Dvoretsky's Endgame Manual Page 8

  The draw is obtained by constructing a stalemate refuge: 1 Kg4! (not 1 g6?? h5!–+) 1…Ke4 2 g6! h6 (2…hg 3 fg f5+ 4 Kg5 f4 5 h5 f3 6 h6=) 3 Kh5!=.

  The game continuation was: 1 gf?? gf 2 Kg4 Ke4 3 Kh3 Kf4 White resigned. Also insufficient was 3 Kh5 K×f5 4 Kh6, when Black’s simplest win is 4…Kg4 5 K×h7 Kh5! (shouldering), but another possible win is 4…Ke6 5 K×h7 f5 6 Kg6 f4 7 h5 f3 8 h6 f2 9 h7 f1Q 10 h8Q Qf5+, with mate soon to follow.

  Aronson – Mednis

  USA 1953

  The exact same position as in the previous diagram, except with all the pieces one rank lower. Here, 1 Kg3? Ke3 2 g5 hg! o would be a mistake; but 1 g5! h5 2 Kg3 Ke4 3 Kh4!= is possible.

  The game actually continued 1 h4?? h5–+.

  Exercises

  “Semi-Stalemate”

  This is what I call the situation when the king is stalemated (on the edge or in the corner of the board), but there are still pawn moves left to make. Instead of stalemate, what we get is zugzwang – usually, a fatal one for the stalemated side.

  Here are two simple examples:

  Marshall – Réti

  New York 1924*

  1 g5! (or 1 Ke5 Kc8 2 g5!) 1…Kc6 2 Ke5 Kd7 3 Kd5! (3 Kf6?? K×d6 4 K×f7 Ke5–+) 3…Kd8 4 Kc6 Kc8 5 d7+ Kd8 6 Kd6+–.

  J. Kling, B. Horwitz, 1851

  1 Bf3+ Kg1 2 Bh1! K×h1 3 Kf1! d5 4 ed e4 5 d6 e3 6 d7 e2+ 7 K×e2 Kg2 8 d8Q h1Q 9 Qg5+ Kh3 10 Qh5+ Kg2 11 Qg4+ Kh2 12 Kf2+–.

  The next example is considerably more difficult and hence more interesting.

  Mandler – Procházka

  Czechoslovakia 1976

  The straightforward 1 b5? leads to a drawn queen endgame with an extra rook’s pawn: 1…K×g2 2 Kb7 K×h3 3 K×a7 Kg4! 4 b6 h3 5 b7 h2 6 b8Q h1Q.

  The other, more promising plan is to squeeze the enemy king into the corner. However, it requires lengthy and accurate calculation.

  1 Kd5!! K×g2 2 Ke4 K×h3 3 Kf3 Kh2 4 Kf2!

  After 4 b5? (or 4 a4?), Black’s king can use Réti’s idea to help him get to the queenside in time: 4…Kg1 5 Kg4 Kg2! 6 K×h4 Kf3=.

  4…h3

  4…a6 is inferior: 5 a4 h3 6 a5! Kh1 7 b5 h2 8 Kf1!+–.

  5 b5!

  5 a4? would be a mistake, in view of 5…a5! 6 ba Kh1 7 a6 h2=.

  5…Kh1

  6 Kf1!

  Again, not 6 a4? a5!=. Now Black must stalemate his own king, since 6…Kh2 allows an easy win by 7 a4 Kh1 8 a5 a6 (8…h2 9 Kf2 a6 10 Kf1) 9 Kf2.

  6…h2 7 b6!!

  The only move! White only gets a draw after 7 a4? a5! or 7 Kf2? a5!.

  7…a5

  7…ab 8 a4 b5 9 a5+–.

  8 b7 a4 9 Ke2! Kg1 10 b8Q h1Q 11 Qb6+

  White wants mate. As can easily be seen, exchanging queens wins also.

  11…Kh2 12 Qd6+ Kg1 13 Qd4+ Kh2 14 Qh4+ Kg2 15 Qg4+ Kh2 16 Kf2 Black resigned.

  Reserve Tempi

  Exploiting Reserve Tempi

  We have already seen more than once how the outcome of a game may hinge on one side’s store of reserve pawn tempi. This is not surprising, considering that zugzwang is the fundamental weapon in the majority of pawn endings.

  The rules involved in the use of reserve tempi are simple and self-evident:

  (1) Use every chance to accumulate reserve tempi and to deprive your opponent of his;

  (2) Hold onto them – do not waste them except when absolutely necessary.

  Let’s observe these rules in action. The first is illustrated in the following two examples.

  1 f5!

  This move secures White two reserve pawn tempi – just enough to squeeze the enemy king at the edge of the board.

  1…K×a4 2 Kc4 Ka3 3 Kc3 a4 4 h4 Ka2 5 Kc2 a3 (5…h5 6 gh a3 7 h6 gh 8 h5 is zugzwang) 6 h5 Ka1 7 Kc1=.

  Despotovic – Dvoretsky

  Moscow tt 1968

  1…g4!

  Now Black has the reserve tempo h6-h5. Here, the game was adjourned; White sealed the move 2 Kc3?, but later resigned without continuing, in view of 2…Ke6 3 Kd4 (3 Kb3 Kd5 4 a4 ba+ 5 K×a4 Kc4–+) 3…Kd6 4 Kc3 Kd5 5 Kd3 h5–+. Thanks to his reserve tempo, Black put his opponent into zugzwang, and his king broke through on the wing.

  But White could have saved himself by playing 2 Ke3! Ke6 3 h3!

  Yes! White loses with either 3 h4? Kd5 4 Kd3 h5 or 3 Kf2? Kd5 4 h3 h5!

  But now what is Black to do? On 3…h5 4 h4 (or 4 hg), he loses his reserve tempo, and the position is now drawn: 4…Kd5 5 Kd3. And in the sharp variation 3…gh 4 Kf2

  4…Kd5, White has time to create kingside counterplay.

  Karsten Müller noted that if, instead of 4…Kd5, Black were to try 4…Kf6 5 Kg1 Kg7!?, then the only move to draw would be 6 Kh1! – the h2-square is mined. The problem is that after 6 Kh2? Kg6 7 K×h3 Kh5 8 g4+, Black does not play 8…fg+? 9 Kg3 Kg6 10 K×g4 Kf6 (10…h5+ 11 Kh4) 11 Kh5 Kf5 12 K×h6 K×f4 13 Kg6, when the white king reaches the queenside in time, but 8…Kg6! 9 Kh4 Kf6! 10 Kh5 fg 11 K×g4 Kg6 – here, it is White who falls into zugzwang.

  5 Kg1 Kc4 6 Kh2 h5 (6…Kb3? 7 g4!) 7 K×h3 Kb3 8 Kh4 K×a3 9 K×h5 (9 Kg5!?) 9…K×b4 10 g4 fg 11 K×g4 Kc3 (11…Kc5 12 Kf3!) 12 f5 b4 13 f6 b3 14 f7 b2 15 f8Q b1Q 16 Qf6+, and the queen endgame is drawn.

  N. Grigoriev, 1931

  White’s king has nothing to do on the kingside (with the kings on the fourth and sixth ranks, it is easy to establish that the opposition is meaningless, and therefore White cannot create zugzwang). The winning plan will be to feint with the king on the queenside, and then march over to the kingside. For this plan to succeed, White will need both of his reserve pawn tempi (with the pawn at b3, the position would be drawn); so it is important not to lose them on the way.

  1 Ke2 Kd7

  1…Kb7 2 Kd3 Ka6 would be senseless, in view of 3 b4 (3 Kc4) 3…Kb5 4 Kc3 Ka6 5 Kc4! Kb7 6 Kd4+–.

  2 Kd3 Ke7! 3 Kc3!

  Both sides must keep in mind that the e6- and c4-squares are mined. After 2…Ke6 3 Kc4, Black cannot play 3…Ke5 in view of 4 b4 Ke6 5 b5; however, if he does not play this, White’s king continues on his way to the queenside. Now on 2…Ke7 3 Kc4? Ke6, the line 4 Kb4(b3) Kd5= is bad; so is 4 Kd4 Kf6! 5 Kc3 Ke5! – therefore, White has to play 4 b3, prematurely using up one of his reserve tempi, which renders the win impossible.

  3…Ke6 4 Kc4 Kd7 5 Kb4 Kc7 6 Ka5 Kb7 7 b3!

  The first tempo is used up, in order to force the black king away from the kingside.

  7…Ka7 8 Kb4 Kb7 (8…Ka6 9 Kc3 Ka5 10 Kc4 Ka6 11 b4) 9 Kc4 Kc7 10 Kd4 Kd7 11 Ke5 Ke7 12 b4+–

  That is where we use the second tempo!

  Tragicomedies

  Kachiani – Maric

  Kishinev izt 1995

  On the kingside – equality (each side has three pawn moves). White has the reserve tempo a3-a4 on the queenside. Unfortunately, she was too eager to use it:

  1 a4??

  White wins by 1 Kc3! Ke6 (1…Ke4 2 Kc4 K×f4 3 Kd5+–) 2 Kc4 Kd6 3 d5 Kd7 4 Kd4 Kd6 5 a4 h6 6 h3 g6 (6…g5 7 fg hg 8 f3 f4 9 Ke4+–) 7 h4 h5 8 f3.

  1…Kd6 2 Kc4 Ke6 3 d5+ Kd7 4 Kd4 Kd6

  Here is where White could have used the reserve tempo – but alas, it is already gone.

  5 Kc4 Kd7 6 Kd3 Ke7

  Of course, Black will not be the first to go to the mined square d6. The game soon ended in a draw.

  Exercises

  Steinitz’s Rule

  Wilhelm Steinitz, the first world champion, put forth the following paradox: that the pawns stand best on their original squares. His explanation: In the endgame, it is useful to have a choice of whether to advance a pawn one or two squares. We shall see the point of his idea to its fullest extent in the following subchapters; so I shall limit myself to just one example of it here. The analysis given below was made by Artur Yusupov when he was still quite young, with the assistance of the author.

  Yusupov – Ionov

  Podolsk 1977*

  We can see at once the idea of a pawn breakthrough on the queenside, after a2-a4 and c4-c5. Obviously, it will have no chance of succeeding unless the black king is taken far enough
away.

  First, it is necessary to put Black on move. It is also important to leave the a-pawn where it is, since from its original square, it has the choice of moving one or two squares forward.

  1 Kf4!

  In playing this, White must be prepared for 1…g6 2 h6 g5+; however, now the pawn breakthrough works: 3 Ke3! (the best square for the king to deal with the black pawns) 3…Kg6 4 a4 K×h6 5 c5! bc (5…dc 6 a5 ba 7 b6 cb 8 d6) 6 a5 c4 7 a6 ba 8 ba c3 9 a7 c2 10 Kd2.

  1…Ke7 2 Kg5

  The king goes inexorably to g6, after which White – thanks to the fact that his pawn is still on a2 – can execute the breakthrough at the ideal moment: when Black’s king is on g8.

  2…Kf8

  2…Kf7 is met by 3 Kf5 Ke7 4 Kg6 Kf8 5 a4! Kg8 6 c5!.

  3 Kg6 Kg8

  4 a3! Kf8 5 a4 Kg8 6 c5! dc 7 a5 ba 8 b6 cb 9 d6 Kf8 10 d7 Ke7 11 K×g7 a4 12 h6 a3 13 h7 a2 14 d8Q+! K×d8 15 h8Q+

  An incautious pawn advance on move one would have let the win slip. Black in response need only be careful which square he picks for his king. For example, after 1 a3?, 1…Kf7? would be a mistake: 2 Kf5 Ke7 (otherwise 3 Ke6) 3 Kg6 Kf8 4 a4+–; but Black could play 1…Ke7! 2 Kg5 (2 Kf5 Kf7!) 2…Kf8! 3 Kg6 Kg8 4 a4 Kf8, and the breakthrough does not work now, and the movelosing maneuver is no longer possible.

  The g- and h-Pawns vs. the h-Pawn

  With Black’s pawn on its starting square, the only winning plan becomes a king invasion at h6. Even the conquest of the h6-square, however, only guarantees victory in the event that one of White’s pawns remains on the second rank, in order to have the choice between moving one or two squares.

  Black to move loses:

  1…h6+ 2 Kf5 Kf7 3 h3 (3 h4 Kg7 4 Ke6 is possible, too) 3…Kg7 4 h4 Kf7 5 h5+–;

  1…Kf7 2 Kh6 Kg8 3 g5 Kh8 4 h4! Kg8 5 h5 Kh8 6 g6 hg 7 hg Kg8 8 g7+–;

  1…Kg8 2 Kh6! (2 Kf6? Kf8=) 2…Kh8 3 g5 Kg8 4 h3! Kh8 5 h4 Kg8 6 h5 Kh8 7 g6 hg 8 hg Kg8 9 g7+–

  But with White to move, the position is drawn. 1 Kf5 Kf7 is useless, and on 1 Kh5 h6! draws. After the h-pawn moves, Black only needs to select the right square for his king to retreat to.

  1 h3 Kg8! 2 Kh6 Kh8 3 g5 Kg8 4 h4 Kh8 5 h5 Kg8 6 g6 hg 7 hg Kh8=

  Clearly, 1 h4 would be met by 1…Kf7! (or 1…Kh8!) 2 Kh6 Kg8, with the same outcome. We can see that the squares g8 and h8 correspond to the position of the pawn (at h3, h4 or h5); and with White’s pawn at g4, there is one correspondence, but with the pawn at g5 – it is the opposite.

  Matters are more complicated when the defending side’s pawn has already left its starting square. Here everything depends on the nuances of king and pawn position. The ideas inherent in such positions are aptly illustrated by the following study.

  R. Réti, A. Mandler, 1921

  1 Kg3!!

  After 1 Kg4? Kf7, the position is lost. For example:

  2 h5 Ke6 (the diagonal opposition);

  2 Kh5 Kf6 3 Kg4 Ke5 4 Kh5 Kf4 5 Kg6 Kg4 6 K×g7 h5;

  2 Kf4 Ke6! 3 Ke4 g6 4 Kf4 Kd5 5 Kf3 Ke5 6 Ke3 Kf5 7 Kf3 h5. Note that with Black’s pawn at g6, the opposition is important for both sides; with the pawn at g7, it is the anti-opposition that is important. This generalization is the mainspring driving this particular endgame.

  2 Kf5 g6+ 3 Ke5 Ke7 4 Kd5 (4 h5 g5 5 Kf5 Kd6) 4…Kf6 5 Ke4 Ke6, etc.;

  2 Kf3 g6! (Black seizes the distant opposition, and then converts it into close opposition, as usual, with an outflanking) 3 Ke3 Ke7! 4 Kf3 Kd6! 5 Ke4 Ke6, etc.

  1…Ke7

  1…Kf7 is met by 2 Kg4! Now 2…Kg6 3 h5+ would be useless; and after 2…Ke6 3 Kf4! White, as should be done in positions with the pawn at g7, gives up the opposition to his opponent (3…Kf6 4 h5= or 3…g6 4 Ke4=); while after 2…g6 he seizes the distant opposition with 3 Kf3! All that remains is 2…Kf6, but then White is saved by the unexpectedly direct 3 Kh5! Ke5 (3…Kf5 is stalemate) 4 Kg6 Kf4 5 K×g7 h5 6 Kf6! Kg4 7 Ke5 K×h4 8 Kf4=.

  2 Kf3!

  Of course not 2 Kf4? Ke6!–+.

  2…Kf6

  2…Ke6 3 Kf4!; 2…g6 3 Ke3!.

  3 Ke4!

  Still the same principle at work – with the pawn on g7, anti-opposition.

  3…Kf7!? 4 Ke3! Ke7 5 Kf3!=.

  Tragicomedies

  Marshall – Schlechter

  San Sebastian 1911

  1…Kg4! 2 Kf2 Kh3 decides: 3 Kg1 (White’s king is on the wrong square: on h1, it would be a draw) 3…g5 4 Kh1 g4 5 Kg1 h4 6 Kh1 g3 7 hg hg 8 Kg1 g2–+.

  Instead, Schlechter played 1…Ke4??, when the position was drawn, because both black pawns have now left their starting rank, and if Black tries to get his king into h3 (there is no other plan), White’s king can always choose the right square on the first rank.

  2 Kf2 Kd3 3 Kf3 g5 4 Kf2 Ke4 5 Ke2 Kf4 6 Kf2 Kg4 7 Kg2 h4 8 h3+ (the simplest, although 8 Kg1 was possible too) Draw.

  Chiburdanidze – Watson

  Brussels 1987

  The position is almost the same as the Réti/Mandler study. White wins with either 1 g4! Ke6 2 Ke2! Kf6 3 Kd3, or 1 Ke3! Ke5 (1…Ke6 2 Kf4! Kf6 3 g4) 2 g4.

  1 Kf3?? Ke7! 2 Kf4 Ke6! 3 g4 Kf6 4 Kf3 Ke7??

  An awful blunder in return. As long as the pawn stood on g3, Black ceded his opponent the right to control the opposition. But now, with the pawn on g4, he cannot give the opposition up! 4…Kf7!= was necessary.

  5 Ke3!Q Kf7 6 Kd4! Kf6 7 Kd5 Ke7 8 Ke5 Kf7 9 Kf5 Kg7 10 Ke6 Kg6 11 h5+ Kg5 12 Kf7 K×g4 13 Kg6 Kf4 14 K×h6 Black resigned.

  The f- and h-Pawns vs. the h-Pawn

  We shall analyze the basic ideas of such positions by using the following study as an example.

  N. Grigoriev, 1920

  If White plays 1 Kh1? (naively hoping for 1…f2?? stalemate), then after 1…Kg4, his position is lost. Black’s plan is elementary: his king goes to e3, and then he advances …f3-f2, forcing the advance of White’s h-pawn. The fact that Black can choose whether to move his h-pawn one or two squares forward allows him to place his opponent in zugzwang. For example: 2 Kg1 Kf4 3 Kf2 Ke4 4 Kf1 (4 Ke1 Ke3 5 Kf1 f2 6 h4 h5!) 4…Ke3 5 Ke1 f2+ 6 Kf1 Kf3! 7 h3 h5! 8 h4 Kg3, or 7 h4 h6! 8 h5 Kg3.

  1 Kf2! Kg4 2 Ke3!

  Thanks to zugzwang, the pawn must leave the h7-square; the position is now a draw. White must only make sure he chooses the right back-rank square for his king (corresponding to the position of Black’s h-pawn).

  2…h6 3 Kf2 Kf4 4 Ke1! Ke3

  Or 4…h5 5 Kf2 Ke4 6 Kf1!.

  5 Kf1 h5

  5…f2 6 h3! Kf3 7 h4 Kg3 8 h5=.

  6 Ke1 f2+

  6…h4 would be met by 7 Kf1! f2 8 h3=. But not the hasty 7 h3?, which would be a terrible blunder here, leading to the Fahrni-Alapin ending we know so well (from Diagram 1-27). Just a reminder: Black wins by triangulating with his king: 7…Ke4 8 Kf1 Ke5 9 Ke1 Kf5! 10 Kf1 Ke4.

  7 Kf1 Kf3 8 h3! Kg3 9 h4=

  So the stronger side wins only if the rook’s pawn is on the starting square. The only exception to this rule was found by Maizelis in 1955 (although it was seen even earlier, in a 1949 study by Valles).

  Here, everything depends on whose turn it is to move. Black to move wins.

  1…Ke4 2 Ke2 h4! (an exceptionally important position – reciprocal zugzwang!) 3 Kf2 Kd3!!

  Control of the opposition is exploited, as usual, by outflanking – although this time, a paradoxical one.

  4 Kf3 h3! 5 Kf2

  There is no help in either 5 Kf4 Ke2 6 K×f5 Kf3! or 5 Kg3 Ke3(e2) 6 K×h3 f4.

  5…Kd2! 6 Kf3 (6 Kf1 Ke3 7 Ke1 Kf3 8 Kf1 f4 9 Kg1 Ke2–+) 6…Ke1 7 Ke3 Kf1 8 Kf3 Kg1 9 Kg3 f4+ 10 Kf3 (10 K×h3 f3) 10…K×h2 11 Kf2 f3–+

  With White to move, there is no win: 1 Ke2 Ke4 2 Kf2 h4 (2…Kd3 3 Kf3 h4 4 Kf4 Ke2 5 K×f5=) 3 Ke2 h3 4 Kf2 Kd3 5 Kf3 Kd2 6 Kf2!=.

  Maizelis’ position serves as a most important guidepost in analyzing situations where one side has an advanced rook’s pawn – the outcome of the battle depends on whether the stronger side can reach Maizelis’ position and whose move it is.

  Maizelis studied the following position, and considered it los
t. His conclusion would appear to be supported by the result of this game.

  Vaganian – Sunye Neto

  Rio de Janeiro izt 1979

  1 Kg3 h5! 2 Kf3 (on 2 Kg2, Black should play 2…Kf4!, and if 2 Kf2, then 2…Kg4! 3 Ke3 Kh3, or 3 Kg2 h4) 2…h4 3 Kg2 Kg4! 4 Kf2 Kf4 5 Ke2 Ke4 (White is in zugzwang) 6 Kf2 Kd3!! 7 Kf3 h3! White resigned.

  I was in Rio de Janeiro. Unfamiliar with Maizelis’ analysis, and astounded that Black could win such a position, I focused on it intensely and quickly found the saving line. Pal Benko (who was Sunye Neto’s second at the Interzonal) came to the same conclusion a bit earlier.

  Since the central problem here is one of reciprocal zugzwang, let’s analyze the corresponding squares. If Black’s king moves to the fourth rank with the pawn at h4 or h6, White’s king must respond by taking the opposition. If the pawn is at h5, the opposite is true – White must take the anti-opposition.

  This means that neither 1 Kg2? Kg4! nor 1 Kf2? Kf4! is good. And we have already seen what happens to 1 Kg3? So:

  1 Ke2!! Kg4

  After 1…Kf4 2 Kf2, White is in fine shape – he has the opposition with the pawn at h6. The game might continue 2…h5 3 Ke2 Ke4 4 Kf2 h4 (with the pawn at h5, the outflanking 4…Kd3 does not work) 5 Ke2=. If 1…h5, then either 2 Ke3 or 2 Kf1.

  2 Ke3! h5 (2…Kh3 3 Kf4) 3 Kf2

  The goal is reached: White has the antiopposition, with the pawn at h5.

  3…Kf4 4 Ke2 Ke4 5 Kf2 h4 6 Ke2=.

  The following exceptionally complex example was first given in Fine’s book (1941), but unfortunately with completely erroneous analysis. Maizelis (1956) did a much better job on the position; and later his conclusions were refined and extended by other authors.