Here is yet another excellent example of corresponding squares from Dvortesky's Endgame Manual and the process one must use to figure them out.

**f4(w) and f6(b)**are in obvious correspondence, for if 1...Kg6 2.e7 Kf7 3.Kxf5 Kxe7 4.Kg6 wins. When White's king is on h4, Black's king must be on g6, and not f6, because of Kh5. So

**h4(w) and g6(b)**correspond. Using these pairs or corresponding squares, we can surmise a third pair by the adjoining squares principle:

White: f4 and h4 adjoin g3

Black: f6 and g6 adjoin g7

Therefore,

**g3(w) and g7(b)**correspond.

Let's examine f3: f3 Adjoins f4 and g3, whose corresponding squares are f6 and g7. f6 and g7 adjoin to g6, therefore

**f3(w) and g6(b)**correspond.

Let's examine h3: h3 adjoins h4(g6 corresponds) and g3(g7 corresponds), and it's corresponding square is f6, which adjoins g6 and g7. Therefore,

**h3(w) and f6(b)**correspond.

So far each square we have examined has a single corresponding square for black. Let's go further back one rank and investigate g2:

From g2, white can move to f3(g6), g3(g7), or h3(f6). The three corresponding squares (g6,g7,f6) connect to f7 - a square guarded by the white e-pawn on e6. Therefore, black cannot occupy f7 when White is on g2. Hence, the solution presents itself. White will retreat his King to g2 and see where Black's king is played in response, and go to the corresponding square of Black's new King position. In summary, the corresponding squares are:

f4(w) and f6(b) - ok

h4(w) and g6(b) - ok

f3(w) and g6(b) - ok

g3(w) and g7(b) - ok

h3(w) and f6(b) - ok

g2(w) and f7(b) - illegal!

**1.Kf3**f3(g6)...

**1...Kg6! 2.Kg2!**Now black must choose his poison. Whichever square he moves to, White will occupy the corresponding square...

**2...Kf6**f6(b) h3(w)...

**3.Kh3! Kg7**g7(b) g3(w)...

**4.Kg3! Kf6**f6(b) f4(w)...

**5.Kf4 Kg6 6.e7 Kf7 7.Kxf5 Kxe7 8.Kg6+-**

## 1 comment:

Corresponding squares are cool.

Post a Comment