.h6 Rxg5=.
10...Kf7 11.g6+ Kf6=.
Back to
D. 1.9 after 3…Kg7.
c) 4.Rb7+ Kf6! 5.g5+ Kf5 6.g6 6.Rb5+ Kf4=.
6...Rc4+ 7.Kg3 Kg5 8.g7 8.Rb5+ Kh6=.
8...Rc8 9.Rxb2 Rg8= (9...Kxh5=).
d) 4.Kg5 Rc5+=.
Back to
D. 1.82) 3.Rb7+
With this move as well as by 3.Rb8 or 3.Rb3 White is able to head to the favourable position of mutual zugzwang (D. 1.3).
3…Kf6 4.Rb6+ Kg7 5.Kg5 D. 1.11 
But not 5.g4? Kf7= (D. 1.6, mutual zugzwang).
5...Rc5+ 6.Kg4 Rc4+ 7.Kh3 Rc2 7...Rc5 8.g4 Rc2 9.Kg3 Kf7 (9...Rd2 10.Kf4 Rd4+ 11.Kg5 Rd5+ 12.Kh4 Rd2+- transposes to D. 1.9 after 3…Kg7 4.Rb6 Rd2? ) 10.Kh4+- (D. 1.3 mutual zugzwang).
8.Kh4 Kf7 9.g4+- (D. 1.3, mutual zugzwang).
B) 2...Rh2+ 3.Kg4 Rc2 4.Rb6+ Kf7 Now 5.Kh4 transposes into the line III.a) 1…Kg7 2.Rb6 Kf7 3.Kh4 Rc4+ 4.g4 Rc2.
Or 5.Kg5 Rc5+ 6.Kh4 Rc4+ 7.g4 Rc2= D. 1. 6 mutual zugzwang).
C) Black can draw also by
2…Rc4+ 3.g4 Rc2, 2...Rd2 3.g4 Rc2 or
2...Rg2 3.g4 Rc2, transposing in D. 1.5 after 3…Kf6.
III. 1...Kg7 As pointed out by Chéron, this move also draws but not in the way he indicated.
2.Kh4 a) 2.Rb6 Kf7 3.g4 (3.Kh4 Rc4+ 4.g4 Rc2 (D. 1.6, mutual zugzwang) 3...Rc3+ 4.Kh4 Rc2= (D. 1.6, mutual zugzwang).
b) 2.g4 Kf7 3.Kh4 Kf6 4.Rb6+ (4.g5+ Kf7) 4...Kf7= (D. 1.6, mutual zugzwang).
1) 2...Kf6? This move of Chéron loses.
D. 1.12
3.Rb6+ Not 3.g4 Rh2+ 4.Kg3 Rc2=.
3...Kg7 4.g4? Strange error by Chéron. 4.Kg5! transposes to D. 1.11.
4...Kf7= (D. 1.6 mutual zugzwang).