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  After 1…Ke4 2 Ke2 f4 3 Kf2 f3, we reach a position which is, indeed, in all endgame books (Fahrni-Alapin), but it is a win. Black wins by triangulation: 4 Kf1 Kf5 5 Ke1 Ke5 6 Kf1 Ke4.

  Exercises

  Other Cases of Correspondence

  Situations with corresponding squares come in all shapes and sizes – from the most elementary to cases so complex that most of the unoccupied squares on the board turn out to be squares of reciprocal zugzwang.

  How is the correspondence between squares determined? There is no special formula. The sensible way is to find key squares, examine the possible plans for both sides, and calculate the simplest variations. This preliminary analysis may uncover some reciprocal zugzwang situations; from there, you may go on to define an entire network of corresponding squares.

  The following examples demonstrate how to make a logical analysis of a position.

  N. Grigoriev, 1921

  Black is obliged to defend the key squares e2 and f2, which he can do either by 1…Ke3 or 1…Kf3. The first appears more natural (the opposition!); but let’s not be too hasty about drawing a conclusion.

  White’s king will attempt to break through on the queenside, by occupying the key square b3 – this too must be prevented. With White’s king at a2, Black‘s king is obliged to occupy the b4-square (a4 would be too far from the kingside). Immediately, we have the whole packet of corresponding squares: a2-b4, b1-c5, c1-d4, d1-e3 and e1-f3. As it turns out, the routine 1… Ke3? loses – after 2 Kd1, Black would be in zugzwang. But 1…Kf3! 2 Kd1 Ke3 draws easily.

  I gave this example a blue diagram, not because it was especially important, but in order to underscore that a system of corresponding squares certainly does not have to always be “straight-line,” as with the opposition. Each case demands concrete analysis. You may only take the opposition after having ensured that this will place your opponent in zugzwang, not yourself.

  And if, as in the present example, you must instead cede the opposition to your opponent, I call such cases of corresponding squares the “anti-opposition.” This term seems more exact than the term “knight’s-move opposition” I have seen used (after all, the entire idea of “opposition” is for the kings to be standing on the same line, not on adjoining lines).

  N. Grigoriev, 1922

  The correspondence of the squares f4 and f6 is obvious (on 1…Kg6, 2 e7 Kf7 3 K×f5 K×e7 4 Kg6 decides). And when White’s king appears on h4, Black must be on the g6-square (but not f6, because of Kh5). The adjoining-squares principle permits us to define yet a third pair of corresponding squares, g3-g7.

  Let’s go further. The square f3 adjoins both f4 and g3 – its obvious correspondent is g6. From h3, the king wants to go to g3 and h4 – thus, the corresponding square for Black is f6.

  Let’s pull back one rank, and look at the g2-square. From here, the king can go to f3 (the corresponding square: g6), g3 (g7), or h3 (f6). The only equivalent square for Black is f7 – but he cannot go there.

  Thus, the solution becomes clear. The g2-square is the key: White must simply retreat his king there, see where Black’s king goes in response, and go to the corresponding square.

  1 Kf3 Kg6! 2 Kg2! Kf6 (2…Kg7 3 Kg3) 3 Kh3! Kg7 4 Kg3! Kf6 (4…Kg6 5 Kh4 Kf6 6 Kh5) 5 Kf4 Kg6 6 e7 Kf7 7 K×f5 K×e7 8 Kg6+–.

  Gulko – Short

  Riga 1995

  First, we must make sure that the direct attempt to force a draw by trading off the e-pawn does not work.

  1 Kf4? Ke7! (it will become clear later why the king goes to this square, and not to f7) 2 Kf5 Kf7 3 e4 (3 Kf4 Ke6 4 Ke4 h4 5 Kf4 Kd5!–+) 3…h4! 4 Kg4 Ke6 5 K×h4 Ke5–+.

  It is also worth noting that if it were White’s move in the position after 3 e4, he would still lose after 4 e5 h4! 5 Kg4 fe 6 K×h4 Ke6 7 Kg3 Kd5 8 Kf3 Kd4. The move e4-e5 only saves the game with Black’s king at g7 or e7 (since the threat is to take on f6 with check).

  Now, what can White do against the black king’s march to the center? The only possibility is to attack the h5-pawn. He can draw, if he can meet Ke6 with Kh4 (with the pawn still on e3).

  But if Black’s king goes to g6, then keeping the king at h4 becomes pointless – here, White must go to f4, with the idea of pushing the e-pawn.

  Note that these paired squares we have found are not corresponding squares, since no zugzwangs exist; but our calculations now allow us to begin the search and analysis of correspondences.

  From f7, Black’s king is ready to move in two directions – to e6 or to g6. White’s king must keep the same possibilities in hand. This clarifies the first, and most important pair of corresponding squares: f7-g3. (And here is why 1 Kf4? is to be met by 1…Ke7! – in order to meet 2 Kg3!? with 2…Kf7!, placing White in zugzwang).

  We are almost ready to make our first move. 1 Kg3? Kf7! is bad; and on 1 Kf2? Ke7! decides – the threat of 2…Ke6 forces White’s king to approach h4 through the mined square g3.

  1 Kg2!! Kg8

  On 1…Kg7, White saves himself by 2 Kf3! The black king can reach e6 only through f7. The white king will then be able to access g3 on its way to h4.

  2 Kf2 (2 Kf3 leads to the same thing) 2…Kf8 3 Kg2! Ke7 4 Kh3! Kf7 5 Kg3! Kg6

  If 5…Ke6 6 Kh4=; and if 5…f5, either 6 e4=, or 6 Kf4 h4 7 e4 h3 8 Kg3 fe 9 K×h3 Ke6 10 Kg3 Kd5 11 Kf2 Kd4 12 Ke2=.

  6 Kf4 Kh6

  On 6…Kf7, the only reply is 7 Kg3! (7 Kf5? Ke7), while on 6…Kg7, it is 7 Kf3! (7 e4? Kg6, and 7 Kf5? Kf7 8 e4 h4–+ are two bad alternatives).

  7 Kf5 Kh7

  If 7…Kg7, then 8 Kf4! (but not 8 e4? Kf7 9 e5 h4–+) 8… Kg6 9 e4! Kh6 10 Kf5 Kg7 11 e5=.

  8 e4 Kh6 (8…Kg7 9 e5) 9 K×f6 h4 Draw.

  Note that the game position is not new – in 1979, C. Costantini composed it as a study. Of course, GM Gulko did not know this – but he was acquainted with the idea of corresponding squares and was able to put the method successfully into practice.

  Exercises

  King vs. Passed Pawns

  The Rule of the Square

  Imagine a square having for one of its sides the path from the pawn to its queening square. If the king stands within the square of the passed pawn, or can reach it on its move, the pawn can be stopped; otherwise, it will queen.

  Black to move gets inside the square and draws (1…Kg4 or 1…Kg3). If it is White’s move, then after 1 b4 the side of the new square becomes the f-file, which Black’s king cannot reach in time.

  If the pawn stood on b2, then because the pawn can move two squares, the square should still be constructed from the b3-square.

  Obstacles in the path of the king: It sometimes happens that even though the king is located within the square, it still cannot stop the passed pawn, because its own pawns get in the way.

  R. Bianchetti, 1925

  1 d5! ed 2 a4 Ke4 (2…d4 3 a5 d3 4 Ke1) 3 a5+–, as Black cannot play 3…Kd5.

  The waiting move 1 Kh3 places Black in zugzwang – now he loses. Without the pawn at c7, the opposite result occurs.

  An analogous zugzwang occurs if you move the pawn at a5 to c5. The only difference is that this time, without the pawn at c7, the position would be drawn.

  Exercises

  Réti’s Idea

  It sometimes happens that a king outside the square of a passed pawn can still catch it. The win of the missing tempo (or even several tempi) is accomplished by the creation of accompanying threats, most often (though not exclusively) involved with supporting one’s own passed pawn.

  R. Réti, 1921

  Black’s king lies within the square of the c6-pawn, while White is short two tempi needed to catch the h5-pawn. Nevertheless, he can save himself – the trick is “to chase two birds at once.” The king’s advance is dual-purpose: he chases after the h-pawn, while simultaneously approaching the queen’s wing.

  1 Kg7! h4 2 Kf6! Kb6

  If 2…h3, then 3 Ke7(or e6), and the pawns queen together.

  3 Ke5! K×c6

  3…h3 4 Kd6 h2 5 c7=.

  4 Kf4=

  A miracle has come
to pass: the king, even though two tempi behind, nevertheless has caught the pawn!

  In 1928, Réti offered a different version of this study: move the white king to h5, and instead of the pawn at h5, put three (!) black pawns at f6, g7 and h6. The solution is similar: 1 Kg6!, and after any Black reply (1…f5, 2…h5, or 1…Kb6) - 2 K×g7!, followed by the well-known “chasing two birds at once.”

  And now, a slightly different version of the same idea.

  L. Prokeš, 1947

  1 Kc8 Kc6 2 Kb8! Kb5 (else 3 a6) 3 Kb7!

  Thanks to the threat of 4 a6, White wins a tempo and gets into the square of the h-pawn. 3 Kc7? h5 is hopeless.

  3…K×a5 4 Kc6=.

  The study we shall now examine shows us that Réti’s idea can be useful in more than just pawn endings.

  A. & K. Sarychev, 1928

  1 c8Q? Bf5+ 2 Kc7 B×c8 is hopeless, as is 1 Kd6? Bf5 2 Kc5 Ke4 3 Kb6 Bc8 4 Ka7 b5. The only saving line starts with a paradoxical move that forces the black pawn to advance.

  1 Kc8!! b5

  As in Réti’s study, White is short two tempi.

  2 Kd7 b4 3 Kd6 Bf5

  Thanks to the threat of 4 c8Q, White wins one tempo; now he wins the other tempo by attacking the black bishop.

  4 Ke5! Bc8 5 Kd4=.

  Tragicomedies

  Yates – Marshall

  Karlsbad 1929

  1 Qc4+ Ka3 2 Qb5(c2)? is a mistake, in view of 2…b1Q! 3 Q×b1 stalemate. But White wins easily after 1 Qc2 a3 2 Kc3 Ka1 3 Kb3 b1Q+ 4 Q×b1+ K×b1 5 K×a3, when the black king can’t reach the square (remember that when the pawn is on the second rank, the square is constructed from f3, not from f2).

  In the game, White chose a less accurate method of transposing into a pawn endgame.

  1 Kc3? b1Q 2 Q×b1+ K×b1 3 Kb4

  This is a situation known to us from the Prokeš study.

  3…Kb2! 4 K×a4 Kc3, with a draw.

  Lasker – Tarrasch

  St. Petersburg 1914

  Black wins without any trouble after 1…Be6+ 2 Kg6 B×g7 3 K×g7 B×b3 4 h4 Bd1. Tarrasch decided that the pawn endgame would be simpler still. However, he overlooked the very same finesse as Yates did in the preceding example.

  1…B×g7? 2 B×f5! K×f5?!

  I leave it to my readers to decide on their own if White could have saved himself after 2…Bh8 or 2…Bf6. Perhaps it would be worthwhile to return to this difficult question after we study the chapter on opposite-color bishops.

  3 K×g7 a5 4 h4 Kg4

  Tarrasch had expected to block White’s king from stopping the passed pawn after 5 Kf6? c4 6 bc bc 7 Ke5 c3 8 bc a4 9 Kd4 a3.

  5 Kg6! K×h4 6 Kf5 Kg3

  6…c4 7 bc bc 8 Ke4 c3 9 bc a4? 10 Kd3 is now useless.

  7 Ke4 Kf2 8 Kd5 Ke3 9 K×c5 Kd3 10 K×b5 Kc2 11 K×a5 K×b3 Draw.

  Exercises

  The Floating Square

  There are cases in which the king must do battle with two separated passed pawns; in these cases, a useful rule is the floating-square rule, suggested by Studenecki in 1939.

  If a square whose two corners are occupied by pawns (on the same rank) reaches the edge of the board, then one of those pawns must queen.

  If the square does not reach the edge of the board, then the king can hold the pawns. If there are two files between the pawns, the king can capture both; if the distance is any greater, he can only prevent their further advance.

  The square having reached the edge of the board, the pawns will queen, regardless of whose move it is.

  1…a4 2 Kb4 e4 3 K×a4 e3+–

  Let’s shift the pawns to a6 and e6. The square now reaches only to the second rank, and the position becomes a draw. In fact, 1…a5? would be bad: 2 Kb5 e5 3 K×a5+–; and so is 1…Kh6? 2 Kd6! a5 3 K×e6 a4 4 Kf7 a3 5 g7 a2 6 g8Q a1Q 7 Qg6#. Black must play 1…Kf6 2 Kc6 (but not 2 Kb6? e5 3 Kc5 a5–+) 2…Kg7 (2…e5 3 Kd5 a5 4 g7 K×g7 5 K×e5= is possible, too) 3 Kc5=.

  This square does not reach the edge of the board, and the distance between the pawns is the most unfavorable: two files. This means the pawns are lost, regardless of who is on move.

  1 Ka4 d4 2 Kb3 Kh6 3 Kc4 a4 4 K×d4+–.

  Let’s examine one more substantive case.

  On the queenside, the square does not reach the edge of the board, so the pawns can be held: 1 Kc3 a3 2 Kc2. On the kingside, however, the pawns are already quite far advanced. True, the king can prevent them from queening – so far; but because of zugzwang, he will soon be forced to let them through.

  Khalifman – Belikov

  Podolsk 1992

  1 h6! gh 2 Kf3 h5 3 Kg3 c5

  There are two files between the black passed pawns; the square does not reach the edge of the board – that means the pawns must be lost. The attempt to defend them with the king is doomed to failure: 3…Kg7 4 c4 c5 5 Kh3 Kh6 6 Kh4 c6 7 Kh3 Kg7 8 Kg3 (triangulation!) 8…Kh6 9 Kh4 e4 10 Kg3 Kg7 11 Kf4+–

  4 Kh4 e4 5 Kg3 Black resigned.

  Tragicomedies

  Stoltz – Nimzovitch

  Berlin 1928

  White would secure the draw by advancing his a-pawn and putting the rook behind it, thus: 1 a5! R×b5 2 Ra3=, or by 1 Ra3! Ke4 2 a5 d3 3 a6 Ke3 4 R×d3+ K×d3 5 a7=. Instead, Stoltz offered to trade rooks:

  1 Rd2?? R×d2! 2 K×d2 f4! 3 gf+ (3 a5 Kd6 4 a6 Kc7) 3…Kd6!

  The square of the d4- and g4-pawns reaches the edge of the board – that means it is impossible to prevent one of them from queening. The same could also be said of White’s pawns – but they are much too late. Note the excellent move of the black king – from d6, it is prepared to stop either white pawn with a minimum of effort.

  4 a5 g3 5 a6 Kc7 6 Ke2 d3+ 7 K×d3 g2 8 Ke4 g1Q 9 Kf5 Qb6 10 Kg5 Kd7 11 f5 Ke7 White resigned.

  We may add to our list of tragicomedies not just White’s gross blunder, but also his stubborn refusal to end resistance in a completely hopeless situation.

  Exercises

  Three Connected Pawns

  It is difficult for the king to fight three connected passed pawns. It has no chance at all if the enemy has any moves in reserve. If not, then a situation of reciprocal zugzwang could arise.

  White to move wins by 1 Kb1! (1…b3 2 Kb2; 1…a3 2 Ka2 c3 3 Kb3; 1…c3 2 Kc2 3 Kb3). Any other first move by White leads to the opposite result.

  Nunn – Friedlander

  Islington 1968

  On the queenside, we have equality: it would be bad for either side to make the first move there. The question is, who will fall into zugzwang when the kingside pawn moves run out?

  White would win by playing 1 Kh2! (or 1 Kg2!?); the important point is to be able to meet 1…h4 with 2 Kh3!, for instance: 2…f5 (2…f6 3 Kg4 f5+ 4 Kh3 f4 5 Kg4) 3 Kh2! g4 (3…f4 4 Kg2) 4 Kg2 f4 (4…h3+ 5 Kg3 f4+ 6 Kh2 f3 7 Kg3; 4…g3 5 Kf3 f4 6 Kg2) 5 Kg1! Kb7 6 Nb4.

  Nothing would be changed by 1…g4 2 Kg3 f5 (2…f6 3 Kf4 f5 4 Kg3) 3 Kg2 f4 4 Kf2(h2); or 1…f5 2 Kg2! (or, with the king at g2 - 2 Kg3 g4 3 Kg2, etc.).

  The actual game took an immediate wrong turn:

  1 Kf2?? h4! 2 Kf3 (2 Kg2 g4) 2…h3 3 Kg3 g4 4 a5

  White has to be the first to upset the queenside equilibrium. He can no longer place his opponent in zugzwang, because the f-pawn retains the right of moving either one or two squares, according to circumstances (an important technique, to which we shall be returning).

  4 Kh2 f6! 5 Kg3 f5 6 Kh2 f4.

  4…f5 5 Nb4+ K×c5 6 a6 Kb6 7 N×d5+ K×a6 8 c4 Kb7 Draw.

  The section which follows is devoted to those cases in which both sides queen simultaneously. In such situations, the game sometimes turns into a “queen versus pawns” endgame – so it makes sense to get to know its theory first.

  Queen vs. Pawns

  The only cases which have significant practical importance are those elementary endings in which a queen plays against a pawn which has reached the next-to-last rank.

  Knight or Center Pawn

  The queen generally wins against either a center or knight pawn.

  The algorithm is simple: the queen uses either checks or at
tacks on the pawn to get closer to the enemy king, and drive it onto the d1-square. This gives White’s king a tempo to get closer to the pawn. This procedure is repeated as often as necessary.

  1 Qc7+ Kb1 2 Qb6+ Kc2 3 Qc5+ Kb1 4 Qb4+ (or 4 Qd4) 4…Kc2 5 Qc4+ Kb2 6 Qd3 Kc1 7 Qc3+ Kd1 8 Kc7 Ke2 9 Qc2 (or 9 Qe5+) 9…Ke1 10 Qe4+ Kf2 11 Qd3 Ke1 12 Qe3+ Kd1 13 Kc6, etc.

  A draw is only very rarely possible – when, for some reason, White is unable to execute this mechanism. An example would be if the white king in our previous diagram were at c7, c6 or c5.

  Sometimes, the queen’s approach is hindered by the presence of additional pawns on the board, as in the following diagram.

  The king cannot be driven to b1, since White is unable to check on the a-file. The most White can achieve is a queen endgame with an extra rook’s pawn by 1 Q×a5!? b1Q; but theory considers that endgame to be drawish. And the pawn endgame is not won either: 1 Qc2 Ka1 2 Ke7 b1Q 3 Q×b1+ K×b1 4 Kd6 Kc2 5 Kc5 Kd3 6 Kb5 Kd4 7 K×a5 Kc5=

  However, with the white king at f7, the exchange of queens leads to victory.

  1 Qc2 Ka1 2 Ke6 b1Q 3 Q×b1+ K×b1 4 Kd5 Kc2 5 Kc4! (the first, but not the last time we shall see “shouldering” used in this book) 5…Kd2 6 Kb5+–.

  Rook or Bishop’s Pawn

  With a rook or bishop’s pawn, the above described winning algorithm does not work – a stalemate defense appears.