# How can the moves from this pawn endgame study be found without a computer?

``````[FEN "8/1p5k/1P1p4/3p4/3Pp2p/2K1P2p/7P/8 w - - 0 0"]
[PlyCount "21"]

1. Kb2 Kg8 2. Ka1 Kf8 3. Ka2 Ke7 4. Kb3 Kd7 5. Kb4 Kc6 6. Ka5
Kd7 7. Kb4 Ke6 8. Kc3 Kf6 9. Kc2 Kg5 10. Kd1 Kg4 11. Ke2
``````

In this great study from Ebersz, white can reach a draw, but only with extreme accurate play. The given variation is one possible. All white moves are forced, any other king move would lose. Suppose, white is in such a position in a practical game, of course without access to electronic topics. Can the correct squares be found out manually ? I know the theory of countersquares, but it seems to be extremely difficult.

How can white save this game without too much efforts ?

• These moves are challenging and great insight is required to find them. That is why it is a study, and not simply, "...and draws." Apr 2, 2015 at 22:38

Some of the corresponding squares are quite easy to work out, especially if you write things down (not allowed in a real game).

First of all, it is clear that

`a5` corresponds to `c6`.

Black can move from `c6` to `g4` in 4 moves. The only square that stops black's invasion and is 4 moves away from `a5` is `e2`, so

`e2` corresponds to `g4`.

The squares on the paths in between need to correspond, so

`b4` corresponds to `d7`,

`c3` corresponds to `e6`,

`d2` corresponds to `f5`.

It then follows that

`b3` corresponds to `e7`,

`c2` corresponds to `f6`,

`d1` corresponds to `g5`.

Hence,

`b2` corresponds to `f7`,

`c1` corresponds to `g6`,

`b1` corresponds to `g7`.

The trickiest part is to show that `a3` corresponds to `e8`. When the black king is on `e8`, it threatens to go to `d7`, `e7` or `f7`, corresponding to `b4`, `b3` and `b2` for white. Therefore the white king needs to be on `a3` or `c3`, but `c3` is the wrong square. Why? Because after a subsequent `Kd8` move form black, white would be in zugzwang. Black can still go to `d7`or `e7`, so white needs to go to a square adjacent to `b4` and `b3`. Such a square is not available from `c3`.

So,

`a3` corresponds to `e8`,

and therefore

`a4` corresponds to `d8`,

`a2` corresponds to `f8`

`a1` corresponds to `g8`.

Now the solution can be explained. In the diagram position Black can go to `g6`, `g7` or `g8`, so white needs to have `c1`, `b1` and `a1` available. Therefore `1. Kb2!`